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\section{Introduction}
Transiting planet systems are valuable because their geometry enables
us to estimate accurate planetary properties. Time-series photometry
during the transit allows us to derive the orbital inclination and the
relative radii of the host star and planet. Combining this with radial
velocity variations and stellar parameters, allows us to derive the
absolute mass of the planet. Hence, the bulk density of the planet can
be estimated with good accuracy, giving us insight into its
composition \citep{Guillot2005, Fortney2007}, thus placing constraints
on planetary structure and formation models. Given the remarkable
diversity in the structure of large planets, it is important to obtain
planetary parameters which are as accurate as possible. However,
obtaining high signal-to-noise transit observations is difficult and
consequently even some of the brightest stars with planets are lacking
good quality light curves and, hence, have poorly determined planetary
parameters.
The RISE (Rapid Imager to Search for Exoplanets) instrument, mounted
on the 2.0m Liverpool telescope \citep{rise2008, Gibson2008} was
designed for exoplanet transit observations. Its main scientific
driver was the detection of transit-timing variations and hence the
search for low-mass companions to ``hot Jupiters''. RISE has a rapid
readout frame transfer CCD and in 2$\times$2 binned mode has a readout
time of less than 1 s. This implies that for exposures longer than 1
s, dead time is negligible substantially increasing the time on
target. However, most exoplanet host stars are relatively bright and
saturate the CCD for 1 second exposure. To avoid dead time losses,
RISE observations are always defocussed
(e.g. \citealt{Gibson2008,Joshi2009}). Defocussed photometry
observations have also the advantage of spreading the PSF over a
larger number of pixels, thereby decreasing flat-fielding errors. RISE
is therefore ideal for obtaining high quality transit light curves for
exoplanets.
WASP-21b is a Saturn-mass planet with $M_p = 0.30 \pm 0.01 $ \hbox{$\mathrm{M}_{\rm Jup}$}\ in
a 4.3 day circular orbit \citep{Bouchy2010}. Its host star is a G3V
type with $M_*=1.01 \pm 0.03\,$\Msun\ , $T_{eff} = 5800 \pm 100\,$K
and a low metallicity, {[M/H]} =$-$0.4 $\pm$ 0.1. It was discovered by
the SuperWASP-North survey \citep{Pollacco2006} in its 2008-2009
observing campaign. \citet{Bouchy2010} argue that WASP-21 is a member
of the galactic thick disc because of its low metal abundances,
velocity relative to the Sun and age $ \sim 12$Gyr, which are similar
to the thick disc population. WASP-21b is among the lowest density
planets, $ \rho_p=0.24\pm 0.05 \rho_J$ \citep{Bouchy2010}, and has one
of the lowest metallicity host stars. Therefore, its properties are
particularly important for irradiation models. The current parameters
of the system \citep{Bouchy2010} are based on the SuperWASP discovery
photometry and a partial transit light curve taken with RISE. However,
the lack of a high precision complete transit light curve required the
assumption of the main sequence mass-radius relation which tends to
bias the estimate of the inclination. Furthermore, the age derived for
WASP-21 is longer than the main-sequence life time of a $1.01 \Msun$
star. This suggests that WASP-21 could be evolved which would
invalidate the main-sequence assumption and bias the parameters of the
system. To test the main-sequence assumption we obtained further
observations of WASP-21.
In this paper, we present transit observations of WASP-21b with RISE
including a full transit light curve. Our high precision light curves
allow us derive the planetary and stellar radii without assuming the
main-sequence mass-radius relation for the host star. We describe our
observations in Section 2. In Section 3, we discuss our transit model
and present the updated parameters of the system in Section
4. Finally, we discuss and summarise our results in Section 5.
\section{Observations}
WASP-21b was observed with RISE \citep{rise2008} mounted at the
auxiliary Cassegrain focus of the robotic 2.0m Liverpool Telescope on
La Palma, Canary Islands. This is a focal reducer system utilizing a
frame transfer e2v CCD sensor. The detector has a pixel scale of 0.54
arcsec/pixel that results in a 9.4 $\times$ 9.4 arcmin field of
view. RISE has a wideband filter covering $\sim 500$--$700\,$nm which
corresponds approximately to V+R. The instrument has no moving parts.
The Liverpool Telescope has a library of flat fields which are taken
manually every couple of months. RISE flats are taken during twilight
at different rotator angles so that there is a uniform illumination of
the CCD. The exposure times of the images are automatically adjusted
so that the peak counts in the individual flats are below the
non-linearity limit of the CCD at 45000 counts. Typically, the
individual flats have between 20000 and 40000 counts. Due to the fast
readout, we can obtain approximately 200 flat frames in a run, these
are combined to create a master flat. For each observation run we use
the master flat that is closest in time, although we note that these
are very stable.
On 2009-09-09 we obtained a full transit of WASP-21b. A total of 6581
exposures in the $2 \times 2$ binning mode with an exposure time of
2.7 seconds were taken. The telescope was defocussed by -1.2mm which
resulted in a FWHM of $\sim 11$ \arcsec. For defocussed photometry,
the star profiles are not Gaussian. However, we found that, in our
case, a Gaussian provided a good fit to the wings of the star profile,
and could be use as a rough estimate of the profile width. Therefore,
we estimated the FHWM in the usual way by cross-correlating a Gaussian
profile with that of the star.
A second full transit observation of WASP-21b was attempted on
2010-11-24. In this case, deteriorating weather terminated the
observations shortly after the mid-transit, by which time, 4008
integrations had been obtained. During these observations, the FWHM
was $\sim 12.5$ \arcsec.
Both data-sets were reduced using the ULTRACAM pipeline
\citep{Ultracam} which is optimized for time-series photometry.
Initially, we bias subtracted the data while we investigated
systematic effects that were introduced by the flat fielding process.
We performed differential photometry relative to five comparison stars
in the field, confirmed to be non variable, and we sampled different
aperture radii and chose the aperture radius that minimised the
noise. For the first night, we used a 22 pixel aperture radius ($\sim
12$\arcsec), and for the second transit, a 32 pixel aperture radius
($\sim 17$\arcsec). The photometric errors include the shot noise,
readout and background noises.
We also included in our analysis, the previously published egress of
WASP-21 taken with RISE \citep{Bouchy2010}. For consistency, we
re-reduced the original data using the same method as for the other
two observations. On 2008-10-07, 2220 exposures of 5 sec duration
were taken. We estimated a FWHM of $\sim 2.7$ \arcsec, therefore, the
level of defocussing was lower than in our observations. The best
aperture radius was found to be 15 pixel ($\sim 8$\arcsec). Our
results agree well with the previous published light curve.
The final high precision photometric light curves are shown in
Figure~\ref{photolc} along with the best-fit model described in
Section~\ref{model}. We overplot the model residuals and the estimated
uncertainties which are discussed in Section~\ref{errors}.
\begin{figure}
\centering
\includegraphics[width=\columnwidth]{figure1.ps}
\caption{Phase-folded RISE light curves for WASP-21. From top to
bottom in chronological order; 2008 October 07, 2009 September 09
and 2010 November 24. We superimpose the best-fit transit model
and also show the residuals for each light curve at the bottom of
the figure. The data are binned into 30 second periods, and bins
displaced vertically for clarity. The individual RISE light curves
plotted here are available in electronic form at CDS.}
\label{photolc}
\end{figure}
\subsection{Optimum exposure time for RISE}
As mentioned above, defocusing is commonly used in exoplanet transit
observations. \citet{Southworth2009} calculated the optimum exposure
time for the DFOSC imager mounted on the 1.54m Danish Telescope. We
follow the same procedure and apply it to RISE mounted on the
Liverpool Telescope and hence, we account for readout noise, photon,
background and scintillation noise. Similar to \citet{Southworth2009},
we do not include flat-fielding noise, assuming that the profile
position is stable.
The key difference is that RISE is a frame transfer CCD whose dead
time is the frame transfer time, 35 milliseconds for observations
longer than 1 second. For the brightest comparison star in our field,
($V \approx 9$), we found the optimum exposure times with RISE are
approximately 2.7, 7.8, 10.8 seconds during bright, gray and dark
time, respectively.
We iterate that the improvement in signal-to-noise for defocussed
observations, reported by \citet{Southworth2009} is only due to
deadtime losses; hence, the defocussing needed is proportional to the
CCD readout time. If the deadtime was zero the best theoretical
signal-to-noise would always be for focused observations, mainly due
to the increase in background noise for wider profiles.
Moreover, in our case, the improvement on signal-to-noise between 1
second and 10.8 seconds exposure times is quite small on the order of
$10\,$ppm per $30\,$sec bin. As we will see below, the strongest
reason for defocussing is to minimise systematic noise which, due to
its nature, is not accounted for in the calculation and can
substantially increase the noise in a transit light
curve. Figure~\ref{complc} shows systematic noise variations larger
than 400 ppm.
\section{Data analysis}
\subsection{Systematic noise}
Exoplanet transit observations are often dominated by systematic
noise. Therefore, to improve the precision of the light curves it is
important to determine and minimise this noise source. For the 2009
September 09 observations, the brightest comparison star (c1) on the
field was affected by systematic noise. This can clearly be seen in
Figure~\ref{complc}, where we show the flux of c1 relative to the
ensemble of comparison stars used in the final 2009 WASP-21 light
curve. This shows a variation of 400 ppm. We found that this
systematic noise was correlated with the star position in the CCD,
which during the transit observation varied by 10 pixels in the $x$
direction and 8 in the $y$. Given that we used an aperture radius of
22 pixels, this implies that only half of the pixels used to perform
aperture photometry were common for the duration of the
observation. Hence, we concluded that the systematic noise was due to
variations in the pixel-to-pixel sensitivity which were not corrected
by flat fielding. In fact, the systematic noise is slightly higher if
we flat field the data. Our master flat is a combination of 150
frames, each with a mean of 35000 counts. The uncertainty in this flat
is 0.5 millimags per pixel which is smaller than the photometric error
($\sim 4.4$ milimags per unbinned point) and the observed systematic
noise. After careful analysis of the data, we found that the c1
comparison star crossed a reflection feature in the CCD that is
rotator dependent (LT is on an alt-azimuth mount) and thus was not
corrected by flat fielding. This experience demonstrates the
importance of good guiding in decreasing the sources of systematic
noise. If the observations were performed in focus and assuming the
seeing was 1 arcsec, the FHWM would have been $\sim 2$ pixels. Using
an aperture radius of $1.5 \times\,$ FWHM $=3\,$ pixels, it would have
implied that there were no common pixels during the
observations. Therefore, we infer, if the observations were focused,
the amount of systematic noise would have doubled. Note that the defocussing does not affect the guiding since the guide camera is always kept in focus.
After this incident the RISE instrument was upgraded. The source of
the reflected feature was identified and removed from the instrument
field of view. We also improved the telescope guiding system's
stability. This led to an improvement in the precision of the light
curves which is evident in the latest light curve of WASP-21 taken
after the upgrades (see Fig. 1). In the November 2010 observations the
variation in position is less than 2 pixels in the $x$ direction and 4
pixels in the $y$.
\begin{figure}
\centering
\includegraphics[width=\columnwidth]{figurec1.ps}
\caption{Light curve of the brightest comparison star for the 2009
September 09 observation relative to the ensemble of comparison
stars used in WASP21b final light curve. It shows systematic noise
with an amplitude of 400 ppm. This comparison was not used in the
final light curve of WASP-21. We also overplot the photometric
errors.}
\label{complc}
\end{figure}
\subsection{Photometric errors}
\label{errors}
An accurate estimate of the photometric errors is important to obtain
reliable system parameters. Our first estimate of errors for each
light curve includes only the shot noise, readout and background
noise, which underestimates the true errors. To obtain a more reliable
estimate we begin by scaling the errors of each light curve so that
the reduced $\chi^2$ of the best fitting model is 1.0. This resulted
in the multiplication of the errors by $1.97$, $1.22$ and $1.44$, for
the 2008, 2009 and 2010 light curves, respectively. We then calculated
the time-correlated noise following the procedure from
\citet{gillon2009}. Using the residuals of the best fit model, we
estimated the amplitude of the red noise, $\sigma_r$ to be $150\,$ppm,
$250\,$ppm and $150\,$ppm, for the 2008, 2009 and 2010 light curves,
respectively. These were added in quadrature to the rescaled
photometric errors and were used in the final Markov Chain Monte Carlo
(MCMC) chains. However, \citet{carter2009} found that this ``time-averaging'' method of estimating the correlated noise can still underestimate the uncertainties by 15-30 per cent.
\subsection{Determination of system parameters}
\label{model}
To determine the planetary and orbital parameters, we fitted the three
RISE light curves of WASP-21b simultaneously. We used the
\citet{Mandel2002} transit model parametrised by the normalised
separation of the planet, $a/R_*$, ratio of planet radius to star
radius, $ R_p/R_* $, orbital inclination, $i$, and the transit epoch,
$T_0$, of each light curve. Our model was originally developed to
measure transit timing variations of exoplanets. Following
\citet{Bouchy2010} that found no evidence for a significant orbital
eccentricity of WASP-21b we adopt a circular orbit. We included the
quadratic limb darkening (LD) coefficients for the RISE filter V+R
from the models of \citet{Howarth2010}: $a=0.45451$ and
$b=0.210172$. These were calculated for $T_{eff} = 5800\,$K, \logg=4.2
and [M/H] =$-$0.5 to match the stellar parameters from
\citet{Bouchy2010}. We initially kept the limb darkening parameters
fixed during the fit. For each light curve, we included two extra
parameters to account for a linear normalization. Therefore, 12
parameters were fitted. Besides the linear normalization, no extra
trends were removed from the light curve.
To obtain the best fit parameters and uncertainties, we used a MCMC
algorithm (e.g. \citealt{Tegmark2004,Cameron2007,Gibson2008}). We
begin by calculating the $\chi^2$ statistic of a set of proposed
parameters,
\begin{equation}
\chi^2=\displaystyle\sum\limits_{j=1}^N {\frac{(f_j-m_j)^2}{\sigma^2_j}},
\end{equation}
where $f_j$ is the flux observed at time $j$, $ m_j$ is the model flux
and $\sigma_j$ is the uncertainty of each $f_j$ as described in
Section~\ref{errors}. At each step in the MCMC chain, each proposed
parameter is perturbed by a random amount which we call a ``jump
function''. Each jump function is proportional to the uncertainty of
each parameter multiplied by a random Gaussian number with mean zero
and unit standard deviation. The new parameter set is accepted with
probability,
\begin{equation}
P=min \left( 1, exp\left(\frac{-\Delta\chi^2}{2}\right) \right),
\end{equation}
where $\Delta\chi^2$ is the difference in the $\chi^2$ of subsequent
parameters sets. Note, the new parameter set is always accepted if its
$\chi^2$ is lower than the previous parameter set ($P=1$). The jump
functions are scaled by a common factor in order to ensure that $25\%$
of the steps are accepted, as suggested by \citet{Tegmark2004}. To
estimate the uncertainty of each parameter and calculate the jump
functions, an initial MCMC fit was performed. With these jump
functions, we computed seven MCMC chains each of 150 000 points and
different initial parameters. The initial 20\% of each chain that
corresponded to the burn in phase were discarded and the remaining
parts merged into a master chain. We estimated the best fit parameter
as the mode of its probability distribution and the 1 $\sigma$ limits
as the value at which the integral of the distribution equals 0.341\%
from both sides of the mode. We computed the \citet{Gelman92}
statistic for each fitted parameter and concluded that chain
convergence was good.
To test how the limb darkening coefficients affect the derived system
parameters, we repeated the MCMC procedure also fitting for the linear
LD ``$a$'' which is the most sensitive to the observing filter. The
quadratic LD coefficient, $b=0.210172$, was kept fixed, because as
reported by \citet{Gibson2008}, the high precision of the RISE light
curves is not enough to fully constrain the LD coefficients (i.e., the
MCMC does not converge when fitting both coefficients). We restricted
the linear LD coefficient to be the same for all the light curves
since they were all taken with the same filter. Therefore, in the
second MCMC procedure we fitted 13 parameters. We estimated $a = 0.337
\pm 0.034$.
\section{Results}
Comparing the fitted and fixed LD solutions, we concluded that
although the fitted linear LD coefficient is statistically
significantly different from the theoretical value, this does not
affect the derived system parameters to any extent. The two solutions
are within $1.5\sigma$ of each other. Contrary to what was found by
other authors (e.g., \citealt{Gibson2008,Southworth2008}), the
uncertainties of the fitted LD solution are slightly smaller than
those of the fixed solution. The $\chi^2$ of the fitted LD solution is
similar to the fixed LD solution which does not justify the addition
of an extra free parameter in the fit. Consequently, we conclude that
our light curve is of insufficient quality to better constrain the
linear LD relative to that achieved by theoretical models and we
choose to present the fixed LD solution.
The estimated transit times, combined the original ephemeris
\citep{Bouchy2010} were used to update the linear ephemeris,
\begin{equation}
T_{t} (HJD) = T(0) + EP.
\end{equation}
We found $P=4.3225060 \pm 0.0000031$ and $T_0 =2455084.51974 \pm
0.00020$ which was set to the mid transit time of the 2009 light
curve. This ephemeris was used in the final MCMC procedures.
For future reference, the time residuals from the linear ephemeris are
given in Table~\ref{ttv}. We conclude that the time residuals of
WASP-21b are consistent with a linear ephemeris.
\begin{table}
\centering
\caption{Time residuals from the linear ephemeris.}
\label{ttv}
\begin{tabular}{ccc}
\hline
\hline
Epoch & Time residuals (sec) & Uncertainty (sec) \\
\hline
-78 & 11 & 40 \\
0 & -7 & 24 \\
102 & 8 & 30 \\
\hline
\end{tabular}
\end{table}
The geometric system parameters of WASP-21 and the 1$\sigma$
uncertainties derived from the MCMC analysis with fixed limb darkening
coefficients are given in Table~\ref{mcmc}. These parameters are
directly measured from the transit light curve and are only weakly
dependent of stellar properties through the limb darkening
coefficients. Note, all the derived parameters presented in
Table~\ref{mcmc} were calculated at each point of the chain.
Therefore, the final derived values and errors were determined from
their probability distribution as done for the fitted values. We
obtain a significantly lower density than was previous reported in the
discovery paper $ \rho_* = 0.84 \pm 0.09 \rho_{\odot} $).
\begin{table}
\centering
\caption{WASP-21 system parameters derived from the mcmc}
\label{mcmc}
\begin{tabular}{lccl}
\hline
\hline
Parameter & Value \\
\hline
Normalised separation $a/R_*$ & $9.68^{+0.19}_{-0.30}$\\
Planet/star radius ratio $ R_p/R_* $ & $0.10705^{+0.00082}_{-0.00086}$ \\
Orbital inclination $ I $ [degrees] & $ 87.34 \pm 0.29$ \\
Impact parameter $ b $ [$R_*$] & $ 0.458^{+0.043} _{-0.036}$ \\
Transit duration $T_T$ [days] & $ 0.1430^{+0.0013}_{-0.0010}$\\
Stellar density $ \rho_* $ [$\rho_{\odot}$] & $ 0.652^{+0.041}_{-0.060} $ \\
\hline
\end{tabular}
\end{table}
\subsection{Stellar mass and age}
To obtain the stellar and planetary physical properties, the geometric
parameters have to be scaled with the stellar mass. The new high
quality transit light curves give a direct estimate of the stellar
density. This allows a more accurate estimation of stellar mass than
log g derived from spectral analysis \citep{Sozzetti2007}. Currently
there are two main methods to derive the stellar mass from the stellar
density. The first uses isochrones and mass tracks from stellar
models \citep{Sozzetti2007} and the second uses an empirical
calibration derived from stellar eclipsing binaries
\citep{Torres2010,Enoch2010}.
\citet{Bouchy2010} derived the stellar mass through the empirical
calibration between $T_{eff}$, $ \rho_* $ and [Fe/H]
\citep{Torres2010} with the parametrisation of \citet{Enoch2010}.
Following the same procedure, with the improved $\rho_*$, we derive a
stellar mass of $1.02 \pm 0.05 \Msun$. In Table~\ref{mcmc2} we
present the mass and radius of WASP-21 and WASP-21b and the 1$\sigma$
uncertainties derived from the MCMC for a stellar mass of $1.02 \pm
0.05 \Msun$. We obtain a significantly larger stellar and planetary
radius than previously reported. This is due to the main-sequence
assumption in the previous analysis which as discussed below is found
to be invalid.
We also estimate the stellar mass from stellar models by interpolating
the Yonsei-Yale stellar evolution tracks by \citet{Demarque2004} using
the metallicity from \citet{Bouchy2010}. These evolution tracks are
plotted in Figure~\ref{isocrones} along with the position of
WASP-21. From the isochrones, we estimate a lower mass of $0.86 \pm
0.04 \Msun$ and an age of $12 \pm 2\,$Gyr for WASP-21. In Figure
\ref{masstrack}, we also show the evolutionary tracks for stellar
masses of 1.0, 0.95, 0.86 and 0.8 \Msun\ adapted from
\citet{Demarque2004}. These suggest that WASP-21 is close to, or is
already in the hydrogen-shell burning phase and hence is evolving off
the main-sequence. This implies that the assumption of a main sequence
mass-radius relationship in the original analysis of
\citet{Bouchy2010} is faulty.
\begin{figure}
\centering
\includegraphics[width=\columnwidth]{isocronesmass.ps}
\caption{ Isochrone models (solid lines) from \citet{Demarque2004}
for WASP-21 using {[Fe/H]} =$-$0.47 and {[M/H]} =$-$0.4 from
\citet{Bouchy2010}. The age in Gyr is marked in the left of the
respective model. We also show the mass tracks (dashed lines) for
a stellar mass of $1.0, 0.9, 0.86$ and $0.8 \Msun$. We overplot
the $T_{eff} = 5800$ value adapted from \citet{Bouchy2010} and the
new $( \rho_*/ \rho_{\odot})^{-1/3} $. }
\label{isocrones}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=\columnwidth]{masstrack.ps}
\caption{Evolutionary mass tracks from \citet{Demarque2004} for the
same stellar parameters as Fig.~\ref{isocrones} from left to right
for stellar masses of 1.0, 0.95, 0.86 and 0.8 \Msun\ . WASP-21
position is in the 0.86 \Msun\ evolutionary mass track for which
we also show the 8, 10, 12, and 14 Gyr points.}
\label{masstrack}
\end{figure}
There is a significant difference between the mass derived from
evolutionary models, $M_*=0.86 \pm 0.04 \Msun$, and the mass derived
from the empirical calibration, $M_* = 1.02 \pm 0.05 \Msun$. In the
past, the \citet{Torres2010} calibration was found to be in agreement
and a more straight-forward alternative to the stellar models
\citep{Torres2010,Enoch2010}. Moreover, it has the advantage that it
can be directly included in a transit fitting procedure
\citep{Enoch2010}. However, recently the same discrepancy between
empirical and isochrone masses was also found for WASP-37
\citep{Simpson2011} and WASP-39 \citep{Faedi2011}. The
\citet{Torres2010} eclipsing binaries sample used for calibrating
their relationship does not contain many low-metallicity systems, in
particular in the low-mass regime. Therefore, this suggests that the
\citet{Torres2010} calibration might not hold for metal poor stars
specially in the low mass regime. For these reasons, for WASP-21, we
favour the lower mass derived from the evolution models. For a stellar
mass of $M_*=0.86 \pm 0.04 \Msun$ we present the stellar and planetary
radii for the WASP-21 system in Table~\ref{mcmc2} along with the
1$\sigma$ uncertainties derived from the MCMC analysis.
To summarise, we derive a lower stellar mass, $0.86 \pm 0.04 \Msun$,
and a lower planetary mass, $0.27 \pm 0.01 \hbox{$\mathrm{M}_{\rm Jup}$}$. We estimate the
inclination of the orbit to be $87.3 \pm 0.3$ degrees. The radius of
the star is found to be $1.10 \pm 0.03\,$ \Rsun\ and the planet radius
is $1.14 \pm 0.04\,$ \Rjup\ , yielding a planetary density of $ 0.18
\pm 0.02\, \rho_J\ $.
\begin{table}
\centering
\caption{WASP-21 system stellar and planetary parameters derived using either the empirical calibration of \citet{Torres2010} or the YY stellar models \citep{Demarque2004}. }
\label{mcmc2}
\begin{tabular}{lccccl}
\hline
\hline
&Torres models & YY models \\
\hline
Stellar mass $ M_* $ [\Msun] & $ 1.02 \pm 0.05 $ & $ 0.86 \pm 0.04 $ \\
Stellar radius $ R_* $ $ R_* $ [\Rsun] &$ 1.161^{+0.037}_{-0.024}$ & $ 1.097^{+0.035}_{-0.022}$\\
Stellar surface gravity $ \log g_* $ [cgs] & $ 4.32 \pm 0.02 $ & $ 4.29 \pm 0.02 $ \\
Orbital semimajor axis $ a $ [AU] & $ 0.052 \pm 0.001 $ & $ 0.0494 \pm 0.0009 $ \\
& & & \\
Planet mass $ M_p $ [\hbox{$\mathrm{M}_{\rm Jup}$}] & $ 0.30 \pm 0.01 $ & $ 0.27 \pm 0.01 $ \\
Planet radius $ R_p $ [\Rjup] & $ 1.210^{+0.048}_{-0.032}$ & $ 1.143^{+0.045}_{-0.030}$ \\
Planet density $ \rho_p $ [$\rho_J$] & $ 0.171^{+0.014}_{-0.018}$ & $ 0.181^{+0.015}_{-0.020}$ \\
Planet surface gravity $ \log g_P $ [cgs] & $ 2.71 \pm 0.02 $ & $ 2.71 \pm 0.02 $ \\
\hline
\end{tabular}
\end{table}
\subsection{Eccentricity}
\citet{Bouchy2010} found the eccentricity to be statistically
indistinguishable from zero, i.e. the $\chi^2$ does not
significantly improve when adding the two additional parameters to
the circular model. For these cases, allowing the eccentricity to
float tends to overestimate the eccentricity \citep{Lucy1971}.
Hence, \citet{Bouchy2010} adopted a circular orbit. Assuming a tidal
dissipation parameter between $10^5$ and $10^6$, the circularisation
timescale for WASP-21b is approximately between $0.017$ and
$0.17\,$Gyr, respectively. Since this is much shorter than the
derived age for the system we expect a circular orbit. However, if
the orbit not circular assuming a zero eccentricity results in
underestimated uncertainties. Therefore, it is interesting to
investigate the effect of a small non-zero eccentricity in the
system parameters and their uncertainties. As an example, we assume
an eccentricity of $0.04\pm 0.04$ which is consistent with the
discovery paper. We repeated the MCMC procedure allowing the
eccentricity to float. Because the transit light curves do not
constrain the eccentricity, we include a prior on the eccentricity
of the form:
\begin{equation}
\frac{(ecc-ecc_{0})^2}{\sigma^2_{ecc}},
\end{equation}
where we assume $ecc_{0}=\sigma_{ecc}=0.04$. This prior is added to
equation 1 at each step of the chain. From the posterior
eccentricity distribution we obtain an eccentricity of $0.038\pm
0.036$ which is close to the input value.
The maximum effect of the eccentricity upon the derived
parameters corresponds to the case where the transit occurs close to
periastron ($\omega=90\hbox{$^\circ$})$ or apastron ($\omega= - 90\hbox{$^\circ$})$.
Hence, in order to investigate the maximum deviation from a circular
orbit we assume $\omega = 90 \hbox{$^\circ$}$. For this particular case by
assuming a circular orbit we would be overestimating $a/R_*$, $inc$
and $\rho_*$, and underestimating the stellar and planetary masses
and radii. The opposite would have happen if we have assumed $\omega
= -90 \hbox{$^\circ$}$.
The derived eccentric solution is within one sigma of the
circular solution and the uncertainties of $a/R_*$, $I$ and $\rho_*$
are $\sim 30\%$ larger. This results in an increased uncertainty of
$\sim 30\%$ on the radii and $\sim 20\%$ on the masses. Hence, we
conclude that if the orbit eccentricity is $ < 0.038$ the system
parameters would be within $\sim 1.3 \sigma$ of the values given in
Tables 2 and 3.
\section{Discussion and Conclusion}
We have presented two high quality transit light curves of WASP-21b
taken with RISE. Together with the previous RISE partial transit,
these were fitted with an MCMC procedure to update the parameters of
the system. We have been conservative in our error estimates by
scaling the $\chi^2$ and by including time correlated noise in our
analysis.
The derived stellar density $ \rho_* = 0.65 \pm 0.05 \rho_{\odot} $
and the estimated age for the system, $12 \pm 2\,$Gyr, suggest that
WASP-21 is in the process of evolving off the main sequence.
Therefore, the main-sequence mass-radius relation assumed for WASP-21
in the discovery paper was invalid which led to a significant
overestimation of the stellar density, thus affecting the derived
planetary properties. Using the stellar models of
\citet{Demarque2004}, we derived a significantly lower stellar, $M_*=
0.86 \pm 0.03 \Msun$, and planetary mass, $ M_p = 0.27 \pm 0.01
\hbox{$\mathrm{M}_{\rm Jup}$}$. This lower host star mass somewhat compensates the lower
stellar density which results in a stellar radius which is within
$1\sigma$ of the one presented by \citet{Bouchy2010}.
We obtained a slightly larger planetary radius, $R_p= 1.14 \pm 0.04\,$
\Rjup, for WASP-21b than previously reported. \citet{Fortney2007}
hydrogen/helium coreless models predict a radius of $\sim 1.06$ \Rsun\
which is consistent within $2\sigma$ with our estimated radius without
the need for any extra heating mechanism. Following
\citet{Laughlin2011} we compute a radius anomaly, $\Re=0.09$, for
WASP-21b. This supports the correlation reported by
\citet{Laughlin2011}, i.e. $\Re =T_{equ}^{1.4}$, where $T_{equ}$ is
the equilibrium temperature of the planet, which is $\sim 1320\,$K for
WASP-21b. \citet{Bouchy2010} argued that the density of WASP-21b
strengthens the correlation between planetary density and host star
metallicity for hot Saturns \citep{Guillot2006}. With the addition of
the latest Saturn-mass planet discoveries (e.g. WASP-39,
\citealt{Faedi2011}; WASP-40, \citealt{Anderson2011}) this correlation
appears weaker. However, if we scale for the equilibrium temperature
with, for example $\Re$, the correlation with metallicity is still
strong (see Figure 6 in \citealt{Faedi2011}). Moreover, the
correlation also holds for the more massive planets (see Figure 3 in
\citealt{Laughlin2011}).
Exoplanet transit light curves are often affected by systematic noise
that can in some cases dominate the photometric noise. Therefore, it
is important to minimise the sources of systematic noise. In Section
3.1, we show an example of systematic noise present in our exoplanet
transit observations and suggest that the first step to decrease this
noise is to maintain the star in the same pixel position in the CCD
during the observations. We confirm that defocused observations can
also help decreasing systematic noise, as well decreasing deadtime
losses and hence improving the signal-to-noise
\citep{Southworth2009}. The systematic noise in our observations was
due to the variation of the stellar position across the CCD.
\section{Acknowledgements}
FPK is grateful to AWE Aldermaston for the award of a William Penny
Fellowship. The RISE instrument mounted at the Liverpool Telescope was
designed and built with resources made available from Queen's
University Belfast, Liverpool John Moores University and the
University of Manchester. The Liverpool Telescope is operated on the
island of La Palma by Liverpool John Moores University in the Spanish
Observatorio del Roque de los Muchachos of the Instituto de
Astrofisica de Canarias with financial support from the UK Science and
Technology Facilities Council. We thank Tom Marsh for the use of the
ULTRACAM pipeline. SCCB is grateful to Catherine Walsh for
proofreading this paper and to Yilen Gómez Maqueo Chew for useful
comments.
\bibliographystyle{mn2e}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,187
|
Frank Aaron (* 30. März 1920) ist ein ehemaliger britischer Langstreckenläufer.
Bei den Leichtathletik-Europameisterschaften 1950 in Brüssel wurde er Vierter über 10.000 m mit seiner persönlichen Bestzeit von 30:31,6 min.
1951 gewann er Silber beim Cross der Nationen.
Einmal wurde er englischer Meister über sechs Meilen (1950) und dreimal im Crosslauf (1949–1951).
Weblinks
10.000-Meter-Läufer (Vereinigtes Königreich)
Crossläufer (Vereinigtes Königreich)
Brite
Geboren 1920
Mann
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,247
|
#ifndef _ITEM_H
#define _ITEM_H
/*
Item -
An Item is a physical/graphical object in the game world
It may have several triggers tied to it for Contact or Use by
Players, Characters, or other Objects (items?)
Examples: Soccer Ball, Crate, Tree, Chest, etc
*/
#include "../../engine/entity/entity.h"
#include "../../engine/entity/loggableObj.h"
#include "../../engine/entity/physicalObj.h"
#include "../../engine/entity/graphicalObj.h"
#include "../../engine/graphics/texture.h"
#define ITEM_OBJ_TYPE 2
class CItem: public CEntity, public IGraphicalObj, public IPhysicalObj
{
public:
CItem();
virtual ~CItem();
virtual void Update( float dt );
virtual void Render();
protected:
CTexture* mTexture;
};
#endif
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,482
|
Good Environmental Choice Australia Limited (GECA) manages a Type 1 Ecolabel program in accordance to ISO 14024 "Environmental Labels and Declarations"and is the owner of the Environmental Choice Australia Ecolabel.
Modern ceiling lighting can be used to create and subtly demarcate any living space. View our range at Ke-Zu Furniture, online or at our Sydney showroom.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,114
|
{"url":"https:\/\/www.gradesaver.com\/textbooks\/math\/precalculus\/precalculus-10th-edition\/chapter-1-graphs-1-3-lines-1-3-assess-your-understanding-page-31\/81","text":"## Precalculus (10th Edition)\n\nPublished by Pearson\n\n# Chapter 1 - Graphs - 1.3 Lines - 1.3 Assess Your Understanding - Page 31: 81\n\n#### Answer\n\nThe slope is $m=\\frac{2}{3}$. The $y$-intercept is $-2$. The graph is attached below.\n\n#### Work Step by Step\n\nIsolate $y$ to obtain: $2x-3y=6$ $-3y=-2x+6$ $y=\\frac{2}{3}x-2$ In a line's equation in slope-intercept form, which is $y=mx+b$, the constant $b$ is the $y$-intercept while the coefficient of $x$, which is $m$, is the slope. Thus, in $y=\\frac{2}{3}x-2$ we have: $y$-intercept is $-2$ slope=$m=\\frac{2}{3}$. Plot the point $(0, -2)$. The slope is $\\frac{2}{3}$ so from $(0, 2)$, move $3$ units to the right and $2$ units upward to have $(3, 0)$. Connect the points using a straight line.\n\nAfter you claim an answer you\u2019ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide\u00a0feedback.","date":"2021-10-25 16:59:46","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7127389907836914, \"perplexity\": 531.3784403631614}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-43\/segments\/1634323587719.64\/warc\/CC-MAIN-20211025154225-20211025184225-00513.warc.gz\"}"}
| null | null |
import { TestOrder } from './enum-testorder';
import { Category } from '../questions/category.model';
import { BrowserTolerance } from './enum-browsertolerance';
import { AllowTestResume } from './enum-allowtestresume';
import { QuestionStatus, AnswerStatus } from '../conduct/question_status.enum';
import { QuestionDisplay } from '../questions/question-display';
import { TestIPAddress } from './test-IPAdddress';
import { CodeSnippetTestCasesDetails } from '../reports/code-snippet-test-cases-details.model';
import { TestCodeSolutionDetails } from '../reports/test-code-solution-details.model';
import { ProgrammingLanguage } from '../reports/programminglanguage.enum';
export class Test {
id: number;
testName: string;
link: string;
startDate: Date | string;
endDate: Date | string;
duration: number;
warningTime: number;
focusLostTime: number;
warningMessage: string;
correctMarks: string;
incorrectMarks: string;
browserTolerance: number;
createdDateTime: Date;
questionOrder: TestOrder;
optionOrder: TestOrder;
allowTestResume: AllowTestResume;
categoryAcList: Category[] = [];
testIpAddress: TestIPAddress[] = [];
isEditTestEnabled: boolean;
isQuestionMissing: boolean;
public isPaused: boolean;
public isLaunched: boolean;
numberOfTestAttendees: number;
numberOfTestSections: number;
numberOfTestQuestions: number;
testCopiedNumber: number;
constructor() {
this.isLaunched = false;
this.isPaused = false;
}
}
export class TestCategory {
public id: number;
public categoryId: number;
public testId: number;
public isSelect: boolean;
}
export class TestQuestion {
public id: number;
public testId: number;
public questionId: number;
public isSelect: boolean;
public question: QuestionDisplay;
public questionStatus: QuestionStatus;
public answerStatus: AnswerStatus;
public codeSnippetQuestionTestCasesDetails: CodeSnippetTestCasesDetails[];
public testCodeSolutionDetails: TestCodeSolutionDetails;
public language: ProgrammingLanguage;
public numberOfSuccessfulAttemptsByAttendee: number;
public totalNumberOfAttemptsMadeByAttendee: number;
public scoreOfCodeSnippetQuestion: string;
public compilationStatus: string;
public codeSolution: string;
public codeToDisplay: string;
public isCodeSolutionDetailsVisible: boolean;
public isCodeSnippetTestCaseDetailsVisible: boolean;
public isCompilationStatusVisible: boolean;
constructor() {
this.codeSnippetQuestionTestCasesDetails = new Array<CodeSnippetTestCasesDetails>();
this.testCodeSolutionDetails = new TestCodeSolutionDetails();
}
}
export class TestQuestionAC {
public id: number;
public categoryID: number;
public isSelect: boolean;
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,123
|
SU Engineering students challenged to think wider
Author: Frieda le Roux
It's not every day that engineering students are challenged with concepts more at home in Social Sciences. But with the Eng. Curriculum of the SU it is the case with a specially developed module, namely Complementary Studies.
And what initially sounds strange is exactly in line with Stellenbosch University's (SU) goal to give students an education and exposure that is as wide as possible – and relevant in our society with its unique challenges.
When Dr Leslie van Rooi, Senior Director of Social Impact and Transformation, and Monica du Toit, Head of the Transformation Office, take their place in front of the class, it is to challenge the engineering students to think even wider than what their lecturers expect from them. The course aims to make students more aware of their role in the wider society – as involved citizens but also as trained engineers who can make a difference with their unique skills.
"In our discussions in class we grapple with the realities on campus and in social media and it is challenging, especially at the start, but we do get to new possibilities," says Monica. "I like the fact that engineers in essence are 'doers'." She says the students have again convinced her that one of our country's greatest assets lies in its engineering students – especially when it comes to the much needed changes in our society. "They have skills to live close to the experiences of regular people while, at the same time, think innovatively about solutions."
For their final assignment the students were divided into groups and asked to tackle different social challenges by looking at it afresh. In this way one group decided to use 'big data' to get access to information that will help universal access for the users of buildings and amenities. With a special app named Enable, users can indicate places where they have difficulties with access. The information on the app, which makes use of GPS coordinates, is in the public domain where all – from building owner to architect, engineer and others with disabilities – can access it and learn from it.
Another group thought about the way engineers can help in creating better integrated communities. And, they found, engineers do not always think of the bigger issue but would rather solve a specific problem as cost-effectively and efficiently as possible. But during their discussions as part of the Complimentary Studies course they realised that the focus – also for engineers – should always include the human aspect and shouldn't only be about technical skills.
Prof Anton Basson, Vice-dean: Teaching & Quality Assurance says students in the Faculty of Engineering are exposed to contemporary societal issues in their third and fourth years. "These modules, which all engineering students must do, expose our graduates to the complex social issues that they will encounter when they enter engineering practice. Our graduates often fulfil leadership roles and these modules raise their awareness of the complexities of gender, racial, access and social responsibilities that challenge leaders."
Leslie, who has been presenting the course for four years, says it is part of the University's goals to equip students with a complete set of skills. "For the University it is important to equip students with skills and the possibility to make an impact on the various levels of our society. This specially developed curricular module gives all engineering students the opportunity to do exactly that." The Faculty of Engineering, together with the Engineering Council of SA, gives the necessary guidance in this regard.
Engineering; Social Impact
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,429
|
My name is Mateusz. I am 22 years old and currently studying at Erasmus University Rotterdam in the Netherlands. I'm following the master's programme Econometrics and Management Science at Erasmus School of Economics. I'm half Polish and half Dutch, so technically I'm not an international student, but I can tell you a lot about what it is like to study here as an international student.
I began my bachelor's degree Econometrics & Operations Research at Erasmus University some years ago. Choosing my program wasn't difficult. I was good at mathematics at my high school but I did not want to end up only knowing and grasping everything about mathematics theoretically, I also wanted to put it into practice.
Since economics was also a field I was interested in, my teacher recommended that I combine these two subjects and study econometrics. Basically, econometrics is the science where you learn mathematics and statistical methods and learn how to apply this into economics and the real world.
Choosing a university was a much more difficult thing to do. Econometrics is a really interesting but not especially known science. The Netherlands is currently the only country with universities offering econometrics at a bachelor level. This is a great reason to study in the Netherlands if you are interested in combining mathematics and statistics with real world applications!
However, there are a lot of universities in the Netherlands and all universities are highly-ranked. Thus, it is not an easy choice to decide where to study, even more so if you are coming from abroad. I eventually chose to study at Erasmus University, and specifically at Erasmus School of Economics since it is a practical school with top-notch, highly ranked research and academics.
Not only the high level theory and internationally-known professors are a draw. The focus at Erasmus School of Economics is also on the practical. Studying in Rotterdam specifically ensures that you'll be practically educated to tackle problems that companies and institutions face in the real world. The school has strong ties to industry with many companies in a stone's throw of the campus. That really stood out to me.
When you're choosing a study abroad destination, my advice is to stay true to yourself and choose a destination that you really have a strong connection with or good feeling about, rather than to be told that it is beautiful or exotic.
I can recommend the Netherlands as a study destination because people here are really open and adapted to international students. Almost everyone speaks English as their second language, they're very open to and interested in people from other countries and cultures, and don't mind making jokes or exchanging funny stories even if they don't know you well.
Finally, when talking about Rotterdam as a destination, I cannot stress enough how internationally-orientated the Erasmus School of economics is in its mindset- not just the institution but also students and professors. Even as someone who did my studies in Dutch. I have a huge number of international friends and have met a tremendous number of international students. You have a lot of ways to feel at home here, but that is something I will talk about in my next blog post.
For now, it is also good to remember that, when studying abroad in Rotterdam, you will soon adapt to the Dutch saying 'Niet lullen, maar poetsen' which translates broadly as ''Don't just say it, also do it.'' In Rotterdam, you will make it happen!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,050
|
package org.kuali.mobility.grades.service;
import java.util.Date;
import java.util.List;
import static junit.framework.Assert.*;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.kuali.mobility.grades.entity.ModuleResults;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.beans.factory.annotation.Qualifier;
import org.springframework.test.annotation.DirtiesContext;
import org.springframework.test.context.ContextConfiguration;
import org.springframework.test.context.junit4.SpringJUnit4ClassRunner;
/**
* Unit test for the Grades Service
*
* @author Kuali Mobility Team <mobility.collab@kuali.org>
*/
@RunWith(SpringJUnit4ClassRunner.class)
@DirtiesContext
@ContextConfiguration("classpath:/GradesSpringBeans.xml")
public class GradesServiceImplTest {
/**
* A reference to the <code>GradesService</code>
*/
@Autowired
@Qualifier("gradesService")
private GradesService gradesService;
/**
* This test will call the service and retrieve a pre-configured set of results.
* This test will simply check that the data is in the order we expected, and also
* provide a sanity check that the services still properly call the DAO,
*/
@Test
public void test() {
List<ModuleResults> results = gradesService.getResults(new Date(), new Date(), "test", "en");
String str;
// Check that we have the expected first mark
str = results.get(0).getExamMark();
assertEquals("76%", str);
// Check the participationMarkComment of the second result
str = results.get(1).getParticipationMarkComment();
assertEquals("Distinction", str);
// Check the module code of the third result
str = results.get(2).getModuleName();
assertEquals("IOPS 1 21 INTRO HUM GEO", str);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,231
|
{"url":"https:\/\/jm.edugain.com\/questions\/Mark-has-zero-marbles-He-wants-to-put-them-in-6-boxes-How-many-marbles-are-there-in-each-box","text":"### Mark has zero marbles. He wants to put them in 6 boxes. How many marbles are there in each box?\n\n0\n\nStep by Step Explanation:\n1. According to the question, Mark has zero marbles and wants to put them in 6 boxes.\n2. Since there are zero marbles, all the boxes remain empty.\n3. Let us prove the above conclusion mathematically.\nAs we know that any number multiplied by 0 gives the result as 0.\ni.e., 6 \u00d7 0 = 0,\nwhich can be written as: 0 \u00f7 6 = 0\nSo, when '0' is divided by a non-zero number, the quotient comes out to be 0.\n4. Therefore, there are zero marbles in each box.","date":"2021-09-22 17:54:50","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8259274363517761, \"perplexity\": 505.7184053757182}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-39\/segments\/1631780057371.69\/warc\/CC-MAIN-20210922163121-20210922193121-00098.warc.gz\"}"}
| null | null |
#food porn
Mine Tonight
June 4, 2017 June 7, 2017 avenger-nerd-mom3 Comments
Warnings: Language, Adult Situations, food porn, NSFW, nipple play, oral sex, face fuck, delayed gratification, edging, denied orgasm
Summary: Chris takes Thalia on a date since their last meeting in public lead to a fight. This time he intends to make it perfect.
The next night, Chris can't believe how beautiful she looks in the wrap style red dress with a simple gold chain at her neck. The little pendant falls just so in her bountiful cleavage and he can't wait to take a closer look at it later. The heels honestly make her the same height as him, and he's so honored to have her on his arm as the whole place seems to watch them walk to their table. Her shoulders are back and her head is held high, but with his hand on the small of her back, he can feel her trembling, a total ball of nerves. Holding her seat out for her, as she sits, he can't help the small whistle of appreciation that escapes his lips. "Christopher! Behave yourself," she admonishes him, her smile growing by the minute.
Pushing her chair in, he leans down over her shoulder, whispering in her ear. "So does that dress have one or two ties holding it closed?"
"Stop it!" She blushes as she swats his arm.
Taking the seat closest to her, she eyes him warily. "I believe it's proper etiquette to sit across from the lady at dinner so you can admire her. You're in the wrong seat, sir."
She halts her teasing when the waiter comes to take their drink orders. Chris watches her bite her lip as his hand runs up her thigh, holding back his own laughter when she clamps her legs shut tight.
Pushing his hand away when the waiter leaves, she rises proper again. "Chris, stop. I mean," she shakes her head and blushes. "Chris. I really don't want you to stop, but," she sighs and scrunches her face up the way she sometimes does when she doesn't want to say what she needs to say that Chris finds so adorable.
"When you do that, and scrunch your nose and pout out your lips, it's adorable," Chris tells her.
Blushing, the pink rose brings out her freckles and he leans back in his seat, placing his hands on the table. "What did you want to tell me, Niña?
She tilts her head back and pushes her hair from her face. This is going to be a long night if he can't stop his thoughts or his hands. "Babe, I rarely get taken on 'fancy' dates like this." She gestures to the menu, "I don't even know what half this stuff is; it's exciting and makes me feel special. I want to enjoy it. I've worked in restaurants before." Her hand rests on his, laying on the table. "The staff always knows when the woman is getting groped under the table; I don't want to be that gi- that woman tonight. Please." She licks her lips nervously waiting for his response.
Nodding his head, he takes a drink of water with his free hand. With his other, he turns it slowly and slides his fingers to entwine with hers. "I understand; tonight is about you." Reaching across he tucks her hair behind her ear and says, "Anything you want, just ask. But I can't control my thoughts and when I get you home with me tonight, you're all mine."
"I wouldn't have it any other way," she grins.
Scouring the menu, they both make their decisions and laugh at the combination of fruits and vegetables in the salad. She tears off a small bite of the seasoned garlic bread and her absolute joy is heavenly. "Carbs," she whispers. "Oh, my God…"
"What are talking about?" Laughing as he dips his bread in the olive oil. "So we don't 'date' but we eat together a lot. I've seen you eat pizza, pancakes and hamburgers and those-"
"You know, as an adult male, those are really unhealthy food options. I should feed you better," she giggles. "A Puerto Rican girl this size? She can cook! Arroz con gandules…. Oh, as cold as it's been I should fix asopao. No. No, you're Italian, I should make pastelón! You'll never want regular lasagna again."
Her eyes gleam and dance, her accent growing thick as the foreign words fall from her lips. "Yes to all of it, whatever you want, considering I have no idea what any of that is, but I love hearing you say it. Growing up, did your family visit Puerto Rico?"
Thalia tells him her adventures as a child visiting her father's family and running through the fields of sugarcane with her cousins who still lived on the island. She pauses long enough for the waiter to serve their meals and she turns her questions to him, asking about his travels and journeys as a historian. Through discussion they even discover at one point they had both been swimming in the caves at Grand Cenote on the Yucatan peninsula within days of one another and had been to many 'off the map' historic stops, leaving them to wonder if they had crossed paths years before, when Thalia was still an undergrad.
The conversation flows freely and lightly. Chris watches in awe as he can see her brain jump from one topic to the next and she shares so many random facts about the places they've both seen, things he didn't even know. Her skin glows in the candle light on the table and as she relaxes, her posture falls a bit, allowing the front of her dress to gape just enough to provide him with a lovely view. Occasionally she rubs her leg against his and smiles innocently. But the damn little lick of her lip lets him know she knows exactly what she's doing.
When the meal winds down, Chris wipes his mouth on his napkin and lays it next to his plate. "So Miss Teaching Assistant; what's my grade? Did I pass our first date? You're not still mad at me like the other night? I wasn't sure who to worry more about- you killing me or Jim?"
A quick pink rises up her chest and across her cheeks. "Sorry about that little display. Oh, but thank you by the way for the heater; you didn't have to do that; you never answered my text? And I really hate presents, just so you know." Tilting her head from side to side, she rubs her lips together. "I was having a really bad day, but you know about that? You reported the creep? I heard about that…" Her eyes drop briefly and the shade of pink grows darker. "I spoke in anger, and I said things I didn't me-"
Putting his hand over hers, "Thalia, you were right. You're too amazing to keep hidden away. I should be taking you out, enjoying things with you. And I sure as hell hope I've never made you feel the way you said. If I did, I'm so, so sorry, and I will do everything I can to fix it. If I've ever made you feel not important or special to me-" Stopping himself he sighs. He scoffs. "Wanna hear something funny? Everyone knew how crazy I was about my wife. Oh, I know girls took my class just to get a look at me, and there were jokes about trading favors for As, but none of that was true. Thalia, you are the first woman to catch my eye since my divorce." He grasps her hands and moves closer to her, looking into her dark eyes. "I wanna make sure you hear this… When I make a commitment, I'm there. I'm all in. I may have sown a few wild oats after the divorce; I'd been with the same woman since I was twenty- two. But revenge sex wasn't all it was cracked up to be and at my age now, I grew tired of it. I hadn't been with anyone in nearly five months till you came along. You are important to me. And too bad, there will be more gifts because I like getting little things for people I care about. It's no different than you making me dinner or bringing me breakfast on days we work together." He winks at her, removing his glasses and putting them in his coat pocket. "And if I haven't said it before, or made you feel it, shame on me. I'd like to make that up to you tonight, if you'll let me."
Her eyes are wide with his confession and he wonders what she thinks. As he's been talking her mouth has slowly dropped open, breathing as though she can't get enough air and her hands are warm in his. "Yes, Chris, I'd like that. Very much." Dropping her eyes, she blushes and says quietly, "You're important to me too. I like the time we share together. It's… different."
Briefly he wonders if she means 'different' as in how the other man in her life treats her,or 'different' in general to past experiences. For a split second, he considers asking her. But this isn't the time or place and secretly he prides himself on keeping his jealousy to himself. He has to make himself look good and doesn't want to tip his hand, and show her how damn envious he really is. "I like hearing that Thalia… I hope you don't mind, but I have other plans for us tonight and I'd like-"
"No dessert? Chris, you can't take a fat girl to a fancy restaurant like this and not have dessert," she laughs, tugging on his arm playfully.
"You're not fat, dammit, and I know you're saying that now as a joke, but I really wish you wouldn't." His hand brushes against her thigh under the table and he feels her tremble as her breath catches and he moves closer to her. Sweeping her hair over her shoulder, his fingers drag gently through her long tresses and her tongue flicks out to lick her lips in desire. "I like when you tremble at my touch, Niña. It shows me you're mine and you want me. Every man in this room has their eyes on you; they want you." Leaning in the smell of sweet orchids on her skin fills his nose and heightens his senses. "They are jealous of me and envision things we might do, and wish it was them. And the women are envious of your beautiful hair and glowing skin. You look positively radiant tonight darling. Red is definitely your color."
Tears rim her eyes and she blinks them away. "I'm beautiful to you, Chris. Thank you, but not everyone else sees me that way."
He wonders again about her other companion and slowly pushes his chair back, signaling to the waiter. "I'm sure someone does," his tone hardens for a moment. To the waiter, he simply hands over his credit card and asks for the rest of their meal to be boxed up. Another server arrives to carry things away. "I have a special treat planned for later, Niña. You'll get more than one type of sweet treat later, I promise."
Giggling she shakes her head and pushes her chair back. Chris rises as she does and she excuses herself to the ladies room. Eyeing her as she saunters away, he worships the sway of her hips, the form fitting dress showing off every curve.
Driving back into town, holding hands tightly in her lap, she watches the city pass through the window. Taking his eyes from the road, he views her strong profile and his heart flips. Turning to him, she winks at catching his gaze. "Where are we? I don't recognize this part of town."
"I wanted to be away from campus; found a duplex housing unit that had a playground for Avery."
"Oh," she whispers quietly. "So we're going back to your house?"
Sensing her hesitancy, he provides an out. "We don't have to; we can go back to your place… Or we can get a hotel?"
"No. No hotels." She fidgets in her seat, looking out the window as she gathers her thoughts. In her quiet husky tone she admits, "It just feels really… intimate. That's your home, with your daughter. Are you sure?"
Raising their hands to his lips, he turns the car down the dark road and gently kisses the back of her hand. Sighing happily, he tells her. "I appreciate your concern, Thalia, but I'm a grown man. I wouldn't bring you here if I didn't want you to be a part of my life, whatever part of it you want. Avery's not here, of course, and to be honest, I haven't decided yet about you two meeting. She'd love you, and no one can help but fall in love with her. It's just-"
"If she's half as amazing as her dad, I'm sure I would adore her. Babe, you don't have to explain it to me." Her eyes glisten in the dark. "Kid of divorce. I get it. It takes a special person to take on additional baggage. Some days I feel like just a kid myself…" Her voice changes. "And no child should be subjected to growing attached to new people who appear in their lives just as suddenly as they leave. You'll know when and if it's right for us to meet…"
In the driveway, he places the car in park. "Thalia Bareo, I don't think there's anything you can't handle. You're so giving and flexible, and you-"
Her giggles stop him and he starts laughing too. "And you have a very filthy mind, and I adore that."
Reaching over to her in the dark he wraps his fingers in her thick mass of curls and she doesn't even wait for him to pull her close. Their lips collide in the middle, meeting over the damn center console. Her hands grip his tie and yanks him to her as their mouths open and their tongues caress against one another and she releases a lusty sigh. "Been waiting on you all night to do that; took you long enough," she teases.
His hands reach into her warm winter coat and he pulls away at the top of her dress, groping at her full breast under the satin and lace. Loudly she moans against his mouth and his cock stirs. God, he wants her. He wastes no time and pushes her back, leaning over the center divide between them to rest against her as he kneads at the tender flesh peeking out of her bra. Shoving the fabric aside he exposes one of her dark nipples to the cold air and pinches it between his fingers. Greedy, his kisses grow with intensity as he alternates between kissing her deeply and nipping at her swollen lips. While their mouths reacquaint with one another, Thalia loosens his tie and makes haste with the buttons on his crisp white dress shirt.
Cold from the winter air her hands dart across his skin in feather light touches, running her fingertips through the soft scattering of hair on his chest. Sliding his mouth down over her jawline, he licks a stripe of wet along her exposed graceful neck before nibbling back up to her ear. She tilts her head back against the cold glass window and she finds her words again. "We're not fourteen and you're gonna hurt yourself there, big boy," she teases as her hands slide down to rest on his waist, his hard cock pressing into her belly. "And it's cold as hell out here. I'm assuming your bed is warmer?"
Climbing off her, he collapses into his seat. "I knew I should have asked for the mini-van in the divorce," he chuckles, squeezing her fingertips.
She simply rolls her eyes. "I'm a lady, kind sir. I don't shag in a vehicle," she teases with the mocked air of royalty.
"Fine, Princess. Stay seated, let me help you. The driveway is slick." Looking over to her beautifully aroused body he reaches across and pulls her dress back in place over her exposed breasts. "Wrap back up. It's cold, and I don't want you getting sick."
Getting out of the car and watching for ice himself, he mumbles aloud. "'I don't want you getting sick.' Hell, I'm not her father. That's just weird, Chris, get it together."
A sound makes him look at her through the front window and he sees her hiding behind her hand, her eyes wide. He raises his arms to shrug a silent question and slips on the ice, catching himself on the grill of the car. "Fuck," he mutters. "Good going. Throw out your back like on old man before you can even get her in bed, Jesus fuckin'-"
Reaching for her door handle, he laughs when he opens it and she has a small pair of flats sitting in her lap. "Is that why your purse is so big? You had shoes in there? Or is it like a closet door to Narnia," he teases.
Spinning around on her ample bottom, her skirt hikes up, exposing her bare thigh as she swings her feet out to the ground. His eyes follow along the long stretch of her leg as she quickly removes her heels and replaces them with the flats. "A Narnia reference, really? I would have gone with Hermione's handbag myself."
"Who?" he laughs but loses his attention when she runs her hand down her leg to put on the other slipper. Damn. The image of her hands caressing over her dark skin makes him hard. He's been holding his thoughts in all evening and now he can allow them to run free. A release of sorts…
"The heels go back on later," he says gruffly, thankful for the long winter wool coat hiding his growing bulge. "Were you laughing at me?"
Nodding her head vigorously her curls shake rapidly. "I was! I could hear you talking to yourself, you fool." She blushes and her smile grins from ear to ear. "Its sweet you worry about me but please, dear God, don't throw out your back," she giggles again, hiding behind her hand.
Blushing, he whispers, "Damn. You weren't supposed to hear all that."
He holds his arm out to her and she rises carefully, making sure the surface isn't slick. Out of the car, he kisses her again, inhaling her warm breath and then blowing it back out to see the bubble of condensation released into the winter night air. Reaching for the left overs in the back seat, he then holds his arm to her, crooked at the elbow for her to link hers in his. "If we fall, I'm grabbing for the steaks and you have to fend for yourself," she tells him and they both nearly fall on the patch of ice when they throw their heads back in laughter.
Walking up the front sidewalk, he's focused on their steps, so they don't get hurt. She makes him feel young and vibrant, and he almost wishes they were dressed better for an impromptu snowball fight. The idea makes him smile, picturing the two of them rolling around in the snow, trying to shove it down each other's shirts and going inside later for a hot shower to warm up and-
"What the hell!" He's startled by the cold against his cheek and neck.
"You squeal like a girl!" Laughing, she drops his arm and moves quickly ahead, more snow from the bushes in her gloved hands. "Chicago, babe. I know how to pack a snowball." She throws it and hits him squarely in the chest.
"Fuck!" He shouts stalking towards her as she grabs the porch rail and moves away from him. "Tomorrow. Tomorrow we will dress for a proper fight, and I will show you how a Boston boy makes snowballs."
"Boston boys are all talk," she taunts him brushing the snow from his coat while he unlocks the front door.
Her playful smile tugs at his heart but he doesn't want to let his guard down now.
The door unlocked, he quickly snakes his free hand around her waist, pulling her close. "Lucky for you we're all action too," he says, kicking the door open and walking her backwards over the threshold. Closing it against the cold air, he tosses the bag of food on the floor and spins her, pinning her back against the wooden door.
"Oh, really? Next thing you know, you'll be trying to tell me Fenway's better than Wrigley," she taunts, her breathing already labored as he pulls at the buttons on her coat and rocks his body against hers, his mouth devouring her neck.
He bites roughly and she sucks in the air deeply, but no sound is released. Her head falls to the side, silent approval of his force and he bites again, his hands finally inside her coat. He growls against her skin, "I'm going to forget you said that. I'm going to make you forget everything tonight. You're mine tonight, Thalia, do you understand me?"
"All yours, babe." The smile is evident in her voice.
His lips continue to lick and suck her neck, but he doesn't leave any marks. Not yet. His hands rub firmly over the front of her body and he grabs the lapels of her coat, pulling her close and away from the door. Quickly he yanks it off of her and then removes his as her hands grasp at his tie- the tie she gave him, and tugs at the other buttons on his shirt. Pulling harder than intended, she gasps when one little disk flies off and her mouth forms a perfectly round "oh."
"Oh, baby, that's so beautiful when your mouth does that." He runs his fingertip across her glossed lips as she pushes his shirt back over his shoulders, and down his arms. Her tongue teases the rough pad of skin and he dips his finger inside her mouth. Her eyes instantaneously darken and the heated sound she makes is very primal, hitting Chris in his gut. Her lips wrap around his finger and her tongue strokes the length. One hand still claws at his bicep, but the other takes hold of his hand at her mouth. Gripping his wrist tightly, she slides his hand in and out, a promise of what's to happen.
Chris can't take it. His voice is dark and deep. "Fuuuck… On your knees. Do it again."
Stepping back slightly, he makes room for her, staring in disbelief as she tugs free the ties from her dress and it falls open to reveal a cream colored satin lingerie set with black lace. The outline of her new tattoo peeks above the lace and he can't wait to feast his eyes and lips upon the ink stain on her skin. She shimmies out of the covering and drops to her knees, grabbing for his belt buckle and pulling on it roughly. "Beautiful, Thalia. Fucking beautiful," he sighs as her manicured nails tease at his bulge hidden in his dress slacks.
His head drops forward to watch her as she tugs free the belt from his waist. He shrugs the shirt from his shoulders, pulling it loose from the waist of his pants. When she she tilts her head back and their eyes connect he is lost in their chocolate depths. Dark and husky she asks, "Can I touch you?"
Asking permission. Fuck, that's so hot. He simply grunts in response, at a loss for words.
Slowly her hand slides the zipper down, feathering her fingers along the exposed fabric of his boxer briefs. Uncontrollably he jumps at the attention and his cock throbs for more of her touch. He just wants to tell her to hurry up, but she deserves the right to some control as well. She's so damn good at it. His precome seeps through the elastic waistband as her hand reaches in his pants and cups under his balls,the fabric rough against his tight skin. Lifting up from her knees, she brings her mouth to his covered cock and blows warmly across him.
Weakened by the open mouthed kisses she places on the cotton fabric, he clutches his hands at his side, bracing himself not to rush her. Her breath hot, her nose brushes from his base to the tip. Tucking her fingers in the waist of his pants, she pulls both garments down tugging over the curve of his toned ass. As he pops free his hard shaft springs into her face and she instantly takes it into her mouth, pushing his pants down his long legs. Her lips tease just his head and he falls forward reaching out to the closed door to catch himself. Tantalizingly her tongue swirls around the swollen tip as she taps his leg to step out of the clothing. Caressing her hands up the backs of his legs she squeezes and pulls his ass cheeks and his heightened breaths fill the air. "So fucking good, Niña."
She begins to take skillful measures with her tongue and teeth, skimming over his veiny ridges as she sucks him into her mouth. His free hand grasps her hair and pulls it tight and she stops, understanding the unspoken command. Stretched, her mouth still holds him. The longer he makes her wait he can feel her salivating, her mouth literally watering for him.
"Do you want it?" he asks gruffly, tugging her hair.
Her teeth clamp on him, but not too rough and her muffled sound is affirmative. Releasing her pressure, he wraps her hair tighter around his hand and grasps the back of her head, holding her still as he begins to face fuck her, reaching the back of her throat and she takes every inch. Her eyes watering, he slows as her hands slide around the front of his thighs and her thumbs push against his taint. His moans fill the air and his sac tightens. The caresses of her hands and mouth are just perfect and he'd love nothing more than to shoot his load into her mouth. Loosening the grip on the back of her head, her hands continue kneading him and slip around to gently hold his balls, rolling them between her fingers. The other hand grasps his shaft, adding a sinful stroking motion to coincide with the sliding of her hot pink lips.
The vision below him is heaven and he's so close to coming. His hand wraps over the top of hers and he aids with the pulling and tugging, adding extra force. The feeling wells up from his toes but he holds back, not wanting the sensation to end. Tilting her head up her eyes are full of lust as he watches her glide across his cock again.
Thalia pulls off with an echoing smack. Still stroking him, eyes connected with his, her smile grows wide and devilish. Her hands still and she rises up to a standing position, confined in the space between him and the door. She wipes the spit from her lips and flicks her tongue out quickly.
Chris can't believe it. She's carried him to the brink and stopped… He chuckles and bites his lip, continuing to pump on his own. Thalia leans her ass against the door, reaching through the pile of clothes for her discarded heels. She lifts one leg and slides a shoe on as he had requested, then repeats the same with the other, running her hands up her curvy legs, stopping at the top of her thick thighs to adjust her panties.
"I'd stop that if I were you. I didn't say you could come yet," she declares as she walks out from under his arm, the confident clicking of her shoes against the hardwood floors. His silk tie is in her hand and trailing the floor behind her. "You promised dessert. Where's the kitchen?"
Pure evil. Pure sass. Pure Thalia.
Click here to read Chapter 28 Just Desserts
AU Fan Fiction, au fiction, Uncategorized#Adult Situations, #avenger-nerd-mom, #chapter 27, #Chris Evans, #Chris Evans AU fic, #Chris Evans fan fic, #delayed graitification, #denied orgasm, #devikafernando, #edging, #educating thalia, #face fuck, #food porn, #Language, #NSFW, #Oral Sex, #plus size, #Plus size romance, #plus-size OFC, #Tom Hiddleston, #Tom Hiddleston AU fic, #tom hiddleston fan fic, au fiction, collaboration, Nipple Play, professors
Word count: 6000 IT'S LONG BUT IT WAS NECESSARY- SORRY!
Warnings: : Language, Adult Situations, stepfamilies, food porn, drinking, NSFW, fingering, exhibitionism, angst, real life discussions
Summary: Thalia enjoys some time out alone with her stepmother, glad she didn't have to choose between her suitors for Valentine's Day, but missing them.
Tired from the long day of classes, the drive into the city, and a few errands, Thalia hides her yawn as she sits at the table. The two women pour over the menu. The older blonde, polished and sophisticated, taps the table with her perfectly manicured nails. Looking over the top of her reading glasses, she asks the young woman what she plans to order.
"Mmm… I can't decide," Thalia admits. "It all looks so good, my mouth is watering just reading the descriptions." She happily sighs. "I think the fish with a baked potato and the house salad? And their garlic bread is to die for!"
"Thalia, that's a lot of food? You really shouldn't be eating all those carbs. I mean, you look great, but…" begins the older woman.
Thalia takes a deep breath, steeling herself as she carefully places the menu on the table. She's used to her stepmother acting this way, and she knows if they can get through the first hour without killing one another, they'll end up having a great visit together.
Shaking her head slightly, she purrs. "Stacey. I'm never gonna be stick thin, like you. I know my limitations and if I decide to indulge there's always the gym tomorrow. Besides. I can't eat too much tonight. Everyone knows tomorrow is half-price chocolate day!" Smiling sweetly, she pats the woman's hand. "Thanks for always worrying about me, but I know my shape. Big and round. And I'm learning to own it, and appreciate it. I'm actually really healthy right now, Moms."
The woman takes the younger woman's hand and squeezes it. "You do look great. Really happy…"
She pauses their conversation as the waiter takes their order, appearing to be miffed they didn't order the dinner special. Thalia looks to her right and left and it's an endless sea of couples with the steak dinner in front of them.
Picking up the conversation where it left off, her stepmother takes a sip of her wine before asking, "Would this have anything to do with the gentleman visitor at Christmas?"
Smiling when the waiter returns with the bread basket, Thalia takes a piece of the bread and smiles up at her mom secretly. "Maybe, yea, a little." Pursing her lips together, she isn't sure how much to give away. "He's… He's really amazing. So sweet and caring, and he…" She stops and shakes her head, realizing she's talking about both men and almost hating how this sentence ends in her mind. Cuz it's fuckin' true of them both, and for the thousandth time, she can't believe she let this happen… Tearing off a piece of the bread, the warm cheese stretches between the two parts and she dips it in the olive oil before popping it in her mouth, chewing carefully.
"Thalia, spit it out. What are you not telling me?" The woman prods.
"Fuck, Stacey. It's kind of a mess, but it's also really wonderful. And I think…" She props her elbow on the table and rubs her fingertips across her brow with exasperation. "I think he loves me." She shakes her head and laughs. Both of them do, and she knows it, and it's just a fucked up mess she's created. "We haven't said it yet, but I think he loves me. And not 'in spite of' my size, but partly because of it." Trying hard to control the grin on her face, she just can't stop herself. It is Valentine's Day afterall… "He's really just… something else, ya know?"
Stacey stares in disbelief for a moment and then releases a happy squeal. "Oh, my God, honey! That's so wonderful! Do you love him back? Is he cute? Is he still in school? Holy shit, I didn't think you'd spill the goods before the food arrived," she chuckles. "It usually takes a pry-bar to get you to open up. He must really be under your skin."
Taking a drink of her wine, she tilts her head slightly to the side while dozens of images race through her mind of both Chris and Tom. Her smile reaches all the way to her eyes and she feels she's positively glowing in the candlelight. Candles on the table. Romance. Damn. She leans forward and blows it out, the smell of smoke filling the air. "In the best ways possible," she hints.
Their salads arrive and Thalia picks off the tomatoes, laying them to the side. The woman across from her raised her since she was nine and she adores her with all her heart, but she can't bring herself to fully open up with all her sordid secrets. Hell, she tries not to think about it herself. Sometimes, she's afraid if she opens up to someone, things are bound to topple over and smack her in the face. As if this is all some magical fairytale and as soon as she breathes a word about it, the bubble will burst. She doesn't want to be judged – least of all in her own head, and she has a pretty good idea she'll start doing that automatically if she shares the details of her secret little love triangle.
She tries her best to answer the questions without actually confessing anything. Scoffing, she replies quickly. "Yes, he is still in school," grinning at her own little joke. "Devilishly handsome. Glasses." She tilts her head back, eyes closed and euphoric, and shakes her whole body happily. God, she's such a nerd. "He's got the most adorable accent and he gets so excited about learning new things, he's like a puppy sometimes, bouncing all over the place." In her mind she wonders again how two men so different can actually be so much alike…
"Thalia María Bareo! You are in love with him!" Stacey drops her fork and bounces with glee. "I never thought I'd see the day, but you are head over heels for this guy. Oh, man! This is amazing, honey! I'm so happy."
Oh, fuck. She's right. Her step-mother is always right. When the hell did that happen?
She quickly downs the rest of her wine and signals the waiter. "Scotch on the rocks, please."
She fends off a few more of Stacey's questions, answering as obliquely as possible. She doesn't want to give her stepmother any details she can nail down. Their conversation dies off as the food arrives. Thalia turns her focus to her mother and asks the required questions about family and adopts an air of interest as Stacey talks about the convention she is attending in the city. Her thoughts begin to drift to 'kill me now, I'd never survive the business world' when her phone begins to buzz, hopping all over the table. 'God Save the Queen' chimes and Stacey looks confused as to the choice of song.
Wiping her mouth on her napkin, she lays it next to her plate. "Stacey? Can I?" She indicates she wants to check her phone and her stepmother frantically waves her approval, mouthing the words 'is that him?'
"It's a text. He can't hear you." She shakes her head at the woman's flightiness. Thalia unlocks the screen to see a selfie of Tom, presumably laying on his bed, wearing the shirt she gave him and the book she found in an old shop lying on his chest. The message reads I miss you, darling. Have fun with your mother. Can we meet for tea and toast in the Commons in the morning?
She runs her fingers over her lips nervously, thinking of the timing and knowing she has to attend a history department meeting at eleven with Chris to take notes about the upcoming exhibit. Who is she kidding? She's juggled them both this long… Of course, Tom! Can't wait to see you SIR.
Good girl. I can't wait either. I have something I wish to ask you.
Thalia's heart jumps into her throat. At least with Chris, she can see where his thinking goes. He sometimes misses the domestic life, and tries to replicate his favorite parts of it with her…
But Tom? She has no idea where his mind wanders off to sometimes. It's like he's truly foreign to her. Some days it's so frustrating to be with him because his British demeanor can be cold and seem harsh, when she knows that underneath it all, he's just a teddy bear that wants to believe in love again, to have someone to love him. Dear God, please don't let me hurt him… She has no clue what he's thinking; what he might want to ask. She finds it unnerving.
Finishing her drink, she pushes her plate away, her appetite lost. Stacey reads her emotions accurately, and asks, "Honey, if you're so crazy about him, why are you here with me and not with him? It's Valentine's Day?"
Nodding, she turns her hands and examines her nails, smiling absently to the waiter as he carries away her plate. "It is. It's Valentine's Day and that's exactly why I couldn't see him today. It's complicated."
Over the loud music and the roar of the revelers, Thalia yells at her stepmother. "How the hell did you talk me into this?"
"I'm your ride home! Now shut up and have another shot. Live a little!"
Stacey turns back to her friends from the convention and Thalia considers calling a cab back to the hotel. The group of older ladies were enjoying the Anti-Valentine's Party atmosphere of the night club. Although most of them were married, that didn't stop them from flirting, accepting drinks or dancing with the scores of good looking men who stopped by their table. The party scene had never been Thalia's style, but she'd danced and was having fun. However her responsibilities are too great and she's exhausted from her long hours. Sliding down from the barstool she reaches across the table to her stepmother to tell her she's going to leave when a strong arm wraps around her waist. A familiar scent fills her nose and a heat radiates through her. She tries to maintain a straight face. Reaching around her, the solid brick wall of a man standing behind her slams another shot of the dark amber liquid down on the table. Rubbing his cock against her curved ass, he growls in her ear, "Another."
Picking up the tiny glass, she turns to face him. She can't hide her smile at the handsome face that floats before her, a sure sight for sore eyes in a sea of strangers. Shooting it down quickly she hands it back to him with a wink. In her husky timbre she boldly says over the loud raucous crowd, "Four."
With his head thrown back in laughter she pushes past him to the dance floor. She can feel his eyes following every move, knowing her high heels add more of a swish to her ass than usual. Realizing he desires her is so stimulating and builds her confidence. Over her shoulder she sees him visibly panting and trying to discretely adjust the already growing bulge in his pants.
Shaking her head she finds a spot in the middle of the crowded dance floor and begins to sway to the music. As one thumping song blends into the next, he shimmies up behind her. His grasp on her is seductive and as she moves the seam of her tight jeans pushes her panties between her aching lips, soaking up her flood. Reaching over her head and behind them, she drapes her arm around his neck, pulling his head down to rest on her shoulder. Turning her head she yells in their close space. "Chris! What the hell are you doing here?"
His hands reach all the way around her and rest on her belly pulling her back to him as they grind to a pulsing Latin beat.
"Stag night," he yells in her ear. "Some of my single friends thought they'd prey on heartbroken women. Looks like they found some at your table." He nibbles on her ear and she lets him. The movements are so sensual it's like fucking in public. As the crowd presses around them, his hands tease down her hips squeezing the tops of her thighs. "Is one of those ladies your stepmom?"
"Shh… Don't talk. Keep doing that." She wants to lose herself in the music with him. In public. Like a goddamn real date on Valentine's Day. It's like Cupid heard her wishes to be with one of her men; she pushes the thought away that maybe it's meant to be a 'sign' that Chris was the one that appeared?
How could she ever hurt Tom? How could she hurt Chris? Her head pounds in a beating pulse and for one more night, she doesn't want to have to come up with an answer.
His hands are rough on her body, tugging and pulling to the beat of the music, unknowingly lifting her from her negative thoughts. His thick fingers grasp at her flimsy top and with each grope higher up her hips and sides, he lifts the silky red fabric until he can drag his fingers along the waist of her jeans. Tickling at her belly button, he traces a path to the snap on her jeans.
Feeling her pupils dilate as another rush of wet releases between her legs she turns her face to his. "What are you doing? We can't…" Her words come out as a rush of air, no real sound to them but the fire returned in his eyes lets her know she heard him.
"Look around, no one's paying attention. I can finger fuck you right here and no one would know."
The sea of dancers swells around them and they are hidden in plain sight. She can't even see the group of Stacey's friends and she doesn't recognize a single face. The ache in her body is intense. Looking down she can see her nipples are hard, obviously peaking under the silk shirt, teased from the lace bra brushing against them. "Aw fuck, Chris. We can't…"
Changing dance positions his other arm comes up over her shoulder and snakes over her breasts, pinching one of her peaks. "Fuck you," she whispers. "I already ache."
"Let me make it better," he offers quietly in return.
She leans back into his body, his cock throbbing and rubbing her ass. If unclothed, he'd fit right into her. No wonder he finds her heeled boots so damn sexy, she thinks. We're the perfect height.
"Niña, don't think." His hands dip lower into her pants, the fabric relaxing and pulling away, allowing him access. "You know if you want me to stop, you just have to tell me. You're so wet, I'm not even to your sweet pussy yet and I can feel your juices on my hand, our dance moves shifting your clothes, teasing you, spreading the wet. The lace is soaked, baby girl. So damn sexy."
The dancers move and swirl around them, new pairings dividing off and others joining in. Chris is her constant, pulling her to the edge with just his hands and his voice. Her heart pounds in her chest. Her head drops back against his shoulder and he bites her neck as he dips into her well, stroking the lace through her lips. "Shit," she moans. Other party-goers jostle against them and his arm is bumped causing him to push deeper. Not expecting the force, her ass pushes back against him and a gasp escapes her lips. He keeps up the pretense of their dance as she melts in his arms.
Pulsating to the beat, Chris holds her tight and fucks her well. "Thalia? Do you want me to stop?" With a barely noticeable turn of her head, she breathes heavily against his skin. "No. No. I need this. I need you."
His hands are magic and she can't say no. His long fingers quickly make work of her as he whispers in her ear, encouraging her to let go, to relax. "Come, come all over my hand and I'll know it the whole drive home." His breath is hot and damp, and the smell of beer as he puffs against her skin seems to add to her own intoxication. The music builds to a crescendo as she comes at his request, spilling over him and the dew seeping down her pants leg.
Removing his hand slowly, he wipes himself clean on the inside of her jeans, tugging her shirt back into place. Keeping up their pretense of a dance, Chris holds her tightly so her limp form doesn't collapse. "Beautiful, so fucking beautiful. Your mind, your body, your spirit, Thalia."
Coming back to earth she can see Stacey's blonde head bobbing up and down over the top of the other dancers, searching for her. Chris spins her quickly, her back to her mother and claims her with a kiss. "Let me take you home," he begs.
Tucking her hair back she tries to clear her mind. "No. We said 'no Valentine's Day'. This is pure coincidence; you got lucky I was here. I'm staying with her at the hotel and I'll see you at lunch tomorrow." She pushes away from his massive chest, ignoring the pounding pain in her head. "I gotta go."
"Been a long time since I had a girl run off 'cause her mother was looking for her," he teases.
Thalia tugs at his shirt collar, catching a glimpse of his tattoo, one of her favorite quotes. "Cradle robber," she jokes, trying to make light of the situation. What kind of magic spell did he just weave? Bastard.
Yanking her close for a moment longer, he shakes his head in total disagreement. "You're all woman, and you're mine."
"Just remember, that makes YOU MINE too, so don't you be flirting with any of these desperate old hags"
Fleetingly, he touches his fingertips to her lips and she smells the proof of her ownership. Pivoting on her heels to walk away, he smacks her on the ass.
When she reaches her table and gathers her coat, she looks around for him and realizes he's gone.
The women decide it's only three blocks back to the hotel so the walk in the cold would be easier than finding a cab. Thalia mentions Uber and most of them look at her like she's speaking an alien tongue, so she just falls back in the group.
The air is cold against her wet jeans and she fears the smell will be detected in the winter breeze. "Jesus fuckin' Christ," she whispers aloud, rubbing her knuckle across her lips like she does when deep in thought.
Stacey steps in next to her and links arms. "I'm sorry I couldn't get out of my meetings today. I can't wait to see your new tattoo. Glad you decided to go ahead and do it, for Amy's memory." The blonde side steps some broken glass on the sidewalk. "That was a fun night, baby girl! I always wanna go out like that, but your father wants to sit at home. I'm glad you came with us!" She takes the end of Thalia's pink scarf and wraps it tighter around her neck. Thalia blanches at the use of her family's nickname for her, having been so long since she'd actually heard family use it. She palms her hand over her mouth and her stomach twists.
Baby girl. Aw, fuck. What have I done? Stacey continues to prattle on, but Thalia hears none of it. All she can think about is how she's broken her own rule. Where did all her determination go? She was the one who didn't want either of them as part of her day, just to be fair to them both. She had wanted to prove to herself she could have fun without them. She shouldn't have given in so easily, to Chris- letting him claim her like that, out in the open. But damn, the man is irresistible… She's betrayed herself, dammit. And to top it all off she let him manipulate her thoughts and he fingered her in a goddamn public place where anyone could have seen them. She feels sick about all the consequences if anyone from campus saw them.
"Hey, I lost ya." Stacey takes her room key out of her purse to gain access to the hotel lobby at the late hour. "You okay?"
The women wait for the buzz of the door to let them in. They all say their goodnights and head off in their individual directions, some staying by the warm fireplace to warm up and chat longer.
Thalia can feel the fakeness to her smile. "Fine. Really. I think the evening just caught up to me and I'm dead on my feet. I need some aspirin and to lie down."
"Can do that, honey." Stacey replies leading them to wait with the group at the bank of elevators. She leans in conspiratorially. "Damn, I know you're in love and all, but that man you were dancing with? Hell, he was hot sex on a stick! Yummy!"
"Stacey!" 'In love.' Her stomach lurches at Stacey's words. Fuck. She is. With both of them, and her wanton public behavior tonight is such an insult to Tom and the privacy of their relationship… Jesus, what the hell am I doing?
"What? Just because I've been married to your father for sixteen years doesn't mean I can't look and appreciate the male form. And believe me, he had a nice one." Stacey bumps her shoulder and giggles.
Girl stuff. Thalia was never good at that. Pulling herself together, she focuses on the now with her stepmother. She giggles too and plays along. "Yes, he did. Solid too."
"A man built like that? He's just right for a girl like you. He could throw you around like a ragdoll," Stacey smiles, her eyes bright from her slight inebriation.
"Oh my God. Hello? Boundaries. You're still my mother, ya know." Thalia laughs for real and shakes her head at the absurdity. She wonders if Stacey will remember this conversation in the morning as she kicks off her heels in the elevator car.
"You need a man," Stacey warrants, bending over to rub the ball of her foot. "If men like that in clubs drool all over you, pick one. And hell, I'll stop worrying about you and food and your weight. I've never had a man look at me like that before, honey. Like he couldn't wait to see you under all those clothes…" Other women from the club chime in their hummed agreement. The blonde tumbles a bit when the elevator stops at their floor. Righting herself she finishes with her audience. "If you can get a man like that one, take him and ride him to the altar and don't let him go."
Echoes of "hell yeahs" reach through the doors as they close and the silence to Thalia is deafening as they are alone and quiet for the first time all night.
Her demeanor changes and she sighs bitterly, really hoping her stepmother doesn't remember her anger in the morning, just her words. "Dammit, Stacey, slow down. Thank you for finally giving me permission to eat whatever the hell I want, that's so kind of you… But grasp your head around this one now: marriage isn't in my cards. At least not for a while… I did not spend all this time and money on an education to give it up for a man and raise his babies. I'm not 'riding' anyone anywhere right now, or for a long time for that matter."
Her headache is growing worse by the minute and she just wants to get to the room and wash the club off her, the smell of smoke out of her hair. She's angry and she knows exactly why and Stacey just happened to say the wrong thing at the wrong time.
"Girl, you're all grown up," Stacey says as she slides the plastic card into the lock. "You don't need me to tell you what to do. And I'm so proud of all the things you do… I just don't want you to be lonely. Find a good man, not a boy, and know the difference. Someone who lets you be you and makes you want to be… well, more."
Stepping into the room, Stacey moves to her bed and flops down face first kicking her feet up in the air. Thalia closes the door and leans against it, banging her head back and closing her eyes. "Maybe that's the problem."
Pushing away from the door the quote from Chris's tat swirls in her brain. When you lose touch with your inner stillness you lose touch with yourself. When you lose touch with yourself, you lose yourself in the world.
Tugging her fingers through her hair, she wonders if she even knows how to be herself without them anymore. It's like being with them has opened a Pandora's box, and now she doesn't know how to close it again. All the new experiences, the self-discoveries. Even though it ties her brain into knots sometimes to deal with the secrecy and onslaught of a dozen different emotions, she wants this, needs this like air.
God, how far gone is she? Will there ever be a point that is too far?
When Thalia arrives to the student Commons for her meeting with Tom, she's ten minutes late and gritting her teeth. She knows how much he values punctuality, but this morning, nothing much has gone her way, traffic was bad leaving the city, and she probably looks as frazzled as her mind is.
He's sitting at the far corner, a little secluded, alternately fumbling with his glasses and rubbing his lips. Surreptitiously straightening her clothes and hair and wondering for the umpteenth time what he wants to talk about, Thalia walks over.
Her heart gives a guilty little lurch when she sees that he's ordered her favorite morning treat – coffee and a blueberry bagel with cream cheese.
He looks up and his jaw tightens as he's watching her approach.
Shit. She's not sure she can handle a pissed-off Tom this morning, with the mix of emotions churning away in her gut and the almost sleepless night making her grumpy.
"Professor Hiddleston."
She stops in front of the table, wondering what to say. He taps his watch and lifts an eyebrow, giving her that stern look that's infuriating and sexy at the same time.
"You're uncharacteristically late, Ms. Bareo."
Ugh. No use making excuses. "I'm sorry. It won't happen again, sir."
His expression softens ever so slightly at the last word and he motions for her to have a seat.
Thalia tosses her bag in an empty seat and sits down, crossing her legs. Tom is wearing one of his hundred nearly identical sweaters today, and the soft burgundy fabric stretches invitingly over his muscles as he folds his arms.
With a swallow, Thalia shifts in her seat. She needs coffee to survive this.
To keep up the pretense, Tom goes through a few project-related things first while they work their way slowly through their breakfast. She keeps having flashbacks of Chris pressed against her on the dance floor, and of the conversation she had with her stepmother.
Why did this have to happen to her? All those years without a real man to catch her attention, and now she has two who couldn't be more different but mean the world to her.
"Ms. Bareo?"
With a start, she realizes Tom has been waiting for an answer from her. Blushing, she takes a last fortifying sip of coffee.
"I'm sorry," she apologizes again.
He looks at her with narrowed eyes, but all of a sudden, his glare gives way to concern. Leaning forward, he lowers his tone. "Are you alright, darling? You look a bit out of it, to be frank."
There it is again, the caring, kind side to him that not many people get to see, although he's always politeness personified. She nods. "Yeah, I'm okay. Just a bit of a rough night."
"I hope your stepmother didn't give you any trouble?"
"No, nothing like that. She and her friends dragged me into a club and we got in late. So I'm a bit hung over, to be honest. I feel as if I'm over forty and they're the party-hungry teenagers or something. They were all laughing and ready to go this morning like it was nothing!"
Tom laughs his characteristic ehehe, but sobers up quickly. Now that the somewhat stern look is gone, she can see that he seems nervous beneath his mask. He keeps adjusting his glasses and pushing the rest of his food around on his plate.
Changing the subject, he asks, "Did you and your stepmother get your tattoos?"
Thalia caresses over the tender spot on her breast carefully. "Yea, I did. Hurts like hell too. But when the redness goes away, it'll be beautiful. Stacey claimed she couldn't get out of her meetings, but I think she punked out at the last minute."
Tom smiles warmly, but his fidgeting hands bely his usual confidence.
Why is he nervous when she's the one who should be feeling like that? It only makes her even more anxious.
"Didn't you want to talk to me about something?" she offers quietly, hoping to alleviate his anxiety.
He swallows hard and starts playing with his empty teacup, long fingers handling the delicate porcelain with utmost care. God, what those fingers can do. They're just as lethal when they're gentle as when they grab her hard enough to leave bruises. She wonders, if given the chance, would he have done the same at the club – driven her crazy with his nimble fingers, leave her panting and wanting more?
Probably so, and his words would have been filthier.
And she'd have loved every goddamn minute of it too.
Who the hell am I becoming?
She closes her eyes briefly and forces herself to focus.
When he speaks, his words are so low she has to lean forward to hear him.
"Would you be my sort-of date at the Alumni Gala next month?"
The napkin she's been twisting slides from her fingers to the table.
Tom runs a hand back through his hair. "Bloody hell, that came out all wrong. I'm sorry."
He takes a breath and continues in a surer tone. "You know that my project is going to be honored, and I want you there by my side because you've played such an important role in it. Without your research, this wouldn't have been possible."
His blue eyes seek out hers, everything about his expression earnest and appealing. She can feel her pulse racing at the thought of accompanying him to the gala. He'll probably wear a three-piece suit or tux and look way too handsome. But…
"Won't I stick out like a sore thumb?" she questions, worrying her lip. "All the staff and dignitaries and VIP guests, and then plain, old, plump me."
He narrows his eyes again, reaching out to her but stopping himself at the last moment and taking a gulp from his water glass instead.
"Nonsense, Thalia. It won't be the first time in history that a grad student has attended the gala for some very valid reason." Looking around the Commons, it's still rather quiet the morning after the holiday, and she sees now she isn't the only dragging, hung over person on campus.
His voice drops and her gaze is drawn to his lovely angled face again. "Darling, there's no way you could ever be 'plain.' You're such a beautiful, charming creature. Everyone will fall at your feet and want to listen to your musical voice."
Blushing, she tries to read him and understand his uncertainty. His gaze softens. "It's perfectly alright for me to invite you. Nobody will think twice about it…and I'm pretty sure the sight of you in a gorgeous gown will make rational thought impossible for anyone, especially me, anyway."
A flirtatious spark darkens his eyes, and she feels her resolve melt away.
"Please accept my invitation, oh fair and gracious lady," Tom adds with a theatrical expression that has her suppress a giggle. "Have mercy on this poor lad who doesn't want to face the crowd alone. I haven't got the faintest idea how Americans handle such events."
Feeling more herself now, she raises a brow at him. "So, you only want me there to save yourself the embarrassment of putting your foot in your British mouth?" she challenges.
Something in his expression shifts from one moment to the other. It's an art he's mastered, and it never fails to throw her off balance.
"Oh, I have a whole list of ideas how you could keep this British mouth of mine busy," he half-growls in a low, deep voice that sends delicious shivers down her spine.
"Tom," she hisses, "not here."
She's had enough with public displays of… lust this week.
A smirk curls his thin lips before he pretends to busy himself with a bite of now cold toast.
"Seriously, though," he goes on, sending her a pleading puppy-dog look. "I'd love to have you by my side and sing your praises, maybe even steal a few hidden touches. You don't necessarily have to stick to my side, though I'd love that. And it will look really good for you, academically speaking, that you've been invited and received some recognition."
Thalia leans back in her chair and sighs. "You're right, of course. As usual."
He smiles. "So you'll be my date?"
"I'll be your guest," she says, stressing the last word and automatically smiling back.
"Marvelous. Don't worry too much about it, you'll fit right in."
"I doubt that," she mutters more to herself than to him. Her eyes widen when she realizes something. "Oh my god, I don't have anything to wear! What sort of gown do you think is expected?"
Tom runs an appreciative gaze over what little of her body is visible, and she feels it like a caress that warms her from the inside.
"I'm sure there are photos or something from previous events to get an idea. Just pick whatever catches your eye, you're going to look more stunning than all the women there put together, no matter what you wear."
Blushing furiously, she wants to say something, but Tom holds up a finger.
"And by the way, I'm paying for the dress. No, don't even think of protesting. You're doing me a huge favor and honor by accompanying me, and a gown for a gala dinner isn't going to cost a couple of bucks. I insist."
"But, but…" she splutters, only to be cut off again when he adds in a low, warning tone, "Are you going to be a good girl and do as I say, or do I need to pick out a dress for you myself?"
Well hell, that doesn't leave her with much of a choice.
"Damn you, Professor, you aren't playing fair," she complains, crossing her arms.
"Stop pouting like this or I'll have to drag you into the next best room and kiss the pout right off your lips."
The sexy threat makes her breath hitch.
Damn, he knows just how to push her buttons.
The look in Tom's eyes is full of promises.
"Glad we've got that settled then. Choose something to show off your lovely legs." He gestures to the meal. "Any more coffee or tea?"
Thalia huffs and shakes her head. Ever the gentleman, Tom rises when she does. He bends to retrieve his leather briefcase and uses the move to whisper into her ear.
"I can't wait to see you bedazzle the crowd, my precious orchid. You're going to make all the other wallflowers wither away."
Click here for Chapter 25 Step Up
AU Fan Fiction, au fiction, Uncategorized#adult situatons, #angst, #avener-nerd-mom, #chapter 24, #Chris Evans, #Chris Evans AU fic, #Chris Evans fan fiction, #devikafernando, #drinking, #educating thalia, #exhibitionism, #fingering, #food porn, #Language, #NSFW, #Plus size romance, #plus sized, #plus-size OFC, #real life discussions, #stepfamilies, #Tom Hiddleston, #Tom Hiddleston AU fic, #Tom Hiddleston fan fiction, collaboration
Chicago: Chapter 11
Warnings: Language, Adult Situations, sightseeing, FOOD PORN, cuteness overload, PLAID PORN, innocent making out
*****SUSPEND REALITY- we know it's not possible to see ALL these attractions in one day. This is what happens when a German girl living in Sri Lanka has never BEEN to Chicago!******
Click here to the beginning of Educating Thalia
Tom feels as if he's entered into a parallel universe or discovered an alter ego of his. It isn't just the clothes, though he does feel different in his red-and-blue plaid shirt and black jeans. He's picked the shirt because it looked awfully comfortable and because he couldn't get the image of Thalia with her cowboy boots out of his head. To brace himself against the biting cold, he's wearing a navy T-shirt beneath the shirt and a quilted navy coat with a zip. He had half a mind to go for cowboy boots himself, but part of him was scared he'd just look like an idiotic wanna-be Yankee mixed with a stiff Englishman. So he opted for his go-to solution, the well-worn grey suede shoes that he loves to combine with basically any casual or even semi-formal outfit.
Thalia has already glanced at the plaid shirt more than once, and the approval in her eyes – turning to quite a lot more than interest when he opens a few buttons in the toasty warm museum – feels like a soothing caress. It makes him feel more at ease, ready to let her play tour guide and boss him around a little.
Since they have embarked on this journey through the city, she's been pointing out landmarks and sharing little snippets that he stores away in a corner of his knowledge-hungry brain for future reference.
The museum is amazing. The building itself caught his attention when he Googled things to see in the city, with its imposing reddish brown façade and half-round side wing full of windows. He listens happily to Thalia sharing some backstory while they make their way inside. Tom chuckles at the way she occasionally bumps him as they move through the holiday crowds. They're walking close enough for their hands to brush, and on an impulse, he laces his fingers with hers.
She shoots him a glance and lowers her lashes, smiling.
He doesn't want to let go of Thalia's hand, and she seems just as happy to let him hold it. Off and on, they look at each other instead of the exhibits, and it feels like a real date, with all the cares in the world a million miles away. She's wearing warm black tights with a denim skirt. To match his outfit–which he'd revealed with a spur-of-the-moment selfie to prove to himself as much as her that he really was in Chicago–she's also opted for a plaid shirt, hers in different shades of blue that match her skirt and coat. And of course, her trusty cowboy boots make today's outfit complete.
Once they've had their history fill in general, Thalia enthusiastically pulls him aside to show him one of the world's largest costume collections. Tom entertains her by imagining little tales for the astonishing historical clothes on display, and they get quite a bit of side eye from other visitors for all their whispering and laughing.
They round off the experience with a hearty brunch at the café on the ground floor.
Pushing his chair back from the table, Tom asks, "What now?" glowing with an overdose of happiness that is partly museum-induced, partly food-induced and most definitely Thalia-induced.
He can't find words for how wonderful it is to share these magical moments with her, away from the whole professor-and-student sword that's usually dangling over their heads.
Thalia drinks the last of her coffee. "Well, I'm pretty sure you did all your touristy homework and have figured out a whole list of things you are planning to tick off. Am I right or am I right?"
Tom throws his head back and guffaws. "You know me entirely too well, darling."
With a flourish, he pulls an actual list out of his pocket, grinning at her exaggerated eye roll. He smoothes the crinkled paper out on the table.
When Thalia leans closer to get a good look, he acts on instinct again and kisses her. It's hardly more than a gentle peck, though he deepens it a little when her lips part, delighting in her sweet sigh. He lets the tip of his tongue brush over her full lips before pulling back.
The look of surprise and joy on her face momentarily lets his mood nose-dive. This, this right there is what she should have. A man who can date her, take her out and spoil her properly. A man who devotes all of his time to her and who will march right into that house and face her family bravely because he intends to stick with her through thick and thin.
Determined not to let the real world burst his giddy bubble, Tom slides a finger down the list.
"Which item can you recommend?"
She concentrates, her forehead puckering in a slight frown. As usual, she's willing to dedicate 100% of herself into whatever needs doing, whether it's studies or showing a secret lover around town.
"The Adler Planetarium is amazing," she says thoughtfully, tugging on a strand of hair until he pulls it out of her grip and curls it around his fingers. "Their Sky Theater offers you virtual-reality trips through space and time which are seriously mind boggling. And if you're really lucky, you can meet one of the top-notch researchers who are responsible for the museum."
"Oh, tempting." Tom can feel himself getting all bouncy in his seat again, which reminds him of his sister Emma always telling him he's somehow managed to trap a five year old in an adult's body.
"The Museum of Science and Industry is another of my favorites," she adds. "It's got a restored U-505 German submarine, a simulated coal mine and a vintage diesel-electric train. Lots of action instead of only dusty exhibits."
"Can't we do both? I promise to keep my enthusiasm mostly bottled up and move quickly through all the halls." Tom gives her his best puppy dog look, which makes her laugh and swat his arm none too gently.
"You're a pain in the ass, Tom."
She hasn't called him Professor Hiddleston once today, and although it's a turn-on to hear her do so, he's rather glad because he wants to be plain Tom here.
Unable to resist temptation, he slides a hand to her knee and toys with the hem of her skirt while leaning close enough to speak into her ear. "As far as I recall, I haven't been allowed close enough to your ass yet to cause any pain, other than the occasional light spanking. But if you feel inclined to change that…"
She makes a squeaky sound and knocks his hand off her leg, wagging a finger at his fit of giggles. "I swear, if you keep that up, I'll happily let my family torture me again and leave you to your own devices in big, mean Chicago."
But her eyes dance merrily, and she holds her hand out to him when she gets up from the chair.
"Now stop acting like a teenager, we've got two items to cross off your list."
"Darling, I have a mental list you should consider sometime as well then."
Her eyes grow wide as his words sink in and he throws his head back in laughter, quickly clearing their table before wrapping his arm around her shoulder to make their exit.
They visit both the Museum of Science and Industry at Hyde Park and the Adler Planetarium with its domed roof that reminds Tom a little of a study trip to Berlin in Germany, several years ago. By the time they are done, Thalia grumbles good-naturedly about wearing holes into the soles of her boots, and Tom is bouncing with another energy boost because the thought of trying out local food is so tempting.
"Don't laugh, love, but I'm hungry again. What do you recommend?" he asks. "I'm starving and I want to try absolutely everything Chicago is famous for."
Thalia gives him another of her eye rolls, hooking her arm through his to pull him to a bus stop.
"Oh my god, that metabolism of yours! I'm jealous!" She shakes her head in disbelief. "Well, Chicago IS famous for food, so brace yourself for the experience of a lifetime."
She holds up her free hand, counting off on her fingers, "We have deep-dish pizza, which is pure heaven and I haven't found any like it out East. There's the Chicago-style hot dog with all the fixings… And all sorts of high-end cuisine stuff if you think it's below your gentleman status to eat what everyone does."
Tom snickers and gives her butt a light slap. "I've heard about the hot dogs, actually. Weren't they a result of the Great Depression?" She nods and he continues. "What's on them?"
"My favorite is just the standard version, with an all-beef hot dog on a steamed poppy seed bun. It's topped with yellow mustard, relish, tomato wedges, chopped onions, pickle, hot peppers and celery salt."
With a groan, he pulls her closer. "Okay, I need one of these like I need air to breathe. Lead the way, oh heroic, merciful tour guide, and prevent my death of starvation."
They take the bus and end up at Portillo's, googling its impressive history and success story while waiting for the food.
Stuffed with hot dogs, but still drooling over their dessert of strawberry shortcake and chocolate éclair cake, they manage small talk between bites.
"So, haven't seen much of the US since you arrived here?" Thalia inquires after he's fed her with a forkful of chocolatey delight.
"No, haven't had the time yet." Tom lets her feed him in return, staring into her eyes while suggestively licking his lips, pleased to see her hand wobble a little. "I've been around the world a bit, though."
"Oh, tell me more."
He shrugs modestly. "Mostly Europe during my youth. Spain, France, Italy, Russia. A couple of years ago, I accompanied some colleagues to Germany."
"What about the rest of Britain? Or do the English make it a point to snobbishly ignore their neighbors?"
He wagged his fork at her and relishes his last bite before answering. "I visited Scotland with my father once. And my mother took my sisters and me to Ireland when we were still young."
His face clouds over momentarily at memories of a childhood that had been anything but easy but was mixed with enough happy incidents to not bother him too much now. At least he hadn't carried any serious scars of his parents' divorce over into adulthood…though he should probably rethink his rules on relationships.
"Oh, and I flew to India for my sister's wedding," he adds with a smile, deliberately pulling himself back to the presence. "That was just…surreal and truly beautiful."
Thalia smiles back at him and entwines her fingers with his when he reaches for her hand.
"And you?" he wants to know. "You've got the whole world waiting for you. Where do you want to go?"
She screws up her face in thought, as if there's too much to consider.
"Everywhere," she answers with a laugh. "I've got to finish my degree, but fortunately it could take me all the places I could never afford to go on my own. I've applied to internships in Paris, Cairo, Athens and Rome for museum work and archeological digs. I'm just waiting each day for the right phone call. There's a box of dusty clothes ready to go in the back of the closet, aching for more dig dirt," she jokes.
Tom leans forward and caresses her cheek with his other hand, scooting his chair closer to hers and resting his leg against hers. "I hate the idea of you being so far away."
Her head drops, her hair falling around her, but her blush is unmistakeable. She seems at a loss for words and it endears her even more to him. Dammit, Thomas. This trip was the best worst idea ever…
She grabs the ticket from the table and the two tussle over who will pay the bill. Tom insists that she's already playing tour guide so he absolutely must pacify his inner gentleman by paying for lunch.
"Fine… But I'd like you to let me take care of you sometimes too, I'm not totally broke you know." His scowl changes her thought and she forges on. "What's next on your list, Mister?" Thalia wants to know, still pouting at her defeat.
"You know what, why don't you suggest something?" He smiles at her. "As I said before, today's in your hands, bills excluded."
She smiles back. "Millennium Park is kind of a no-brainer, despite the weather. Want to give it a go?"
"Sure, that's where the Bean is, right?!"
She laughs at his enthusiasm, bundling up to brave the cold Chicago air. They make it there in no time, discovering that a lot of people are bracing the cold to get photos in front of the iconic Cloud Gate sculpture with its metallic bean form and cloud-reflecting surface. Tom takes a few selfies with Thalia, debating with himself whether to share these lovely memories with his family or not, and deciding against it with a heavy heart.
As happy as he is today, he keeps realizing one thing that obscures his inner sunshine momentarily: In a world with less prejudice, Thalia and he would make a great couple. But as things are, a normal relationship is out of the question and not something either of them is ready for anyway. He knows he should wish for her to have someone else to share such joyous moments with—but he doesn't. He wants her to himself for a bit longer at least. They take their own sweet time, walking the city streets, giggling and window shopping, dragging out their stolen moment together.
Close to evening, Thalia takes him up on the 360 Chicago, formerly known as the John Hancock Observatory where the 94th floor – 1,000 feet up – offers a view for miles and miles, across four states. They dine up there at the restaurant, and again Tom can't resist doing all the little things proper couples should. He feeds her and teases her, touches her as much as he can, asks her personal questions and stores away each morsel of information as if his life might one day depend on the right answer.
"Tom, I've had a wonderful day. I hate to see it end, but I really should be getting back home. There's a train switch and I don't want to miss it." Her eyes glisten in the low light of the restaurant. Her tone is wistful and tells him what he wishes to know.
Taking her hand in his across the table he runs his thumb over the back of her hand. "Thalia, darling, if you're willing to risk it with your family, I'd love for you to stay with me tonight."
Her eyes search his, search his face, looking for what, he doesn't know. She brushes a floppy curl from his forehead and runs her thumb across his scarf, and he leans into her touch, craving more. Slowly her grin turns up. "Let me make a call to my stepmother. She'll know how to soothe things over with Dad."
Nestled in the warmth of the back seat of the cab, Tom is pleasantly surprised when Thalia takes the initiative to kiss him, at first unsure and timid. He tugs on her scarf, pulling her closer and acknowledging his need. Her hand grips his thigh as the kisses grow more heated. He bites back a chuckle at the cab driver watching in the rearview mirror while they make out like two lusty teenagers. The ride is entirely too short and by the time they make it to the hotel, he wants nothing more than to lose himself inside her and forget the rest of the world.
Click here for Chapter 12 Tutoring
AU Fan Fiction, au fiction, Uncategorized#avenger-nerd-mom, #chapter 11, #chicago, #Chris Evans, #Chris Evans AU fic, #Chris Evans fan fiction, #devikafernando, #educating thalia, #Fluff, #food porn, #innocent making out, #Language, #PLAID, #plaid porn, #Plus size romance, #plus sized fiction, #plus sized OFC, #plus sized romance, #real life, #Tom Hiddleston, #Tom Hiddleston AU fic, #Tom Hiddleston fan fiction, #vacation, au fiction, collaboration, professors
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 14
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Q: how to find out open file description for linux process? While runing my go program, i find out so many error such as "too many open files" in the logs, and i just to find out which process run out of the fds, and i run this command:
lsof -n |awk '{print $2}'|sort|uniq -c |sort -nr
it returns the result such as
279605 20341
62748 19861
10310 19712
5434 21318
3484 27344
2842 19781
2400 20372
2346 24153
2123 5214
1540 21123
process which pid is 20341 is a mongod process, and i'm surprised about that. So i try another way:
lsof -p 20341 | wc -l
but something make me trouble is that it's result is:567.
After that, i try another way:ll /proc/20341/fd | wc -l which result is 496。
And i am so confusion now,which one is right,and what the different between them?
thanks.
updated at:2018-05-31 10:35:33
*
*Get the mongodb PID
[root@node26 10:34:54 ~]$ps aux | grep mongo
mongodb 20341 2.4 1.9 25419812 1257420 ? Sl May28 107:58 /usr/bin/mongod --quiet -f /etc/mongod.conf run
*Command lsof -p
[root@node26 10:36:12 ~]$lsof -p 20341 | wc -l
570
*Directory
[root@node26 10:36:33 ~]$ll /proc/20341/fd/ | wc -l
499
*Command lsof + grep
[root@node26 10:37:33 ~]$lsof | grep 20341 | wc -l
282223
*
*front 10
mongod 20341 mongodb cwd DIR 9,127 4096 2 /
mongod 20341 mongodb rtd DIR 9,127 4096 2 /
mongod 20341 mongodb txt REG 9,127 12238320 2499177 /usr/bin/mongod
mongod 20341 mongodb mem REG 9,127 67108864 1969114 /var/lib/mongodb/a_dev.0
mongod 20341 mongodb mem REG 9,127 536870912 1968852 /var/lib/mongodb/a_dev.ns
mongod 20341 mongodb mem REG 9,127 67108864 1968447 /var/lib/mongodb/a.0
mongod 20341 mongodb mem REG 9,127 536870912 1968347 /var/lib/mongodb/a.ns
mongod 20341 mongodb mem REG 9,127 67108864 1968453 /var/lib/mongodb/b.0
mongod 20341 mongodb mem REG 9,127 536870912 1968449 /var/lib/mongodb/b.ns
mongod 20341 mongodb mem REG 9,127 67108864 1968590 /var/lib/mongodb/c.0
*middle 10
mongod 20341 27018 mongodb 490u IPv4 143223380 0t0 TCP node26:27017->node24:59172 (ESTABLISHED)
mongod 20341 27018 mongodb 491u IPv4 143758325 0t0 TCP node26:27017->node25:43016 (ESTABLISHED)
mongod 20341 27018 mongodb 492u IPv4 143762443 0t0 TCP node26:27017->node24:60602 (ESTABLISHED)
mongod 20341 27018 mongodb 493u IPv4 154865226 0t0 TCP node26:27017->node26:54800 (ESTABLISHED)
mongod 20341 27018 mongodb 494u IPv4 164046515 0t0 TCP node26:27017->node24:42952 (ESTABLISHED)
mongod 20341 27018 mongodb 495u IPv4 164046516 0t0 TCP node26:27017->node24:42960 (ESTABLISHED)
mongod 20341 27018 mongodb 497u IPv4 154865844 0t0 TCP node26:27017->node25:41976 (ESTABLISHED)
mongod 20341 27018 mongodb 500u IPv4 164046517 0t0 TCP node26:27017->node24:42968 (ESTABLISHED)
mongod 20341 27018 mongodb 502u IPv4 164046518 0t0 TCP node26:27017->node26:60306 (ESTABLISHED)
mongod 20341 27018 mongodb 503u IPv4 164046519 0t0 TCP node26:27017->node26:60314 (ESTABLISHED)
*tail 10
mongod 20341 32608 mongodb 492u IPv4 143762443 0t0 TCP node26:27017->node24:60602 (ESTABLISHED)
mongod 20341 32608 mongodb 493u IPv4 154865226 0t0 TCP node26:27017->node26:54800 (ESTABLISHED)
mongod 20341 32608 mongodb 494u IPv4 164046515 0t0 TCP node26:27017->node24:42952 (ESTABLISHED)
mongod 20341 32608 mongodb 495u IPv4 164046516 0t0 TCP node26:27017->node24:42960 (ESTABLISHED)
mongod 20341 32608 mongodb 497u IPv4 154865844 0t0 TCP node26:27017->node25:41976 (ESTABLISHED)
mongod 20341 32608 mongodb 500u IPv4 164046517 0t0 TCP node26:27017->node24:42968 (ESTABLISHED)
mongod 20341 32608 mongodb 502u IPv4 164046518 0t0 TCP node26:27017->node26:60306 (ESTABLISHED)
mongod 20341 32608 mongodb 503u IPv4 164046519 0t0 TCP node26:27017->node26:60314 (ESTABLISHED)
mongod 20341 32608 mongodb 505u IPv4 164046523 0t0 TCP node26:27017->node26:60322 (ESTABLISHED)
mongod 20341 32608 mongodb 730u IPv4 117137926 0t0 TCP node26:27017->node25:54730 (ESTABLISHED)
A: /proc/${pid}/fd contains file descriptors connected to the shell, which show up as a number followed by a u in lsof:
$ la /proc/$$/fd
total 0
lrwx------ 1 username users 64 May 30 20:08 0 -> /dev/pts/0
lrwx------ 1 username users 64 May 30 20:08 1 -> /dev/pts/0
lrwx------ 1 username users 64 May 30 20:08 2 -> /dev/pts/0
lrwx------ 1 username users 64 May 30 20:08 255 -> /dev/pts/0
$ lsof -p $$
COMMAND PID USER FD TYPE DEVICE SIZE/OFF NODE NAME
bash 3720 username cwd DIR 254,3 12288 1835009 /home/username
bash 3720 username rtd DIR 254,2 4096 2 /
bash 3720 username txt REG 254,2 859688 2890163 /usr/bin/bash
bash 3720 username mem REG 254,2 46912 2885785 /usr/lib/libnss_files-2.27.so
bash 3720 username mem REG 254,2 2942480 2930144 /usr/lib/locale/locale-archive
bash 3720 username mem REG 254,2 457800 2890072 /usr/lib/libncursesw.so.6.1
bash 3720 username mem REG 254,2 2105608 2885835 /usr/lib/libc-2.27.so
bash 3720 username mem REG 254,2 14144 2885777 /usr/lib/libdl-2.27.so
bash 3720 username mem REG 254,2 363064 2890132 /usr/lib/libreadline.so.7.0
bash 3720 username mem REG 254,2 177680 2885836 /usr/lib/ld-2.27.so
bash 3720 username 0u CHR 136,0 0t0 3 /dev/pts/0
bash 3720 username 1u CHR 136,0 0t0 3 /dev/pts/0
bash 3720 username 2u CHR 136,0 0t0 3 /dev/pts/0
bash 3720 username 255u CHR 136,0 0t0 3 /dev/pts/0
They are both "right," but the count from lsof is the one relevant for running out of open files.
To find the relevant open files limit use ulimit -n.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,892
|
Prerequsites
----------------------------------------------------------------
Install the Aerospike C Client from the download page:
http://www.aerospike.com/download/client/c/3.1.24/
Make sure the C Client library is in your search path.
You need the readosm library:
https://www.gaia-gis.it/fossil/readosm/index
On RedHat/CentOS:
sudo yum install -y readosm-devel
sudo yum install -u jansson-devel
On MacOS:
brew install readosm
brew install jansson
export CPATH=/usr/local/include
Building
----------------------------------------------------------------
make
Running
----------------------------------------------------------------
Usage:
OBJS/osm_load --usage
Execute the program, argument is path to osm pbf data file:
OBJS/osm_load san-francisco-bay_california.osm.pbf
Docker
----------------------------------------------------------------
A Docker file is included that packages up the code and dependencies to run the load,
Usage:
docker build -t <myuser>/osm-load:cplusplus .
docker run --rm -v ~/Downloads:/data <myuser>/osm-load:cplusplus -h localhost -p 3000 /data/san-francisco-bay_california.osm.pbf
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,808
|
Very often we can become afraid that if we listen to our heart we might lose our head.
Years ago, I went to study in person the Institute of HeartMath in California.
Their research shows that the electromagnetic field of the heart is 60 times more powerful than the energy field of your brain.
"The heart is a sensory organ and acts as a sophisticated information encoding and processing center that enables it to learn, remember, and make independent functional decisions," HeartMath Institute Director of Research Rollin McCraty reports.
Your vagus nerve – your tenth cranial nerve – connects your brain, your heart and your digestive tract.
That is partly why when you are emotionally upset, you may experience high blood pressure, increased heart rate and a multitude of gastrointestinal problems.
When you relax, your brain and heart come into entrainment, which is a phenomenon in which your brain waves and the electrical rhythm of your heart come into energetic coherence.
When this happens, you access a high degree of intelligence.
Given the scientific support for following your heart, you can rest assured that listening to your heart is a smart thing to do.
Get comfortable. Either sit or lie down. Make sure you are warm enough. Prop yourself with pillows so that your physical body can completely let go.
Practice breathing exercises for a few minutes until you feel deeply calm. I recommend yoga pranayama exercises for this purpose. You can learn how for FREE at this link on my website Unlimited Energy Now. Deep breathing is one of the quickest ways for you to come into entrainment because your breath controls your heart rate.
Focus on your heart area. Relax!
Imagine placing the person, situation or dilemma you have been wanting guidance about in your heart area.
Allow yourself to feel the shift in energy.
When you have finished receiving your guidance, go back to your deep breathing exercises. Finish by calming your heart as much as possible.
When you are complete, write down your observations. Writing down the guidance you receive from your heart will help you integrate the information because you bring it back into language your brain can accept and understand.
From an energetic perspective, your heart may be either open or closed.
This happens when you are either punishing yourself for mistakes you feel you have made or if you are protecting yourself from further emotional pain.
If you close your heart down, however, you may be shutting down not only your emotions but also subtle degrees of intelligence.
It takes courage to feel.
When I practiced this exercise recently to receive my own guidance, my heart literally shuddered as I grasped the totality of the situation.
That is why you will want to begin and end with pranayama, allowing your breath to guide your heart back to a peaceful state.
Because your heart is so much more powerful than your brain, you may be surprised to discover the depth of information you receive this way.
Open your heart and have the courage to listen to what it has to say!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 29
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{"url":"https:\/\/twhl.info\/thread\/view\/18410?page=30","text":"# Sledge (Hammer Alternative) Alpha Build Created 4 years ago2013-07-27 02:24:39 UTC by Penguinboy\n\nCreated 4 years ago2013-07-27 02:24:39 UTC by\n\nPosted 2 years ago2016-02-20 00:52:01 UTC Post #328952\ngot your game setup locations set properly?\nTetsu0Original Cowboy\nPosted 2 years ago2016-02-20 14:47:20 UTC Post #328953\nI think so.\nBy the way, it doesn't launch the game nor does it ask me to run the game.\nPosted 2 years ago2016-02-20 19:40:00 UTC Post #328954\nI've never checked \"steam install\" maybe that's the cause.\nTetsu0Original Cowboy\nPosted 2 years ago2016-03-22 23:36:59 UTC Post #329520\nI have a question. Is there any way to save as an OBJ and keep the texture coords? I would absolutely love a good BSP map editor for use in more modern engines that use 3d model based maps. Its so hard to find a decent way to make BSP style maps. Exporting from Sledge works awesomely but having to redo the UVs and re texture everything is a bummer\nPosted 2 years ago2016-03-23 02:51:37 UTC Post #329522\nI don't know a lot about the OBJ format, so I'm not sure. It's probably possible, but I've never worked with OBJ textures before.\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-03 07:52:35 UTC Post #329641\nHello guys, does sledge editor work with floating point?\n\nI am hopeful because I want know because JackHammer has bug of floating points because after compiler than maybe textures are not visibled or shaped-holes. That is why I want know if sledge 1:1 like Valve Hammer Editor too\nPosted 2 years ago2016-04-03 08:16:11 UTC Post #329643\nIt should! If you find any cases of Sledge producing different results to VHE (plus HLFix or a floating point hack), let me know and I'll try and fix it!\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-04 21:02:20 UTC Post #329665\nOh @Penguinboy, thanks for support!\n\nBut I am using JackHammer but JackHammer has problem with floating point because saved map was generated by JackHammer and vhlt compiled too slow because vhlt doesn't know about wrong map format ( just it used only Quake map format because it hasn't reader for floating points.\n\nPS: HLFix doesn't work for me because I am using Windows 7 Pro x64\n\nVHE with hack of floating points yes and 32K yes I have here... from Sven Coop Forum\nPosted 2 years ago2016-04-05 04:53:38 UTC Post #329668\nI could never get VLuzacn floating points to work with Sledge.\nError: Not hammer.exe from Valve Editor etc..\nPosted 2 years ago2016-04-05 19:14:17 UTC Post #329677\n@Kachito\nHi You know \"floating point\" need if you draw brush very small and it compiles generating brush like my window of nice hallway. ( Did you see pictures from Show your screenshots from WIP )\nFor example it is happens without floating points and vhlt builds very slow because map file hasn't recognizable floating points. It means issues of map structure and it looks like Quake Map format and it sees wrong to generate to bsp-file with hidden hole-able shapes. That is why floating point is very important for GoldSource Engine. It is fixing to hidden shapes.\nExample for screenshot: Used from JackHammer or Sledge Editor:\nUsed with valve Hammer Editor with floating points:\nThat is proof of floating point.\nI hope your crab hugs sledge developer\nPosted 2 years ago2016-04-05 20:21:01 UTC Post #329678\nHi SourceSkyBoxer, thanks for showing those examples.\nCould you use Jpegs for screenshots, though? 1MB for a 1024 x 768 image is pretty unnecessary.\nI know 1MB isn't massive, but I view the site on my phone quite often and it all adds up!\nArchieGoodbye Moonmen\nPosted 2 years ago2016-04-05 21:26:11 UTC Post #329679\n@Archie:\nOh sorry! I will convert to beloved jpeg all screenshots. I never forget that. You're right because I don't know what is big compressed image format. jpeg is smaller than png...\nPosted 2 years ago2016-04-05 21:29:36 UTC Post #329680\nArchieGoodbye Moonmen\nPosted 2 years ago2016-04-05 22:12:13 UTC Post #329681\nI know 1MB isn't massive, but I view the site on my phone quite often and it all adds up!\nWhen you pay 3 euros\/month for 5GB of mobile internet those problems cease to exist.\nStrikerI seriously doubt myself\nPosted 2 years ago2016-04-05 22:12:45 UTC Post #329682\nThanks for the screenshots, can you upload an RMF so I can test it for myself?\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-06 02:13:39 UTC Post #329685\n@Striker:\nMobile data 5 GB pro 3 Euro\/Month.\n\nSo arrogant!! My mobile data 5 GB from Germany, o2 costs 25 Euro\/Month. But it is too expensive to you. You're living in Greek, right?\n\n@Penguinboy:n\nSorry my rmf was after my working progress. Find my http:\/\/twhl.info\/forums.php?thread=14851&page=last with pictures andmy username! Because I am continued to drawing map and snap many screenshots...\nPosted 2 years ago2016-04-06 02:53:01 UTC Post #329686\nThat's fine, but without a sample map to reproduce the issue, I wont be able to fix it... can you make a sample map that I can use to compare VHE to Sledge?\n\nDoes anyone else know about this issue? I'd like to be able to reproduce it if possible.\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-06 17:18:09 UTC Post #329696\n@PenguinBoy,\nI send you zip with 3 different version of vhe, jh and se\n3 different format:\nvhe = Valve Hammer Editor with Floating Point\njh = JackHammer\nse = Sledge Editor\n\nPS: jh and se have made by me black hollowed outside-box\nAnd vhe has red holdbox like you want check without leak.\nPlease do not open rmf just you should open _vhe.map than you see with red hollowed outside-box drawing by Valve Hammer Editor\n\nPlease promise me! Do not injury my copy right of my map!\n\nIf you're serious and honest than I will make safe. Because I am scared who steals my nice map than somebody hurts my architecture!\n\nThanks! File was uploaded to private message.\nPosted 2 years ago2016-04-06 17:42:01 UTC Post #329697\nWe employed* Penguinboy as a programmer on our mod The Core. Turns out he was just doing it to get access to our nice map and hurt all the architecture. Don't fall for the same trick, SSB!\n\n\n\n*non-monetarily\nArchieGoodbye Moonmen\nPosted 2 years ago2016-04-06 18:42:37 UTC Post #329698\n@Archie:\nOh PenguinBoy is programmer.. Yes I saw because he uses C# good work!\n\nI want believe this because you are good like other real good people. But I don't know that. That is why I will publish sometimes things if you customize your The Core of my great brush architecture - If you decide.\n\nI am working as my own mod. But I don't know what is correct game title of my own mod. I accept to publish my files if you need. Because I appreciate safe with TWHL\n\nDon't worry!\nPosted 2 years ago2016-04-06 20:35:47 UTC Post #329700\nHaha, don't worry, nobody will steal your work here.\nArchieGoodbye Moonmen\nPosted 2 years ago2016-04-06 23:29:09 UTC Post #329704\nThanks for uploading those files, SSB. I actually couldn't get it looking nicely on any editor I used, see examples:\n\nSledge: https:\/\/i.imgur.com\/Npg2id1.png\nJackhammer: https:\/\/i.imgur.com\/jeOiPAs.png\nHammer (normal): https:\/\/i.imgur.com\/HHP2m6u.png\nHammer (floating point hack): https:\/\/i.imgur.com\/WKxUrp8.png\n\nAs you can see the normal Hammer is closest (because everything is rounded to the closest grid point) but there are still a few gaps in the geometry. What's your process to get it looking correct?\n\nWith all the off-grid geometry it's not hard to see why each editor behaves differently. The MAP format has trouble with off-grid points at the best of times.\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-06 23:54:32 UTC Post #329705\nHi @Penguinboy, I have a problem with the sledge, when I put the spirit.fgd file it gave me a error saying:\n\nUnable to parse FGD. Unexpected At On line 1296, character 1.\n\nIn hammer and jackhammer doesn't gave me errors so I don't know what's happening, could yo help me? :V\nPosted 2 years ago2016-04-07 00:11:05 UTC Post #329706\nCan you post the FGD on dropbox\/pastebin? I'm not getting that error on the SOHL 1.8 FGD.\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-07 02:17:34 UTC Post #329707\nPosted 2 years ago2016-04-07 03:07:14 UTC Post #329708\nLooks like you've got a badly formatted FGD file. I made some fixes to it, this one will work: http:\/\/pastebin.com\/raw\/3jvcW5dz\n\nThe problems were:\nLine 1295, an empty class with no brackets, I added [ and ]\nLine 1348, the close quote was missing, I added \" to fix it\n\nThe first problem could potentially be a feature of FGD that Sledge doesn't support, so I'll fix it so it works. But the second issue is definitely a problem in the FGD itself, not Sledge.\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-07 03:12:35 UTC Post #329709\nThanks, it work, keep working on sledge its a very nice program, the only thing it needs its a source suport :V\nPosted 2 years ago2016-04-07 03:25:19 UTC Post #329710\nSledge will support Source just as soon as The Core is released.\nPosted 2 years ago2016-04-07 03:33:58 UTC Post #329711\nI suspect it may take longer than that >_>\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-07 09:41:19 UTC Post #329712\nHi @Penguinboy correct!\n\njust my mistake because \"normal\" or \"floating-used\" hammer. It is okay my process.\n\nThanks for understanding! I am sorry for my mistake because I don't know because hammer.exe by me I have 2 patchers by Vluzacn and i have applied 2 patchers to hammer. Than hammer still patched and I used it. That is why I am not using floating. I don't know because I already installed Valve Hammer Editor with -\/+32768 and fresh. If you like fix normal version of valve hammer editor to Sledge Editor, why not?\n\nPlease please give me! I need spirit.fgd for Valve Hammer Editor 3.5.3 because fgd doesn't show models... How do I fix it?\n\nThanks\n\n@DiscoStu , really you want The Core released than Source Engine moves to SLedge Editor? Wow Nice to listen! I hope Sledge Editor wins again JackHammer. I would like to support Sledge Editor because it is good and has built-in universal patch compiler.\n\n\/\/ Edit:\nI have big idea for sledge:\nFix normal points like normal valve hammer 3.5.x\nSphere and trows should to get 2 triangles like my map curved brushwork and do not use cube-like brush from Source Engine. You know I created trows-like tunnel for subways from Sven Coop Forum I think post from 2012 - 2014.\n\nThat is why sphere and trows need many triangles better than cube-like brushwork.\n\nAnd we would like to add function choose for aurora if you use spirit of half life than sledge will visible a built-in button ... Like models and sprite and creator for detail textures and manage with own was if you have customized textures than you just add or update texture to own wad package better than we don't need to use wally or makewad. Why not?\n\nImport\/export supports 3ds format because it is very cleaning better than wavefront obj\n\nIf you like my suggestions\nPosted 2 years ago2016-04-09 07:33:48 UTC Post #329740\nCould someone please confirm that you can only nudge vertices with the arrow keys if you drag-select them (and cannot if you click-select them)? It's kinda annoying.\nPosted 2 years ago2016-04-09 08:37:14 UTC Post #329742\nYou're right, what an odd bug. The VM tool is rewritten in the next release, so it shouldn't be an issue, but I'll double check to make sure it's fixed.\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-10 00:30:27 UTC Post #329748\nSo I can't compile any maps. Sledge doesn't seem to be including any of my .wad files during compile. I have the necessary .wad files in my mod folder and the valve folder. I have set my max texture memory to a huge number. I am not using too many\/large textures or too many wads. I have all wad names entered both as \"additional wads\" and as \"whitelisted wads\" in the texture section.\n\nHere's a typical HLCSG error log that I get...\n\n[i]\nWadfiles not in use by the map will be excluded\n\nCreateBrush:\n(0.00 seconds)\nWorld bounds: (-384 -448 -32) to (224 160 160)\n124 map planes\nCSGBrush:\n(0.01 seconds)\n240 csg faces\n120 used faces\n0 tiny faces\n0 tiny clips\nWad files required to run the map: (None)\n\nFindTextures elapsed time = 0ms\nqsort(miptex) elapsed time = 0ms\n\n--- END hlcsg ---\n[\/i]\n\nI can compile my simple test map perfectly fine in Hammer, so the problem is only occurring in Sledge. Am I missing a step somewhere?\nPosted 2 years ago2016-04-10 03:33:11 UTC Post #329750\nHmm, I'd have to see your settings to make sure, but there's a few things you can try. First of all, there's no need to have so many copies of the wads, it's best to remove them from 'additional wads' and only have one copy in your game or mod directory, Sledge will automatically load them from there. Also make sure you're using the VHLT tools because the original HL tools have issues finding the correct files when they all aren't on the same drive.\n\nThe compile will set the detected list of wads as a property of your map, you can see it in the 'map properties' dialog. The list should be set to the list of wads that the compile tools will use during compile.\nIf that list is wrong, then it's likely a problem with Sledge. If the list looks right, it might be an issue with your settings or compile tools. If you can provide more info about your settings and compile tools it would help with finding the issue.\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-10 05:52:36 UTC Post #329751\nSorry, I should have mentioned I'm using Vluzcan's ZHLT. Thanks for the quick reply!\n\nThe issue is definitely in the Map Properties! The \"wad\" value is blank! You mentioned a list of wads and the screenshot makes it look like it's located somewhere in the Half-Life directory. Is this just a text file of some sort or a folder or...?\n\nEDIT: Compiling my map seems to set the \"wad\" value to blank. Values that I enter in that field stay until I attempt to compile.\nPosted 2 years ago2016-04-10 07:34:19 UTC Post #329752\nYep, the setting is auto-set by Sledge every time you compile to the list of wad files that are being used by the map. For some reason, it thinks that no textures are in use.\n\nAfter a compile it's supposed to look something like: \"C:\\path\\to\\halflife.wad; C:\\path\\to\\decals.wad\", and so on. Setting it manually won't work because it'll get overwritten every time.\n\nIt's a strange one because I don't know how it would happen. Are you using WON or Steam Half-Life? What region\/language is your Windows set to? Is Sledge showing the textures correctly when you open the map?\nThis might be a possible cause, can you see if this helps:\n\nTake a look at the extensions of your wad files - See if the file extension is \"WAD\" (upper case) or \"wad\" (lower case). I suspect that Sledge might not work correctly when the extension isn't exactly \"wad\" in lower case, try renaming them if that's the case.\n\n(Note: In Windows XP you need to rename to something else first or it won't work:\n\nLet me know if this works! If not, it might help if you took a screenshot of the wads in your file explorer so I can see if anything looks strange. Make sure 'hide file extensions' is turned off!\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-10 08:02:11 UTC Post #329753\nWhy, hello guys. Peter Brev here, author of a Half-Life mod called Hard Duty.\n\nI installed many Hammer alternatives on my computer and I have a few things to say about Sledge.\n\nI started my mod using VHE. I, then, heard about Sledge and Jackhammer being two alternatives to mapping. I gave both editors a go. Sorry if I offense you, Penguinboy, but I chose Jackhammer over Sledge.\n\nWhile Sledge is very good and user-friendly, at least for me, I encountered too many problems as it is way too unstable on my computer and crashed each time I either alt+entered, saved the map, pressed apply on the entity's property, and so on... Most of the time, I lost all the development progress and that was very frustrating.\n\nI understand it is an alpha build, but as soon as there is a more stable version, I will gladly give it a new try.\n\nI wish you good luck, though, and keep up the good work,\n\nPeter\nPosted 2 years ago2016-04-10 08:14:26 UTC Post #329754\nThanks for the feedback. When did this happen, and did you submit the bugs when the error happened? It sounds like your particular issues might be caused by something in your FGD so it might be easy to fix.\n\nOf course Sledge shouldn't crash even if your FGD is strange so it needs to be fixed in the program itself. Do you mind sending me a copy of your FGD and a small RMF that is problematic (e.g. one entity or whatever triggers the error) so I can debug the issue?\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-10 08:41:27 UTC Post #329755\nI have sent you the necessary elements in a PM. Have you received them?\nPosted 2 years ago2016-04-10 09:14:18 UTC Post #329758\nYes, thanks. I'll see if they cause any issues for me.\n\nThe best way for me to find an issue is through the bug submission system. Did you ever click 'submit bug' when you got an error? If you could let me know what day you did that, I'd be able to find the error detail in the error logs. Or even better, could you open Sledge and try and get the error to happen, and then submit the error details?\n\nEdit: sorry, those FGDs are Jackhammer ones which Sledge doesn't support yet. Can you upload the ones you were using with Sledge? Though if you haven't made any changes to them from the standard HL1 FGD then it might be a different issue entirely.\nPenguinboyHaha, I died again!\nPosted 2 years ago2016-04-10 09:43:18 UTC Post #329760\nUnfortunately, I did not. I only got that box once, but did not send anything. I shall remember to do it next time. However, most of the time, Sledge just stops responding and I have to kill the process anyway, so I do not get the error report box most of the time.\n\nI will upload the Sledge's FGD I was using in a PM.\n\nEdit: I have sent the FGDs.\nPosted 2 years ago2016-04-10 09:55:49 UTC Post #329761\nhi @PeterBrev\nPlease check my screenshots - If Sledge doesn't generate clean map because it has bug since compiling and you see bad in-game.\n\nHey my dear @penguin, iek iek, How is your fix? sometimes, I have readden your messages. Please fix bug of map version of Valve Hammer Editor. I am hopeful to you. If you find solutions.\n\nThanks\nPosted 2 years ago2016-04-10 13:55:32 UTC Post #329767\n@SSB\n\nThanks for the taking the time to respond but the problem is entirely different. I am talking about random crashes when performing the same things (see my first post above).\nPosted 2 years ago2016-04-10 17:02:25 UTC Post #329768\n@PeterBrev\nYes you know Valve Hammer Editor still is almost best program for Half-Life and Mods ever.\n\nOther like JackHammer and Sledge Editor are bugged to write bad map structure because both apps have \"negative\" floating points and Valve Hammer Editor hasn't floating points because vhe can write clean x, y, z and scale, texture and entities. Good work. Did you see my screenhots from current page number ( just scroll up and find my posts ) hallway-window picture sees bad because it was written by JackHammer or Sledge Editor. So sadly to other apps you can not be happy if your map structure messed up since writing process and generated bad shapes or holes to in-game.\n\nBut Q3Radiant and Valve Hammer Editor are winner again Sledge Editor and JackHammer.\n\n@all I think who want develop map editor like Valve Hammer Editor. It sees not good. So sad because Valve Software is great world champion again Russian and Australian.\n\nThat is bad for both countries like this:\nFor Australia: kangaroos hit with boxing to cowboys - No chance because cowboys have weapons. Peng, peng kangaroos are died.\nFor Russia: arrogant armies watch hard to cowboys - No chance because our horses trample scats to armies. Cowboys laugh to armies and they vomit since bad smelling of scats by horses.\n\nWho are cowboys? Answer it is Valve Software LLC, from the United States of America.\n\nThat is why you should believe me. I am not crazy I want explain only because I feel that Americans are very very arrogant and stubborn. That is why they won't release next version of Valve Hammer Editor because it is enough to use. Why do Americans not develop expand of map range from -\/+8192 to -\/+32768. I can not understand because they are sensitive. But Id Software , Quake Mapping hasn't problem to expand map range upto -\/+131072.\n\nIf you want continue your development of Sledge Editor. but map format still incorrectly. How? Maybe developers outside from the United States of America want be hero again Americans...\n\nBut you can't because map format is \"Valve220\" just it is not same of Quake map because it is different structure of Goldsource and Quake are very differently. I know you say me reasons....\n\nI want tell truth because typical Americans...\n\n\/\/ Edit:\nPosted 2 years ago2016-04-10 17:21:24 UTC Post #329769\nThat is why they won't release next version of Valve Hammer Editor because it is enough to use.\nThat version already exist and is called Hammer 4, it is designed for Source 2004 to Source 2013 and the \"next-gen\" version of Hammer is already developed and it's for Source 2 (see Dota 2 Workshop Tools).\nWhy do Americans not develop expand of map range from -\/+8192 to -\/+32768.\nValve has stopped the support of GoldSource for years. After they released the 2013 update and fixed the majority of problems, they were done with GoldSource, they moved to Source 2013, Source 2, Steam, HTC Vive, CS:GO, Dota 2 and insert the rest here.\nI can not understand because they are sensitive. But Id Software , Quake Mapping hasn't problem to expand map range upto -\/+131072.\nDo you want to know why we can extend the map limits on id Tech engines ? It's because id Software always released the FULL source code of their engines\/games and the only people to get their hand on the FULL GoldSource\/Half-Life engine are people who either paid for it or earned it (Sven-Coop). It's a shame that id Software stopped that tradition since id Tech 4 (RAGE\/Wolfenstein The New Order\/Wolfenstein The Old Blood\/The Evil Within).\nBut you can't because map format is \"Valve220\" just it is not same of Quake map because it is different structure of Goldsource and Quake are very differently. I know you say me reasons....\nThe format isn't a barrier, how Linux is able to read\/write on NTFS (Windows) partitions which Microsoft never released the specs for it ? Reverse engineering and testing. How can Jackhammer export to RMF ? Reverse engineering and testing. This how mods for \"un-moddable games\" are born.\nShepard62700FRHalf-Cat is watching...\nPosted 2 years ago2016-04-10 17:44:03 UTC Post #329770\n@Penguinboy\nYep! Renaming my wad files from \".WAD\" to \".wad\" fixed the issue! My wad-creating program exports them with a capitalized extension, so I'll have to remember to rename them for now.\nPosted 2 years ago2016-04-10 19:14:08 UTC Post #329773\n@Shepard62700FR\n\nBut \"Valve Hammer Editor\" doesn't release more yet for Goldsource Engine because only. I know World Hammer Editor is for Source Engine 1\nAnd other dark Hammer 5 for Source Engine 2 ( Dota )\nValve has stopped the support of GoldSource for years. After they released the 2013 update and fixed the majority of problems, they were done with GoldSource, they moved to Source 2013, Source 2, Steam, HTC Vive, CS:GO, Dota 2 and insert the rest here.\nOh no Urby is right because I can not sell Goldsource Mod, Shit that is why I must convert from old Goldsource into Source. I feel because Source Engine can to sell and to make happy and to get success. I am shy that. Grr I must \"remake\" Thanks for help and support. I will move to Source Engine now. But I can't understand why Valve Software lies us since Half-Life has big update since Trinity Renderer doesn't work - if you play map than you are going to \"changelevel\" than Half-Life stopped changelevel. But Sven Coop has \"old version\" of Half-Life Engine and it works fine with Trinity Renderer check my video. But gun from weapon will be crashed but it is maybe okay maybe not work?\n\n@McSqueaky\nYeah I hope you fix it...\nPosted 2 years ago2016-04-10 20:01:36 UTC Post #329776\n@SSB Floating points are not too much of a big deal for me as I tend to rarely have such issue. If it were to happen, a simple brush recreation fixes the issue. Floating points happens mostly on terrain creating since it implies alot of vertex point movement which often lead to some points being off. However, again, this rarely happen.\nPosted 2 years ago2016-04-10 20:13:06 UTC Post #329777\nOh no Urby is right because I can not sell Goldsource Mod, Shit that is why I must convert from old Goldsource into Source.\nIf you want to stick to GoldSource. You can send an e-mail to Valve to see if they still allow you to purchase a full GoldSource licence. If you want to sell a Source engine game, you will need to buy the Havok license which is \\$25k and if you are using Bink\/Miles Sound System buy those licenses too.\nBut Sven Coop has \"old version\" of Half-Life Engine\nI thought they were using the latest version. I don't remember SC 5.0 having Direct3D support or scrapped off the SDL2 binaries. Are you sure it's really the case ?\nShepard62700FRHalf-Cat is watching...\nPosted 2 years ago2016-04-10 22:43:25 UTC Post #329794\nI don't know what the hell this conversation is even about anymore. Anywho...\n\nPengy, have you considered implementing skewing textures?\nI'm vaguely aware that you can do it with something like \"align to view\" and certain camera angles \u2014 I'll have to find the tutorial or whatever it was that detailed that \u2014 but a more standardized method could be helpful to some.","date":"2018-06-24 17:01:20","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.29727232456207275, \"perplexity\": 5218.42676526672}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-26\/segments\/1529267866984.71\/warc\/CC-MAIN-20180624160817-20180624180817-00489.warc.gz\"}"}
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Q: Update Computed column in entity framework I have a DateTime column in my app that gets auto calculated by SQLServer. It works great when creating the row. However, I would like to be able to updated later. I have the property marked with [DatabaseGenerated(DatabaseGeneratedOption.Computed)].
Is it possible to updated a property like that in EF?
A: No, with DatabaseGeneratedOption.Computed EF will never include the property in update and insert statements. In stead, it will always read its value after such statements.
If you have to update the property in client code you have no other option than removing the data annotation an make it an ordinary updateable property. You can set a default value in the owning class's constructor.
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Hillary Scott's Former School Destroyed in Nashville Tornado
Donelson Christian Academy, the school at which Lady Antebellum member Hillary Scott spent her middle and high school years, was destroyed in a tornado that tore through Nashville on Tuesday morning (March 3). Video and photos on social media show debris from the building and its classrooms scattered where the school once stood.
Donelson Christian Academy was located on Danyacrest Drive in the Donelson area of Nashville. The site of the school is almost directly east from the Five Points area of East Nashville, which also sustained major damage. Scott spent fifth grade through her senior year of high school at the school.
"Praying for our city. I can't believe the devastation ... People's homes are gone. Lives lost," Scott writes on Instagram Stories. "Please pray for Nashville, its people and all of the first responders who are trying to keep everyone safe as they assess the damage."
Tweets from Shelby Sansone, a reporter with Nashville's WSMV-TV, show the devastating damage the Donelson Christian Academy campus sustained:
At least 21 people, including two in East Nashville specifically, died in the Tuesday morning tornado, which swept through Nashville's Germantown and East Nashville neighborhoods before moving east. More than 40 buildings collapsed, the Tennessean reports, and others sustained everything from minor to major damage. The tornado affected Wilson County, Benton County and Putnam County in addition to Davidson County, in which Nashville is located.
"There's a really good possibility that there may be more," Gov. Bill Lee said at a morning press conference. "It's early yet."
The Community Foundation of Middle Tennessee has set up a fund for victims of the tornado; to donate, visit CFMT.org. Hands on Nashville, the Red Cross and other organizations are also organizing donations and volunteers, according to the Tennessean.
More From the 2020 Nashville Tornado
Source: Hillary Scott's Former School Destroyed in Nashville Tornado
Filed Under: 2020 nashville tornado, hillary scott, lady antebellum
Categories: Country, Music News, Tri-State Weather
Cold Weather's a Beach - Panama City Beach Trip Winner
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David Oyetokunbo Oyelowo (Oxford, Oxfordshire, 1 maart 1976) is een Brits acteur.
Biografie
Oyelowo werd geboren in Oxford als zoon van Nigeriaanse ouders. Hij volgde theaterstudies aan het City and Islington College waar zijn leraar hem aanraadde acteur te worden. Hij verkreeg een studiebeurs aan de London Academy of Music and Dramatic Art waar hij na drie jaar opleiding afstudeerde in 1998. Hij begon zijn theatercarrière in 1999 bij de Royal Shakespeare Company en speelde verschillende rollen in televisieseries. Vanaf 2001 speelde hij ook enkele kleinere rollen in speelfilms. In 2012 kreeg hij een rol in Middle of Nowhere van Ava DuVernay (première op het Sundance Film Festival) en in The Paperboy van Lee Daniels, die meedeed in de competitie van het Filmfestival van Cannes. Hij speelde opnieuw in een film van Lee Daniels, The Butler (2013) en in een film van Ava DuVernay, Selma (2014). Voor deze laatste vertolking kreeg hij verschillende nominaties.
Filmografie
*Exclusief televisiefilms
The Midnight Sky (2020)
A Wrinkle in Time (2018)
The Cloverfield Paradox (2018)
A United Kingdom (2016)
A Most Violent Year (2014)
Selma (2014)
A Most Violent Year (2014)
Default (2014)
Interstellar (2014)
The Butler (2013)
Jack Reacher (2012)
Lincoln (2012)
Middle of Nowhere (2012)
The Paperboy (2012)
Red Tails (2012)
96 Minutes (2011)
The Help (2011)
Rise of the Planet of the Apes (2011)
Rage (2009)
A Raisin in the Sun (2008)
Who Do You Love? (2008)
The Last King of Scotland (2006)
As You Like It (2006)
Derailed (2005)
The Best Man (2005)
A Sound of Thunder (2005)
Dog Eat Dog (2001)
Televisie
Les Misérables (2018)
The Lion Guard (2016-2019)
Star Wars Rebels (2014-2018)
Complicit (2013)
Small Island (2009)
Five Days (2007)
Spooks (2002-2004)
Prijzen & nominaties
Oyelowo ontving een tiental filmprijzen en nominaties waarvan de belangrijkste:
Externe link
Brits acteur
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\section{Introduction}
\label{sec:introduction}
Recent works have shown how low-dimensional representations captured
by generative models can be successfully exploited in reinforcement
learning (RL) settings. Among others, these generative models have
been used to learn low-dimensional latent representations of the state
space to improve the learning efficiency of RL
algorithms~\cite{zhang2018arxiv,gelada2019arxiv}, or to allow the
generalization of policies learned on a source domain to other target
domains~\cite{finn2016icra,higgins2017icml}. The DisentAngled
Representation Learning Agent (DARLA) approach~\cite{higgins2017icml},
in particular, builds such latent representations using variational
autoencoder (VAE) methods~\cite{kingma2013arxiv,rezende2014arxiv}, and
shows how learning disentangled features of the observed environment
can allow an RL agent to learn a policy robust to some shifts in the
original domain.
In this work, we explore the application of these latent
representations in capturing different input sensory modalities to be
considered in the context of RL tasks. We build upon recent work that
extends VAE methods to learn joint distributions of multiple
modalities, by forcing the individual latent representations of each
modality to be similar~\cite{suzuki2016arxiv,yin2017aaai}. These
multimodal VAEs allow for \emph{cross-modality inference}, replicating
more closely what seems to be the nature of the multimodal
representation learning performed by
humans~\cite{damasio1989cognition,meyer2009tns}. Inspired by these
advances, we explore the impact of such multimodal latent
representations in allowing a reinforcement learning agent to learn
and exploit policies over different input modalities. Among others, we
envision, for example, scenarios where reinforcement learning agents
are provided the ability of learning a visual policy (a policy learned
over image inputs), and then (re-)using such policy at test time when
only sound inputs are available. Figure~\ref{fig:example} instantiates
such example to the case of video games---a policy is learned over
images and then re-used when only the game sounds are available,
\emph{i.e.,} when playing ``in the dark''.
\begin{figure}[t]
\centering
\includegraphics[width=8.5cm]{images/intro/eps/intro_image.eps}
\caption{Concrete scenario where a policy trained over
one input modality (game videoframes) is transferred to a
different modality (game sound).}
\label{fig:example}
\end{figure}
To achieve this, we contribute an approach for multimodal transfer
reinforcement learning, which effectively allows an RL agent to learn
robust policies over input modalities, achieving better out-of-the-box
performance when compared to different baselines. We start by first
learning a generalized latent space over the different input
modalities that the agent has access to. This latent space is
constructed using a multimodal generative model, allowing
the agent to establish mappings between the different modalities---for
example, \emph{``which sounds do I typically associate with this
visual sensory information''}. Then, in the second step, the RL
agent learns a policy directly on top of this latent
space. Importantly, during this training step, the agent may only have
access to a subset of the input modalities (say, images but not
sound). In practice, this translates in the RL agent learning a policy
over a latent space constructed relying only on some
modalities. Finally, the transfer occurs in the third step, where, at
test time, the agent may have access to a different subset of
modalities, but still perform the task using the same policy. These
results hold consistently across different OpenAI
Gym~\cite{brockman2016arxiv} and Atari-like~\cite{bellemare2013jair}
environments. This is the case even when using different multimodal
generative models~\cite{yin2017aaai} and reinforcement
learning algorithms~\cite{mnih2015nature,lillicrap2015arxiv}.
\begin{figure*}[t]
\centering
\begin{subfigure}[t]{0.16\linewidth}
\centering
\includegraphics[height=4.3cm]{images/deep_generative_models/eps/singlemod_representation_learning.eps}
\caption{}
\label{fig:vae_model}
\end{subfigure}
%
%
\begin{subfigure}[t]{0.31\linewidth}
\centering
\includegraphics[height=4.3cm]{images/deep_generative_models/eps/hang_representation_learning.eps}
\caption{}
\label{fig:avae_model}
\end{subfigure}
%
%
\begin{subfigure}[t]{0.52\linewidth}
\centering
\includegraphics[height=4.3cm]{images/deep_generative_models/eps/suzuki_representation_learning.eps}
\caption{}
\label{fig:jmvae_model}
\end{subfigure}
%
\caption{Networks of different generative models, highlighting the
models' data encoding (orange) and decoding (green) pipelines. The
similarity constraints imposed by their training procedures are
presented in dashed lines. \ref{fig:vae_model}) The VAE model
learns a latent representation of the data distribution of a
single modality. \ref{fig:avae_model}) On the other hand, the AVAE
model extends the previous framework to account for multiple
modalities, allowing for cross-modality
inference. \ref{fig:jmvae_model}) Finally, the JMVAE model learns
a representation of both modalities, allowing for both single and
joint modality reconstruction, and cross-modality inference.}
\label{fig:deep_generative_models}
\end{figure*}
The third and last step unveils what sets our work apart from the
existing literature. By using (single-modality) VAE methods, the
current state-of-art approaches implicitly assume that the source and
target domains are characterized by similar inputs, such as raw
observations of a camera. In these approaches, the latent space is
used to capture isolated properties (such as colors or shapes) that
may vary throughout the tasks. This is in contrast with our approach,
where the latent space is seen as a mechanism to create a mapping
between different input modalities.
The remainder of the paper is structured as follows. We start in
Section~\ref{sec:preliminaries} by introducing relevant background and
related work on generative models and reinforcement learning. Then, in
Section~\ref{sec:approach} we introduce our approach to multimodal
transfer reinforcement learning, and evaluate it in
Section~\ref{sec:experimental evaluation}. We finish with some final
considerations in Section~\ref{sec:conclusions}.
\section{Preliminaries}
\label{sec:preliminaries}
This section introduces required background on deep generative models
and deep reinforcement learning.
\subsection{Deep Generative Models}
\subsubsection{Variational Autoencoders}
Deep generative models have shown great promise in learning
generalized representations of data. For single-modality data, the
variational autoencoder model (VAE) is widely used. The VAE
model~\cite{kingma2013arxiv} learns a joint distribution
$p_{\theta}(\vect{x}, \vect{z})$ of data $\vect{x}$, which is
generated by a latent variable $\vect{z}$. Figure~\ref{fig:vae_model}
depicts this model. The latent variable is often of lower
dimensionality in comparison with the modality itself, and acts as the
representation vector in which data is encoded.
The joint distribution takes the form
$p_{\theta}(\vect{x}, \vect{z}) = p_{\theta}(\vect{x}\, \vert\,
\vect{z}) \, p(\vect{z})$, where $p(\vect{z})$ (the
\textit{prior} distribution) is often a unitary Gaussian
($\vect{z} \sim \mathcal{N}(\vect{0}, \mat{I})$). The
generative distribution $p_{\theta}(\vect{x}\, \vert\, \vect{z})$,
parameterized by $\theta$, is usually composed with a simple
likelihood term (e.g. Bernoulli or Gaussian).
The training procedure of the VAE model involves the maximization of
the evidence likelihood $p(\vect{x})$, by marginalizing over the
latent variable and resorting to an inference network
$q_{\phi}(\vect{z}\vert\vect{x})$ to approximate the posterior
distribution. We obtain a lower-bound on the log-likelihood of the
evidence (ELBO)
$\log p(\vect{x}) \geq \mathcal{L}_{\text{VAE}}(\vect{x})$, with
\begin{equation*}
\mathcal{L}_{\text{VAE}}(\vect{x})
=
\lambda\, \EE[q_{\phi}(\vect{z}\vert\vect{x})]
{\log p_{\theta} (\vect{x}\vert\vect{z})} - \, \beta\, \KL \left[q_{\phi}(\vect{z} \vert \vect{x}) \, \| \, p(\vect{z})\right],
\end{equation*}
where the Kullback-Leibler divergence term
$\KL \left[q_{\phi}(\vect{z} \vert \vect{x}) \,\|\,
p(\vect{z})\right]$ promotes a balance between the latent channel's
capacity and the encoding process of data. Moreover, in the model's
training procedure, the hyperparameters $\lambda$ and $\beta$ weight
the importance of reconstruction quality and latent space
independence, respectively. The optimization of the ELBO is performed
resorting to gradient-based methods.
\subsubsection{Multimodal Variational Autoencoders}
VAE models have been extended in order to perform inference across
different modalities. The Associative Variational Autoencoder (AVAE)
model~\cite{yin2017aaai}, depicted in Figure~\ref{fig:avae_model}, is
able to learn a common latent representation of two modalities
($\vect{x}, \vect{y}$). It does so by imposing a similarity
restriction on the separate single-modality latent representations
($\vect{z}_{x}, \vect{z}_{y}$), employing a KL divergence term on the
ELBO of the model:
\begin{equation*}
\mathcal{L}_{\text{AVAE}}(\vect{x}, \vect{y})
=
\mathcal{L}_{\text{VAE}}(\vect{x}) + \mathcal{L}_{\text{VAE}}(\vect{y}) - \,\alpha \KL^{\star} \left[q_{\phi}(\vect{z}_{x} \vert \vect{x}) \, \| \, q_{\phi}(\vect{z}_{y} \vert \vect{y})\right]
\end{equation*}
where $\KL^{\star} \left[p \, \| \,q \right]$ is the symmetrical
Kullback-Leibler between two distributions $p$ and $q$, and $\alpha$
is a constant that weights the importance of keeping similar latent
spaces in the training procedure~\cite{yin2017aaai}. We note that each
modality is associated with a different encoder-decoder
pair. Moreover, the encoder and the decoder can be implemented as
neural networks with different architectures.
Other models aim at learning a joint distribution of both modalities
$p_\theta(\vect{x}, \vect{y})$. Examples include the Joint Multimodal
Variational Autoencoder (JMVAE)~\cite{suzuki2016arxiv} or the
Multi-Modal Variational Autoencoder
(M$^2$VAE)~\cite{korthals2019arxiv}. These generative models are able
to build a representation space of both modalities simultaneously
while maintaining similarity restrictions with the single-modality
representations, as shown in the JMVAE model presented in
Figure~\ref{fig:jmvae_model}.
However, a fundamental feature of all multimodal generative models is
the ability to perform \emph{cross-modality inference}, that is the
ability to input modality-specific data, encode the corresponding
latent representation, and, from that representation, generate data
of a different modality. This is possible due to the forced
approximation of the latent representations of each modality, and the
process follows the orange and green arrows in
Figure~\ref{fig:deep_generative_models}.
\subsection{Reinforcement Learning}
Reinforcement learning (RL) is a framework for optimizing the
behaviour of an agent operating in a given environment. This framework
is formalized as a Markov decision process (MDP)---a tuple
$\mathcal{M} = (\mathcal{X},\mathcal{A},P,r,\gamma)$ that describes a sequential decision problem
under uncertainty. $\mathcal{X}$ and $\mathcal{A}$ are the state and action spaces,
respectively, and both are known by the agent. When the agent takes an
action $a\in\mathcal{A}$ while in state $x\in\mathcal{X}$, the world transitions to
state $y\in\mathcal{X}$ with probability $P(y\mid x,a)$ and the agent receives
an immediate reward $r(x,a)$. Typically, functions $P$ and $r$ are
unknown to the agent. Finally, the discount factor
$\gamma \in [0, 1)$ sets the relative importance of present and future
rewards.
Solving the MDP consists in finding an optimal policy $\pi^*$---a
mapping from states to actions---which ensures that the agent collects
as much reward as possible. Such policy can be found from the optimal
$Q$-function, which is defined recursively for every state action pair
$\left( x, a \right) \in \mathcal{X} \times \mathcal{A}$ as
\begin{equation*}
Q^*(x,a)
=
r(x, a)
+
\gamma
\sum_{y\in\mathcal{X}}
P(y \mid x, a)
\max_{a'\in\mathcal{A}} Q^*(y, a').
\end{equation*}
Multiple methods can be used in computing this
function~\cite{sutton1998reinforcement}, for example
$Q$-learning~\cite{watkins89phd}.
More recently, research has geared towards applying deep learning
methods in RL problems, leading to new methods. For example, Deep $Q$
Network (DQN) is a variant of the $Q$-learning algorithm that uses a
deep neural network to parameterize an approximation of the
$Q$-functions $Q(x, a; \theta)$, with parameters $\theta$. DQN assumes
discrete action spaces $\mathcal{A}$, and has been proved suitable for learning
policies that beat Atari games~\cite{mnih2015nature}. Continuous
action spaces require specialized algorithms. For example, Deep
Deterministic Policy Gradient (DDPG) is an actor-critic, policy
gradient algorithm that can deal with continuous action spaces, and
has been shown to perform well in complex control
tasks~\cite{lillicrap2015arxiv}.
\section{Multimodal Transfer Reinforcement Learning}
\label{sec:approach}
Consider an agent facing a sequential decision problem described as an
MDP $\mathcal{M} = \left( \mathcal{X}, \mathcal{A}, P, r, \gamma \right)$. This agent is
endowed with a set $\left\lbrace I_1, I_2, \dots, I_N \right\rbrace$
of $N$ different input modalities, which can be used in perceiving the
world and building a possibly partial observation of the current state
$x \in \mathcal{X}$. Different modalities may provide more, or less, perceptual
information than others. Some modalities may be redundant
(\emph{i.e.}, provide the same perceptual information) or complement
each other (\emph{i.e.}, jointly provide more
information). Figure~\ref{fig:perceptual information} provides an
abstract illustration of the connection between different input
modalities, and corresponding impact in the state space that can be
perceived by the agent.
Our goal is for the agent to learn a policy while observing only a
subset of input modalities $\boldsymbol{I}_\mathrm{train}$, and then use that same
policy when observing a possibly different subset of modalities,
$\boldsymbol{I}_\mathrm{test}$, with as minimal performance degradation as
possible.
Our approach consists of a three-stages pipeline:
\begin{enumerate}
\item
\emph{Learn a perceptual model of the world.}
\item
\emph{Learn to act in the world.}
\item
\emph{Transfer policy.}
\end{enumerate}
We now discuss each step in further detail.
\begin{figure}[t]
\centering
\includegraphics[width=4cm]{images/multimodal_transfer_reinforcement_learning/eps/state_space_v2.eps}
\caption{Connection between two abstract input modalities $I_1$ and
$I_2$, and corresponding impact in the agent's perceptual
information. The elliptic surface $\mathcal{I}$ depicts the complete
perceptual space the agent can perceive with both modalities. The
colored projections in the axes depict the (reduced) perceptions
of the agent when using single modalities.}
\label{fig:perceptual information}
\end{figure}
\subsection{Learn a perceptual model of the world}
\label{subsec:generative model}
Let $\mathcal{I}$ denote the Cartesian product of input modalities,
$\mathcal{I} = I_1 \times I_2 \times \dots \times I_N$. Intuitively, we can
think of $\mathcal{I}$ as the complete perceptual space of the agent. We write
$\vect{i}$ to denote an element of $\mathcal{I}$. Figure~\ref{fig:map
modalities} depicts an example on a game, where the agent can have
access to two modalities, $I_\mathrm{image}$ and $I_\mathrm{sound}$,
corresponding to visual and sound information.
At each moment $t$, the agent may not have access to the complete
perception $\vect{i}(t)\in\mathcal{I}$, but only to a partial view
thereof. Following our discussion in Section~\ref{sec:introduction},
we are interested in learning a multimodal latent representation of
the perceptions in $\mathcal{I}$. Such representation amounts to a set of
latent mappings $\mathcal{F}=\{F_1,\ldots,F_L\}$. Each map $F_\ell$ takes the
form $F_\ell:\mathrm{proj}_\ell\to\mathcal{Z}$, where $\mathcal{Z}$ is a common latent
space and $\mathrm{proj}_\ell$ projects $\mathcal{I}$ to some subspace of $K$
modalities,
$\mathcal{I}_\ell=I_{\ell_1}\times I_{\ell_2}\times\ldots\times I_{\ell_K}$. In
Figure~\ref{fig:map modalities} the set of mappings $\mathcal{F}$ is used to
compute a latent representation $\vect{z}$ from sound and image data.
To learn such mappings, we start by collecting a dataset
of $M$ examples of simultaneous sensorial information:
\begin{equation*}
\mathcal{D}(\mathcal{I})
=
\big\{
\vect{i}^{(1)},\ldots,\vect{i}^{(M)}
\big\}.
\end{equation*}
We then follow an unsupervised learning approach, and train a
multimodal VAE on dataset $\mathcal{D}(\mathcal{I})$ to learn a generalized latent space
over the agent's input modalities. The latent mappings in $\mathcal{F}$
correspond to the encoders of the VAE model, while the decoders can be
seen as a set of inverse latent mappings,
$\mathcal{F}^{-1}=\{F^{-1}_1,\ldots,F^{-1}_L\}$ that allow for modality
reconstruction and cross-modality inference. Figure~\ref{fig:cross
modality} depicts an example of how the multimodal latent space can
be used for performing cross-modality inference of sound data given an
image input using the modality-specific maps.
The collection of the initial data needed to generate $\mathcal{D}(\mathcal{I})$ can be
easier/harder depending on the complexity of the task. In
Section~\ref{sec:experimental evaluation} we discuss mechanisms to
perform this.
\begin{figure*}[t]
\centering
\begin{subfigure}[t]{0.49\textwidth}
\centering
\includegraphics[width=8.20cm]{images/multimodal_transfer_reinforcement_learning/eps/learning_to_see.eps}
\caption{}
\label{fig:map modalities}
\end{subfigure}
%
\hfill
%
\begin{subfigure}[t]{0.49\textwidth}
\centering
\includegraphics[width=8.10cm]{images/multimodal_transfer_reinforcement_learning/eps/cross_modality_inference.eps}
\caption{}
\label{fig:cross modality}
\end{subfigure}
\caption{\ref{fig:map modalities}) Each time step of a game includes
visual and sound components that are intrinsically coupled. This
coupling can be encoded in a latent representation using the
family of latent maps $\mathcal{F}$.
%
\ref{fig:cross modality}) shows how the multimodal latent
representation can be used in inferring the sound associated with
a given image, using the image latent map $F_\mathrm{image}$ and
sound inverse latent map $F_{\mathrm{sound}}^{-1}$.}
\label{fig:map and cross modality}
\end{figure*}
\subsection{Learn to act in the world}
\label{subsec:reinforcement learning}
After learning a perceptual model of the world, the agent then learns
how to perform the task. We follow a reinforcement learning approach
to learn an optimal policy for the task described by MDP $\mathcal{M}$. During
this learning phase, we assume the agent may only have access to a
subset of input modalities $\boldsymbol{I}_\mathrm{train}$.
As a result, during its interaction with the environment, the agent
collects a sequence of triplets
\begin{equation*}
\left\lbrace
\left(
\vect{i}_\mathrm{train}^{(0)},
a^{(0)},
r^{(0)}
\right),
%
\left(
\vect{i}_\mathrm{train}^{(1)},
a^{(1)},
r^{(1)}
\right),
%
\dots
\right\rbrace,
\end{equation*}
where $\vect{i}_\mathrm{train}^{(t)}$, $a^{(t)}$, $r^{(t)}$ correspond
to the perceptual observations, action executed, and rewards obtained
at timestep $t$, respectively.
However, our reinforcement learning agent does not use this sequence
of triplets directly. Instead, it pre-processes the perceptual
observations using the previously learned latent maps
$\mathcal{F}$ to encode the multimodal latent state at each
time step as
$
\vect{z}^{(t)}
=
F_\mathrm{train}
\left(
\vect{i}_\mathrm{train}^{(t)}
\right)
$,
where $F_\mathrm{train}\in\mathcal{F}$ maps $I_\mathrm{train}$ into $\mathcal{Z}$. In practice, the RL agent uses a sequence of triplets
\begin{equation*}
\left\lbrace
\left(
\vect{z}^{(0)},
a^{(0)},
r^{(0)}
\right),
%
\left(
\vect{z}^{(1)},
a^{(1)},
r^{(1)}
\right),
%
\dots
\right\rbrace
\end{equation*}
to learn a policy $\pi : \mathcal{Z} \to \mathcal{A}$, that maps the latent states to
actions. Any continuous-state space reinforcement learning algorithm
can be used to learn this policy $\pi$ over the latent states. These
latent states are encoded using the generative model trained in the
previous section, and as such, the weights of this model should remain
frozen during the RL training.
\subsection{Transfer policy}
\label{subsec:transfer}
The transfer of policies happens once the agent has learned how to
perceive and act in the world. At this time, we assume the agent may
now have access to a subset of input modalities
$\boldsymbol{I}_\mathrm{test}$, potentially different from
$\boldsymbol{I}_\mathrm{train}$, \emph{i.e.}, the set of modalities it
used in learning the task policy $\pi$. As a result, during its
interaction with the environment, at each time step $t$, the agent will now
observe perceptual information $\vect{i}_\mathrm{test}^{(t)}$.
In order to reuse the policy $\pi$, the agent starts by pre-processing
this perceptual observation, again using the set of maps $\mathcal{F}$
previously trained, but now generating a latent state
$\vect{z}^{(t)} = F_\mathrm{test} \left( \vect{i}_\mathrm{test}^{(t)}
\right)$, where $F_\mathrm{test}\in\mathcal{F}$ now maps $I_\mathrm{test}$ into $\mathcal{Z}$. Since policy $\pi$ maps the latent space $\mathcal{Z}$ to the action
space $\mathcal{A}$, it can now be used directly to select the optimal action at the new
state $\vect{z}^{(t)}$.
Effectively, the agent is reusing a policy $\pi$ that was learned over
a (possibly) different set of input modalities, with no additional
training. This corresponds to a zero-shot transfer of
policies. Figure~\ref{fig:learn_to_act} summarizes the three-steps
pipeline hereby described.
\begin{figure}[b]
\centering
\includegraphics[width=8cm]{images/multimodal_transfer_reinforcement_learning/eps/learn_to_act_v2.eps}
\caption{Summary of the three-steps approach for cross-modality
transfer in reinforcement learning. The first step learns a
perceptual model of the world, described by the latent mappings
$\mathcal{F}$ (and corresponding inverses), which map perceptions to a
common latent space $\mathcal{Z}$. In the second step, the agent learns a
policy $\pi$ that maps the latent space to actions, with an RL
approach using observations from a given subset of input
modalities. The third step concerns the reuse off the same policy
$\pi$, assuming new observations from a potentially different
subset of modalities. This is possible by first encoding the new
observations in the multimodal latent space.}
\label{fig:learn_to_act}
\end{figure}
\section{Experimental Evaluation}
\label{sec:experimental evaluation}
We evaluate and analyze the performance of our approach on different
scenarios of increasing complexity, not only on the task but also on
the input modalities. We start by considering a modified version of
the \textsc{pendulum} environment from OpenAI gym, with a simple sound
source. Then, we consider \textsc{hyperhot}, a \textsc{space
invaders}-like game that assesses the performance of our approach in
scenarios with more complex and realistic generation of sounds.
\begin{figure}[t]
\centering
\includegraphics[width=3cm]{images/pendulum/eps/pendulum_description.eps}
\caption{Visual and sound perceptual information in the
\textsc{pendulum} scenario. The tip of the pendulum emits a
frequency that is received by three microphones placed at the
bottom left and right ($bl,br$) and middle top $(mt)$.}
\label{fig:pendulum}
\end{figure}
\subsection{\textsc{pendulum}}
We consider a modified version of the \textsc{pendulum} environment
from OpenAI gym---a classic control problem, where the goal is to
swing the pendulum up so it stays upright. We modify this environment
so that the observations include both an image and a sound
component. For the sound component, we assume that the tip of the
pendulum emits a constant frequency $f_0$, which is received by a set
of $S$ sound receivers
$\left\lbrace \rho_1, \dots, \rho_S
\right\rbrace$. Figure~\ref{fig:pendulum} depicts this scenario, where
the pendulum and its sound are in red, and the sound receivers
correspond to the circles.
Formally, we let $\mathcal{I} = I_{\mathrm{image}} \times I_{\mathrm{sound}} $
denote the complete perceptual space of the agent. The visual input
modality of the agent, $I_{\mathrm{image}}$, consists of the raw image
observation of the environment. On the other hand, the sound input
modality, $I_{\mathrm{sound}}$, consists of the frequency and
amplitude received by each of the $S$ microphones of the
agent. Moreover, both image and sound observations may be stacked to
account for the dynamics of the scenario.
In this scenario, we assume a simple model for the sound
generation. Specifically, we assume that, at each timestep, the
frequency $f'_i$ heard by each sound receiver $\rho_i$ follows the
Doppler effect. The Doppler effect measures the change in frequency
heard by an observer as it moves towards or away from the
source. Slightly abusing our notation, we let $\boldsymbol{\rho}_i$
denote the position of sound receiver $\rho_i$ and $\boldsymbol{e}$
the position of the sound emitter. Formally,
\begin{equation*}
f'_i
=
\left(
\frac
{c + \dot{\vect{\rho}}_i \cdot \left( \vect{e} - \vect{\rho}_i \right)}
{c - \dot{\vect{e}} \cdot \left( \vect{\rho}_i - \vect{e} \right)}
\right)
f_0,
\end{equation*}
where $c$ is the speed of sound and we use the dot notation to
represent velocities. Figure~\ref{fig:doppler effect} depicts the
Doppler effect in the pendulum scenario.
We then let the amplitude $a_i$ heard by receiver $\rho_i$ follow the
inverse square law
\begin{equation*}
a_i = \frac{K}{\|\vect{e} - \vect{\rho}_i\|^2},
\end{equation*}
where $K$ is a scaling constant. Figure~\ref{fig:inverse square law}
depicts the inverse square law applied to the pendulum scenario,
showing how the amplitude of the sound generated decreases with the
distance to the source.
We now provide details on how our approach was set up. All constants
and training hyper-parameters used are summarized in
Appendix~\ref{subsec:constants and params}.
\subsubsection{Learn a perceptual model of the world}
\label{subsubsec:pendulum learn to map inputs}
For this task, we adopted the Associative Variational AutoEncoder
(AVAE) to learn the family of latent mappings $\mathcal{F}$. The AVAE was
trained over a dataset $\mathcal{D}(\mathcal{I})$ with $M$ observations of images and
sounds
$\vect{i}^m = \left(i_\mathrm{image}^m, i_\mathrm{sound}^m\right)$,
collected using a random controller. The random controller proved to
be enough to cover the state space. Before training, the images were
preprocessed to black and white and resized to $60 \times 60$
pixels. The sounds were normalized to the range $[0, 1]$, assuming the
minimum and maximum values found in the $M$ samples.
For the image-specific encoder we adopted an architecture with two
convolutional layers and two fully connected layers. The two
convolutional layers learned $32$ and $64$ filters, respectively, each
with kernel size $4$, stride $2$ and padding $1$. The two fully
connected layers had $256$ neurons each. Swish
activations~\cite{ramachandran2017arxiv} were used. For the
sound-specific encoder, we adopted an architecture with two fully
connected layers, each with $50$ neurons. One dimension batch
normalization was used between the two layers. The decoders followed
similar architectures. The optimization used Adam gradient with
\textsc{pytorch}'s default parameters, and learning rate
$\eta_{\textsc{avae}}$.
The AVAE loss function penalized poor reconstruction of the image and
sound. Image reconstruction loss was measured by binary cross entropy
scaled by constant $\lambda_{\mathrm{image}}$, and sound
reconstruction loss was measured by mean squared error scaled by
constant $\lambda_{\mathrm{sound}}$. The prior divergence loss terms
were scaled by $\beta$, and the symmetrical KL divergence term by
$\alpha$.
\begin{figure}[t]
\centering
\begin{subfigure}[t]{0.49\linewidth}
\centering
\includegraphics[width=3.57cm]{images/pendulum/eps/doppler_4.eps}
\caption{}
\label{fig:doppler effect}
\end{subfigure}
%
\hfill
%
\begin{subfigure}[t]{0.49\linewidth}
\centering
\includegraphics[width=3.57cm]{images/pendulum/eps/amplitude_2.eps}
\caption{}
\label{fig:inverse square law}
\end{subfigure}
%
\caption{Different sound properties in the pendulum scenario.
\ref{fig:doppler effect}) Depicts the Doppler effect. As the sound
source moves near (away from) the observer, the arrival time of
the emitted waves decreases (increases), thus increasing
(decreasing) the frequency. \ref{fig:inverse square law}) Depicts
the how the amplitude of the sound decreases with the distance
from the source. Fading semi-circles denote smaller intensities.}
\label{fig:doppler and amplitude}
\end{figure}
\subsubsection{Learn to act in the world}
\label{subsubsec:pendulum learn to act}
The agent learned how to perform the task using the DDPG algorithm,
while only having access to the image input modality---that is
$\boldsymbol{I}_\mathrm{train} = I_{\mathrm{image}}$. These image
observations are encoded into the latent space using
$\mathcal{F}_\mathrm{train} = \mathcal{F}_\mathrm{image}$---the image-specific encoder
of the AVAE trained in~\ref{subsubsec:pendulum learn to map
inputs}. Thus, the agent learns a policy $\pi$ that maps latent
states to actions.
The actor and critic networks consisted of two fully connected layers
of $256$ neurons each. The replay buffer was initially filled with
samples obtained using a controller based on the Ornstein-Uhlenbeck
process, with the parameters suggested by~\citet{lillicrap2015arxiv}.
The Adam gradient was used for optimizing both networks, with learning
rates $\eta_{\textsc{critic}}$ and $\eta_{\textsc{actor}}$.
\subsubsection{Transfer policy}
We evaluated the performance of the policy trained
in~\ref{subsubsec:pendulum learn to act}, when the agent only has
access to the sound input modality, \emph{i.e.},
$\boldsymbol{I}_\mathrm{test} = I_{\mathrm{sound}}$.
Given a sound observation, the agent first preprocesses it using the
latent map $\mathcal{F}_\mathrm{test} = \mathcal{F}_\mathrm{sound}$, generating a
multimodal latent state $\vect{z}$---we denote this process as
\textsc{avae\textsubscript{s}}. The agent then uses the policy to
select the optimal action in this latent state.
As a result, we are measuring the zero-shot transfer performance of
policy $\pi$---that is, the ability of the agent to perform its task
while being provided perceptual information that is completely
different from what it saw during the reinforcement learning step,
without any further training. Table~\ref{tab:pendulum crossmodality
performance} summarizes the transfer performance in terms of average
reward observations throughout an episode of $300$ frames. Our
approach \textsc{avae\textsubscript{s} + ddpg} is compared with two
baselines:
\begin{itemize}
\item \textsc{random} baseline, which depicts the performance of an
untrained agent. This effectively simulates the performance one
would expect from a non-transferable policy trained over image
inputs, and later tested over sound inputs.
\item \textsc{sound ddpg} baseline, a DDPG agent trained directly over
sound inputs (\emph{i.e.} the sounds correspond to the
states). Provides an estimate on the performance an agent trained
directly over the test input modality may achieve.
\end{itemize}
From Table~\ref{tab:pendulum crossmodality performance}, we conclude
our approach provides the agent with an out-of-box performance
improvement of over $300\%$, when compared to the untrained agent
(non-transferable policy). It is also interesting to observe that the
difference in performance between our agent and \textsc{sound ddpg}
seems small, supporting our empirical observation that the transfer
policy succeeds very often in the task: swinging the pole
up\footnote{We also note that the performance achieved by the
\textsc{sound ddpg} agent is similar to that reported in the OpenAI
gym leaderboard for the \textsc{pendulum} scenario with state
observations as the position and velocity of the pendulum.}.
\subsection{\textsc{hyperhot}}
We consider the \textsc{hyperhot} scenario, a novel top-down shooter
game scenario inspired by the \textsc{space invaders} Atari
game\footnote{We opted to use a custom environment implemented in
\textsc{pygame}, since the \textsc{space invaders} environment in
OpenAI gym does not provide access to game state, making it hard to
generate simulated sounds.}, where the goal of the agent is to shoot
the enemies above, while avoiding their bullets by moving left and
right.
\begin{figure}[t]
\centering
\includegraphics[width=2.75cm]{images/hyperhot/eps/hyperhot_description_3.eps}
\caption{Visual and sound perceptual information in the
\textsc{hyperhot} scenario. All enemies and bullets emit sounds
that are received by four microphones at bottom left and right
($bl, br$) and paddle left and right ($pl, pr$).}
\label{fig:hyperhot}
\end{figure}
\begin{table}[b]
\caption{Zero-shot performance of the policy trained over the image
input modality, when using sound inputs only. Presents the average
reward per episode, over $75$ episodes. Results averaged over $10$
randomly seeded runs.}
\label{tab:pendulum crossmodality performance}
\begin{tabular}{@{}m{2.5cm}c@{}}
\multicolumn{2}{c}{\textsc{pendulum}} \\ \toprule
& \ \ \textbf{Rewards} \\
\textbf{Approach} & \ \ \textbf{avg} ${\boldsymbol \pm}$ \textbf{std} \\ \midrule
\textsc{avae\textsubscript{s} + ddpg} & $-2.00 \pm 0.97$ \\ \midrule
\textsc{random} & $-6.30 \pm 0.29$ \\
\textsc{sound ddpg} & $-1.41 \pm 0.91$ \\ \bottomrule
\end{tabular}
\end{table}
Similarly to the \textsc{pendulum}, in this scenario, the observations
of the environment include both image and sound components. In
\textsc{hyperhot}, however, the environmental sound is generated by
multiple entities $e_i$ emitting a predefined frequency
$f_0^{(i)}$:
\begin{itemize}
\item
Left-side enemy units, $e_0$, and right-side enemy units, $e_1$,
emit sounds with frequencies $f_0^{(0)}$ and $f_0^{(1)}$,
respectively.
\item
Enemy bullets. $e_2$, emit sounds with frequency
$f_0^{(2)}$.
\item
The agent's bullets, $e_3$, emit sounds with frequency
$f_0^{(3)}$.
\end{itemize}
The sounds produced by these entities are received by a set of $S$
sound receivers $\left\lbrace \rho_1, \dots, \rho_S
\right\rbrace$. Figure~\ref{fig:hyperhot} depicts the scenario, where
the yellow circles are the enemies; the green and blue bullets are
friendly and enemy fire, respectively; the the agent is in red; and
the sound receivers correspond to the white circles. The agent is
rewarded for shooting the enemies, with the following reward function:
\begin{equation*}
r =
\begin{cases}
10 & \text{if all enemies are killed, \emph{i.e.}, win} \\
-1 & \text{if player is killed or time is up, \emph{i.e.}, lose} \\
0 & \text{otherwise}
\end{cases}
\end{equation*}
The environment resets whenever the agent collects a non-zero reward,
be it due to winning or losing the game.
We assume the perceptual space of the agent as
$\mathcal{I} = I_{\mathrm{image}} \times I_{\mathrm{sound}}$, with the visual
input modality of the agent, $I_{\mathrm{vision}}$, consisting in the
raw image observation of the environment. The sound, however, is
generated in a more complex and realistic way. We model the sinusoidal
wave of each sound-emitter $e_i$ considering its specific frequency
$f_0^{(i)}$ and amplitude $a_0^{(i)}$. At every frame, we consider the
sound waves of every emitter present in the screen, according to their
distance to each sound receiver in $S$. The sound wave generated by
emitter $e_i$ is observed by receiver $\rho_j$ as
\begin{equation*}
a^{(i)}
=
a_0^{(i)} \exp{\left(-\delta \| \vect{e}_i - \vect{\rho}_j \|^2 \right)}
\end{equation*}
where $\delta$ is a scaling constant, $\vect{e}_i$ and $\vect{\rho}_j$
denote the positions of sound emmitter $e_i$ and sound receiver
$\rho_j$, respectively. We generate each sinusoidal sound wave for a
total of \num{1047} discrete time steps, considering an audio sample
rate of \num{31400} Hz and a video frame-rate of 30 fps. As such, each
sinusoidal sound wave represents the sound heard for the duration of a
single video-frame of the game\footnote{This is similar to what is
performed in Atari videogames.}. Finally, for each sound receiver,
we sum all emitted waves and encode the amplitude values in 16-bit
audio depth, considering a maximum amplitude value of $a_M$ and a
minimum value of $-a_M$.
We now provide details on how our approach was set up. All constants
and training hyper-parameters used are again summarized in
Appendix~\ref{subsec:constants and params}.
\subsubsection{Learn a perceptual model of the world}
We trained an AVAE model to learn the family of latent mapping $\mathcal{F}$,
with a dataset $\mathcal{D}(\mathcal{I})$ with $M$ observations of images and sounds
collected using a random controller. Before training, the images were
preprocessed to black and white and resized to $80 \times 80$ pixels,
and the sounds normalized to the range $\left[ 0, 1 \right]$.
For the image-specific encoder we adopted an architecture with three
convolutional layers and two fully connected layers. The three
convolutional layers learned $32, 64$ and $64$ filters,
respectively. The filters were parameterized by kernel sizes $8, 4$
and $2$; strides $4, 2$ and $1$; and paddings $2, 1$ and
$1$. \textsc{ReLU} activations were used throughout. For the
sound-specific component, we used two fully connected layers of $512$
neurons each, with one dimension batch normalization between the
layers. The decoders followed similar architectures. The increase in
size of these layers when compared to the \textsc{pendulum} task is
due to more complex nature of the sounds considered in this scenario.
The optimizer and loss function were configured in the same way as in
the previous scenario.
\subsubsection{Learn to act in the world}
The agent learned how to play the game using the DQN algorithm, while
having access only to image observations,
$\boldsymbol{I}_\mathrm{train} = I_\mathrm{image}$, corresponding to
the video game frames. The image observations are encoded into the
latent space using $\mathcal{F}_\mathrm{train} = \mathcal{F}_\mathrm{image}$---the
image-specific encoder of the AVAE model trained in the previous
step. As such, the policy learned to play the game, maps these latent
states to actions.
The policy and target networks consisted of two fully connected layers
of $512$ neurons each, and we adopted a decaying $\epsilon$-greedy
policy.
\subsubsection{Transfer policy}
We then evaluated the performance of the policy learned with image
inputs, when the agent only has access to the sound modality,
\emph{i.e.}, $\boldsymbol{I}_\mathrm{test} =
I_{\mathrm{sound}}$. Given a sound observation, the agent preprocesses
it using the latent map $\mathcal{F}_\mathrm{test} = \mathcal{F}_\mathrm{sound}$, thus
generating a multimodal latent state $\vect{z}$---this process is
denoted as \textsc{avae\textsubscript{s}}. The agent then uses the
policy to select the optimal action in this latent state.
Table~\ref{tab:hyperhot crossmodality performance} summarizes the
transfer performance of the policy produced by our approach
\textsc{avae\textsubscript{s} + dqn}, in terms of average discounted
rewards and game win rates over $100$ episodes. We compare the
performance of our approach with additional baselines:
\begin{itemize}
\item \textsc{avae\textsubscript{v} + dqn}, an agent similar to ours,
but encodes the latent space with visual observations (as
opposed to sounds).
\item \textsc{image dqn}, a DQN agent trained directly over the visual
inputs.
\end{itemize}
Considering the results in Table~\ref{tab:hyperhot crossmodality
performance}, we observe:
\begin{itemize}
\item
A considerable performance improvement of our approach over the
untrained agent. The average discounted reward of the
\textsc{random} baseline is negative, meaning this agent tends to
get shot often, and rather quickly. This is in contrast with the
positive rewards achieved by our approach. Moreover, the win rates
achieved by our approach surpass those of the untrained agent by
$5$-fold.
\item A performance comparable to that of the agent trained directly
on the sound, \textsc{sound dqn}. In fact, the average discounted
rewards achieved by our approach are slightly high\-er. However, we
note that the \textsc{sound dqn} agent followed the same DQN
architecture and number of training steps used in our approach. It
is plausible that with further parameter tuning, the \textsc{sound
dqn} agent could achieve better performances.
\item The approach that could fine-tune to the most informative
perceptual modality, \textsc{image dqn}, achieved the highest
performances. Our approach, while achieving lower rewards, is the
only able to perform cross-modality policy transfer, that is, being
able to reuse a policy trained on a different modality. One may
argue that this trade-off is worthwhile.
\end{itemize}
The DQN networks of all approaches followed similar architectures and
were trained for the same number of iterations.
\begin{table}[b]
\caption{Zero-shot performance of the policy trained over the image
modality, when using sound inputs only. Provides a comparison with
different baselines. Middle column is the average discounted reward
per episode. Right column is the win rate of the agent. Both
averaged over $100$ episodes. Results averaged over $10$ randomly
seeded runs.}
\label{tab:hyperhot crossmodality performance}
\begin{tabular}{
l
S[separate-uncertainty]
S[separate-uncertainty]
}
\multicolumn{3}{c}{\textsc{hyperhot}} \\ \toprule
& {\qquad \quad \textbf{Rewards}} & {\qquad \quad \textbf{Win pct}} \\
{\textbf{Approach}} & {\qquad \quad avg $\pm$ std} & {\qquad \quad avg $\pm$ std} \\
\midrule
\textsc{avae\textsubscript{s} + dqn} & 0.15 \pm 0.16 & 36.1
\pm
10.38
\\ \midrule
\textsc{avae\textsubscript{v} + dqn} & 0.21 \pm 0.11 &
43.20 \pm 7.03 \\
\textsc{random} & -0.33 \pm 0.16 & 8.3
\pm 5.75 \\
\textsc{sound dqn} & 0.1 \pm 0.22 &
27.3\pm 21.44
\\
\textsc{image dqn} & 1.54 \pm 0.20 & 75.00 \pm 5.33
\\ \bottomrule
\end{tabular}
\end{table}
\subsection{Discussion}
The experimental evaluation performed shows the efficacy and
applicability of our approach. The results show that this approach
effectively enables an agent to learn and exploit policies over
different subsets of input modalities. This sets our work apart from
existing ideas in the literature. For example, DARLA follows a similar
three-stages architecture to allow RL agents to learn policies that
are robust to some shifts in the original
domains~\cite{higgins2017icml}. However, that approach implicitly
assumes that the source and target domains are characterized by
similar inputs, such as raw observations of a camera. This is in
contrast with our work, which allows agents to transfer policies
across different input modalities.
Our approach achieves this by first learning a shared latent
representation that captures the different input modalities. In our
experimental evaluation, for this first step, we used the AVAE model,
which approximates modality-specific latent representations, as
discussed in Section~\ref{subsec:generative model}. This model is
well-suited to the scenarios considered, since these focused on the
transfer of policies trained and reused over distinct input
modalities. We envision other scenarios where training could
potentially take into account multiple input modalities at the same
time. Our approach supports these scenarios as well, when considering
a generative model such as JMVAE~\cite{suzuki2016arxiv}, which can
learn joint modality distributions and encode/decode both modalities
simultaneously.
Furthermore, our approach also supports scenarios where the agent has
access to more than two input modalities. The AVAE model can be
extended to approximate additional modalities, by introducing extra
loss terms that compute the divergence of the new modality specific
latent spaces. However, it may be beneficial to employ generative
models specialized on larger number of modalities, such as the
M$^2$VAE~\cite{korthals2019arxiv}.
\section{Conclusions}
\label{sec:conclusions}
In this paper we explored the use of multimodal latent representations
for capturing multiple input modalities, in order to allow agents to
learn and reuse policies over different modalities. We were
particularly motivated by scenarios of RL agents that learn visual
policies to perform their tasks, and which afterwards, at test time,
may only have access to sound inputs.
To this end, we formalized the multimodal transfer reinforcement
learning problem, and contributed a three stages approach that
effectively allows RL agents to learn robust policies over input
modalities. The first step builds upon recent advances in multimodal
variational autoencoders, to create a generalized latent space that
captures the dependencies between the different input modalities of
the agent, and allow for cross-modality inference. In the second step,
the agent learns how to perform its task over this latent
space. During this training step, the agent may only have access to a
subset of input modalities, with the latent space being encoded
accordingly. Finally, at test time, the agent may execute its task
while having access to a possibly different subset of modalities.
We assessed the applicability and efficacy of our approach in
different domains of increasing complexity. We extended well-known
scenarios in the reinforcement learning literature to include, both
the typical raw image observations, but also the novel sound
components. The results show that the policies learned by our approach
were robust to these different input modalities, effectively enabling
reinforcement learning agents \emph{to play games in the dark}.
\section*{Acknowledgments}
This work was partially supported by national funds through the
Portuguese Funda\c{c}\~{a}o para a Ci\^{e}ncia e a Tecnologia under
project UID/CEC/50021/2019 (INESC-ID multi annual funding) and the
Car\-negie Mellon Portugal Program and its Information and
Communications Technologies Institute, under project
CMUP-ERI/HCI/0051/ 2013. Rui Silva acknowledges the PhD grant
SFRH/BD/113695/2015. Miguel Vasco acknowledges the PhD grant
SFRH/BD/139362/2018.
\newpage
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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\section{Introduction}
Conjunctive queries are the most basic and arguably most important class of database queries~\cite{ChandraM77}. As such, they have been one of the major objects of study of database theory and are by now well understood in many respects.
In this short note, we close a minor gap in the literature of the counting complexity of conjunctive queries. This area has seen intensive study in the last few years, see e.g.~\cite{PichlerS13,BraultBaron13,DurandM14,DurandM15,GrecoS14,ChenM15,DellRW19} for a sample. The main difference between the counting complexity of CQs and the decision problem for Boolean CQs is that projection, also called existential quantification, matters for the complexity of queries: it was already observed by Pichler and Skritek in~\cite{PichlerS13} that there are classes of acyclic CQs for which counting answers is $\mathsf{\#P}$-hard whereas Yannakakis' algorithm shows that Boolean CQs can be decided efficiently. The understanding of counting was refined in~\cite{DurandM14} by introducing a notion called \emph{quantified star size} that is shown for acyclic queries to determine the (parameterized) complexity of answer counting. Note that quantified star size can be seen as a quantitative generalization of free-connex acyclicity, previously introduced in the setting of enumeration algorithms~\cite{BaganDG07}, in the sense that the free-connex acyclic queries are exactly the acyclic queries of quantified star size $1$.
The work of Dell, Roth and Wellnitz~\cite{DellRW19} gave a sweeping generalization of~\cite{DurandM14}, in particular introducing the perspective of fine-grained complexity in the setting which allows to give tight results on the exponent of the polynomial runtimes. This in particular allows considering the complexity of individual queries instead of whole classes of them as they have to be considered in the parameterized complexity approach. Concretely, \cite{DellRW19} shows that a parameter called \emph{dominating star size} lower bounds the exponent of counting algorithms for CQs.
The very interesting results of~\cite{DellRW19} have two slight shortcomings: first, the complexity is measured in the size of the domain of the input database. This is very common in many areas of theoretical computer science, in particular most work on graph algorithms does this. However, in database theory it is common to measure complexity with respect to the size of the database instead. It is not hard to see that the dominating star size of~\cite{DellRW19} does not determine the counting complexity of CQs when measured in the size of the input database which is a slight mismatch between~\cite{DellRW19} and most of the literature in database theory. Second, the results of~\cite{DellRW19} only consider cases in which the dominating star size of the query is at least $3$. This in particular makes it impossible to determine the queries in which linear time counting is possible. This is unfortunate since, due to very big input databases, differentiating between problems that have linear time algorithms and such that have not is often particularly interesting in database theory, see e.g.~the great amount of work in this direction on enumeration algorithms~\cite{BerkholzGS20}.
In this short note, we slightly improve some of the results of~\cite{DellRW19} in the directions pointed out above: we show that for acyclic CQs the original star size as introduced in~\cite{DurandM14} is a lower bound for the exponent of the runtime for counting when measuring in the size of the database. In particular, this is also true for quantified star size $1$ and $2$ which allows us to show that the free-connex CQs are the only self-join free acyclic CQs which allow linear time counting. Let us stress that the the techniques used to show these results are neither new nor surprising; essentially, we tie together several techniques from the literature here.
\section{Preliminaries}
We assume that the reader is familiar with the basics of database theory~\cite{AbiteboulHV95}. We only consider conjunctive queries which we generally assume to be self-join free. The hypergraph of a conjunctive query $q$ has as vertices the variables of $q$ and an edge $e$ for every atom which contains exactly the variables appearing in the atom.
A \emph{join tree} of a hypergraph $H$ is a tree $T$ whose vertices are the edges of $H$ and which has the following property: for every $v\in V(H)$ the set $\{e\in E(H) \mid v\in e\}$ is connected in $T$.
A hypergraph $H$ is called \emph{acyclic} if it has a join tree. A CQ is called acyclic if its hypergraph is acyclic.
All graphs in this note are simple and undirected.
A dominating set $S$ of a graph $G=(V,E)$ is a set $S\subseteq V$ such that every vertex not in $S$ has a neighbor in $S$. The problem $k$-Dominating Set (short $k$-DS) is to decide, given a graph $G$, if $G$ has a dominating set of size at most $k$. We will use the following result from~\cite{PatrascuW10}.
\begin{theorem}\label{thm:PatrascuW10}
If SAT has no algorithm with runtime $O(2^{n(1-\epsilon)})$ for any $\epsilon>0$, then there is no constant $\epsilon'$ such that there is a constant $k$ and an algorithm for $k$-DS with runtime $O(n^{k-\epsilon'}) $ on graphs with $n$ vertices.
\end{theorem}
We remark that the assumption on SAT in Theorem~\ref{thm:PatrascuW10} is in particular implied by the strong exponential time hypothesis (SETH)~\cite{ImpagliazzoP01}. As a consequence, our results below could also be with SETH. However, we keep the formulation as in~\cite{PatrascuW10}.
\section{Star Queries}
In this section, we will refine the lower bounds for the star queries that have already been a crucial building block of the lower bounds in~\cite{DurandM14,DurandM15,DellRW19}.
\begin{lemma}\label{lem:stars}
Let $k\in \mathbb{N}$ with $k\ge 2$. If there is an algorithm that counts answers to \[q^\star_k(x_1, \ldots, x_k) := \exists z \bigwedge_{i\in [k]} R(x_i,z)\] in time $O(m^{k-\epsilon})$ on databases with $m$ tuples for some $\epsilon>0$, then there is a $k'\in \mathbb{N}$ such that $k'$-DS can be decided in time $O(n^{k'-\epsilon})$ on graphs with $n$ vertices.
\end{lemma}
We remark that a very similar lower bound has already been shown in~\cite{DellRW19} but there the lower bound is for $n^{k-\epsilon}$ where $n$ is the size of the domain and not $m^{k-\epsilon}$. Moreover, there is was assumed that $k\ge 3$. We slightly improve the bound here by encoding several vertices of the graph in each variable of the query instead of just encoding a single vertex per variable.
\begin{proof}[Proof of Lemma~\ref{lem:stars}]
Choose $k'$ as a fixed integer such that $k'>k^2/\epsilon$ and $k'$ is divisible by $k$. We will encode $k'$-DS into the query $q^\star_k$. To this end, let $G=(V, E)$ be an input for $k'$-DS. Set
\[R:=\{(\vec{u}, v)\mid v\in V, \vec{u} = (u_1,\ldots, u_{k'/k}), \forall i \in [k'/k]: u_iv\notin E, u_i \ne v\}.\]
Then any assignment $\vec{u}^1, \ldots, \vec{u}^k$ to $x_1, \ldots, x_k$ in $q(x,y)$ corresponds to a choice $S$ of at most $k'$ vertices in $G$. The set $S$ is a dominating set if and only if there is no vertex $v$ in $V$ that is not in $S$ and has no neighbor in $S$. This is the case if and only if there is no $v\in V$ that assigned to $z$ makes $q^\star_k$ true. Thus the answers to $q^\star_k(x_1, \ldots, x_k)$ are exactly the assignments $\vec{u}^1, \ldots, \vec{u}^k$ that do \emph{not} correspond to dominating sets in $G$. Hence, any algorithm counting the answers to $q^\star_k(x_1, \ldots, x_k)$ directly yields an algorithm for $k'$-DS.
We now analyze the runtime of the above algorithm. Note that the time for the construction of $R$ is negligible, so the runtime is essentially that of the counting algorithm for $q^\star_k(x_1, \ldots, x_k)$. First observe that the relation $R$ has at most $n^{\frac{k'}{k}+1}$ tuples. The exponent of the runtime of the counting algorithm is thus
\begin{align*}
\left(\frac{k'}{k}+1\right)(k-\epsilon) &= k'+k-\frac{k' \epsilon}{k} - \epsilon\\
& < k'+k - \frac{k^2 \epsilon}{\epsilon k} -\epsilon\\
& = k'-\epsilon
\end{align*}
where the inequality comes from the choice of $k'$ satisfying $k'> k^2/\epsilon$.
\end{proof}
\begin{corollary}
If SAT has no algorithm with runtime $O(2^{n(1-\epsilon)})$ for any $\epsilon>0$, then there is no constant $k\in \mathbb{N}$, $k\ge 2$ and no $\epsilon' > 0$ such that there is an algorithm that counts answers to \[q^\star_k(x_1,\ldots, x_k) := \exists z \bigwedge_{i\in [k]}R(x_i,z)\] in time $O(m^{k-\epsilon'})$ on databases with $m$ tuples.
\end{corollary}
\section{Tight Bounds for Acyclic Queries}
In this section, we will lift Lemma~\ref{lem:stars} to self-join free acyclic queries. The approach is rather standard, see e.g.~\cite{BraultBaron13,DurandM14,DellRW19}, but we here consider somewhat tighter runtime bounds so we give it here for completeness. We closely follow the theory as developed in~\cite{BraultBaron13}. The argument is a minimal adaption of one found in the (unfortunately unpublished) full version of~\cite{BaganDG07} for enumeration. An acyclic conjunctive query with hypergraph $\mathcal{H}$ and free variables $S$ is called \emph{free-connex} if the hypergraph $\mathcal{H}\cup \{S\}$ that we get from $\mathcal{H}$ by adding $S$ as an edge is acyclic as well.
\begin{theorem}\label{thm:embedding}
Let $q$ be a self-join free conjunctive query that is acyclic but not free-connex. Then, assuming that SAT has no algorithm with runtime $O(2^{n(1-\epsilon)})$ for any $\epsilon>0$, there is no algorithm that counts the solutions of $q$ on a database with $m$ tuples in time $O(m^{2-\epsilon'})$ for any $\epsilon'$.
\end{theorem}
\begin{proof}
The idea is to embed $q^\star_2$ into $q$. Let $S$ be the set of free variables of $q$ and let $\mathcal{H}$ be the hypergraph of $q$. Since $q$ is acyclic but not free-connex, we get by \cite[Theorem 13]{BraultBaron13} that there is a set $S'$ such that $(\mathcal{H}\cup \{S\})[S']$ is a cycle plus potentially some unary edges. Note that acyclic hypergraphs cannot contain such an induced cycle, so we it follows that $(\mathcal{H}\cup \{S\})[S']$ contains at least one edge $xx'$ which is not in $\mathcal{H}[S']$. Moreover, since the cycle is induced, it must contain a vertex not in $S$. It follows that we can choose a path $P$ in $\mathcal{H}[S']$ such that the endpoints of $P$ are in $S$ but none of the internal vertices are. Let $x_1, x_2$ be the endpoints and let $z_1, \ldots, z_\ell$ be the internal vertices.
Let $\mathbb{D}$ be an input database for $q^\star_2$. We construct a database $\mathbb{D}'$ for $q$ as follows: all vertices not on $P$ are forced to take a fixed value $d$ in all relations. If there are any unary atoms $U$ on any variable on $P$, we make sure that those do not constrain the variables: if the atom is in the variable $x_i$, we set $U^{\mathbb{D}'}$ to the active domain of $x_i$ in $\mathbb{D}$. If the atom is in any $z_i$, we set $U^{\mathbb{D}'}$ to the active domain of $z$ in $\mathbb D$. Moreover, we construct the following:
\begin{itemize}
\item For every atom $R_1$ containing $x_1$ and $z_1$, we construct a relation $R_1$ such that the projection $\Pi_{x_1, y_1}(R_1)$ is $R^{\mathbb{D}}$, the relation for $R(x_1, z)$ in $q^\star_2$.
\item We set the relations for atoms containing $x_2$ and $z_\ell$ analogously.
\item Finally, for every atom of $q$ that contains a pair $z_i, z_{i+1}$, we construct a relation $R$ such that the projection $\Pi_{z_i, z_{i+1}}(R)= \{(v,v)\mid v\in D_z\}$ where $D_z$ is the active domain of $z$ in $\mathbb{D}$.
\end{itemize}
Note that, for every atom of $q$ containing a variable on $P$, exactly one of the cases applies. It follows that $\mathbb{D}'$ has linear size in $\|\mathbb D\|$ and that it can be constructed in quasilinear time. Also, observe that $\Pi_{x_1, x_2}(q(\mathbb{D}')) = q^\star_2(\mathbb{D})$ and that this projection is a bijection. It follows that $|q(\mathbb{D}')| = |q^\star_2(\mathbb{D})|$, so counting for $q$ directly allows counting for $q^\star_2$.
\end{proof}
Note that it is well known that for every free-connex acyclic CQ the answers can be counted in linear time, see e.g.~\cite{BraultBaron13,CarmeliZBKS20}, so we get that being free-connex completely characterizes the linear time case for acyclic CQs in the following sense.
\begin{corollary}
Assume that SAT has no has no algorithm with runtime $O(2^{n(1-\epsilon)})$ for any $\epsilon>0$. Let $q$ be a self-join free acyclic CQ. Then there is an algorithm that counts the answers of $q$ for every database in linear time if and only if $q$ is free-connex.
\end{corollary}
We remark in passing that Theorem~\ref{thm:embedding} can be generalized beyond quadratic runtime lower bounds. There is a quantitative generalization of being non-free-connex called \emph{quantified star size} that essentially measures under which conditions $q^\star_k$ can be embedded into a query. Inspection of the proof of Lemma~4.3 from~\cite{DurandM15} shows that the following is true (we refer to~\cite{DurandM15} for definitions and all details).
\begin{corollary}
Let $q$ be a self-join free conjunctive query of quantified star size $k$. Then, assuming that SAT has no algorithm with runtime $2^{n(1-\epsilon)}$ for any $\epsilon>0$, there is no algorithm that counts the solutions of $q$ on a database with $m$ tuples in time $m^{k-\epsilon'}$ for any $\epsilon'$.
\end{corollary}
\section{Conclusion}
We have slightly improved some of the bounds from~\cite{DellRW19} here by improving the dependence of the runtime bounds from the size of the domain to the size of the database. This makes these bounds connect better to other results in database theory. Let us stress that we have only considered a very small portion of the results in the very rich paper~\cite{DellRW19}. It would be interesting to see which other results from there can be adapted in this way.
One compelling question that we leave open is fully determining the queries for which---under a standard complexity assumption---answers can be counted in linear time. Note that, combining our result with the work of Brault-Baron in~\cite{BraultBaron13}, this question reduces to showing lower bounds for the so-called Loomis-Whitney joins. Brault-Baron conjectured that those queries do not have a linear time counting algorithms and from this and our result it would follow that the queries that allow linear time counting are exactly the free-connex acyclic queries of~\cite{BaganDG07}. We note that there are reasons to consider Brault-Baron's conjecture as plausible, see the discussion in~\cite[Section~6]{BerkholzGS20}. However, it would still be interesting to find more evidence for it.
Unfortunately, doing so looks quite challenging: even for the question of counting triangles in a graph, the smallest Loomis-Whitney join, there are no known conditional lower bounds under standard assumptions, despite the fact that triangles play a major role in fine-grained complexity, see e.g.~\cite{williams2018some}.
\paragraph{Acknowledgement.} This note would not have been written without a question Nicole Schweikardt and it would not have been published without another question by Nofar Carmeli. The author thanks Nofar Carmeli and Nikolaos Tziavelis for helpful discussions and their generous advice that helped greatly improve upon an earlier version of this note.
\bibliographystyle{plain}
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package org.apereo.cas.support.oauth.web;
import org.apereo.cas.support.oauth.OAuthConstants;
import org.apereo.cas.support.oauth.util.OAuthUtils;
import org.apereo.cas.support.oauth.validator.OAuthValidator;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.beans.factory.annotation.Qualifier;
import org.springframework.cloud.context.config.annotation.RefreshScope;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.AbstractController;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
/**
* This controller returns an access token according to the given OAuth code and client credentials (authorization code grant type)
* or according to the refresh token and client credentials (refresh token grant type) or according to the user identity
* (resource owner password grant type).
*
* @author Jerome Leleu
* @since 3.5.0
*/
@RefreshScope
@Controller("accessTokenController")
public class OAuth20AccessTokenController extends AbstractController {
/**
* The logger.
*/
protected transient Logger logger = LoggerFactory.getLogger(getClass());
/**
* The OAuth validator.
*/
@Autowired
@Qualifier("oAuthValidator")
protected OAuthValidator validator;
@Autowired
private OAuth20Grant[] oAuth20Grants;
@RequestMapping(path = OAuthConstants.BASE_OAUTH20_URL + '/' + OAuthConstants.ACCESS_TOKEN_URL, method = RequestMethod.POST)
@Override
protected ModelAndView handleRequestInternal(final HttpServletRequest request, final HttpServletResponse response) throws Exception {
if (!this.validator.checkParameterExist(request, OAuthConstants.GRANT_TYPE)) {
logger.error("Authorize request verification fails");
return OAuthUtils.writeTextError(response, OAuthConstants.INVALID_REQUEST);
}
final String grantType = request.getParameter(OAuthConstants.GRANT_TYPE);
try {
for (OAuth20Grant oAuth20Grant : oAuth20Grants) {
if (grantType.equals(oAuth20Grant.accessTokenGrantType())) {
return oAuth20Grant.accessToken(request, response);
}
}
} catch (RuntimeException e) {
return OAuthUtils.writeTextError(response, e);
}
// Unsupported grant_type.
logger.error("Unsupported grant type: {}", grantType);
return OAuthUtils.writeTextError(response, OAuthConstants.INVALID_REQUEST);
}
}
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"redpajama_set_name": "RedPajamaGithub"
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Basil Spence
Title: Basil Spence
Subject: Grade I listed buildings in Brighton and Hove, DoCoMoMo Key Scottish Monuments, University of Southampton, Buildings and architecture of Brighton and Hove, Coventry Cathedral
Collection: 1907 Births, 1976 Deaths, 20Th-Century Architects, 20Th-Century British Architects, Academics of the University of Leeds, Alumni of the Edinburgh College of Art, Architects of Cathedrals, British Army Personnel of World War II, Brutalist Architects, Camoufleurs, English Ecclesiastical Architects, People Educated at George Watson's College, People from Edinburgh, Presidents of the Royal Institute of British Architects, Scottish Architects, Scottish Knights
Sir Basil Urwin Spence
(1907-08-13)13 August 1907
19 November 1976(1976-11-19) (aged 69)
Yaxley, Suffolk
Basil Spence & Partners
Coventry Cathedral
Hyde Park Barracks
New Zealand parliament extension
Sir Basil Urwin Spence, OM, OBE, RA (13 August 1907 – 19 November 1976) was a Scottish architect, most notably associated with Coventry Cathedral in England and the Beehive in New Zealand, but also responsible for numerous other buildings in the Modernist/Brutalist style.
Early career 2
Army service 3
Post-war career 4
Coventry Cathedral 4.1
Later work 4.2
List of projects 6
Spence was born in Edinburgh College of Art (ECA) in 1925, studying architecture, where he secured a maintenance scholarship on the strength of the "unusual brilliance" of his work. He won several prizes at the college, and meanwhile carried out paid work drawing architectural perspectives for practising architects including Leslie Grahame-Thomson and Reginald Fairlie.
In 1929–1930 he spent a year as an assistant, along with William Kininmonth, in the London office of Sir Edwin Lutyens, whose work was to have a profound influence on Spence's style, where he worked on designs for the Viceroy's House in New Delhi, India. While in London he attended evening classes at the Bartlett School of Architecture under A. E. Richardson. Returning to ECA in 1930 for his final year of studies, he was appointed a junior lecturer, despite the fact that he was still a student. He continued to teach at ECA until 1939.
Southside Garage in the art deco style
After graduating in 1931, Kininmonth and Spence set up in practice together, based in a room within the office of Rowand Anderson & Paul (at that time having Arthur Forman Balfour Paul as sole partner), in Rutland Square, Edinburgh. The practice was founded on two residential commissions which Kininmonth had obtained that year. Spence also received commissions to illustrate other architect's work, including the Southside Garage, on Causewayside, Edinburgh, in an Art Deco style (although credited to Spence his name appears nowhere on the official warrant drawings and only appears as a signature on the artist's perspective).[1]
In 1934 Spence married, and the Kininmonth & Spence practice merged with Rowand Anderson & Paul. Balfour Paul died in 1938, leaving Kininmonth and Spence in charge of the renamed Rowand Anderson & Paul & Partners. Spence's work was now concentrated on exhibition design, including three pavilions for the 1938 Empire Exhibition in Glasgow, and country houses. The first two of these, Broughton Place at Broughton near Biggar, and Quothquhan in Lanarkshire, were executed in traditional Scottish styles at the client's request. The third was entirely modern. Gribloch was designed for John Colville, grandson of the founder of Colville's Iron Works, and his American wife. It was designed in a modernist Regency style, with assistance from Perry Duncan, an American architect hired by the Colvilles when Spence was too busy with exhibition work to progress the project.
Army service
In 1939, Spence was commissioned as a 2nd Lieutenant into the Camouflage Training and Development Unit of the British Army.[2] He was initially based at Farnham in Surrey, and took part in the D-Day landings in 1944. He was demobilised in September 1945, having reached the rank of major and been mentioned in despatches twice.[3]
Post-war career
Spence returned to Rowand Anderson & Paul & Partners briefly, before setting up his own practice, Basil Spence & Partners, with Bruce Robertson. He was awarded an OBE in 1948 for his work in exhibition design, work which he continued with the Sea and Ships Pavilion for the 1951 Festival of Britain.[4] That year he opened a London office, moving there permanently from 1953. A second office was opened in 1956 at Canonbury, which became the creative hub of the practice. Spence was External Professor of Architecture at the University of Leeds from 1955 to 1957 and from 1958 to 1960 he was the President of the Royal Institute of British Architects.
Coventry Cathedral (1956–1962)
On 14 November 1940, Coventry's Anglican Cathedral was extensively damaged by German bombing, a year into World War II.
In 1944, Sir Giles Gilbert Scott submitted a design proposal to rebuild the cathedral but this was rejected by the Royal Fine Arts Commission. In 1950, a competition was launched to find the most suitable design from a Commonwealth of Nations architect. Over 200 entries were received, and Spence's radical design was chosen. Work began in 1956 and the structure was completed in 1962. Spence was knighted in 1960 for his work at Coventry, while the cathedral was still being built.
On 23 February 2012 the Royal Mail released a stamp featuring Coventry Cathedral as part its "Britons of Distinction" series.[5]
Later work
The New Zealand Parliament's executive wing, the Beehive
In 1959 Spence secured two important commissions, for the British Embassy in Rome (completed 1971), and for the Hyde Park Cavalry Barracks in London (completed 1970). He was also responsible for designing the high-rise Hutchesontown C housing in Glasgow. These were intended to replace the notorious slum tenements in the Gorbals area of the city. A combination of social deprivation and exclusion in the relevant areas, coupled to poor execution of his designs meant that the developments created as many problems as they solved, and led to their demolition in 1993. He was also responsible for modernist buildings on The Canongate in Edinburgh, opposite the new Scottish Parliament and in view of Holyrood Palace. This area is named Brown's Close and was listed in 2008. These buildings are privately owned and are currently undergoing repairs to the roofs and spallings (Nov 2010) to preserve this early example of his work. Other work in the 1960s included the executive wing of the New Zealand Parliament Buildings in Wellington, nicknamed "The Beehive", Edinburgh University Library, and Abbotsinch Airport (now Glasgow Airport). In 1960, Spence designed Mortonhall Crematorium in Edinburgh's Braid Hills area (based on the same angled fin concept as found at Coventry Cathedral). He also designed Trawsfynydd nuclear power station, which was unveiled in Snowdonia, north Wales, in 1964.
The Spence practice was rearranged in 1964, with the Canonbury office being renamed Sir Basil Spence OM RA, and the second London office Spence Bonnington & Collins. The Edinburgh office was also renamed for its partners, Spence Glover & Ferguson. From 1961 to 1968, Spence was Professor of Architecture at the Royal Academy. Through the 1970s, Spence continued to work on public and private commissions, universities and offices including Aston University Library and Management Centre. His last work was for an unexecuted cultural centre for Bahrain, which he worked on during illness in 1976. Some of his final commissions were built after his death; for example, his design for the new Glasgow Royal Infirmary was completed in 1981.
Spence died in November 1976 at his home at Yaxley, Suffolk and was buried at nearby Thornham Parva. His practice – Spence, Ferguson and Glover continued until 1992 before being disbanded.
The British Embassy in Rome
In 2006 he was the subject of a DoCoMoMo as one of Scotland's sixty key monuments of the post-war years, in the same year as it was demolished.
In August 2010 English Heritage recommended that the Spence-designed Sydenham School be given Grade II listed status: the building was due to be demolished to make way for a new building.
50 Queen Anne's Gate, completed 1976
Falmer House (grade 1 listed), part of the University of Sussex campus, 1962
Broughton Place (a private house in the style of a 17th-century Scottish tower house in Broughton, Scottish Borders with decorative reliefs by architectural sculptor Hew Lorimer) (1938)
Gribloch (a house near Kippen, Stirling) (1938–39)
Kilsyth Academy, Kilsyth (opened 1954, designed 1930)
Sea and Ships Pavilions for Festival of Britain (1951)
Duncanrig Secondary School, East Kilbride, Greater Glasgow (1953)
St Paul's Church, Wordsworth Avenue, Sheffield
Sydenham School, Sydenham, London (1956)
Agricultural Science Building, University of Nottingham, Sutton Bonington campus (1956–58)
The churches of St Oswald, Tile Hill – St Chad, Wood End – St John the Divine, Willenhall. Built simultaneously in Coventry (1957).
Thurso High School Thurso, Scotland (1957)
The Chadwick Physics Laboratory (1957–9), University of Liverpool
Campus development plan at the University of Nottingham (1957–60) including Chemistry Building, Physics and Mathematics Building, Mining and Fluid Mechanics Laboratory, Pope Building, Coates Building
Thorn EMI House, 5 Upper St. Martin's Lane, London (1959) (Spence's original exterior was demolished in the 1990s; reborn as Orion House with a full-height floor plate addition and re-skinned elevations. A 60-foot-tall (18 m) metal sculpture by Geoffrey Clarke for the original façade (incorporating allusions to electric lamp filaments) has been remounted onto the added lift and service riser.)
Great Michael Rise and Laverockbank Crescent, social housing developments in Newhaven, Edinburgh
Erasmus Building, Friars Court, Queens' College, Cambridge (1959–1960)
Swiss Cottage Leisure Centre (originally 'Swimming Baths'), London (1962–4)
Spence House, near Beaulieu, Hampshire (designed 1961, for Spence's own use and listed Grade II)
Coventry Cathedral, completed 1962
The initial campus design at the University of Sussex (1960s) including Falmer House (1962, now a Grade I listed building)
Hutchesontown C flats, Gorbals, Glasgow (1962 – demolished in 1993)
Physics Building, Streatham Campus, University of Exeter.
Herschel Building, Newcastle University (1962)[7]
Nuffield Theatre, Highfield Campus, University of Southampton (1964)
The "Beehive", the executive wing of the New Zealand Parliament Buildings Wellington, New Zealand (1964)
Trawsfyndd Nuclear Power Station (1965)
St Aidan's College, University of Durham
Edinburgh University Main Library
Glasgow Airport (1966) (Spence's original façade was covered over in 1989 when an extension was built to house new check-in desks. The original structure can now only be seen from the departure gates and runway.)
British pavilion, Expo 67 (1967).[8]
Newcastle Central Library (1968)[9] - demolished in 2007
65 – 103 Canongate, social housing developments in The Canongate, Edinburgh
Civic Centre, Sunderland (1970)
Hyde Park Barracks, London (1970)
British Embassy, Rome (1971)
Glasgow Royal Infirmary redevelopment – Phases 1 & 2 (1971–82) – now known as the Queen Elizabeth Building and University Block
Kensington and Chelsea Town Hall (1972-6), including chambers, offices, and public areas[10][11]
Aston University Library (1975) (Sir Basil Spence, Glover and Ferguson). Extended and remodelled in 2010
50 Queen Anne's Gate (the former Home Office building), London (1976)
The Sydney Jones Library (1976) at the University of Liverpool
Buildings and architecture of Brighton and Hove
Category:Basil Spence buildings
^ City of Edinburgh Council: Building Warrant Archive
^ "Edinburgh, Ravelston Dykes Road, Ravelston House, Garden". A set of oblique aerial photographs of Ravelston House garden with military vehicles, Edinburgh taken as a camouflage test. Sir Basil Spence Archive. Royal Commission on the Ancient and Historical Monuments of Scotland. 14 March 1944. pp. Canmore ID 273364. Retrieved 2 August 2012.
^ http://www.theguardian.com/society/2007/oct/16/communities
^ http://canmore.org.uk/collection/1028251
^ "Coventry Cathedral architect in Royal Mail stamp set". BBC. 23 February 2012. Retrieved 2012-01-02.
^ http://www.stuartbrown.info/#ArtworksScotland_Spence
^ http://www.pmsa.org.uk/pmsa-database/9567/
^ "Ex. 4". Expo 67 press kit. Citynoise. Retrieved 13 December 2009.
^ http://canmore.rcahms.gov.uk/en/286495/details/tyne+and+wear+newcastle+upon+tyne+city+library/
^ Twentieth Century Society: Kensington and Chelsea Town Hall
^ Canmore database: Kensington and Chelsea civic centre
Long, Philip and Thomas, Jane, (eds.) Basil Spence: Architect, National Galleries of Scotland/RCAHMS, 2007
Stringer, Michael (2010-08-06). "Heritage listing threatens Sydenham School rebuild".
"Site record for Greater London, Southwark, Camberwell And Dulwich, Dartmouth Road, Sydenham School".
Article published by WalesHome about Trawsfynydd, October 2009
Last-ditch attempt to save Sir Basil's Trawsfynydd from demolition
"Sir Basil Spence Archive Project". Homepage for the archive of nearly 40,000 items held by the Royal Commission on the Ancient and Historical Monuments of Scotland (RCAHMS).
"Basil Urwin Spence". Dictionary of Scottish Architects. Architect biography.
"Basil's Bairns". Royal Scottish Academy. Exhibition held from 1 January 2008 at the Royal Scottish Academy, looking at careers of the architects who worked in the studio of Sir Basil Spence.
Rudden, Liam (2006-09-08). "Rebuilding Sir Basil Spence's battered reputation". The Scotsman. UK.
"Basil Spence archive exclusive". Extract from the Sir Basil Spence archive, featuring 57 images of his work.
Camoufleurs
1st World War
Lucien-Victor Guirand de Scévola
Loyd A. Jones
John Graham Kerr
Alister Hardy
André Mare
Solomon Joseph Solomon
Abbott Handerson Thayer
Maximilian Toch
Leon Underwood
Edward Wadsworth
Everett Warner
Norman Wilkinson
2nd World War
Tony Ayrton
Geoffrey Barkas
Hugh Casson
John Codner
Edward Bainbridge Copnall
Hugh B. Cott
Victorine Foot
Frederick Gore
Stanley William Hayter
Ivan Konev
Jasper Maskelyne
Oliver Messel
Colin Moss
Roland Penrose
Peter Proud
Fred Pusey
Edward Seago
Alan Sorrell
Steven Sykes
Ernest Townsend
Julian Trevelyan
Wilfred Clement Von Berg
20th-century architects
Academics of the University of Leeds
Alumni of the Edinburgh College of Art
Architects of cathedrals
British Army personnel of World War II
Brutalist architects
English ecclesiastical architects
People from Edinburgh
Presidents of the Royal Institute of British Architects
Scottish architects
Scottish knights
20th-century British architects
University of Cambridge, England, Imperial College London, Durham University, University of Edinburgh
Diocese of Coventry, Berlin, Coventry, Church of England, John Witcombe
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,344
|
Biden marks first full year in office
WRAL News at 7PM
FDA cracks down on claims that cannabis can cure cancer
Posted November 2, 2017 9:33 a.m. EDT
Updated November 2, 2017 8:47 p.m. EDT
By Ben Tinker (CNN)
(CNN) — The US Food and Drug Administration is cracking down on companies it says are making unsubstantiated claims that certain products made from marijuana can cure cancer.
This week, the agency responsible for policing the American food and drug market issued warning letters to four companies that are "illegally selling products online that claim to prevent, diagnose, treat, or cure cancer without evidence to support these outcomes." It said in a statement, "The illegally sold products allegedly contain cannabidiol (CBD), a component of the marijuana plant that is not FDA approved in any drug product for any indication."
FDA Commissioner Scott Gottlieb said in the statement, "We don't let companies market products that deliberately prey on sick people with baseless claims that their substance can shrink or cure cancer and we're not going to look the other way on enforcing these principles when it comes to marijuana-containing products."
The companies that received the warning letters are required by law to respond within 15 working days, indicating what steps they have -- or will -- take to address the FDA's concerns.
"Failure to promptly correct these violations may result in legal action without further notice, including, without limitation, seizure and injunction," each of the four letters warned.
What products were targeted?
The FDA said the 25-plus products that are part of this crackdown include oil drops, capsules, syrups, teas, topical lotions and creams.
The four companies that received warning letters on Wednesday are Greenroads Health, Natural Alchemist, That's Natural! Marketing & Consulting and Stanley Brothers Social Enterprises LLC.
Greenroads Health did not respond to a request for comment.
Natural Alchemist declined a request for comment.
In a statement, That's Natural! said, "We are excited about the opportunity to address the FDA's Warning letter by complying to their warning letter and taking down the requested links and information, including that to peer-reviewed journal articles, on our site and social media."
Stanley Brothers said in a statement, "We take regulatory compliance very seriously. Our customers love to share their very personal stories about how our products helped improve their lives or those of their loved ones. ... We will work with the FDA to ensure that we better monitor how we share third-party testimonials."
CNN featured the Stanley brothers in a 2013 documentary, "Weed," about a young girl who struggled with severe seizures.
"There are a growing number of effective therapies for many cancers," Gottlieb said. "When people are allowed to illegally market agents that deliver no established benefit they may steer patients away from products that have proven, anti-tumor effects that could extend lives."
When issuing warning letters about other "illegal" cancer treatments in April, the FDA said consumers should not use them "because they may be unsafe and could prevent a person from seeking an appropriate and potentially life-saving cancer diagnosis or treatment," according to Douglas W. Stearn, director of the FDA's Office of Enforcement and Import Operations."We encourage people to remain vigilant whether online or in a store, and avoid purchasing products marketed to treat cancer without any proof they will work. Patients should consult a health care professional about proper prevention, diagnosis and treatment of cancer."
What should consumers look out for?
"I think the biggest red flag would be that any product that hasn't undergone FDA review is making a claim that it can treat or cure cancer," Jason Humbert, a regulatory operations officer in the FDA's Office of Regulatory Affairs, told CNN in April. "Only products that have been evaluated -- approved FDA drugs -- can make those claims. So if a consumer happens upon a website or a social media site and they see that this product is marketed as a natural cure for cancer or a natural treatment for cancer, they should be very skeptical, because unless that product has been evaluated by FDA, there's no reason to believe it's safe or effective for that use."
Although claims vary from product to product, the FDA says fraudulent cancer products "often use a particular vocabulary." The agency identified these phrases as the most common red flags:
Treats all forms of cancer Miraculously kills cancer cells and tumors Shrinks malignant tumors Selectively kills cancer cells More effective than chemotherapy Attacks cancer cells, leaving healthy cells intact Cures cancer
"The overarching point is that these products are untested, and some of the ingredients may present direct risk to the consumer's health or interact with any medications they might be taking," Humbert said in April. "They're not a substitute for appropriate treatment, and using these products can not only endanger consumers' health but waste their money and waste their time, as well."
In this week's statement, Gottleib said, "We have an obligation to provide caregivers and patients with the confidence that drugs making cancer treatment claims have been carefully evaluated for safety, efficacy, and quality, and are monitored by the FDA once they're on the market.
"We recognize that there's interest in developing therapies from marijuana and its components, but the safest way for this to occur is through the drug approval process -- not through unsubstantiated claims made on a website. We support sound, scientifically-based research using components derived from marijuana, and we'll continue to work with product developers who are interested in bringing safe, effective, and quality products to market."
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,147
|
Seraphon: What We'd Like to See in the New Battletome
captberk February 2, 2020 1 Comment
In my opinion, the Seraphon are quite a unique faction in Age of Sigmar. At first glance one would perhaps expect these savage reptilian creatures to belong to the Destruction Grand Alliance, but no these beasts fight for the forces of Order.
In addition, the Seraphon units themselves possess the 'Daemon' keyword, a unique occurance outside of the Chaos Grand Alliance – they really are bit of an oddball faction within the Order Grand Alliance.
With the recent announcement of an imminent Battletome (read all about it here), I thought I'd take a closer look at this most mysterious of factions.
Prior to 'Age of Sigmar', the Seraphon were known simply as the 'Lizardmen' – a race of reptilian humanoids 'created' by the enigmatic 'Old Ones' to rid the 'Old World' of those races whose existence were deemed a detriment to the 'Great Plan'.
The Lizardmen consisted of several sub-races – the Saurus warrior breed, the Skink workforce and the mighty Kroxigor who served as beasts of burden (and were also quite handy in a scrap too). The different lizardmen entered the world via great 'Spawning Pools' deep within the 'Lustrian' jungles, emerging fully formed and able to wage war or perform their preordained duties from the get go. But the Lizardmen themselves were just servants to much greater powers – the Slann and their 'Old One' masters.
Not much is know about the enigmatic Old Ones. They were a race of super intelligent space faring aliens diametrically opposed to Chaos and by whose hand much of the 'Old World' was shaped.
Serving the Old Ones, and distinct from the Lizardmen were the mighty Slann. The Slann were powerful Mage-Priests, able to shift continents and topple cities with their amazing magical powers. They served as leaders to the Lizardmen and, when the Old Ones vanished, worked hard to enact the 'Great Plan' in their masters' absence.
The Lizardmen had a rough time of it during the 'End Times'. Their jungle homelands were invaded by large numbers of Skaven and were ultimately destroyed by massive walls of fire and sunk beneath huge tidal waves.
However not all was lost and we learnt that before the end, several Slann escaped the confines of the 'Old World' before it's ending in a great exodus. We must assume these Slann are the very same that now exist in the game world today.
Which brings us back to the Age of Sigmar. Much of the Lizardmen range now exists as the 'Seraphon' – no longer a race of mortal reptiles spawned into the world but instead an army of Daemons imagined into being by the mighty Slann Starmasters.
It seems those Slann that escaped the destruction of the 'Old World' were led to the 'Realm of Azyr' by the Godbeast 'Dracothian'. Here in the utmost reaches of the Realm of Heavens do the Slann now dwell and contemplate their eternal battle against the forces of the Chaos Gods.
Now, when the Slann deign to march to war they summon legions of daemons, made manifest in the image of the Lizardmen of old, formed of celestial magic.
In the beginning of the Age of Sigmar it was said that these summoned armies could only exist in the presence of a Slann Starmaster, and were quick to dissipate once battle was over or the Slann who summoned them perished. However it now seems that this is no longer always the case and there are now reports of Seraphon continuing to exist in the absence of a Slann for long periods of time.
So how do Seraphon play on the table top?
Well it is worth remembering that they are currently one of the oldest Battletomes in the game. They have received regular updates in the different versions of the 'General's Handbook' but I have to admit they do feel a bit lacking compared to the more recent Battletomes.
They do have access to a lot of different units and it is possible to build an army to suit a variety of different playstyles, some much more effective than others it has to be said. You can build an army that leans into the magical game, with powerful casters in the Slann and Skink mages.
You can go infantry heavy with blocks of Saurus Warriors and Saurus Guard, buffed with Saurus characters like the Oldbloods and Scar-Veterans. They also have access to several mobile units such as Teradons, Ripperdactyls and Cold One Knights and finally they have access to several behemoths in the form of the Carnosaurs, Stegadons and Bastilidons.
Several of the units have access to ranged weapons, so it is even possible to build quite a shooty (albeit short ranged) army too, perhaps worth bearing in mind in today's meta. As you can see their is a wide range of units to choose from, but I have to admit the warscrolls themselves do leave a bit to be desired, especially when you compare them to the more recent releases.
Perhaps the biggest strength of the Seraphon, and the one that enables to them to compete effectively on the tabletop today is their summoning and board control. The Seraphon Allegiance abilities allow you to summon new units of Seraphon to the battlefield by spending 'Celestial Conjuration Points' (CCP). The new units must arrive on the battlefield next to Slann Starpriests or Astrolith Bearers. CCP are accrued by a Slann General in place of casting spells and a small number of additional CCP can be aquired throughout the game by the presence of the Slann General or Astrolith Bearer.
In addition to this summoning mechanic, the Engine of the Gods unit can also summon a selection of units onto the tabletop using it's own special rules. Using the right combination of command traits, artifacts and units it is quite possible to summon a large number of Seraphon units to the battlefield over the course of a game. Which is just as well as most of the Seraphon's units are not quite as durable as one would like.
I also mentioned board control as a strength of the army. Another of the Seraphon's Allegiance abilities allows you to 'teleport' a single unit in your hero phase anywhere else on the battlefield (outside of 9″ of an enemy of course). In addition to this teleport mechanic there are some battalions which allow you to hold units back in reserve and teleport them onto the battlefield during the course of the game.
So, what would I like to see in the new Seraphon Battletome?
I'll start with the lore and background. I've always liked the Old One/Slann/Lizardmen lore from previous editions. The idea of the Old Ones traveling to different worlds (perhaps even those that exist in the 41st Millennium) in their great war against the Chaos Gods really fuels my imagination.
Then of course tragedy occurred and the Old Ones disappeared, leaving the Slann to battle on, attempting to interpret the will of their missing masters as best they could.
The Slann were a dying race in the Old World but it seems in the Age of Sigmar they have had a resurgence of sort, now no longer servants of the Old Ones but masters themselves.
I would like to learn more of their purpose, of their ambitions and of their own 'Great Plan'. It is mentioned in the lore that some mortals believe they travel between worlds on great space vessels and it would be interesting to learn more about the great battle against Chaos that must surely exist beyond the confines of the Mortal Realms as we know them.
In this age of Gods such as Nagash, Allarielle and Sigmar I would like to see the Slann raised in stature. They are alien beings that seem to originate from outside the bounds of the Mortal Realms, fighting the great enemy in other places within the Universe.
In my mind I see the Slann as alien and outsiders, they are not bound to the rules of the Mortal Realms that perhaps constrain the other Gods that we know of and therefore their power should also be of a different type and scale.
Some of the recent background that we have seen indicate that some Seraphon now appear to exist 'for real' within the Mortal Realms, no longer the stuff of celestial magic but of flesh and blood.
I would like to see this developed substantially, maybe there are now Seraphon cities and conclaves within the Moral Realms in the image of those 'Temple Cities' that once existed in the Old World. The new Serapon scenery piece makes me optimistic this may be the case 🙂
What about on the table top?
Well, aside from a general update of the existing warscrolls to bring them in line with more recent Battletomes I would like to see some big changes to the one of their main mechanics – summoning.
Don't get me wrong, I like the idea of summoning units of Seraphon to the battlefield and teleporting them around. What I don't like is the fact that I have to include some of the greatest spellcasters in the lore…but not actually cast any spells :(. To me it is just not fun that I have this fantastic looking Slann model on the table top, with access to a really fun spell lore and superb casting talents, who just sort of sits there all game, outside of line of sight try to avoid being engaged or shot at.
Instead I'd like to see Games Workshop come up with a really interesting summoning mechanic that feels exciting and unique and allows me to use my awesome Slann Starmaster to wreak havoc amongst the enemy army with their spells whilst summoning in Seraphon reinforcements from the realm of Azyr.
Perhaps the most obvious 'fix' is to adopt a mechanic similar to that of the 'Disciples of Tzeentch' (check out Podcast for a review of this particular Battletome), accruing CCP when a spell is successfully cast, but this does not feel unique enough to me. Maybe they could be given a new spell lore, where each spell has a different 'tier' of power. The player could choose to cast the more powerful version of a spell but this would grant fewer CCP. Instead the player could cast a less powerful version of the spell but this would instead allow the Slann to accrue more CCP. Command traits, artifacts or battalions could perhaps modify this mechanic and allow Slann to benefit from both the CCP generation and the more powerful spell effects in certain scenarios? I really hope they come up with something cool though as I'd love to be able to cast spells AND summon at the same time 😀
I think it is safe to assume that there will be some variation on the 'Stormhost' theme in the new Seraphon Battletome. One guess is that these will be based on the different 'Constellations' as they exist today, allowing you to build armies that could be Skink heavy, Saurus heavy, possibly even behemoth heavy (Stegadon Battleline perhaps?). This seems most likely to me and could be just what the Seraphon army needs to turn it from a 'jack of all trades, masters of none' type of Battletome into a more lethal specialised force.
Anyway, hopefully we will not have much longer to wait and I'm really looking forward to the release of the new Battletome. Let us know what you're hoping for in the new book below or on social media.
Published by captberk
View all posts by captberk
News, Warhammer Age of Sigmar
Age of Sigmar, aos, Battletome, Games Workshop, preview, Seraphon
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Pingback: Sons of Behemat: What We Would Like To See In The New Battletome – Sprues & Brews
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,465
|
require 'to_csv'
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,698
|
Q: How to invoke a listener or interceptor in @CacheEvict I have a requirement to invoke some functionality when @CacheEvict is being called. Is there a way to call a listener or interceptor to be invoked in Spring @CacheEvict?
A: Typically, this is very "cache provider" specific since no 2 cache providers have the same capabilities.
For instance, I primarily work with In-Memory Data Grid (IMDG) technology like Pivotal GemFire and the OSS version Apache Geode. Both, of which, can be used as a "caching provider" in Spring's Cache Abstraction. With GemFire/Geode, you can register a o.a.g.cache.CacheListener callback on the GemFire/Geode Region (essentially a java.util.Map) that is backing the Spring Cache interface, and used in Spring's Caching infrastructure as the "Adapter" to the backing store. As you can see with SD GemFire/Geode provider implementation, the "eviction" triggers a GemFire/Geode Region.remove(key). This eviction can subsequently be captures and handled in the Region's registered CacheListener.afterDestroy(:EntryEvent) callback method.
However, this is just 1 way to handle notifications on evictions in your application.
Of course, as @Borino pointed out, you can leverage Spring's AOP support to "intercept" the cache eviction operation. The advantage of this approach is that it is more generic and reusable across different caching providers.
Although, I would say that you should not be developing a AOP Pointcut expression based on the underlying "caching provider", as @Borino instructed, i.e. ...
execution(* org.springframework.cache.concurrent.ConcurrentMapCache.evict(..))
This expression ties your AOP Aspect to the ConcurrentMapCache "provider", the default in Spring's Cache Abstraction (and in Spring Boot).
What happens when you use Ehcache, Hazelcast, Redis, GemFire/Geode or multiple combinations of these "providers" in your application?
Rather, you can slightly adjust the AOP Pointcut expression to this...
execution(* org.springframework.cache.Cache.evict(..))
See here. This is safe because all "caching providers" must supply 2 things: a CacheManager implementation and a Cache implementation for each cache specified in the application. Again, the Cache interface is a "Adapter" to the backing store. Again, see the docs for more details.
There are tradeoffs with either approach. Provider specific solutions generally give your more control capabilities, but using the AOP approach is more reusable. Do what is right for your UC.
Hope this helps.
Cheers!
-John
A: John's answer is on the right track, but it's important to know that the Cache class is not a Spring managed bean, the CacheManager is.
For this reason, you'd have to pull in additional AspectJ dependencies and do some sort of compile or post-compile weaving to target the Cache.evict method.
A: i tried injecting cache manager to an aspect and set up my point cuts on methods that cache eviction was necessary like here:
package com.example.aspectdemo.aop;
import com.example.aspectdemo.domain.Customer;
import com.example.aspectdemo.service.CustomerService;
import org.aspectj.lang.JoinPoint;
import org.aspectj.lang.annotation.*;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.cache.CacheManager;
import org.springframework.stereotype.Component;
@Component
@Aspect
public class CacheEvictHandler {
public static Logger logger = LoggerFactory.getLogger(CacheEvictHandler.class);
private final CacheManager cacheManager;
private final CustomerService customerService;
public CacheEvictHandler(CacheManager cacheManager, CustomerService customerService) {
this.cacheManager = cacheManager;
this.customerService = customerService;
}
@Pointcut(value = "execution(* com.example.aspectdemo.service.impl.CustomerServiceImpl.save(..))")
public void loggingPointCut() {
}
@AfterReturning(value = "loggingPointCut()", returning = "customer")
public void LoggAfterEviction(JoinPoint joinPoint, Customer customer) throws Throwable {
cacheManager.getCache("customer-cache").clear();// remove cache
logger.info("*** saved customer id : {}", customer.getId());// do what you like here, i added some logs
logger.info("*** after eviction : {}", customerService.findAll());
logger.info("*** cache evicted ***");
}
}
here is where i save:
@Transactional
@Override
public Customer save(Customer customer) {
log.debug("Request to save Customer : {}", customer);
return customerRepository.save(customer);
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 267
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About a year ago, I setup an apache + suexec + mod_fastcgi + php-fpm setup with multiple virtual host, in which each virtual hosts has it's own user, and it's own php-fpm pool running. This all worked fine (also over previous updates) until last week when I updated the setup from php 5.5.10 to php 5.5.12.
[Tue Jun 10 11:26:42 2014] [error] [client 192.168.1.2] FastCGI: incomplete headers (0 bytes) received from server "/var/www/cgi-bin.d/cgi-control/php-fpm"
So it looks that the FastCGI process tries to access the socket using the apache user, and not as the pool user (control). Is there something changed in the new php-version? Did Suexec or mod_fastcgi change something? Or did I miss something else?
APACHE2_OPTS="-D DEFAULT_VHOST -D INFO -D SSL -D SSL_DEFAULT_VHOST -D SUEXEC -D LANGUAGE -D FASTCGI -D PHP_FPM"
drwxrwxr-x 2 control control 4.0K Jul 29 2013 .
drwxr-xr-x 9 root root 4.0K Aug 4 2013 ..
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,884
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package alexsmirnov
import alexsmirnov.scalafx.ObservableImplicits
package object pbconsole extends ObservableImplicits {
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,843
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Q: Programmatically refresh layer usage information In AutoCAD, through the Layer Properties Manager UI, a user can refresh the usage information for layers in a drawing. The information sometimes get stale when you have layers from xrefs. In your code, when you access the LayerTable for a drawing, its records will not include those layers that it deems as unused (in the Layer Properties Manager UI grey status = unused, blue status = used). In order for me to get to those layers, my program first needs to refresh the usage information. Is there a way to do this through the .NET API, COM API, or the command line?
I'm also pursing an answer in parallel in the AutoCAD forums
http://forums.autodesk.com/t5/NET/Programmatically-refresh-layer-usage-information-NET/m-p/2794756
A: It seems that the problem was that one of the entities in the particular drawing was corrupt. When one of the users recreated the drawing (including re-adding some xrefs) and ran the program it worked just fine. I also double checked the code and it was definitely hitting the correct layers this time. I am however open to suggestions or solutions on how to detect corruption problems in AutoCAD drawings.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,173
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Sebastião Barros Vale has joined the Future of Privacy Forum (FPF) as EU Policy Fellow in 2021, where he is following and analyzing privacy and data protection European and national case law, recent academic research, guidelines, and decisions from the European Data Protection Board and national Data Protection Authorities. He actively monitors the activity of EU institutions around privacy and data protection, including Communications and Proposals of the European Commission and legislative reports from the European Parliament and the Council of the EU.
He completed a traineeship at the Cabinet of EU Commissioner Věra Jourová, where he contributed to the 1st Annual Review of the EU-US Privacy Shield and to the Commission Guidance on the applicability of the General Data Protection Regulation (GDPR). As a qualified lawyer in Portugal, he practiced at the ICT practice area of Vieira de Almeida e Associados, assisting clients from the healthcare, retail, energy and financial sectors in complying with the GDPR and other privacy-related laws. Just before joining FPF, he worked as an in-house privacy expert at Johnson & Johnson.
Sebastião holds an LL.M in EU Law from the College of Europe (2016, Belgium), where he wrote his thesis on alternative international data transfer mechanisms after the invalidation of the EU-US Safe Harbour Decision by the Court of Justice of the European Union. Two notable papers he published focused on the interplay between the GDPR and the EU's DIrective on digital payment services (PSD2), and on how European anti-discrimination, consumer, and data protection rules shield consumers against pervasive online price personalization.
Posts by Sebastião
"Are crumbles all that remains of the cookies?" A conversation on the future of ad tech at the Nordic Privacy Arena 2021
Event Report: From "Consent-Centric" Frameworks to Responsible Data Practices and Privacy Accountability in Asia Pacific
Upcoming data protection rulings in the EU: an overview of CJEU pending cases
Blogs by Daniel
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 5,502
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\section{Introduction}
\input{figures/teaser.tex}
\input{sections/intro.tex}
\section{Related Work}
\input{sections/related.tex}
\section{The \NetName{} Model}
\input{sections/method.tex}
\section{Experiments}
\input{sections/experiments.tex}
\section{Conclusion}
\input{sections/conclusion.tex}
\section*{Acknowledgments}
This work was partially supported by the National Key R\&D Program of China (No. 2017YFA0700804) and the National Natural Science Foundation of China (No. 61876171).
\begin{Appendix}
\section{AVA-ActiveSpeaker Dataset Statistics}
Table~\ref{table:ava-statistics} provides some statistics about the AVA-ActiveSpeaker dataset. Note that the number of entities is a rough estimate of the upper bound, which is the number of merged face tracks that are obtained through the procedure described in Sec.~\ref{sec:setup} of the main paper.
\section{Cross-Dataset Evaluation Results}
In this section, we provide detailed results for the cross-dataset evaluations in Sec.~\ref{sec:cross-dataset-eval} of the main paper. Results for Columbia, RealVAD and AVDIAR are presented in Table~\ref{table:columbia}, \ref{table:realvad}, and \ref{table:avdiar}, respectively. For an analysis of these results, please refer to Sec.~\ref{sec:cross-dataset-eval} of the main paper. Note that we find if the sampling rate of the videos are changed to 25fps (the value used during training), the results will be different. In these tables, we report the results at the original sampling rate. For fine-tuned results on RealVAD, we report the numbers from the best models on the validation set.
\input{tables/ava_statistics.tex}
\input{tables/columbia.tex}
\input{tables/realvad.tex}
\input{tables/avdiar.tex}
\end{Appendix}
\bibliographystyle{ACM-Reference-Format}
\subsection{Experimental Setup}\label{sec:setup}
\paragraph{Datasets:} The \textbf{AVA-ActiveSpeaker} dataset \cite{DBLP:conf/icassp/RothCKMGKRSSXP20} consists of $262$ YouTube movies from film industries around the world. Each video is annotated from minutes $15$ to $30$. The dataset provides face bounding boxes that are linked into face tracks and labeled for both speech activity and whether the speech is audible. As shown in Fig.~\ref{fig:dataset-teaser}a, the dataset is highly challenging as it contains occlusions, low-resolution faces, low-quality audio, and varied lighting conditions. It has become a mainstream benchmark for the ASD task.
The \textbf{Columbia} dataset \cite{DBLP:conf/eccv/ChakravartyT16} annotates $35$ minutes of a one-hour panel discussion video featuring $6$ unique panelists (Bell, Bollinger, Lieberman, Long, Sick, Abbas). The panelists directly face the camera, but occasionally look elsewhere, or perform spontaneous activities (e.g. drinking water). Upper body bounding boxes and voice activity ground truth labels are provided, as shown in Fig.~\ref{fig:dataset-teaser}b.
The \textbf{RealVAD} dataset~\cite{9133504} (Fig.~\ref{fig:dataset-teaser}c) is also constructed from a panel discussion lasting approximately $83$ minutes. The video is recorded using a static, mounted camera, capturing the $9$ panelists in a full shot. The panelists sit in two rows, and at any given time can be looking anywhere. In addition, the panelists perform natural, spontaneous actions that may result in partial occlusion of the face and mouth (e.g. touching faces, cupping chins). Similar to Columbia, upper body bounding boxes and voice activity labels are provided. These two datasets are often adopted by works on visual voice activity detection (V-VAD) and ASD~\cite{DBLP:conf/iccvw/ShahidBM19,Shahid_2021_WACV,DBLP:conf/accv/ChungZ16a,DBLP:conf/eccv/AfourasOCZ20}.
The \textbf{AVDIAR} dataset \cite{DBLP:journals/pami/GebruBLH18} (Fig.~\ref{fig:dataset-teaser}d) provides $27$ recordings of informal conversations involving one to four participants. The dataset is extremely challenging due to varying levels of speaker movement and speech overlap, occlusions, profile faces, and even back-facing speakers. The dataset was originally used to evaluate speaker diarization and tracking systems. We re-purpose the dataset to evaluate our ASD model by aggregating per-frame ASD results into diarization results.
\vspace{-2ex}
\paragraph{Data Preprocessing:} We preprocess each dataset by cropping the face tracks of each visible candidate. For the \textbf{AVA-ActiveSpeaker} dataset, to group face tracks into ``scenes" for training and testing, each video is first segmented into shots with \texttt{ffmpeg}~\cite{DBLP:journals/ijig/Lienhart01}. Within each shot, disconnected tracks belonging to the same candidate are then merged based on bounding box overlap with an IoU threshold of $0.8$. All videos are resampled to $25$fps, and ground truth labels are computed according to frame timestamps via nearest-neighbor interpolation. The \textbf{Columbia} and \textbf{RealVAD} datasets do not provide face bounding boxes, so face detection is performed with an off-the-shelf RetinaFace detector~\cite{DBLP:conf/cvpr/DengGVKZ20}. The resulting bounding boxes are expanded by $1.3$x to mimic AVA-style detections. For the \textbf{AVDIAR} dataset, the provided bounding boxes are used directly.
\vspace{-1.5ex}
\paragraph{Train/test Split and Evaluation Metric:}
For \textbf{AVA-ActiveSpeaker}, the official train/validation/test split is adopted. The test set is held out for the ActivityNet challenge and unavailable, so we perform our analysis on the validation set instead, as several previous methods do~\cite{DBLP:conf/cvpr/AlcazarCMPLAG20,zhangmulti2019,DBLP:journals/corr/abs-1906-10555}. The \textbf{Columbia} and \textbf{AVDIAR} datasets do not provide training sets, so we only perform zero-shot testing. For Columbia, following previous practice \cite{DBLP:conf/eccv/ChakravartyT16,DBLP:conf/accv/ChungZ16a}, we report results on all but one speaker (Abbas) which is usually held out for validation. For \textbf{RealVAD}, in addition to zero-shot results, we also report leave-one-out cross-validation results for each of the $9$ panelists to compare with previous work. We only train on scenes in which the test speaker does not co-occur with the training speakers.
Finally, we report the metric that was commonly adopted by previous work for each dataset: mean Average Precision (mAP) and Area Under Receiver Operating Characteristic Curve (AUROC) for AVA-ActiveSpeaker, frame-wise $F$-1 score for Columbia and RealVAD, and Diarization Error Rate (DER) for AVDIAR, which is defined as the sum of false alarm rate, missed speech rate, and speaker confusion rate.
\vspace{-2ex}
\paragraph{Implementation Details:} We implement our model with PyTorch and the \texttt{pytorch-lightning} package. All models are trained using the AdamW optimizer~\cite{DBLP:conf/iclr/LoshchilovH19}. The network parameters are initialized using He initialization~\cite{DBLP:conf/iccv/HeZRS15}. During training, we augment the data via random horizontal flipping and uniform corner cropping along the input face tracks, followed by random adjustments to brightness, contrast, and saturation. All cropped face tracks are resized to $144\times 144$, and randomly cropped to $128\times 128$ for training. We use a central $128\times 128$ patch for testing. We only run one trial with a fixed random seed for all experiments to ensure the results are comparable, but find the results to be stable under different network initializations and data augmentations. During inference, occasional long segments that do not entirely fit into GPU memory are split into shorter, fixed-size chunks and the predictions are re-combined later.
To facilitate training, we apply curriculum learning~\cite{DBLP:conf/icml/BengioLCW09} by first training a single-candidate model without relational context according to Eq.~\eqref{eq:total_loss}, and then continuing to train on up to $3$ candidates, \textit{i.e.} $N=3$ in Eq.~\eqref{eq:equivariant-gamma}, but without the $\mathcal{L}_{\mathrm{a}}$ term in Eq.~\eqref{eq:total_loss}. This is theoretically and empirically enough to learn the relevant parameters since $R_\mathrm{V}$ captures \textit{pairwise} interactions, and we observe diminishing returns by sampling $4$ or more candidates (note that we can still test on $N>3$ candidates due to parameter sharing). In addition, we follow the sampling strategy in \cite{DBLP:conf/cvpr/AlcazarCMPLAG20} during training, and randomly sample a $1.12$s segment ($28$ frames at $25$fps)\footnote{The mean speech segment duration is $1.11$s on the AVA-ActiveSpeaker dataset~\cite{DBLP:conf/icassp/RothCKMGKRSSXP20}.} from every training example in the dataset for each epoch. Therefore, our epoch size correlates with the number of identities or scenes rather than face detections, which prevents over-fitting.
\vspace{-2ex}
\subsection{Ablation Studies}\label{sec:ablations}
In this subsection, we provide a thorough ablation study on the modeling components on the challenging AVA-ActiveSpeaker dataset.
\vspace{-4ex}
\paragraph{Spatial Context:} We compare two different ways to incorporate spatial context. The first approach applies \textit{early fusion} and concatenates speaker-centric spatial embeddings $h_i$ with the $i$-th candidate's face track features before dimensionality reduction. In contrast, the second approach applies \textit{late fusion} and concatenates $h_i$ with the dimension-reduced face track features $v_i$. As shown in Table~\ref{table:head-maps}, both improve upon the simplest baseline that only uses convolutional temporal context (see Sec.~\ref{sec:temporal-context}) on audio and face track features, but early fusion performs $0.8$\% worse than late fusion, likely because information is lost too early as features pass the $128$-dim bottleneck. Thus, we adopt late fusion hereafter. As shown in Table~\ref{table:ablations}, simply concatenating the spatial context embeddings to the face track features (denoted by \textbf{+S}) already yields stronger visual representations that improve the baseline ($+1.7$\% mAP).
\vspace{-2.5ex}
\input{tables/head_maps.tex}
\vspace{-1ex}
\paragraph{Relational Context:}
We assess the efficacy of our relational context component. As shown in Table~\ref{table:ablations}, incorporating the relational context module yields noticeable improvements of $1.7$\% absolute over the 1D CNN baseline when used alone (\textbf{+R}), and works in synergy with spatial context ($+2.6$\% mAP for \textbf{+S+R}). We observe that this improvement is especially pronounced in multi-speaker and small face scenarios. This is because \NetName{} can model the relationships among the candidates in the scene, which leads to a better holistic understanding under these challenging settings. Moreover, we introduce weight sharing strategies within a permutation-equivariant formulation, which enables our model to process a theoretically unlimited number of speakers at test time.
\vspace{-1.5ex}
\paragraph{Temporal Context:}
Finally, incorporating temporal context with Bi-GRUs yields the most substantial improvement, which is reasonable because Bi-GRUs have access to each candidate's complete track history, and can thus make more reliable predictions. By leveraging long-term temporal context, our model ultimately achieves $92.0$\% mAP, improving upon our initial baseline by as much as $8.0$\%.
\input{tables/ablations.tex}
\vspace{-4ex}
\paragraph{Non-active Speaker Suppression:}
In this part, we discuss the role of non-active speaker suppression in ASD. As shown in Table~\ref{table:suppression}, we compare two ways of integrating global information against a baseline that does not apply non-active speaker suppression. One uses mean-pooling to obtain $\eta_{\mathrm{global}}$, and the other uses max-pooling. Both improve over the model that only introduces visual relational context, and max-pooling performs slightly better, yielding fewer false negatives. One possibility is that max-pooling back-propagates more gradients to the most salient speaker, while mean-pooling results in weaker gradients for the true active speaker, making the model less confident.
\input{tables/speaker_pooling.tex}
\vspace{-2ex}
\subsection{Comparison with the State-of-the-Art}
As shown in Table~\ref{table:ava-sota}, our full \NetName{} outperforms previous audio-visual ASD methods including the state-of-the-art by a large margin. Remarkably, we achieve $92.0$\% mAP without any pre-training, surpassing $90$\% for the first time on the AVA-ActiveSpeaker validation set at the time of submission. This overall performance strongly supports the effectiveness and superiority of our \NetName{}. It is also worth noting that the \textbf{+S+T} model in Table~\ref{table:ablations} provides a very strong baseline ($90.3$\% mAP) that already exceeds all methods available for comparison here, which further confirms the benefits of our model. As a side note, applying ImageNet pre-training to the encoders further boosts performance by $0.2$\% to $92.2$\%.
\vspace{-2.5ex}
\input{tables/ava_sota.tex}
\vspace{-1ex}
\subsection{Cross-dataset Evaluation}\label{sec:cross-dataset-eval}
Apart from evaluating on the large-scale AVA-ActiveSpeaker dataset, we also conduct cross-dataset evaluation on three other datasets: Columbia, RealVAD and AVDIAR. These datasets contain challenges that are previously rare, if not unseen in AVA-ActiveSpeaker, such as overlapped speech, reverberation, extreme poses, heavy face occlusion, and different spatial distributions of candidates. Our cross-dataset evaluation protocols include zero-shot testing on all three datasets using the model trained on AVA-ActiveSpeaker, and leave-one-out testing after fine-tuning on RealVAD. We report both visual-only and audio-visual performance. The former is obtained using only the auxiliary visual prediction layer $V_{\mathrm{aux}}$ in Sec.~\ref{sec:losses}.
\vspace{-1.5ex}
\paragraph{Zero-shot Testing:}
Overall, our \NetName{} model achieves good zero-shot generalization performance on all three datasets. In particular, it achieves state-of-the-art performance on both Columbia ($90.9\%$ average $F$-1 score) and RealVAD ($80.30\%$ average $F$-1 score). On AVDIAR, we are the first to report diarization performance using single-channel audio and a single camera viewpoint.
On the \textbf{Columbia} dataset,
we outperform all previous state-of-the-art under both visual-only and audio-visual settings. It is worth noting that the listed visual-only methods~\cite{DBLP:conf/eccv/ChakravartyT16,DBLP:conf/iccvw/ShahidBM19,Shahid_2021_WACV} use the entire upper body, while we only use cropped face tracks. We also outperform previous audio-visual models~\cite{DBLP:conf/eccv/AfourasOCZ20,DBLP:conf/accv/ChungZ16a} that are pre-trained on datasets a magnitude larger ($224$ and $606$ hours, respectively).
Similarly, on \textbf{RealVAD}, our zero-shot visual-only and audio-visual results outperform the only existing method available for equal comparison by a large margin ($87.22$\% V, $80.30$\% AV vs. $53.04$\%). An interesting observation is that on RealVAD and Columbia, our visual-only model outperforms the audio-visual model in most cases. This is because labels provided with the two datasets are not frame-accurate; our audio-visual model successfully detects short inter-sentence pauses that are not annotated as "not speaking" by the dataset curators. Due to the page limit, please refer to the supplementary video for details.
On \textbf{AVDIAR}, our model provides the first baseline
that only uses mono microphone input, and a single camera viewpoint -- a highly challenging setup. Nevertheless, our audio-visual model still achieves $55.16\%$ DER on average, which is a very promising zero-shot result. Moreover, our full \NetName{} model reduces DER by $4$ to $5$\% over the \textbf{+S+T} model that does not model relationships among the candidates, which once again supports the efficacy of our relational context in multi-speaker scenarios.
\vspace{-2ex}
\paragraph{Fine-tuned Results:} We also evaluate fine-tuned leave-one-out performance on RealVAD (Columbia and AVDIAR do not provide training splits). After
fine-tuning on homogeneous data, our model quickly adapts to the unseen scenario (panel discussion) and noises (e.g. spontaneous actions near the face). The average audio-visual $F$-1 score is improved by $5.34$\% from $80.30$\% to $85.64$\%, and the average visual $F$-1 score is improved by $1.47$\% from $87.22$\% to $88.69$\%. More detailed results can be found in the appendix.
\vspace{-1.5ex}
\subsection{Performance Breakdown}
In this part, we provide a more in-depth analysis of three known challenging scenarios on AVA-ActiveSpeaker. We provide a comparison with the existing state-of-the-art and highlight some advantages and appealing properties of the \NetName{} model.
\vspace{-1ex}
\paragraph{Low-resolution Faces.} We summarize performance for different face sizes in Fig.~\ref{fig:breakdown}a. Following previous evaluation procedures \cite{DBLP:conf/icassp/RothCKMGKRSSXP20,DBLP:conf/cvpr/AlcazarCMPLAG20}, we partition the dataset into three bins by the widths of the detected faces: small (width $\le$ 64 pixels), medium (width between $64$ and $128$ pixels), and large (width $>128$ pixels). While the \textbf{+S+T} model provides a strong baseline with $90.3$\%mAP that already outperforms the previous state-of-the-art \cite{DBLP:conf/cvpr/AlcazarCMPLAG20}, our full model which exploits relational context yields further performance gains of around $2$\% mAP absolute on all subsets. In particular, on the ``small" subset, we outperform the previous state-of-the-art ($56.2$\% mAP) by a large margin of $15.1$\%.
\input{figures/perf_breakdown.tex}
\vspace{-1ex}
\paragraph{Multiple Candidates.} Fig.~\ref{fig:breakdown}b shows the model performance according to the number of detected faces in a frame. We consistently improve over the previous state-of-the-art for different numbers of detected faces, with the most significant gain being in the subset with $3$ faces ($+14.7$\% mAP). Moreover, we surpass $90\%$ on the two-faces subset for the first time. The improvements prove the effectiveness of spatial and relational context modeling, which equip the model with the ability to discern background actors from main actors, as well as an enhanced understanding of the relationships between the visible speakers.
We refer interested readers to the supplemental video for more interesting qualitative results.
\vspace{-1ex}
\paragraph{Out-of-sync Audio and Video.}\label{sec:out-of-sync}
Many real-world videos suffer from poor synchronization between audio and video, caused by transmission or re-encoding. To evaluate our model's performance on out-of-sync data, we assume perfect synchronization in the source videos and artificially shift the audio stream by up to $10$ frames to mimic out-of-sync videos. We then assess the performance of our model on manually de-synchronized videos. As shown in Fig.~\ref{fig:sync}, our model is fairly resilient to A-V sync errors. Remarkably, even on videos that are shifted by $10$ frames ($0.4$sec), our full \NetName{} model only loses $0.67$\% mAP, and still outperforms a simple Bi-GRU baseline by $1.64$\% absolute, which does not model spatial and relational context (\textbf{+T} in Table~\ref{table:ablations}), and receives \textit{synchronized} inputs. This shows that our model is robust to synchronization error.
\input{figures/sync.tex}
\vspace{-1.5ex}
\subsection{Encoders}
Given an input video clip, we first crop face tracks for each candidate speaker and transform them into low-dimensional speaker descriptors for further analysis. Likewise, we encode each audio frame into a low-dimensional audio descriptor.
\vspace{-2ex}
\paragraph{Face Track Encoder:}
To encode short-term temporal dynamics, at each time step $t = 1,2,\dots, T$, where $T$ is the total number of time steps (i.e. frames), the $i$-th candidate's input ($i=1,2,\dots,N$ where $N$ is the number of visible candidates at time step $t$) is a stack of $k$ consecutive face crops $\mathbf x_i^t = \big(x_i^{t-[k/2]}, \dots, x_i^{t+[k/2]}\big)$ between step $t-[k/2]$ and $t+[k/2]$. Then an encoder $\phi$, which is a ResNet-18~\cite{DBLP:conf/cvpr/HeZRS16} is used to produce an average-pooled feature vector $\phi \big(\mathbf x_i^t\big)\in\mathbb{R}^{d}$ over the $k$ face crops. To keep the computational cost manageable, we choose $k=5$ for our experiments and reduce the dimensionality of each $\phi \big(\mathbf x_i^t\big)$ to $d'=128$ using a single shared fully-connected layer to obtain each candidate's final face track features:
\[v_i=\big(\mathrm{FC}\big(\phi\big(\mathbf x_i^1\big)\big),\mathrm{FC}\big(\phi\big(\mathbf x_i^2\big)\big),\dots,\mathrm{FC}\big(\phi\big(\mathbf x_i^T\big)\big)\in\mathbb{R}^{T\times d'}.\]
\paragraph{Audio Encoder:}
At each time step $t$ ($t=1,2,\dots, T$), we obtain audio representations from a $400$ms window preceding $t$. A ResNet-18 encoder $\psi$ takes $13$-dimensional MFCCs of the window as input and outputs a $512$-dimensional average-pooled feature. These features are also dimension-reduced to $128$ using a single fully-connected layer. We denote the final audio features by $a = (a_1, a_2,\dots, a_T)$.
\vspace{-2ex}
\subsection{Spatial Context}
The purpose of modeling spatial context is twofold. First, active speakers usually occupy a central position in the scene and depict higher visual saliency, especially in movies (commonly known as "the language of the lens"). Hence knowing the scale, position, and trajectory of a face in the video can help eliminate unlikely candidates. Second, people tend to look at the active speaker as they listen. Therefore, we wish to reflect such gaze-related information to some degree, by providing the model with both facial features and the relative positions of the candidates in the scene.
Specifically, we encode head positions of all candidates in the scene using $64\times 64$ coordinate-normalized maps of 2D Gaussians, which is motivated by \cite{DBLP:conf/cvpr/Marin-JimenezKM19}. The position and radius of each Gaussian represent the relative position and size of each candidate's face. Next, as shown in Fig.~\ref{fig:head-maps}, to further indicate the candidates' relationships, for each candidate $i$ ($i=1,2,\dots, N$) we construct his/her person-specific head map $H_i$ by generating a color-coded version of the initial Gaussian map in the following principle: yellow denotes candidate $i$, and blue denotes other candidates in the scene. To further facilitate the subsequent modeling of relationships between candidate $i$ and every other candidate $j$ ($j=1,2,\dots,N$, $j\ne i$), we tweak the color coding scheme to construct paired variants $H_{ij}$ as follows: red denotes candidate $i$, green denotes candidate $j$ and blue denotes the rest. Alternatively, $H_{ij}$ can be thought of an RGB image: the red channel shows the "subject", i.e. candidate $i$; the green channel shows the "object", i.e. candidate $j$; and the blue channel shows other "context" candidates, i.e. those other than $i$ and $j$ ($H_i$ is identified with $H_{ii}$). Then a VGG-M-inspired network with four 2D convolutional layers~\cite{DBLP:conf/cvpr/Marin-JimenezKM19} is used to embed these colored head maps $H_i$ and $H_{ij}$ into $64$-dimensional vectors per frame, $h_i$ and $h_{ij}$ for each candidate $i$ and candidate pair $(i, j)$ with $i\ne j$. We term the resulting embeddings $h_i$ and $h_{ij}$ candidate $i$'s \textbf{spatial context}.
\vspace{-4.5ex}
\input{figures/head_maps.tex}
\subsection{Relational Context}\label{sec:relational-context}
After obtaining face track and audio descriptors as well as the spatial context embeddings, we jointly model all candidates in the scene and refine each candidate's representations for robust ASD. Our motivation for modeling relational context is the inevitable presence of local ambiguities which make directly matching face motion (e.g. lip movements) and audio non-trivial. Therefore, we integrate different types of contextual clues and consider all candidates in the scene holistically to facilitate the ASD task. For example, a person who is speaking tends to receive more attention from others and is usually portrayed with higher visual saliency (especially in movies).
Specifically, our relational context component is designed to complete two natural sub-tasks for ASD: learning a contextual visual representation for \textit{visual voice activity detection}, and a contextual audio-visual representation for \textit{audio-visual affinity modeling}. The two resulting representations will then be fused for joint analysis to produce the final prediction. As shown in Fig.~\ref{fig:architecture}, our relational context contains two parts: \textbf{visual} ($R_\mathrm{V}$) and \textbf{audio-visual relational context} ($R_\mathrm{AV}$). This special decoupled design strengthens the model's robustness to synchronization errors between audio and video. We now elaborate on the design of $R_\mathrm{V}$ and $R_\mathrm{AV}$.
\paragraph{Visual Relational Context:}
\iffalse
When jointly optimizing all candidates in the scene, the order in which the candidates are arranged within the context should be immaterial. To simultaneously process all candidates, we introduce a \textit{permutation-equivariant} layer $\gamma$ which produces exactly one output per candidate, while preserving the order they are provided to the network. In other words, given any permutation of the $N$ candidate speakers $\sigma\in S_N$, where $S_N$ is the symmetry group in $N$ letters, the following holds:
\begin{equation}
\sigma\big(\gamma(v_1),\dots,\gamma(v_N)\big) = \big(\gamma\big(v_{\sigma(1)}\big),\dots,\gamma\big(v_{\sigma(N)}\big)\big).
\end{equation}
\fi
Our key idea is to represent each candidate's visual activity by aggregating his/her locally perceived activity and all pairwise interactions with other candidates in the scene. Specifically, we design a \textit{permutation-equivariant} layer $\gamma$ to process all candidates simultaneously, while preserving the order in which they are provided to the network. Denote the input visual feature stack by $\mathbf v=(v_1,v_2,\dots, v_N)\in\mathbb{R}^{N\times T\times d'}$. For each candidate $i=1,2,\dots,N$, the output in the $i$-th position of $R_\mathrm{V}(\mathbf v)$ is
\begin{equation}\label{eq:equivariant-gamma}
R_{\mathrm{V},i} = \gamma(v_i) = \frac 1N\bigg[\alpha(v_i, h_i) + \sum_{j\ne i}\beta(v_i, v_j, h_{ij})\bigg],
\end{equation}
where $\alpha(\cdot,\cdot)$ and $\beta(\cdot, \cdot, \cdot)$ are two networks that respectively model visual activity and pairwise visual interactions, $v_i$ denotes candidate $i$'s visual features, $h_i$ is the head map embedding for candidate $i$, and $h_{ij}$ is the head map embedding for the candidate pair $(i, j)$.
Here, the parameters of $\alpha$ are shared across all candidates, and the parameters of $\beta$ are shared across all candidate pairs, so Eq.~\eqref{eq:equivariant-gamma} is directly applicable to any number of co-occurring candidates $N$. This in turn means that our network can analyze a theoretically unlimited number of candidates at test time. On the surface, Eq.~\eqref{eq:equivariant-gamma} appears to suggest that obtaining $R_\mathrm{V}(\mathbf v)$ requires $N^2+N$ evaluations in total. In our implementation, we reduce this number by half to $N(N+1)/2$ by parameterizing $\beta(\cdot,\cdot, \cdot)$ as a \textit{skew-symmetric} function
\begin{equation}\label{eq:skew-symmetry}
\beta_{ji}=\beta(v_j,v_i,h_{ji}) = -\beta(v_i,v_j,h_{ij})=-\beta_{ij}.
\end{equation}
Our intuition is that $\beta_{ij}$ should learn to represent the amount of attention that candidate $i$ directs at candidate $j$, and rank the relative ``activeness" and visual saliency of candidate $i$ against candidate $j$. In this case, it suffices to express the direction of such relationships using a single positive or negative sign for each component.
\vspace{-1.5ex}
\paragraph{Audio-Visual Relational Context:}
We model the affinity between the audio and each speaker's face track in terms of both synchrony and cross-modal biometrics. We fuse per-frame face track features with the audio features via concatenation, and then pass them through a shared network $\eta$ to compute local A-V affinity features $\eta(v_i, a)$. Considering that local A-V affinity estimation is vulnerable to signal ambiguity (e.g. low-resolution faces, profile faces, noisy audio etc.), we aggregate evidence from all candidates to make more reliable predictions: if one or some candidates display strong A-V agreement, then the network can accordingly lower its affinity predictions for other candidates to mitigate false positives. We therefore design a learnable module to adaptively suppress only the non-active candidates whose A-V affinities are of significantly lower magnitudes, while leaving the active candidates' features "as-is". To this end, we introduce an element-wise max-pooling operation over all A-V affinity features to obtain a global representation. Specifically, the pooled feature is computed as $\eta_{\mathrm{global}} = \max_{1\le i\le N}\eta(v_i, a)$,
\iffalse
\begin{equation}\label{eq:eta_global}
\eta_{\mathrm{global}} = \max_{1\le i\le N}\eta(v_i, a),
\end{equation}
\fi
and then concatenated to each candidate's initial features $\eta(v_i, a)$. The concatenated features are further processed with two fully connected layers to generate each candidate's final contextual audio-visual representation $R_{\mathrm{AV},i}$. We name this operation \textit{non-active speaker suppression} after its effect.
\vspace{-2ex}
\subsection{Temporal Context}\label{sec:temporal-context}
As shown in Fig.~\ref{fig:architecture}, we incorporate temporal context in the networks $\alpha$, $\beta$, and $\eta$ inside $R_\mathrm{V}$ and $R_\mathrm{AV}$. This benefits ASD in two ways: first, it improves the consistency of the relational context modeling process and smoothes out local, instantaneous noises; second, it helps to alleviate synchronization errors between the audio and the video stream, an issue that is ubiquitous with in-the-wild videos.
We choose two basic architectures as our temporal modeling back-end: temporal convolutions (1D CNNs), and bi-directional Gated Recurrent Units (Bi-GRUs). The former is favorable since it has a fixed temporal receptive field, and can be easily adapted for online settings. The latter is more reliable since it has access to both past and future information. Hence, for the rest of the paper, by mentioning \textbf{temporal context} we refer to the usage of a one-layer Bi-GRU backend with $256$ cells. Otherwise, we apply a stack of two 1D convolutional layers of kernel size $3$ interleaved with Batch Normalization~\cite{DBLP:conf/icml/IoffeS15} and ReLU activation by default.
Finally, the refined contextual representations $R_\mathrm{V}$ and $R_\mathrm{AV}$ are concatenated and fed to a fully-connected layer that is shared across candidates. The outputs are passed through sigmoid activation, resulting in real values between $0$ and $1$ that indicate each candidate's probability of being the active speaker.
\vspace{-2ex}
\subsection{Losses}\label{sec:losses}
The model is trained end-to-end with a multi-task loss formulation. Since the goal is to predict a binary speaking/not speaking label, for each loss term we apply the standard binary cross-entropy (BCE) loss, averaged over all time steps. The BCE loss is defined as
\begin{equation}
\mathcal{L}_{\mathrm{BCE}}(y, \hat y) = -\hat y\log y - (1-\hat y)\log(1-y),
\end{equation}
where $y$ and $\hat y$ are the predictions and ground truth labels, respectively. For a scene with $T$ frames and $N$ candidates, let $A_{\mathrm{aux}}$ and $V_{\mathrm{aux}}$ denote the audio and visual auxiliary prediction layer, $AV_{\mathrm{pred}}$ the final audio-visual prediction layer, and $\hat{y}_{\mathrm{v},i}^{1,2,\dots,T}$ and $\hat{y}_{\mathrm{av},i}^{1,2,\dots,T}$ the corresponding visual and audio-visual ground truths of the $i$-th candidate. We add two auxiliary losses $\mathcal{L}_\mathrm{a}$ and $\mathcal{L}_\mathrm{v}$ to learn discriminative features for both visual and audio streams. The \textit{audio} prediction loss is
\begin{equation}
\mathcal{L}_\mathrm{a} = \frac 1T\sum_{t=1}^T \mathcal{L}_{\mathrm{BCE}}\big(\sigma(A_{\text{aux}}(a^t)), \hat{y}_{\mathrm a}^t\big),
\end{equation}
where $\hat{y}_{\mathrm a}^t = \max_{1\le i\le N}\hat{y}_{\mathrm{av},i}^t$ is the audio ground truth indicating whether at least one of the candidates is speaking, and $\sigma(\cdot)$ is the sigmoid function.
When training on a single candidate (\textit{i.e.} without employing relational context),
the \textit{visual} prediction loss and main \textit{audio-visual} prediction loss are given by
\begin{align}
\begin{split}
\mathcal{L}_\text{v} &= \frac 1T\sum_{t=1}^T \mathcal{L}_{\mathrm{BCE}}\big(\sigma(V_{\text{aux}}(v_i^t)), \hat{y}_{\mathrm v,i}^t\big),\\
\mathcal{L}_\text{av} &= \frac 1T\sum_{t=1}^T \mathcal{L}_{\mathrm{BCE}}\bigg(\sigma\big(AV_{\text{pred}}\big(a^t\oplus v_i^t\big)\big), \hat{y}_{\mathrm{av},i}^t\bigg),
\end{split}
\end{align}
where $\oplus$ denotes feature concatenation. When training with relational context on multiple candidates, the losses are slightly different. To compute the visual and audio-visual losses, we replace the inputs to the auxiliary prediction layers with (concatenated) contextual representations, and aggregate losses from all candidates:
\begin{align}\label{eq:loss-nspk}
\begin{split}
\mathcal{L}_\text{v} &= \frac 1N\sum_{i=1}^N\frac 1T\sum_{t=1}^T \mathcal{L}_{\mathrm{BCE}}\bigg(\sigma\big(V_{\text{aux}}\big(R_{\mathrm{V}, i}^t\big)\big), \hat{y}_{\mathrm v,i}^t\bigg),\\
\mathcal{L}_\text{av} &= \frac 1N\sum_{i=1}^N\frac 1T\sum_{t=1}^T \mathcal{L}_{\mathrm{BCE}}\bigg(\sigma\big(AV_{\text{pred}}\big(R_{\mathrm{V}, i}^t\oplus R_{\mathrm{AV}, i}^t\big)\big), \hat{y}_{\mathrm{av},i}^t\bigg).
\end{split}
\end{align}
Finally, the \textit{total} loss is defined as follows:
\begin{equation}\label{eq:total_loss}
\mathcal{L} = \mathcal{L}_\text{a} + \mathcal{L}_\text{v} + \mathcal{L}_\text{av}.
\end{equation}
Note that in practice, $N$ may vary over time as candidates enter and leave the scene.
\vspace{-1.5ex}
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{
"redpajama_set_name": "RedPajamaArXiv"
}
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Q: Upload and insert into text area I have wrote a very simple php upload. But I want to insert image in text area with "click". I want to do this with jquery.
Is there any jquery plugin for do this? (or an application for upload like in stackoverflow)?
Thanks in advance
A: Check http://elrte.org/. It's a jQuery WYSIWYG editor with files manager.
A: Look up TinyMCE - it's not part of jQuery, but, it's a JS library that gives you a WYSIWYG editor with image control.
A: Try CKEditor, it's even working with jQuery too.
A: After submit (some help about that in here) call to the rich text area and insert the img tag, something like this:
$('#imageform').bind('submit', function(e) {
e.preventDefault(); // <-- important
$(this).ajaxSubmit({
target: '#preview',
success: function() {
$('#richtext_area').html('<img id"newimg" src="" alt="your file"/>');
$('#newimg').attr("src",$("#filePath_text").val())
}
});
}).submit();
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"redpajama_set_name": "RedPajamaStackExchange"
}
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\section{Introduction}
The intrinsic spectral energy distribution (SED) of active galactic nuclei
(AGNs), which extends from the radio up to $\gamma$-rays, cannot be observed
from the Lyman limit and up to several hundred eV due to Galactic and
intrinsic absorption. It is widely assumed that the intrinsic SED in this
range is a superposition of a bump and one or more power-law components
(Figure~\ref{f:sed+bump}). The hypothetical bump is known as the `Big Blue
Bump'. The deviation from the hypothetical power-law at the high energy
wing of the bump is known as the `soft excess'. This latter feature
receives special attention because it may be within the observable soft
X-ray range.
There are theoretical and observational motivations for the bump plus
power-law hypothesis. Generically, thermal radiation from accretion on a
black hole has a typical effective temperature $T_a$ that is related to the
typical size of the emitting region $r_a$ and to the accretion luminosity
$L_a$ by $L_a \propto 4\pi r_a^2 T_a^4$. Order of magnitude estimates, as
well as more detailed accretion disk models, suggest that typical AGN
accretion temperatures should be of order ${\rm few}\times10^5$\,K,
corresponding to a Wien bump peaking at $\sim 40$\,eV (see e.g. Blandford
\cite{BNW} for a review). However, thermal bumps may be absent if the
accretion is advection-dominated (Narayan \& Yi
\cite{NY}; Narayan, Kato \& Fumio \cite{NKH}).
The power-law components are suggested by the observed shape of the SED at
the optical-UV and soft to hard X-ray continua, which in many cases have
different spectral indices. The power-law continua are thought to be
produced by inverse Compton scattering of accretion disk
emission. Observationally, there are indications that the mean optical--UV
slope tends to be flatter than the mean soft X-ray slope (e.g. Puchnarewicz
et al. \cite{Puchnarewicz}). This suggests a break somewhere in between
these two spectral ranges, on top of which a bump may be superimposed.
Two major issues motivate the efforts to uncover the ionizing SEDs of
AGNs. The AGN paradigm postulates that AGNs are powered by an accreting
super-massive black hole. Major advances have been made in recent years in
establishing the existence of super-massive black holes in the nuclei of
normal and active galaxies (Kormendy \& Richstone \cite{KR}; Miyoshi et al
\cite{Miyoshi}; Eckart \& Genzel \cite{EG}). On the other hand, there is
little direct evidence to support the accretion hypothesis. A detection of
a signature of the accretion mechanism in the SED will have important
implications both for validating the AGN paradigm as well as for
discriminating between various accretion models. The ionizing SED is also
important for modeling the effects of the AGN on its environment, since the
SED is a critical input for photoionization models. Such modeling is a
major tool in the study of AGN emission lines and in attempts to
disentangle the contribution of the nuclear emission from that of
starbursting regions in ultra-luminous IR galaxies (Lutz et
al. \cite{Lutz1}; Genzel et al. \cite{Genzel}).
The spectral range of the SED that is accessible to direct observations can
be maximized in the soft X-ray regime by observing nearby Seyfert galaxies,
and in the UV by observing high redshift AGNs (e.g. Zheng et
al. \cite{Zheng}), although this is limited by inter-galactic
absorption. Even so, a large gap remains. Almost all the current
information on the SED in this gap comes from combining UV and X-ray data,
following two basic approaches. The first is to interpolate, assuming some
model, between the UV and X-rays. The second is to look for correlations
between the continuum slopes at the optical-UV and X-rays.
A survey of results from recent X-ray studies (Walter \& Fink \cite{WF};
Laor et al. \cite{LFEWM1}; Walter et al. \cite{WOCFMOW}; Puchnarewicz et
al. \cite{Puchnarewicz}; Laor et al. \cite{LFEWM2}; Brunner et
al. \cite{Brunner}) reveals a confusing picture. Answers to the key
observational questions: Is there a bump? Is there a soft excess? What is
the object-to-object scatter in bump temperature? Is intrinsic absorption
significant?, vary greatly. As a consequence, answers to key theoretical
questions, such as the consistency of the SEDs with accretion models and
the origin of the object-to-object scatter, also remain inconclusive. There
are several possible reasons for this. One is the strong and rapid
variability of AGNs in the X-rays. To date, very few AGNs have been observed
simultaneously in the the UV and X-rays (NGC\,4151 is one of them). Another
problem is the very low spectral resolution of current X-ray spectroscopy,
which requires a-priori models of the SED and of the galactic absorption to
reconstruct the spectrum. As a result, the conclusions are strongly model
dependent and many different models fit the data equally well (see
e.g. Walter et al. \cite{WOCFMOW}). Because of this ambiguity, the
interpolation approach has been primarily applied to samples of AGNs rather
than to individual objects. The correlation approach is, by definition,
statistical and applicable only to AGN samples. Therefore, both approaches
are vulnerable to the statistical pitfalls of sample analysis, such as
selection effects and small sample fluctuations. It is widely suspected
that selection effects play an important role in biasing the results of
X-ray sample studies (e.g. Puchnarewicz et al. \cite{Puchnarewicz}). If
this is true, then there must be a large variety of AGN SEDs. This raises
doubts whether the sample mean is a useful quantity for describing such a
heterogeneous population. In this situation, a method that can be applied
to an {\em individual object} rather than to AGN {\em samples} can be
extremely useful. It is in this that IR spectroscopy can play a role.
Following the first detection of $\bFeXb$ in NGC\,4151 (Oke \& Sargent
\cite{OS}), it became apparent that AGNs emit optical lines from highly
ionized species (`coronal lines'). More recently, IR coronal lines have
been detected with ground based telescopes (Oliva \& Moorwood \cite{OM},
Oliva et al. \cite{Oliva}, Reconditi \& Oliva \cite{RO}), and
photoionization models predicted that many additional lines were likely to
be observed in the spectral range covered by ISO (e.g. Spinoglio \& Malkan
\cite{SM}, Greenhouse et al. \cite{Greenhouse}, Voit \cite{Voit}). The
observed line ratios point to photoionization rather than collisional
ionization as the dominant ionization mechanism (Oliva et
al. \cite{Oliva}). In this case, each of these lines probes the SED at
energies $\ge \ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$, where we designate by $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$ the ionization
energy required to produce the emitting ion from the preceding
ionization stage. Figure~\ref{f:eion} demonstrates that the gap in
the SED of NGC\,4151 is well covered by the observed optical and IR
lines. Therefore, by fitting a photoionization model to the observed
lines, it should be possible to constrain the SED.
With this in mind, our ISO-SWS program on bright galactic nuclei
included detailed spectroscopic observations of several AGNs. The
results from a pilot study of the nearby Seyfert 2 Circinus Galaxy
(Moorwood et al. \cite{Moorwood}) demonstrated the potential of this
approach to the reconstruction of the ionizing SED. Here we extend
this method to NGC\,4151, one of the brightest, closest and most
extensively studied Seyfert galaxies. The ISO-SWS observations of
NGC\,4151 are presented in a companion paper (Sturm et
al. \cite{Sturm}, Paper I), which compares the IR emission line
profiles to optical line profiles and derives constraints on models of
profile asymmetry. In this work, we add observed UV, optical, near-IR
and far-IR lines from the literature to the mid-IR SWS lines, and
perform an extensive search in parameter space to find the best
fitting SED.
The rest of this paper is organized as follows. The observed properties of
Seyfert galaxy NGC\,4151 and the inferences about its physical conditions
are discussed in Sect.~\ref{s:n4151}. The observed emission line fluxes
compilation and the criteria for selecting the lines to use in the modeling
are described in Sect.~\ref{s:lines}. The photoionization models and the
way they are compared to the data are described in
Sect.~\ref{s:models}. The results are presented in Sect.~\ref{s:results},
discussed in Sect.~\ref{s:discuss} and summarized in Sect.~\ref{s:summary}.
\section{The physical properties of NGC\,4151}
\label{s:n4151}
Although NGC\,4151 is often called the `classical' Seyfert galaxy, it
displays a variety of properties typical of different AGN classes. It
may therefore be in fact more complex than average. We present here a
brief overview of its properties that are relevant to this work.
NGC\,4151 is a barred spiral galaxy, which is seen almost face on
(Pedlar et al. \cite{Pedlar92}) at redshift $z=0.0033$ (distance
$D=9.9h^{-1}$\,Mpc) and magnitude $m_V = 11.5$ (Brinkmann et
al. \cite{Brinkmann}). Its spectrum is dominated by the non-stellar
component (Kaspi et al. \cite{Kaspi}). NGC\,4151 is highly variable in
both the continuum and the broad emission lines. Originally classified
as an intermediate, Seyfert 1.5 galaxy (Osterbrock \& Koski
\cite{OK}), it went through an extreme low state in 1984, where it
took on the characteristics of a Seyfert 2 galaxy (Penston \& P\'erez
\cite{PP}). Currently, NGC\,4151 is usually grouped in the Seyfert 1 class.
\subsection{The ionizing continuum}
\label{s:n4151-SED}
The December 1993 multi-wavelength campaign to monitor this variability
(Crenshaw et al. \cite{Crenshaw}; Kaspi et al. \cite{Kaspi}; Warwick et
al. \cite{Warwick}; Edelson et al. \cite{Edelson}) yielded a simultaneous
SED extending from the optical to $\gamma$-rays. NGC\,4151 was observed
in the optical from Wise Observatory and Lowell Observatory, in the UV by
IUE, in X-rays by ROSAT and ASCA and in $\gamma$-rays by
CGRO. Warwick et al. (\cite{Warwick}) deconvolved the X-rays to $\gamma$-rays
spectrum by assuming an $F_\nu\propto\nu^{-0.5}$ powerlaw and a thermal
bremsstrahlung component with a temperature of 0.5 keV, absorbed by a two
component intrinsic absorber (`partial covering model') and by galactic
neutral hydrogen.
In order to extend the spectral coverage up to the Lyman limit, we
used the 1995 HUT best-fit power-law continuum (Kriss et
al. \cite{KDZKE}), which we added to the 1993 SED after normalizing
them to the IUE flux at the line-free wavelength of 1806\AA\,. A large
uncertainty is associated with this flux value due to variability and
reddening. The 1995 UV continuum was $\sim5$ times more luminous and
much bluer than it was in 1990 (Kriss et al. \cite{Kriss}). By fitting
a power-law continuum to the HUT data, Kriss et al. (\cite{Kriss})
derive $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi=0.04$, which they use to deredden the SED. This amounts
to a factor of 1.3 increase in the flux at 1806\AA\,. However,
estimates of the reddening vary greatly (Sect. \ref{s:n4151-gas}). In
view of these problems, we treat the flux at 1 Ryd as another source
of uncertainty in the model, which is investigated by the model
fitting procedure described in Sect.~\ref{s:fit}.
Figure \ref{f:sed+bump} shows the composite SED, where for compatibility
with Edelson et al. (\cite{Edelson}), we plot the SED as $L_\nu = 4\pi D^2
f_\nu$, where $f_\nu$ is the observed flux density in
erg\,s$^{-1}$\,cm$^{-2}$\,Hz$^{-1}$ and $D=20$\,Mpc. The SED displays the
typical strong optical-UV emission lines of a Seyfert galaxy and
significant intrinsic absorption below 4 keV. Above 4 keV, a hard power-law
extends up to a cutoff at $\sim90$\,keV. Also visible are X-rays emission
lines at $\sim 6.4$\,keV. Figure \ref{f:sed+bump} also shows a schematic
sketch of a superimposed Big Blue Bump and a powerlaw spectrum, constructed
by superimposing a power law spectrum, $L_\nu\propto\nu^{-1}$ and 5
blackbody spectra normalized to have equal total luminosities,
$B_\nu(T)/T^4$, with $T=10^5, 2\,10^5,\ldots,5\,10^5$\,K. Note that the
shape of the excess emission above the power-law depends in part on the way
the SED is presented. A bump in $\nu L_\nu$ may appear only as a `shoulder'
or a flattening in $L_\nu$. Here, we use the term `bump' irrespective of
the presentation of the SED.
\subsection{The galactic nucleus}
\label{s:n4151-nucleus}
The line spectrum of NGC\,4151 points to the existence of the three
kinematically distinct nuclear emission regions that are observed in many
Seyfert galaxies: an unresolved Broad Line Region (BLR) with typical line
widths of $> 5000$\,km s$^{-1}$, densities above $10^8$\,cm$^{-3}$ and an
ionization parameter (see below) $U \sim 0.03$--$1$, a Narrow Line Region
(NLR) with typical line widths of $\la 500$\,km s$^{-1}$, densities in the
range $10^3$ to $10^6$\,cm$^{-3}$ and $U \sim 0.001$--$0.01$, and an
Extended Emission Line Region (EELR) with typical line widths of $\la 50$
km s$^{-1}$, densities below $10^3$\,cm$^{-3}$ and $U \la 0.005$ (Schulz \&
Komossa \cite{SK}). The line widths of the SWS forbidden IR lines point to
an origin in the NLR. This is supported by estimates of the gas density
from the line ratios (Sect. \ref{s:n4151-gas}) and is consistent with the
small critical densities ($<{\rm few}\times10^5$\,cm$^{-3}$) of the mid-IR
fine-structure line transitions, which are incompatible with the
BLR. However, the very large aperture of the SWS includes also the EELR,
and some contribution from the EELR cannot be ruled out, especially for the
lower ionization lines.
Narrow band images in the [O\,{\sc ii}], [O\,{\sc iii}] and Balmer lines
(Heckman \& Balick \cite{HB}; P\'erez et al. \cite{Perez}; Unger et
al. \cite{Unger}; P\'erez-Fournon \& Wilson \cite{PW}; Yoshida \& Ohtani
\cite{YO}; Hutchings et al. \cite{Hutchings}) show that NGC\,4151 has an
elongated, knotty EELR which extends along PA $\sim 50^\circ$ up to $\sim
30^{\prime\prime}$ to the SW of the nucleus, but only up to $\sim 15^{\prime\prime}$ to the
NE. The NLR symmetry axis coincides with the radio jet at PA
$\sim77^\circ$ (Pedlar el al. \cite{Pedlar93}), but is misaligned with the
EELR. This misalignment is consistent with a very wide ionization cone
with an opening angle $\sim 120^\circ$, intercepting the galactic disk at
grazing incidence (Pedlar et al. \cite{Pedlar92},
\cite{Pedlar93}). Images show that the opening angle of the line
emitting conical section of the galactic plane is $\sim 80^\circ$
(Hutchings et al. \cite{Hutchings}). The $\bOIIIbB$ profile, which traces
the velocity field (Schulz \cite{Schulz90}) and the $\bOIIIbB/\ifmmode {\rm H}\alpha \else H$\alpha$\fi$ line
ratio, which traces $U$ (Robinson et al. \cite{Robinson}), both indicate a
NLR size of $\sim5^{\prime\prime}$. An $\bOIIIbB$ map of the inner nucleus reveals
NLR emission from $\la 0.5^{\prime\prime}$ of the nucleus up to $2-3^{\prime\prime}$
(Hutchings et al. \cite{Hutchings}). These size estimates are unlikely to
be affected significantly by projection, since the inclination angle of the
galactic disk at the inner few arcsec is estimated at
$12^\circ<i<21^\circ$, and the NLR symmetry axis is within a few degrees of
the line of nodes (Boksenberg et al. \cite{Boksenberg}).
The volume filling factor of the NLR gas, $F$, which is
defined as the fraction of the volume actually filled with gas\footnote{By
definition, $F^{1/2} = n_{\rm rms}/n_{\rm loc}$, the ratio of the volume
averaged rms electron density to that of the local one. $n_{\rm loc}$ is
estimated from emission line ratios, while $n_{\rm rms}$ is estimated from
the observed flux of a recombination line, $\ifmmode f_\ell \else $f_\ell$ \fi$, and the angular size of
the line emitting region,$\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$, from $4\pi D^2 \ifmmode f_\ell \else $f_\ell$ \fi \propto n_{\rm rms}^2
(D\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi)^3$. This introduces an $h$ dependence to the estimate of the
filling factor $F=F_1h$.}, is very small. Yoshida \& Ohtani (\cite{YO})
estimate $F \sim 10^{-3}h$ based on the low excitation $\ifmmode {\rm H}\alpha \else H$\alpha$\fi$, $\bOIIb$ and
$\bOIIIbB$ maps. Similarly, Robinson et al. (\cite{Robinson}) find $F \sim
{\rm few}\times 10^{-4}h$ in an $\ifmmode {\rm H}\alpha \else H$\alpha$\fi$ knot at the edge of the NLR. Thus the
low excitation NLR lines are emitted from a very clumpy distribution of
gas. Unfortunately, narrow band images or position-resolved long-slit
spectroscopy of coronal lines are not yet available for NGC\,4151, so their
NLR size and filling factor cannot be estimated directly. However, an upper
limit on the size of the coronal line emitting region can be estimated from
the low ionization NLR lines with the plausible assumption that the high
ionization emission is more compact. This is indeed observed in the spatial
distribution of the $\CIV$ line relative the $\bOIIIbB$ line (Hutchings et
al. \cite{Hutchings}), and is observed in the coronal lines of the Circinus
galaxy (Maiolino et al. \cite{Maiolino}).
Spatially resolved maps of the line emission in the NLR and EELR make it
possible to estimate the anisotropy parameter $A$, defined as the ratio
between the ionizing luminosity that the gas must see in order to emit the
observed line flux, and that directed at the observer. There are
indications that the anisotropic bi-conic structure of the NLR and EELR
traces an anisotropy in the photoionizing luminosity (as opposed to an
anisotropy in the matter distribution). Estimates of $A$ vary greatly since
they strongly depend on the assumed intrinsic continuum (usually
interpolated over the gap in the SED by a power-law) and the
photoionization model of the NLR /EELR. Schulz \& Komossa (\cite{SK})
derive $1\le A \le 6$, Yoshida \& Ohtani (\cite{YO})) derive $1<A\la 3$,
Penston et al. (\cite{Penston90}) derive $A\sim13$, and Robinson et
al. (\cite{Robinson}) obtain $A\sim1$ and $A\sim10$ by two different
methods.
Strong absorption on the line of sight in the optical, UV and X-ray
bands implies the existence of a multi-component system of absorbers
that may also modify the ionizing SED in other directions. Warwick et
al. (\cite{Warwick}) fit the X-ray SED with a two component absorber
of total hydrogen column density in the range $10^{22}$ to $10^{23}$
cm$^{-2}$. The location of this absorber along the line of sight is
unknown. Kriss et al. (\cite{Kriss}) fit the Lyman absorption lines
with an outflowing clumpy and dense absorber ($n > 10^{9.5}$
cm$^{-3}$) with a neutral hydrogen column density in the range
$6\,10^{17}$ to $6\,10^{20}$\,cm$^{-2}$, which covers a large fraction
of the continuum source and the BLR along the line of sight. Both the
high density and the outflow velocity suggest that the absorber is
between the BLR and NLR. The properties of this absorber are
incompatible with that of the X-ray absorber, and it must be assumed
that these are two different components. HUT observations of NGC\,4151
at a later epoch (Kriss et al. \cite{KDZKE}) show that the properties
of the absorber appear to change in time, possibly in response to
changes in the continuum luminosity, and indicate the possible
presence of an additional component with neutral hydrogen column
density of up to $5\,10^{20}$\,cm$^{-2}$.
\subsection{The line emitting gas}
\label{s:n4151-gas}
The ISO spectra can be used to estimate the gas density and
temperature, given estimates of the reddening and starburst
contribution to the line emission.
Estimates of the reddening of the NLR in NGC\,4151 range from almost
negligible, $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi = 0.04$ (Kriss et al. \cite{KDZKE}), 0.05 (Penston et
al. \cite{Penston81}; Wu \& Weedman \cite{WW}; Boksenberg et
al. \cite{Boksenberg78}), to considerable, $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi = 0.12$ (Kaler
\cite{Kaler}), 0.13 (Malkan \cite{Malkan}), and 0.09 to 0.28 (Ward et
al. \cite{Ward}). The elemental abundances and the internal dust content of
the line emitting gas are even less certain. In a study of the Circinus
galaxy, Oliva et al (\cite{Oliva}) find that the coronal line emitting gas
is dust free and has solar abundances to within a factor of $\sim 2$.
Similarly, Ferguson et al. (\cite{FKF}) compare an extensive grid of
coronal line photoionization models to observed line ratios in several
Seyfert galaxies and conclude that dust is unlikely to exist in coronal
line emitting gas, and that the Si and Ca abundances are solar to within a
factor of $\sim2$. The NGC\,4151 EELR models of Schulz \& Komossa
(\cite{SK}) point to a $\sim 1/2$ solar metallicity for all metals
(consistent with metal poor gas, but not with dust depletion). In our
models, we will assume that the abundances are solar and that the gas is
dust free.
The stellar population in the nucleus of NGC\,4151 can affect the
observed line fluxes in two ways. First, the presence of
star-formation regions within the large ISO-SWS aperture can introduce
line emission due to radiation fields peaking at $\sim1$\,Ryd, which
are typical of hot stars (if these regions are shielded from the
central continuum). An upper limit on the starburst contribution to
the lower ionization lines can be estimated from the upper limits on
the PAH emission, which are tracers of star forming activity (Roche et
al. \cite{RASW}; Genzel et al. \cite{Genzel}). The mean observed
ratios of the integrated $\ifmmode {\rm [Ne}\,{\sc ii]}\,\lambda 12.8 \mu{\rm m$ line flux to that of the PAH
feature in starburst galaxies M83, NGC\,4945, NGC\,3256 and NGC\,7552
(see also Genzel et al. \cite{Genzel}) are $\ifmmode {\rm [Ne}\,{\sc ii]}\,\lambda 12.8 \mu{\rm m$/PAH\,$\lambda
7.7\mu$m = 0.07 $\ifmmode {\rm [Ne}\,{\sc ii]}\,\lambda 12.8 \mu{\rm m$/PAH\,$\lambda 6.2\mu$m = 0.22, as compared
to the NGC\,4151 lower limits of $\ifmmode {\rm [Ne}\,{\sc ii]}\,\lambda 12.8 \mu{\rm m$/PAH\,$\lambda 7.7\mu{\rm m}
> 0.32$ and $\ifmmode {\rm [Ne}\,{\sc ii]}\,\lambda 12.8 \mu{\rm m$/PAH\,$\lambda 6.2\mu{\rm m} > 0.47$, the latter
value being much less certain. The more reliable upper limit on the
PAH\,$\lambda 7.7\mu$m flux leads to an upper limit of $\sim 20$\% of
starburst contribution to $\ifmmode {\rm [Ne}\,{\sc ii]}\,\lambda 12.8 \mu{\rm m$ in NGC\,4151. This is supported
by the [O\,{\sc iii}], $\ifmmode {\rm H}\alpha \else H$\alpha$\fi$ and $\ifmmode {\rm H}\beta \else H$\beta$\fi$ line ratios and maps in the
bright knots of the EELR, which point to photoionization by the
central source rather than photoionization by hot stars (Schulz
\cite{Schulz88}; P\'erez et al. \cite{Perez}; P\'erez-Fournon \&
Wilson \cite{PW}). Second, a strong underlying stellar continuum may
affect the flux measurements of the optical lines. Robinson et
al. \cite{Robinson} fit the off-nuclear continuum ($5^{\prime\prime}$ SW of
the center) by an old stellar population and conclude that the effect
of absorption features on the emission line flux measurements is
negligible. Thus, stars are unlikely to affect the reconstruction of
the SED.
Three density-sensitive pairs of bright lines of the same ion are found
among the NGC\,4151 emission lines detected in the ISO-SWS wavelength
range: [Ne\,{\rm v}]\,$\lambda14.32,24.32\mu$m, [Ne\,{\sc
iii}]\,$\lambda15.55,36.04\mu$m, and [S\,{\sc
iii}]\,$\lambda18.71,33.48\mu$m. We also consider the [O\,{\sc
iii}]\,$\lambda51.81,88.35\mu$m pair observed with ISO-LWS by Spinoglio et
al. (\cite{Spinoglio}). Among these observations, the [Ne\,{\sc v}] ratio
is perhaps most interesting since it samples a NLR species with an
ionization energy near 100\,eV. However, for NGC\,4151 with its apparently
very small contribution of star forming activity to the emission line
spectrum, the [Ne\,{\sc iii}], [S\,{\sc iii}], and [O\,{\sc iii}] ratios
are valuable NLR diagnostics as well. Figure \ref{f:density} shows the
density dependence for these ratios, computed by solving the rate equations
for five level systems. Transition probabilities and collision strengths
have been taken from Mendoza \& Zeippen (\cite{MZ}) ([S\,{\sc iii}]),
Galavis et al. (\cite{GMZ95}) ([S\,{\sc iii}]), Galavis et
al. (\cite{GMZ97}) ([Ne\,{\sc iii}], [O\,{\sc iii}], [Ne\,{\sc v}]), Butler
\& Zeippen (\cite{BZ}) ([Ne\,{\sc iii}]), and Lennon and Burke (\cite{LB})
([Ne\,{\sc v}], [O\,{\sc iii}]). ISO observations of the planetary nebula
NGC\,6302 (Pottasch et al. \cite{Pottasch}) support the Lennon \& Burke
collision strengths for [Ne\,{\sc v}] that were previously questioned by
Oliva et al. (\cite{OPR}).
The [O\,{\sc iii}] ratio of 1.5 measured by Spinoglio et
al. (\cite{Spinoglio}) for NGC\,4151 corresponds to a density of about
450\,cm$^{-3}$. The other ratios should be near their low density limits at
such densities and hence of little use to further constrain the density of
the NLR gas in NGC\,4151. Indeed, the measured ratios [Ne\,{\sc v}],
[Ne\,{\sc iii}], [S\,{\sc iii}] of 0.98, 5.9 and 0.67, respectively, are
close to their low density limits, and consistent with the [O\,{\sc iii}]
density. For simplicity and since [O\, {\sc iii}] may tend to sample
slightly lower density extended regions when compared to higher ionization
species, we adopt an electron density of 1000\,cm$^{-3}$ for our
modelling. This is still well below the critical densities of the lines we
model.
The infrared ground state fine-structure line fluxes are insensitive to the
electron gas temperature for $T\ga 2\,10^3$\,K. However, the optical
forbidden line intensities, which originate from higher energy levels, are
sensitive to the gas temperature of $\sim 10^4$\,K in photoionized gas. The
ratios of optical and infrared transition may therefore be useful
diagnostics of the gas temperature. The dominant uncertainty in such an
analysis is often the reddening correction to be applied to the optical
fluxes. Because of the low extinction to the NLR of NGC\,4151, for which
we adopt $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi=0.05-0.15$ (see above), a temperature analysis based on such
an optical/infrared comparison appears promising. The four species
discussed above emit optical lines which are in principle suitable. The
optical/IR ratios of [S\,{\sc iii}]\,$\lambda 9096$\AA\, and [S\,{\sc
iii}]$\lambda 9531$\AA\, to $\ifmmode {\rm [S}\,{\sc iii]}\,\lambda 18.7 \mu{\rm m$ and $\ifmmode {\rm [S}\,{\sc iii]}\,\lambda 33.5 \mu{\rm m$ and of
$\bOIIIbA$\AA\, and $\bOIIIbB$\AA\, to [O\,{\sc iii}]\,$\lambda 51.81\mu$m
and [O\,{\sc iii}$\lambda 88.35\mu$m, however, are of limited diagnostic
value at the density of the NGC\,4151 NLR and electron temperatures in the
range 10000-20000\,K, since the temperature dependence is small compared to
the density dependence and compared to the effects of possible problems in
inter-calibration of optical and infrared measurements. In case of
[Ne\,{\sc v}], a suitable large aperture measurement of the 3426\AA\/
transition is not available. It is hence not possible to test claims of
very high $T_e$ in the highly ionized region that are based on analysis of
optical [Fe\,{\sc vii}] lines in several Seyferts (Erkens et
al. \cite{EAW}).
A robust diagnostic is the [Ne\,{\sc iii}]$\lambda3868$\AA\/ /
$15.55\mu$m ratio which is basically density-insensitive (Figure
\ref{f:temp}). We adopt the 3868\AA\/ flux of Oke \& Sargent
(\cite{OS}) (Table \ref{t:flux}) and correct for
$\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi=0.05-0.15$. Taking into acount this uncertainty in extinction,
and an additional 30\% calibration uncertainty which we have added
directly, we use the [Ne\,{\sc iii}] ratio to derive an electron
temperature of 13000$\pm$2500\,K, with lower electron temperature
obtained for lower adopted extinction. It is instructive to compare
this value to the classical NLR temperature diagnostic that uses the
optical [O\,{\sc iii}]$\lambda4363$\AA\/ / 5007\AA\/ ratio, since the
two species require similar ionization energies for their creation and
should sample a similar part of the NLR. Correcting the 4363\AA\/ /
5007\AA\/ ratio of 0.03 (Oke \& Sargent \cite{OS}) for
$\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi=0.05-0.15$, we infer an [O\,{\sc iii}] electron temperature of
about 19000\,K. We caution however that this number may be very
uncertain due to the faintness of the 4363\AA\/ line. Similarly high
[O\,{\sc iii}] electron temperatures in NLRs and EELRs
(e.g. Storchi-Bergman et al. (\cite{SWMB}), using a more careful
treatment of the [O\,{\sc iii}]$\lambda4363$\AA\/ line) have been used
to put fairly strong constraints on photoionization scenarios
(e.g. Binette et al. \cite{BWS}), or to reject them entirely. An
electron temperature of 13000\,K as obtained from the optical/IR
[Ne\,{\sc iii}] ratio is however reproduced by fairly standard
photoionization models, thus avoiding the need for more complex
scenarios. Clearly, an extension of this analysis to other AGNs is
highly desirable to investigate this discrepancy.
\section{The observed line flux compilation}
\label{s:lines}
The advantage of the IR lines as diagnostics of the gas and SED lies in
their insensitivity to the gas temperature and to external
reddening. However, this also means that they cannot constrain these
physical parameters. In particular, UV and optical lines are needed to fix
$\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$. In order to maximize the observational constraints on the models,
we added to the measured mid-IR lines observed UV, optical and NIR lines from
the literature, as well as recently measured far-IR lines.
We initially compiled a list of about 120 observed emission lines: UV
lines from Kriss et al. (\cite{Kriss}), optical lines from Oke \&
Sargent (\cite{OS}), Anderson (\cite{Anderson}), Netzer
(\cite{Netzer74}), Boksenberg et al. (\cite{BSAFPS}), Osterbrock \&
Koski (\cite{OK}); NIR lines from Osterbrock, Shaw \& Veilleux
(\cite{OSV}), Thompson (\cite{Thompson}). In addition, we included
some far-IR lines observed by the ISO Long Wavelength Spectrometer
(LWS) (Spinoglio et al. \cite{Spinoglio}).
Various problems stand in the way of merging these different
observations into a self-consistent data set that can be compared with
the model results. The observations span some thirty years, and the
spectra were taken with a variety of instruments and reduction
techniques, at different resolutions and with different apertures. The
UV lines were measured with an aperture of $18^{\prime\prime} \oslash$ (Kriss
et al. \cite{Kriss}). The apertures of the optical and near-IR spectra
(in those papers where quoted), range from
$1^{{\prime\prime}}\times3^{{\prime\prime}}$ (Boksenberg et
al. \cite{BSAFPS}) to $8^{\prime\prime} \oslash$ (Anderson
\cite{Anderson}). The NIR spectra were taken with long slits, ranging from
$2^{\prime\prime}\times10^{\prime\prime}$ (Thompson \cite{Thompson}) to $3^{\prime\prime}\times72^{\prime\prime}$
(Osterbrock, Shaw \& Veilleux \cite{OSV}). The apertures of the SWS
($14^{\prime\prime}\times20^{\prime\prime}$ to $20^{\prime\prime}\times33^{\prime\prime}$), and LWS ($80^{\prime\prime}
\oslash$) are much larger and cover the entire nucleus (See Figure 1 in Sturm et al.
\cite{Sturm}). Since the EELR emission in NGC\,4151 is very extended and
inhomogeneous, these aperture differences cannot be corrected by simply
normalizing the fluxes to the aperture size. Here we attempt no such
correction. This can be justified by noting that Yoshida \& Ohtani
(\cite{YO}) find that 83\% and 78\% of the $\bOIIIbB$ and $\bOIIb$ emission
in the central $28^{\prime\prime}\times10^{\prime\prime}$ of the nucleus, respectively, is
emitted from the NLR in a $4.3^{\prime\prime}\times3.9^{\prime\prime}$ region centered on
the continuum source. We expect that the higher ionization species will be
even more centrally concentrated and conclude that the overall error in the
line flux measurement that is introduced by the aperture differences is
smaller than 20\% for all but the smallest apertures.
Broad / narrow component decomposition, variability and absolute flux
normalization are three other inter-related problems. The narrow
component was separated only in the strongest broad lines, while the
quoted values of the weaker permitted lines include both
components. Profile decomposition is an uncertain procedure even under
ideal conditions. In the case of NGC\,4151, it is further complicated
by its high variability. In particular, the 1984 low state of
NGC\,4151 occurred at the middle of the epoch covered by the
observations cited above. Most of the optical lines are given only
relative to the narrow or full $\ifmmode {\rm H}\beta \else H$\beta$\fi$ line. The absolute flux value is
given only in a few cases, so that the inter-calibration of these
measurements is uncertain. Finally, in most cases no measurement
errors are given.
Our choice of the lines to include in the modeling is guided by four
criteria. First, we want to consider only lines that are primarly formed by
photoionization by the extreme-ultraviolet and soft X-ray continuum of the
central source. We exclude all lines emitted from ions with $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi <
13.6$\,eV, since these can be easily ionized by other processes. In
particular, this criterion leads to the exclusion of the far-IR O\,{\sc
i}\,$\lambda 63.2 \mu$m, O\,{\sc i}\,$\lambda 145.5\mu$m and C\,{\sc
ii}\,$\lambda 157.7\mu$m lines, which are produced mainly in neutral
photon-dominated regions (PDRs) (Tielens \& Hollenbach \cite{TH}; Sternberg
\& Dalgarno \cite{SD}) and in X-ray dissociation regions (Maloney,
Hollenbach \& Tielens \cite{MHT}). Second, we want to consider only lines
with reliable flux measurements. This leads to the exclusion of all optical
lines taken with spectral resolution worse than 20\AA\/, those with fluxes
lower than 1/4 of the narrow $\ifmmode {\rm H}\beta \else H$\beta$\fi$ flux, or permitted lines whose narrow
component was not decomposed from the broad one. The HUT narrow UV lines
(Kriss et al. \cite{Kriss}) are somewhat problematic in this respect, since
their FWHM is 1.5 to 3 times larger than that of the NLR forbidden lines
(c.f. Paper I). Although this may indicate a problem in their decomposition
we choose to retain these measurements because of the importance of UV data
for constraining the reddening. We exclude the two weakest ISO-SWS lines,
$\ifmmode {\rm [Si}\,{\sc ix]}\,\lambda 2.6 \mu{\rm m$ and $\ifmmode {\rm [Mg}\,{\sc iv]}\,\lambda 4.5 \mu{\rm m$, which have a low S/N. Third, we require that the
line measurements not be too inconsistent with the large SWS and LWS
apertures. This criterion leads to the exclusion of all optical lines taken
with apertures whose smaller dimension is less than $3^{\prime\prime}$, which
unfortunately include the important He\,{\sc ii}\,$\lambda4686$ line. The
$\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi < 13.6$\,eV criterion also helps in this respect by decreasing the
contamination of EELR line emission. Fourth, we need to take into account
the limitations of the photoionization models. This leads to the exclusion
of the $\bFeXb$\AA\/ and $\bFeXIb$\AA\/ lines, whose collision strengths
values are highly uncertain. The final, much shortened line list is given
in Table~\ref{t:flux}. Whenever more than one measurement of the line
exists, we quote the average flux and use the rms scatter as an error
estimate. We list also some of the lines and upper limits that are not used
in the model fits, so that their consistency with the model results can be
checked.
\section{The photoionization models}
\label{s:models}
\subsection{The SED / cloud models}
\label{s:SEDcloud}
Two basic ingredients have to be specified for calculating the
photoionization models: the gas, or `cloud' properties and the ionizing
SED. The main objective of this work is to reconstruct the ionizing SED of
NGC\,4151 as model-independently as possible. However, cloud models must be
adopted in the analysis, and these introduce some uncertainties in the
reconstructed SED. The cloud models we consider here (Figure~\ref{f:geom})
are probably too simple to fully reproduce the rich observed nuclear
structure, but they allow us to assess the robustness of the reconstructed
SED to the uncertainties of modeling the gas.
In the cloud models that are described below, we distinguish between
compact and clumpy gas models. In a compact gas model the entire cloud
volume is filled with gas, that is, its filling factor $F$ is unity. In a
clumpy gas model, $F<1$, and the cloud can be described as composed of many
`cloudlets', or alternatively as having a porous, sponge-like
structure. Only the {\em total} column density of the cloud is specified.
The photoionization calculations assume infinitesimal cloudlets (or holes)
that are randomly and uniformly distributed in the cloud volume. In this
limit, $F$ only enters the photoionization calculations by modifying the
volume emissivity and the optical depths (Ferland \cite{Ferland}). Although
the cloud as a whole may be optically thick, the cloudlets can be thought
of as a system of optically thin clouds that filter, but do not entirely
absorb, the ionizing radiation, and then transmit it to cloudlets further
to the back of the cloud. This picture is physically relevant as long as
the illuminated cross-section of the cloud is not too small relative to its
depth, since it is unlikely that a thin filament of cloudlets can remain
perfectly aligned so as to have the specified total column density. A
clumpy gas distribution `stretches' a given column density over a larger
physical distance relative to that of a compact cloud. Therefore, the
geometrical dilution of the radiation is larger, and hence the ionization
structure of the gas and its emission line spectrum are modified.
We consider four types of clouds.
\begin{enumerate}
\item Constant density, optically thick clumpy clouds (i.e. $F<1$),
distributed on a spherical shell of radius $r$ centered on the
continuum source.
\item Constant density, optically thick compact clouds (i.e. $F=1$),
similarly distributed.
\item Optically thick compact clouds, similarly distributed, with a density
gradient that increases linearly into the cloud (i.e. away from the
illuminated surface),
\begin{equation}
n(d) = n_0(1+d/r_0)\,,
\end{equation}
where $n_0$ is the gas density at the irradiated surface, $d$ is the
distance into the cloud and $r_0^{-1}$ the gradient. Such models may
approximate clouds with evaporating irradiated surfaces (Binette,
Wilson \& Storchi-Bergmann \cite{BWS}) or a density stratification due
to radiation pressure (Binette et al. \cite{BWRS}; Binette
\cite{Binette}).
\item Two components of constant density, optically thick compact clouds,
distributed on two spherical shells of radii $r_1 < r_2$, where it is
assumed that the SED seen by component 2 is the same as that seen by
component 1, with no obscuration but with geometrical dilution.
\end{enumerate}
Realistically, the clouds are likely to have a distribution of
densities and positions relative to the central continuum
source. However, these simple models may be justified by noting that
the large aperture of the SWS (and to a lesser degree the smaller
apertures of the optical and NIR spectrometers) integrates over this
population. The validity of these models then rests on the assumption
that the population average can be approximated by a single,
representative cloud type.
We model here only ionization bounded (IB) clouds (i.e. optically thick
clouds). Matter bounded (MB) cloud models (i.e. optically thin clouds) have
been proposed in response to discrepancies between the observations and the
predictions of NLR / EELR single component models, which assumed a simple
power law SED (Viegas \& Prieto
\cite{VP}; Binette et al. \cite{BWS}). Those models typically
under-predict the high ionization lines, under-predict the electron
temperature and under-predict the scatter in the ${\rm He}\,{\sc
ii}\,\lambda4686/\ifmmode {\rm H}\beta \else H$\beta$\fi$ ratio among Seyfert 2 galaxies. An additional
MB component comes at the price of three additional free parameters:
the column density of the matter bounded clouds, their fraction
relative to the IB clouds and the fraction of IB clouds that see the
filtered radiation from the back of the MB clouds relative to those
that see the unfiltered radiation. In view of the considerable
complexity introduced by these additional free parameters, and
because, as will be shown below, the high ionization lines are well
modeled by the IB clouds while the gas temperature may not, in fact,
be problematic in NGC\,4151 (Sect.~\ref{s:n4151-gas}), we do not
consider these types of models here.
In addition to the cloud geometry and density structure, it is necessary
to specify the ionization parameter, which for isotropic emission is
defined as
\begin{equation}
U = \frac{Q_{\rm ion}}{4\pi r^2 n_0 c}\,,
\end{equation}
where $Q_{\rm ion}$ is the ionizing photon emission rate (s$^{-1}$)
and $c$ is the speed of light. It is also necessary to specify the
cloud column density $N_c$, which we assume to be effectively
infinite, the element abundances, which we assume to be solar, and the
amount of dust, which we assume to be negligible.
The ionizing SED model we use here is based on the observed
multi-wavelength SED of the 1993 campaign (Edelson et al. \cite{Edelson}).
In using it, we are making the assumption that this SED can be used to
approximate the time averaged SED over time scales comparable to the light
crossing time of the NLR, which is of the order of 1000\,yr.
The enumeration on the SED assumes a four-segment broken power-law
(Figure \ref{f:sed_temp}). The observed SED is adopted longward of 1 Ryd
and shortward of 50 keV. In between these two limits, the enumeration
proceeds by choosing all possible combinations of the three luminosity
densities (erg\,s$^{-1}$\,Hz$^{-1}$) at 4, 8, and 30 Ryd ($L_4$, $L_8$
and $L_{30}$, respectively) and connecting them by straight lines in
the $\log E$--$\log L_\nu$ plane. This yields a total of
$13\times13\times3 = 507$ possible combinations. The break energies of
1 and 4 Ryd were chosen because they correspond to the H and He\,{\sc
ii} ionization edges. The break energy of 30 Ryd roughly brackets the
highest ionization potential of the observed lines. The break energy
of 8 Ryd is arbitrary and is intended to allow additional flexibility
in the SED shape. The quantity $\log L_{30}$ varies between 25.4 and
26.0 in increments of 0.3. The lower limit for the $\log L_{30}$ range
is obtained by extrapolating down to 30 Ryd a power-law using
observations at 50 to 80 keV (longward of the power-law break at
$\sim90$\,keV) and at $\sim 4.1$\,keV (shortward of the absorption
feature but still longward of the X-rays Fe emission lines). The upper
limit on $L_{30}$ allows for possible errors in the determination of
the continuum at $\sim 5$\,keV. The values of $\log L_4$ and $\log L_8$
vary between 25.4 and 27.8 in increments of 0.2, thereby reaching
values as high as that at 1 Ryd and as low as that at 30 Ryd. This
allows the enumeration on the SED to explore the possibility of a very
prominent bump, as well as that of a steeply falling continuum in the
1-30 Ryd range.
The photoionization calculations were carried out using the numerical
photoionization code ION97, the 1997 version of the code ION described in
Netzer (\cite{Netzer96}).
\subsection{The fit procedure}
\label{s:fit}
The very large parameter space we investigate in this work makes it
necessary to adopt a goodness-of-fit score to rank the models. There are,
however, some difficulties in applying standard methods to the problem at
hand. Both the measurement error estimates and those of the model are
highly uncertain, and are probably non-Gaussian and dominated by
systematic errors. Therefore, the commonly used $\chi^2$ score, which
relies on the accuracy of the errors estimates, can be highly
misleading. Furthermore, the model lines and the observed lines are not
directly comparable, since the model calculations yield the line luminosity
per unit area on the cloud's cross-section, whereas the observations are of
the line flux on earth. A standard procedure for comparing the two is to
normalize all the line fluxes relative to the flux of some strong reference
line, such as $\ifmmode {\rm H}\beta \else H$\beta$\fi$, which is roughly proportional to the total ionizing
luminosity. However, in the case of NGC\,4151, this is problematic because
the narrow $\ifmmode {\rm H}\beta \else H$\beta$\fi$ flux measurement is very uncertain due to the difficulty
of decomposing the narrow and broad components. The observed forbidden
optical and IR lines are also not suited for this purpose because of the
large measurement errors. An error in the flux of the chosen reference line will
bias all the line ratios, and because of the highly non-linear behavior of
the photoionization calculations, this is likely to have complicated
effects on the best-fit SED. Moreover, the freedom of choosing the reference
line adds ambiguity to the results, since there is no compelling reason for
preferring one reference line over another.
Here we formulate and adopt a different fit procedure that does not
rely on the accuracy of the measurement error estimates and treats all
the lines on an equal footing. We assume that for the correct model
and for ideal, error-free observations, the observed and model line
fluxes are related by
\begin{equation}
\exp(\ifmmode k_\ell \else $k_\ell$ \fi\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi)\ifmmode f_\ell \else $f_\ell$ \fi = \alpha \ifmmode m_\ell \else $m_\ell$ \fi\,,
\end{equation}
where $\ifmmode f_\ell \else $f_\ell$ \fi$ is the observed flux in line $\ell$, $\ifmmode m_\ell \else $m_\ell$ \fi$ is the model
luminosity per unit surface area of the irradiated face of the cloud,
and the exponent is the dereddening factor with $\ifmmode k_\ell \else $k_\ell$ \fi$ given by the
extinction curve at the line's rest wavelength. The parameter $\alpha$
is a proportionality factor common to all the lines. It has a simple
geometrical interpretation in the framework of the cloud models we
consider here. These models assume an isotropic line emission from
clouds distributed on the surface of a sphere of radius $r$, which is
centered on the continuum source. In this case
\begin{equation}
\exp(\ifmmode k_\ell \else $k_\ell$ \fi\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi)\ifmmode f_\ell \else $f_\ell$ \fi = C 4\pi r^2 \ifmmode m_\ell \else $m_\ell$ \fi / 4\pi D^2\,,
\end{equation}
where the covering factor, $C$, is the fraction of the spherical surface
covered by the clouds and $D$ is the distance of the AGN from the
observer. Therefore,
\begin{equation}
\alpha = C \left(\frac{r}{D}\right)^2\,.
\end{equation}
The distance $r$ and the corresponding angular distance $\theta$ are fixed
by the ionization parameter and the hydrogen gas density,
\begin{equation}
\label{e:r2}
r^2 = \frac{D^2\ifmmode q_{\rm ion} \else $q_{\rm ion}$ \fi A}{Un_0c}\,,
\end{equation}
\begin{equation}
\label{e:theta}
\theta^2 = \frac{\ifmmode q_{\rm ion} \else $q_{\rm ion}$ \fi A}{Un_0c}\,,
\end{equation}
where $A$ is the anisotropy parameter of the ionizing luminosity (Sect.
\ref{s:n4151-nucleus}), and $\ifmmode q_{\rm ion} \else $q_{\rm ion}$ \fi$ is the ionizing photon flux that would
be observed if all the ionizing photons reached the observer. The parameter
$\ifmmode q_{\rm ion} \else $q_{\rm ion}$ \fi$ is a function of $L_4$, $L_8$ and $L_{30}$, which parameterize the
SED. Equation \ref{e:theta} can be used to estimate $A$ if the other
parameters are known independently. Here, this information is unavailable,
and $A$ remains a free parameter. The covering factor is related to
$\alpha$ by
\begin{equation}
\label{e:cf}
C = \frac{\alpha}{\theta^2}\,.
\end{equation}
The depth of the ionized fraction of the gas distribution, $d_{\rm ion}$,
which is calculated by the photoionization code, can be used to estimate
the angular extent of the NLR, $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$,
\begin{equation}
\label{e:tion}
\theta_{\rm ion} = d_{\rm ion}/D = d_{\rm ion}h/cz\,.
\end{equation}
$\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$ corresponds to the angular extent of the Balmer lines
emission. Emission from higher ionization lines is more concentrated
towards the inner part of the NLR.
We note that there exists a scaling relation between $A$ and $F=F_1h$
(observations fix $F$ only up to a factor of $h$) that allows models, which
differ only in these parameters, to have identical best-fit line fluxes and
SEDs. This can happen when both $U$ and the geometrical dilution of the
ionizing flux,
\begin{equation}
g(d) = (1+d/r)^{-2}\,,
\label{e:geom_dilute}
\end{equation}
are unaffected by the changed parameters. Since $d$ scales as $1/F$,
$g$ remains constant if $r$ is also scaled as $1/F$, which in turn
implies that $U$ should scale as $A F_1^2$. Therefore, if $A$ scales
as $1/F_1^2$, the line ratios remain constant, while $\theta$ and
$\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$ scale as $A^{1/2}$ and $C$ as $1/A$.
The search for the best combination of cloud model and SED proceeds by
enumerating over various possible cloud models, whose input parameters are
$U$, $n_0(d)$, $F$, $A$, the number of gas components and their elemental
abundances, and the `boundary conditions' of the SED template, $L_1$ and
$L_{4134}$. In addition, $h$ is required for calculating $r$ and $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$
and also for estimating the input value of $F$ from the observations. For
each cloud model, we then enumerate over $L_4$, $L_8$ and $L_{30}$, and use
a `goodness-of-fit' score, which is described below, to find the values of
$L_4$, $L_8$, $L_{30}$, $\alpha$ and $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$ that best fit the observed
line fluxes. The observational constraints on $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$ and the derived
quantities $C$, $\theta$ and $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$ can then provide additional checks on
the best fit values of these parameters.
Lacking reliable measurement and model error estimates, a natural measure
of the match or mismatch between model and observations is simply the line
ratios, $\ifmmode m_\ell \else $m_\ell$ \fi/\ifmmode f_\ell \else $f_\ell$ \fi$. For this reason, photoionization models are commonly
judged by their ability to fit the data to within a factor of $S$, on
average, where $S<2$ is considered to be a reasonable fit. We formalize
this intuitive concept of a logarithmic scale by defining a score function
\begin{equation}
\ifmmode \log^2S \else $\log^2S$ \fi = \frac{1}{n}\sum_{\ell=1}^n\left(\log\frac{\alpha \ifmmode m_\ell \else $m_\ell$ \fi}
{\exp(\ifmmode k_\ell \else $k_\ell$ \fi\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi)\ifmmode f_\ell \else $f_\ell$ \fi}\right)^2\,,
\end{equation}
where $n$ is the number of lines used in the fit. The $\ifmmode \log^2S \else $\log^2S$ \fi$ score is an
explicit function of $\alpha$ and $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$ and an implicit function of $L_4$,
$L_8$, $L_{30}$ and the cloud model through the model line fluxes $\ifmmode m_\ell \else $m_\ell$ \fi$. An
average fit factor $S$, in the rms sense, is defined as
\begin{equation}
S = \exp\left(\sqrt{\ifmmode \log^2S \else $\log^2S$ \fi}\right)\,,
\end{equation}
with the worst fitting line off by a factor of
\begin{equation}
\max S_\ell = \max_\ell \exp\left(\left|\log \frac{\alpha^0 m_\ell^0}{\exp(\ifmmode k_\ell \else $k_\ell$ \fi
E^0_{\sc b-v})\ifmmode f_\ell \else $f_\ell$ \fi}\right|\right)\,.
\end{equation}
It is straightforward to show that for a given set of model fluxes,
$\{\ifmmode m_\ell \else $m_\ell$ \fi\}$, $\ifmmode \log^2S \else $\log^2S$ \fi$ is minimized when $\alpha$ and \ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi take the values
\begin{equation}
\log\alpha^0 =
\frac{\langle k\rangle\langle k \log m/f\rangle-
\langle k^2\rangle\langle\log m/f\rangle}
{\langle(k-\langle k\rangle)^2\rangle}\,,
\end{equation}
and
\begin{equation}
E_{\sc b-v}^0 =
\frac{\langle(k-\langle k\rangle)(\log m/f-\langle\log m/f\rangle)\rangle}
{\langle(k-\langle k\rangle)^2\rangle}\,,
\end{equation}
where the notation $\langle\ldots\rangle$ designates the average of the
bracketed quantities over the $n$ lines, and it is assumed that not all the
\ifmmode k_\ell \else $k_\ell$ \fi are equal. $E_{\sc b-v}^0$ can formally take negative, non-physical
values. In such cases we set it to zero, and re-minimize \ifmmode \log^2S \else $\log^2S$ \fi as a function
of $\alpha$ only, in which case
\begin{equation}
\log\alpha^0 = -\langle\log m/f\rangle\,.
\end{equation}
The search for the minimum of \ifmmode \log^2S \else $\log^2S$ \fi in $L_4$, $L_8$, $L_{30}$, \ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi and
$\alpha$ proceeds by first calculating numerically $\{\ifmmode m_\ell \else $m_\ell$ \fi\}$ for each point
in ($L_4$, $L_8$, $L_{30}$) space and then analytically evaluating \ifmmode \log^2S \else $\log^2S$ \fi at
each point using $\alpha^0$ and $E^0_{\sc b-v}$.
\ifmmode \log^2S \else $\log^2S$ \fi is easily generalized to the case of two cloud components,
\begin{equation}
\ifmmode \log^2S \else $\log^2S$ \fi = \frac{1}{n}\sum_{\ell}\left(\log\frac{
\alpha (w_1 m_{\ell1}+w_2 m_{\ell2})}
{\exp(\ifmmode k_\ell \else $k_\ell$ \fi\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi)\ifmmode f_\ell \else $f_\ell$ \fi}\right)^2\,,
\end{equation}
where $w_1+w_2 = 1$ are the mixture weights of the two components,
which emit a line luminosity $m_{\ell1}$, $m_{\ell2}$ per unit surface
area, respectively. The covering factor of each component is given by
Eq.~(\ref{e:cf}) with $\alpha w_1$ or $\alpha w_2$ in place of
$\alpha$. The \ifmmode \log^2S \else $\log^2S$ \fi score is minimized by numerically enumerating on
possible values of $w_1$, calculating $\ifmmode m_\ell \else $m_\ell$ \fi = w_1 m_{\ell1}+w_2
m_{\ell2}$ and then proceeding as in the case of a single cloud
component.
The errors on the best-fit parameters of a given cloud model, which are
introduced by the measurement errors in the line fluxes, are estimated by
Monte Carlo simulations (see e.g. Press et al. \cite{PTVF}). A set of
simulated observations $\{f^\prime_\ell\}$ are drawn using the best-fit
parameters and the given measurement error estimates (or a conservative
guess of a 50\% error if such an estimate is unavailable),
\begin{equation}
f^\prime_\ell = \alpha^0 m^0_\ell
\exp(-k_\ell E^0_{\sc b-v})(1+\epsilon)\,,
\end{equation}
where $\epsilon$ is a Gaussian deviate of zero mean and standard deviation
equal to the quoted fractional measurement error of the line. The
minimization procedure is then repeated with the given cloud model for each
simulated set, and the best-fit parameters are recorded. The 99.9\%
confidence limits, which are quoted below, lie between the minimal and
maximal values that the parameters take on the contour of constant $S$ that
encloses 99.9\% of the simulated results. The errors are given for each
parameter separately and are {\em not} statistically independent. Note that
this is the only aspect of the modeling where the measurement error
estimates affect the results. Note also that the confidence limits on the
SED are confined to the range covered by the SED enumeration (see
Figure~\ref{f:sed_temp}). The quoted confidence interval may in fact be
smaller than the true one in places where it extends right up to the edge
of the enumerated region. It should also be emphasized that the confidence
limits are conditional, in the sense that they are calculated under the
assumption of a given gas model and SED template. We present below errors
only on the parameters $L_4$, $L_8$, $L_{30}$, $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$, $\theta$, $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$
and $C$. The Monte Carlo procedure can be easily generalized so as to
obtain confidence limits on all the free parameters, including those of the
gas model, $U$, $F$, $n_0$ and $r_0$. However, because we test only a small
number of values for these parameters, this is not attempted here.
The weighting strategy is a major concern in any fit procedure, including
this one. The $\ifmmode \log^2S \else $\log^2S$ \fi$ score assigns equal weights to the different lines. In
doing so, we are ignoring the fact that some lines carry more physical
information than others, and that the unequal spacing of the lines in
$\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$ over-emphasizes some energy ranges in the SED at the expense of
others. Nevertheless, because it is unclear how to deal in a satisfactory
and general way with these problems, and because of the simplicity and
elegance of the $\ifmmode \log^2S \else $\log^2S$ \fi$ fit, we choose to use the simplest option of equal
weights. As will be shown in Sect.~\ref{s:discuss}, we test the sensitivity
of the best-fit SED to the weighting by dropping a subset of the lines from
the fit (i.e. setting their weight to zero) and re-fitting. We find that
the best-fit SED is insensitive to these changes.
A combined SED and cloud model is considered successful if it fulfills the
following criteria.
\begin{enumerate}
\item
The model fits the lines to within a factor of 2, on average ($S < 2$).
\item
The worst fitting line is off by no more than a factor of 3 ($\max
S_\ell< 3$).
\item
The covering factor is less than 0.25, which corresponds to a
bi-cone with an opening angle of $\sim 80^\circ$ ($C \la 0.25$).
\item
The angular distance of the illuminated face is consistent with the
inner radius of the NLR ($\theta \la 0.5^{\prime\prime}$).
\item
The angular extent of the line emitting gas is consistent with the size
of the NLR ($\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi \sim 5^{\prime\prime}$).
\item
The extinction is small ($\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi < 0.15$).
\end{enumerate}
We also attempt to detect systematic trends in the residuals, or remaining
discrepancies, of the best-fit model. This is done by calculating the
correlation between the best-fit ratios, $\alpha^0 m_\ell^0/\exp(k_\ell
E_{\sc b-v}^0)f_\ell$, and the line properties $\lambda_0$, $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$ and the
element's tendency for depletion. These test, respectively, for problems
with the assumed extinction law, problems with the assumed SED
parameterization, and problems with the elemental abundances (assumed
solar), which may be depleted by dust. In order to minimize the sensitivity
of these tests to numeric uncertainties and to assumptions on the nature of
the correlations (e.g. linear in the logarithm of the quantity vs. linear
in the quantity itself), we use the non-parametric Kendall's $\tau$
correlation coefficient (see e.g. Press et al. \cite{PTVF}). Using this
estimator, the possibility of depletion is simply tested by correlating the
best-fit ratios (after summing line fluxes from the same element) with the
depleted abundances, which we take to be those of the Galactic interstellar
medium given by Spitzer (\cite{Spitzer}). A negative correlation is
expected if the true abundances are depleted. The residuals of the correct
model should not display any such correlations.
\section{Results}
\label{s:results}
The cloud models that we investigate in this work (Table~\ref{t:score})
include clumpy and compact single component models with constant density,
compact single component models with a density gradient, and two component
models that result from mixing pairs of compact, single component constant
density models. The ionization parameters lie in the range $U=0.005$ to
$0.4$ and the filling factors in the range $10^{-3}h$ to $1$. We assume
density profiles of the form $n (r) = n_0(1+d/r_0)$, where
$n_0=1000$\,cm$^{-3}$ and $r_0 = 3\,10^{17}$\,cm or $r_0 = \infty$, the
latter corresponding to the constant density case. We assume throughout
$h=0.65$, isotropic source emission ($A=1$), solar abundances, no dust, and
use a column density that is large enough to absorb all the ionizing
radiation. For each of these cloud models, we enumerate on the SED template
(Figure~\ref{f:sed_temp}), calculate the resulting line fluxes with the
photoionization code and perform the fit procedure. The extinction curve
used for dereddening is based on Seaton (\cite{Seaton}) for the UV and
optical and Lutz et al. (\cite{Lutz2}) for the IR. Because of the
uncertainty involved in inter-calibrating the optical and ISO flux
measurements, we repeat the fit procedure twice, once using only the ISO
line list (`ISO fit') and once with the full ISO and optical line list
(`full fit'). This also allows us to assess the sensitivity of the equal
weighting scheme that underlies the $\ifmmode \log^2S \else $\log^2S$ \fi$ score.
The range of ionization parameters was chosen after spot-checks indicated
that lower $U$ models severely under-produce the high ionization lines
emission, while higher $U$ models over-produce them. The filling factor
$F=0.065$ corresponds to that used to model the Circinus SED by Moorwood
et al. \cite{Moorwood}, and that of $6.5\,10^{-4}$ is the value inferred
for the low excitation clouds of the NLR in NGC\,4151 (Yoshida \& Ohtani
\cite{YO}); Robinson et al. \cite{Robinson}). We also made some
spot-checks with models of higher and lower densities ($n_0 = 300$ and
$3000$\,cm$^{-3}$) and larger and smaller density gradients ($r_0 = 10^{17}$
and $10^{18}$\,cm). These modified models gave qualitatively similar
results to the models included in the grid, and will therefore not be
further discussed.
\subsection{Single component models}
The best-fit scores of the single component models are listed in
Table~\ref{t:score}. As expected, the ISO fits, which are less
constrained, have better $S$ scores than the full fits. Of the six fit
criteria ($S$, $\ifmmode \max S_\ell \else $\max S_\ell$ \fi$, $C$, $\theta$, $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$ and $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$), $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$ is
the least restrictive (Some high $U$ models have ISO fits with very
large $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$ values, but these have very large errors). The covering
factor $C$ points to models with low ionization parameter or small
filling factor. The angular distance $\theta$ is uncomfortably large
for the lowest $U$ models, $S$ and $\ifmmode \max S_\ell \else $\max S_\ell$ \fi$ disfavor models with the
lowest ionization parameters and constant density models with high
filling factors. The most restrictive criterion turns out to be
$\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$. Only models with the lowest filling factor have the large
observed NLR size.
Taking into account all the criteria, the best model, for both the
ISO and full fits, is that with $U=0.025$, $F=6.5\,10^{-4}$ and a
constant density. This particular choice of $U$ is somewhat arbitrary,
as the higher $U$ models in the $F=6.5\,10^{-4}$ sequence fit almost
equally well (see Sect.~\ref{s:discuss}). The Balmer lines NLR emission
in this model extends between $\sim 0.5^{\prime\prime}$ to $\sim 5^{\prime\prime}$.
The $\bOIIIbB$ emission is more centrally concentrated, and extends
only up to $\sim 3^{\prime\prime}$. This is consistent with the images of
Hutchings et al. (\cite{Hutchings}). The covering factor of this
model, $C=0.24$, implies that the gas fills almost the entire NLR
bi-cone. With such a large cloud cross-section, the clumpy gas
description is physically reasonable (Sect.~\ref{s:SEDcloud}).
The properties of the best-fit model are summarized in Table~\ref{t:fit},
the best fitting SEDs are shown in Figure~\ref{f:best_SED} and the best-fit
line ratios are shown in Figures \ref{f:ratio_ISO} and
\ref{f:ratio_full}. Apart for a somewhat large $\ifmmode \max S_\ell \else $\max S_\ell$ \fi$ in the full fit,
both fits are successful in the sense defined in
Sect.~\ref{s:fit}. The results of the two fit procedures are very
similar. In particular, in both the SED falls steeply beyond 1 Ryd
towards 4 Ryd and then rises again at around 8 Ryd. As can be
expected, the ISO fit does not constrain $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$ very well and also
does not reproduce the optical lines as well as the full fit. Although
the confidence limit on $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi$ are improved in the full fit, they are
still quite large. This probably reflects the fact that the fit is
still dominated by the IR lines, and that the optical reddening
indicators give conflicting estimates (Sect.~\ref{s:n4151-gas}). The
best fit $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi = 0.03$ is consistent with the lower range of these
reddening estimates. In both fits, even the lines that were excluded
from the $\ifmmode \log^2S \else $\log^2S$ \fi$ score are reproduced reasonably well, including the
He\,{\sc ii}\,$\lambda4686$ line, for which wide aperture measurements
are unavailable.
\subsection{Two component models}
The fitting procedure of the two component models treats the mixture
weight, $w_1$, as an additional free parameter of the fit. We find that in
most cases, the best-fit two component model was one with the trivial
mixture weight of $w_1=0$ or $1$, meaning that the fit of the better model
of the two could not be improved by any amount of mixing with the other
model. In those cases where a non-trivial mixture did improve the fit, the
score was still worse than that of the best fitting single component model,
for both the full and ISO fits. We therefore conclude that there is no
compelling reason to prefer the mixed two-component constant density models
over the simpler single component models.
\subsection{Dependence on the UV template SED}
\label{s:UV}
The template SED has two points fixed by the observations, one at 1 Ryd and
one at 56.2\,keV (4134 Ryd). The hard photons at 56.2\,keV are well beyond
the ionization potentials of the lines and affect only the low ionization
lines by maintaining partially ionized regions at the back of the
clouds. Therefore, the model line fluxes should not depend strongly on the
exact SED value at 56.2\,keV. This is not necessarily the case for the
fixed point at 1 Ryd, as these photons interact very strongly with the gas.
There are two sources of uncertainty in the value adopted for the SED at 1
Ryd. One is the strong variability of NGC\,4151, which may result in a big
difference between the template SED and the effective, time-averaged SED
that is relevant for the NLR. The other is the accuracy of dereddening
correction that was used to derive the intrinsic SED from the observed one.
In order to check the sensitivity of the results to the shape of the UV
SED, we repeated the calculations for the clumpy, $F=6.5\,10^{-4}$, model,
this time varying also the flux at 1 Ryd. Instead of the fixed value of
$\log L_1 = 27.88$, we enumerated on the values $\log L_1=27.8$, 28.0 and
28.2. These correspond to the dereddened IUE flux at 1806\AA\/ with
$\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi=0.03$, $0.08$ and $0.11$, respectively, or to any combined effect of
reddening and variability of order $\Delta\log L_1 = 0.4$. For reasons of
computational economy, we enumerated only on every second point in $L_4$
and $L_8$. The best-fit results of these models and those of the fixed 1
Ryd luminosity models are almost identical. We therefore conclude that
uncertainties of the order of 1 mag in the determination of the luminosity
at the Lyman limit do not have a significant effect on the best-fit
results.
\section{Discussion}
\label{s:discuss}
\subsection{Properties of the best fit model}
The best-fit model succeeds in reproducing the observed line fluxes to
within a factor of 2, on average. It is also consistent with low
extinction and with the geometrical constraints on $\theta$, $C$ and
$\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$. We find no indication of a Big Blue Bump rising beyond the
Lyman limit (c.f. Figure~\ref{f:sed+bump}). On the contrary, the SED
declines steeply from the Lyman limit towards 4 Ryd and then rises
sharply and peaks at around 8 Ryd. The exact shape of the SED varies
among the various successful models, but generally, we get very
similar results from the different cloud models we considered here.
The best-fit model is a single component, clumpy cloud with constant
density. We find no compelling evidence pointing towards multi-component
clouds, non-solar elemental abundances or significant continuum anisotropy.
The $A/F$ scaling relation (Sect.~\ref{s:fit}) cannot be applied in this
case without increasing $\theta$ and $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$ beyond their observed values
of $\la 0.5^{\prime\prime}$ and $\sim 5^{\prime\prime}$, respectively. Note, however, that
$A$ is not well constrained because of the uncertainty in $\ifmmode q_{\rm ion} \else $q_{\rm ion}$ \fi$ (cf
Eq.~\ref{e:theta}) that follows from the uncertainty in $L_1$
(section~\ref{s:UV}). A clumpy gas distribution enhances the geometrical
dilution of the ionizing radiation relative to a compact distribution by
`stretching' a given gas column density over a larger physical distance. It
appears that geometrical dilution plays an important role in establishing
the ionization structure that is required for reproducing the observed line
ratios. This can be seen by noting that models with a density gradient
(e.g. $U=0.01$, $\log r_0 = 17.5$) also reproduce the line ratios very
well, and result in the same best-fit SED. In such models, geometrical
dilution is qualitatively mimicked by the increase in gas density with
depth into the cloud. However, such models fail to reproduce the large
extent of the NLR since only the thin illuminated skin of the cloud emits
lines. It is possible that an extended system of such clouds with suitably
adjusted ionization parameters may be consistent with the observations as
well, and may be natural in the context of radiatively accelerated clouds
(Binette
\cite{Binette}). However, the dynamics of the NLR are outside the
scope of this work, and in view of the success of simpler models, this
possibility was not investigated here.
The insensitivity of the low-$F$ results to the value of $U$
(Sect.~\ref{s:results}) is also related to the geometrical dilution. The
volume averaged ionization parameter $\bar{U}$ (taking into account
only geometrical dilution but ignoring true absorption) is
\begin{eqnarray}
\bar{U} & = & U\left.\int_{r_1}^{r_2}r^2 g(r)\,{\rm d}r \right/
\int_{r_1}^{r_2}r^2\,{\rm d}r \nonumber\\
& = & 3U\left/\left[1+r_2/r_1+(r_2/r_1)^2\right]\right. \nonumber\\
& \stackrel{\textstyle \longrightarrow}{_{r_2\gg r_1}} & 3g(r_2)U\,,
\end{eqnarray}
where $r_1$ and $r_2$ are the inner and outer radius of the NLR,
respectively, and $g(r) = (r_1/r)^2$ is the geometrical dilution (c.f.
Eq.~\ref{e:geom_dilute}). When the geometrical dilution is significant,
$\bar{U}$ can be much smaller than $U$, which is defined at the
irradiated face of the cloud, since the volume average is dominated by
the outer shells. The inner and outer radii of the NLR are free
parameters in our fit procedure. Inspection of the $F=6.5\,10^{-4}$,
$U\ge 0.01$ full fit results in Table~\ref{t:score} shows that the
best fit value of $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi/\theta$ varies with $U$ in such a way that
$g(r_2) \propto U^{-1}$ approximately, and therefore $\bar{U}
\sim 10^{-3} \ll U$ irrespective of $U$. In addition, the best fit
SEDs of all these models are almost identical to that of the $U=0.025$
model (Figure~\ref{f:best_SED}). Thus, while the emission line data
cannot precisely constrain $U$, $\theta$ or $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi$ individually, they
clearly favor models with a low $\bar{U}$, a low $F$ and a very hard
SED.
The filling factor of the best-fit model, $F=6.5\,10^{-4}$, was deduced
from observations of the low excitation NLR lines. The high ionization
lines of the best-fit model are produced closer to the central source,
while the lower ionization lines are produced deeper in the cloud in less
ionized regions. The fact that this model reproduces both the high and low
excitation lines supports a simple picture that the clumps emitting the
coronal lines and those emitting the optical narrow lines form one
continuous and homogeneous system.
The most significant result of this work is that we find no evidence for a
Big Blue Bump just beyond the Lyman limit of NGC\,4151. Such a bump is too
soft to be consistent with the flux ratios of the high and low ionization
lines. This is demonstrated in Figure~\ref{f:gray}, which shows the model
to data ratio ($\alpha\ifmmode m_\ell \else $m_\ell$ \fi/\exp(\ifmmode k_\ell \else $k_\ell$ \fi\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi)\ifmmode f_\ell \else $f_\ell$ \fi$) of the $\ifmmode {\rm [Ar}\,{\sc iii]}\,\lambda 9.0 \mu{\rm m$ line
($\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi=27.6$\,eV) and $\ifmmode {\rm [Si}\,{\sc ix]}\,\lambda 3.9 \mu{\rm m$ line ($\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi=303.2$\,eV) as function of
$L_4$ and $L_8$ for the best-fit model. Both lines are very well modeled
without a Big Blue Bump (Figure~\ref{f:ratio_full}). The right hand side of
the gray scale plots corresponds to high luminosity at 4 Ryd, and the lower
right hand side corner to the case of an extreme Big Blue Bump, which peaks
at around 4 Ryd ($\sim 50$\,eV) and then falls off steeply. Such a bump
clearly over-produces the low ionization line and under-produces the high
ionization line.
A single power-law is also not favored by the observations. As an example,
we calculated the fit for a model similar to the best fit model, but
with $\log L_4= 26.6$, $\log L_8 = 26.4$ and $\log L_{30} = 25.7$, which is
the SED that most closely resembles a single power-law and is still
contained within the 99.9\% confidence interval shown in
Figure~\ref{f:best_SED}. This SED obtains a fit score of $S = 2.8$,
under-predicts the $\ifmmode {\rm [Mg}\,{\sc viii]}\,\lambda 3.0 \mu{\rm m$ line by a factor of 8.5, and under-predicts
the other high $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$ lines as well. The fit of this model to the observed
NLR geometry is also clearly worse than that of the best fit model. The
softer SED results in a smaller ionization depth, $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi = 3.6^{\prime\prime}$, a
higher covering factor, $C = 0.35$, and a higher photon flux $\ifmmode q_{\rm ion} \else $q_{\rm ion}$ \fi$, so
that $\theta = 0.83^{\prime\prime}$ (Eq.~\ref{e:theta}).
\subsection{Comparison with the Circinus SED}
It is instructive to discuss the SED of NGC\,4151 by comparing it to
the SED of the Seyfert 2 Circinus galaxy. Unlike NGC\,4151, the
reconstructed SED of the Circinus galaxy does appear to have a Big
Blue Bump (Moorwood et al. \cite{Moorwood}). As a check on the
Circinus results and a test of the $\ifmmode \log^2S \else $\log^2S$ \fi$ method, we applied the
fitting procedure described here to the observed optical and IR line
fluxes of Circinus (Oliva et al. \cite{Oliva}; Moorwood et
al. \cite{Moorwood}). We assume, as in Moorwood et
al. (\cite{Moorwood}), a single component constant density clumpy gas
distribution of solar abundances with $n_0 = 5000$\,cm$^{-3}$, an
ionization parameter $U=0.45$, a filling factor $F=0.1h$, isotropic
emission ($A=1$) and $h=0.65$. Figure~\ref{f:best_cir} shows the
best-fit SED for this model. Only lines with $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi>30$\,eV were used
in the fit to avoid contamination from lines excited by hot stars,
which contribute significantly to the nuclear emission (Moorwood \&
Oliva \cite{MO}; Marconi et al. \cite{Marconi}; Moorwood et
al. \cite{Moorwood}). This model fits 17 observed lines with $S=1.7$,
$\ifmmode \max S_\ell \else $\max S_\ell$ \fi = 2.8^{-1}$, $\ifmmode E_{\sc b-v} \else $E_{\sc b-v}$ \fi=2.0$, $C=0.05$, $\theta=0.3^{\prime\prime}$,
$\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi=1.9^{\prime\prime}$ and $\log Q_{\rm ion}= 53.73$ (for $h=0.65$).
These values are consistent with the observed properties of the
Circinus galaxy (Oliva et al. \cite{Oliva}). A Big Blue Bump is seen
in our best $\ifmmode \log^2S \else $\log^2S$ \fi$ fit, in qualitative agreement with the Moorwood et
al. (\cite{Moorwood}) result. However, this result cannot be
established with a high confidence level due to the lack of
constraints on the SED below 30\,eV (note that Figure~\ref{f:best_cir}
shows only a 90\% confidence region). Although the confidence limits
on the best-fit Circinus SED can accommodate a NGC\,4151-like trough,
this is not required by the narrow emission line data. In the
following, we adopt the simplest working hypothesis, namely that the
Circinus SED has the UV bump that is seen in the best-fit SED. We note
that careful separation of the nuclear and stellar contributions is
crucial in studies of this type. When we also include in the fit
star-contaminated lines with $13.6<\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi<30$\,eV, we find that the
best-fit SED no longer exhibits a bump, but rather falls off sharply
from 1 Ryd to 4 Ryd and then remains flat.
The possibility that the SEDs of Seyfert 1 and Seyfert 2 galaxies are
intrinsically different may have far reaching consequences for unification
schemes of AGNs (Antonucci \cite{Antonucci}). This cannot be reconciled
with the idea that the observational differences between these two AGN
types are due to orientation effects alone. Bearing in mind the large
uncertainties in the reconstruction of the Circinus SED, we make a
tentative attempt to understand the differences between the NGC\,4151 and
Circinus results. We consider first the narrow emission line spectra. A
simple comparison of the NGC\,4151 to Circinus emission line ratio as
function of $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$ (Figure~\ref{f:cir2ngc}) reveals a strong decrease with
$\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$. This trend implies that the NLR of NGC\,4151 is not as highly
ionized as that of Circinus. However, our analysis shows that the SED of
NGC\,4151 is as hard or harder than that of Circinus. The mean ionizing
photon energy (between 1 to 30 Ryd) is 3.2\,Ryd (ISO fit) to 4.1\,Ryd (full
fit) for NGC\,4151 as compared to 3.2\,Ryd for Circinus. This is in the
opposite sense needed to explain the trend illustrated in
Figure~\ref{f:cir2ngc}. It is possible, in principle, that the paucity of
UV photons immediately beyond the Lyman edge in the harder NGC\,4151 SED is
a bottle-neck for the ionization sequence that leads to the high
ionizations levels. However, this is not supported by the trends apparent
in Figure~\ref{f:gray}, which show that modifying the SED by replacing hard
photons by soft photons only further decreases the ionization level. The
geometrical dilution in the NLR of both AGNs is quite similar
($\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi/\theta=6.6$ in Circinus as compared to $\ifmmode \theta_{\rm ion} \else $\theta_{\rm ion}$ \fi/\theta=7.2$ in
NGC\,4151), so that the difference in $\bar{U}$ simply reflect that in $U$,
with $\bar{U}=0.02$ in Circinus being 20 times larger than $\bar{U}=0.001$
in NGC\,4151.
We investigate the relative contribution of these factors to the NLR
ionization by calculating the ratios between the ISO line fluxes in the
best full fit NGC\,4151 model and those in a sequence of four NLR models:
(i) the best fit Circinus model; (ii) a model identical to the best full
fit NGC\,4151 model, apart for having the high Circinus $U$; (iii) a model
identical to the best full fit NGC\,4151 model, apart for having the
Circinus-like bumpy SED shown in Figure~\ref{f:abs_SED}; and (iv) a model
identical to the best full fit NGC\,4151 model, apart for having both the
Circinus $U$ and a bumpy SED. The ratios are plotted in
Figure~\ref{f:ngc_models} as function of $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$. The run of ratio (i) with
$\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$ is generally similar but not the same as that based on the observed
line fluxes (Figure~\ref{f:cir2ngc}) because the models don't exactly
reproduce the observations (in particular, the discrepancies in reproducing
the $\ifmmode {\rm [S}\,{\sc iv]}\,\lambda 10.5 \mu{\rm m$ and $\ifmmode {\rm [O}\,{\sc iv]}\,\lambda 25.9 \mu{\rm m$ lines mask the observed trend below
$\sim50$\,eV). We use these model ratios as the standard for comparison
with the other three models. We find that ratio (ii) falls off with $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$
even more sharply than (i). This shows that the large $\bar{U}$ difference
can more than account for the much higher ionization in Circinus. Ratio
(iii) rises with $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$. This demonstrates that all other factors being
equal, the harder SED of NGC\,4151 can ionize the NLR to levels higher than
in Circinus. Ratio (iv) does fall off with $\ifmmode E_{\rm ion} \else $E_{\rm ion}$ \fi$, but less than ratio
(ii). We therefore conclude that the $\bar{U}$ difference dominates the
relative line strengths in the spectra of the two AGN. Although the
relative hardness of the two SEDs affects the line ratios in the opposite
sense, the effect is smaller and it does not reverse the trend.
Next, we attempt to interpret the difference in the SEDs. We begin by
assuming that they are the intrinsic SEDs, i.e. that they directly
reflect the accretion process. There is evidence that the two AGNs
have different radiative efficiencies. Edelson et al. (\cite{Edelson})
estimate that the black hole mass in NGC\,4151 is $4-10\,10^7\,\ifmmode M_\odot \else $M_\odot$ \fi$,
and that it is emitting at a rate of $L/L_{\sc e}\la 0.01$, where
$L_{\sc e}$ is the Eddington luminosity. Maiolino et
al. (\cite{Maiolino}) place an upper bound of $4\,10^6\,\ifmmode M_\odot \else $M_\odot$ \fi$ on the
black hole mass in Circinus, which together with a non-stellar
luminosity estimate of $L\sim10^{10}\,\ifmmode L_\odot \else $L_\odot$ \fi$ (Moorwood et
al. \cite{Moorwood}), translates into $L/L_{\sc e}> 0.1$. There are
two known families of {\em steady-state} accretion solutions: hot thin
disks (Shakura \& Sunyaev \cite{SS}), and advection-dominated
accretion flows (ADAFs) (Narayan \& Yi \cite{NY}; Narayan et
al. \cite{NKH}). Naked thin disks (i.e. without additional components)
emit thermally and are therefore efficient compared to ADAFs, where
most of the energy is advected into the black hole by the ions, and
only a small fraction of it is emitted through synchrotron radiation
and inverse Compton scattering. The high efficiency of Circinus and
the bump in its SED are naturally explained by thin disk
accretion. However, the reconstructed SED of NGC\,4151 does not fit
our pre-conceived ideas about accretion flow spectra. Neither the high
luminosity of NGC\,4151 nor the sharp features in its SED can be
easily reconciled with ADAF spectra (c.f. Lasota et
al. \cite{Lasota}). Hybrid ADAF / thin disk models such as considered
by Lasota et al. (\cite{Lasota}) can do better in matching the
observed high luminosity, but would re-introduce a significant thermal
bump component. The mismatch between the spectra of these accretion
models and NGC\,4151 may be related to the AGN's extreme
variability. It is conceivable that such high variability can no
longer be described as a small perturbation on top of a steady-state
accretion flow, but is related to some other transient flow geometry.
\subsection{Internally absorbed ionizing continuum}
The alternative assumption is that the reconstructed SED of NGC\,4151 {\em
is not} the intrinsic one, but only the one that photoionizes the NLR. This
can be the case if the NLR is partially shielded from the continuum source
by an absorbing medium (Halpern \& Steiner (1983) suggested that varying
amounts of shielding can naturally explain the Seyfert to LINER sequence of
NLR spectra). The relatively high NLR covering factor of the best fit
model, together with the large broad to narrow $\ifmmode {\rm H}\beta \else H$\beta$\fi$ flux ratio, suggest
that this is indeed the case in NGC\,4151. The measured broad to narrow
$\ifmmode {\rm H}\beta \else H$\beta$\fi$ ratio varies from 1.5 (DeRobertis \cite{DeRobertis}) during a low
state epoch of the AGN (Penston \& P\'erez \cite{PP}), through $\sim 6$
(Osterbrock \& Koski \cite{OK}), up to $\sim 10$ (Kaspi et
al. \cite{Kaspi}) during a high state epoch\footnote{Based on the low
resolution Wise spectrum of the full $\ifmmode {\rm H}\beta \else H$\beta$\fi$ line and the narrow $\ifmmode {\rm H}\beta \else H$\beta$\fi$ flux
given in Table~\ref{t:flux}.} (Kriss et al. \cite{KDZKE}). We adopt a
broad to narrow $\ifmmode {\rm H}\beta \else H$\beta$\fi$ flux ratio of 5 as a representative value. The $\ifmmode {\rm H}\beta \else H$\beta$\fi$
line emission in the best fit model is well modeled by case B recombination
with $T=10^4$\,K, which is close to the maximal efficiency for $\ifmmode {\rm H}\beta \else H$\beta$\fi$
production by recombination per unit covering factor. The covering factor
of the best fit model is $C=0.24$, which implies an unphysical covering
factor for the BLR, $C_{\rm BLR} = C\times5> 1$. It is well known (Netzer
\cite{BNW}) that the uncertainty in calculating the broad Balmer line
intensities is large, due to complicated radiative transfer effects. It is
likely that these effects, combined with collisional excitation of $\ifmmode {\rm H}\alpha \else H$\alpha$\fi$
and $\ifmmode {\rm H}\beta \else H$\beta$\fi$, can enhance the $\ifmmode {\rm H}\beta \else H$\beta$\fi$ intensity over its recombination value, by
a factor of a few. However, this is unlikely to lower $C_{\rm BLR}$ to an
acceptably low value, especially if the broad to narrow $\ifmmode {\rm H}\beta \else H$\beta$\fi$ flux ratio is
in fact as large as $\sim 10$.
A possible solution to this problem is to assume that the BLR gas sees more
ionizing photons than the NLR. For example, the unabsorbed Circinus-like
bumpy SED, shown in Figure~\ref{f:abs_SED}, has 10 times as many ionizing
photons as the best fit SED. In this case, $C_{\rm BLR} = C\times5/10 \sim
0.1$, which is the canonical value for $C_{\rm
BLR}$. Figure~\ref{f:abs_SED} shows that the rise towards 8 Ryd in the best
fit SED is well reproduced by the recovery slope of an intrinsically bumpy
SED absorbed by $\sim 5\,10^{19}$\, cm$^{-2}$ of neutral hydrogen, located
between the BLR and the NLR. The covering factor of the absorber relative
to that of the NLR, $C_{\rm abs}$, has to be close to unity, but some
leakage is still consistent with the best fit SED.
It is remarkable that such an absorber was detected with completely
independent methods by absorption lines (Kriss et al. \cite{Kriss,KDZKE})
(Sect.~\ref{s:n4151-nucleus}). The absorber's H\,{\sc i} column density and
location , although poorly determined, are consistent with what we deduce
from the narrow emission line data. The observations also imply that the
absorber is very dense and that it covers a large fraction of the continuum
source and the BLR along the line of sight. The only additional assumption
that is required is that it also covers the line of sights to the NLR,
which is consistent with our assumption of $A=1$. The high density of the
absorber, $n > 10^{9.5}$\,cm$^{-3}$, implies that it does not contribute to
the forbidden line emission. Therefore, our gas models, which do not
include possible narrow line emission from this component, remain valid
since they rely predominantly on the forbidden lines. The analysis of Kriss
et al. (\cite{Kriss,KDZKE}) suggests that the absorber has a complex,
possibly time-variable multi-zone structure. Since the energy resolution of
the enumerated SED is too coarse, and additional free parameters will make
the enumeration too large, such detailed modeling is outside the scope of
this work. We note that the evidence for an intrinsic UV bump comes only
from the BLR covering factor problem. The absorption beyond the Lyman edge
is so high that there are almost no direct constraints on the intrinsic SED
in that energy range.
To summarize, we have to consider two possible interpretations of the
reconstructed SED of NGC\,4151.
\begin{enumerate}
\item
It is an intrinsic SED, which does not fit easily with present-day
accretion theories, which challenges AGN unification schemes, and
which photoionizes a BLR with an unusually high (perhaps even an
impossibly high) covering factor.
\item
It is an intrinsically bumpy SED (consistent with a hot thin accretion
disk), similar to the one we find in the Seyfert 2 Circinus galaxy,
which is modified by an absorber between the BLR and NLR. The
existence of an absorber with the required properties was discovered
independently by UV absorption lines.
\end{enumerate}
Both simplicity and plausibility arguments lead us to adopt the second
interpretation. While the intrinsic SED of NGC\,4151 cannot be
reconstructed from narrow emission line data because of the internal
absorption, it is encouraging that our method has succeeded in detecting
this feature in the SED. The absorption in NGC\,4151 is very complex and it
is not at all clear, due to the lack of similar quality data, how typical
is this among AGN. It is quite possible that the narrow emission lines of
other AGNs will provide us with direct information on the intrinsic
ionizing continuum. A program of ISO observations of additional AGNs,
currently in progress, will hopefully shed more light on this issue.
\section{Summary}
\label{s:summary}
In this paper we inferred the spectral shape of the obscured
photoionizing continuum source in the nucleus of the Seyfert 1 galaxy
NGC 4151. We constrained the SED by fitting the observed intensities
of NLR emission lines, particularly the high-ionization IR lines, to a
large grid of photoionization models. The advantage of this approach
over methods that attempt to relate the observed UV and soft X-ray
spectra is that it can be used to study AGNs individually rather than
statistically, thus avoiding the problems of selection effects and
small sample statistics.
We used the available information on the line emitting gas to
construct a large grid of gas models. For each of these models, we
enumerated extensively on possible ways to bridge the UV--X-ray gap in
a SED based on simultaneous multi-band observations of the AGN
continuum, and calculated the line emission with a photoionization
code. We then employed a new method to fit the model predictions to a
compilation of observed line fluxes ranging from the UV to the IR. In
addition to minimizing over the SED, this method also minimizes over
the gas geometry and the extinction, thus making it possible to use
observational information on these quantities to further constrain the
best-fit model. Our results suggest that the filling factor of the NLR
plays an important role in determining the properties of the observed
line spectrum. We find that the best-fit SED for NGC\,4151 {\em does
not} have a Big Blue Bump, but rather falls steeply beyond the Lyman
limit towards 4 Ryd, and then rises sharply again towards 8 Ryd. Using
our new method we confirm our previous conclusion that a Big Blue Bump
is present in the SED of the Seyfert 2 galaxy Circinus.
We consider the possibility that the reconstructed NGC\,4151 SED is
the intrinsic SED produced by the accretion mechanism. However, such a
SED does not have enough UV photons to reproduce the BLR recombination
emission. This leads us to adopt the interpretation that the BLR is
photoionized by the intrinsic continuum source, which does contain a
strong UV component (perhaps a Big Blue Bump), but that this UV
component is absorbed by material located between the NLR and BLR. Our
analysis suggests that the absorber consists of $\sim
5\,10^{19}$\,cm$^{-2}$ of neutral hydrogen. Such an absorber was
detected independently by UV absorption lines.
\acknowledgments
We thank Insu Yi for helpful discussions of advection dominated
accretion flows. This work was supported by DARA under grants 50-QI-8610-8
and 50-QI-9492-3, and by the German-Israeli Foundation under grant
I-196-137.07/91.
\clearpage
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{
"redpajama_set_name": "RedPajamaArXiv"
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Jump to navigation Jump to search Jump to content
LafargeHolcim Worldwide
Contribute to better cities
Contribute to more housing cities
Contribute to more compact cities
Contribute to more durable cities
Contribute to more beautiful cities
Contribute to more connected cities
Cities in pictures
The Group at a Glance
All About Cement
LafargeHolcim Awards
Latest News / Press Releases
Working for Hima Cement Limited
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Geocycle
Cement »
All About Cement »
Step 1: extraction of raw materials
The raw materials needed to produce cement (calcium carbonate, silica, alumina and iron ore) are generally extracted from limestone rock, chalk, clayey schist or clay. Suitable reserves can be found in most countries.
These raw materials are extracted from the quarry by blasting. They are then crushed and transported to the plant where they are stored and homogenized.
Step 2: raw grinding and burning
Very fine grinding produces a fine powder, known as raw meal, which is preheated and then sent to the kiln. The material is heated to 1,500°C before being suddenly and dramatically cooled by bursts of air.
This produces clinker, the basic material required for the production of all cements.
Clinker is the main ingredient in cement. These hardened granules are obtained by firing a mixture of approximately 80% limestone and 20% clay to a high temperature. Cement is obtained by grinding clinker, in some cases supplementing it with additives.
Step 3: cement grinding and shipping
A small amount of gypsum (3-5%) is added to the clinker to regulate how the cement will set. The mixture is then very finely ground to obtain "pure cement". During this phase, different mineral materials, called "cement additives", may be added alongside the gypsum.
Used in varying proportions, these additives, which are of natural or industrial origin, give the cement specific properties such as reduced permeability, greater resistance to sulfates and aggressive environments, improved workability, or higher-quality finishes.
Finally, the cement is stored in silos before being shipped in bulk or in bags to the sites where it will be used.
CO2 and cement
Why does manufacturing cement produce CO2?
Cement manufacturing is the source of 5% of global carbon dioxide (CO2) emissions. The cement industry is a natural producer of CO2:
60% of emissions are due to the transformation of raw materials at high temperatures (decarbonation of limestone),
40% result from the combustion needed to heat the cement kilns to 1500°C.
Lafarge: building better cities
In 2050, 70% of the world population will live in cities, twice as many as there were in 1970. Whether large, medium or small, whether in mature or emerging countries, cities are central to the challenges facing the planet. To improve cities, Lafarge contributes to the construction of cities around the world, through its innovative solutions providing them with more housing and making them more compact, more durable, more beautiful, and better connected.
© Lafarge 2021
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"redpajama_set_name": "RedPajamaCommonCrawl"
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Le théâtre aux États-Unis plonge ses racines dans la tradition occidentale et emprunte ses bases au théâtre européen. La comédie musicale est la forme de théâtre la plus appréciée aux États-Unis, avec pour foyer principal Broadway à New York.
Histoire
Débuts
Les plus anciens témoignages sur le théâtre américain remontent à l'époque coloniale anglaise. Un théâtre existait déjà à Williamsburg en 1715, et on sait que Thomas Kean avait joué Richard III à New York en 1750. Le théâtre amateur s'est développé dès le . On sait que la troupe de Lewis Hallam, venant de Londres, se produisit à Williamsburg en 1752. C'est Lewis Hallam, Jr. qui fonda la "compagnie américaine" et ouvrit un théâtre à New York. Les puritains les plus radicaux condamnèrent l'expression théâtrale : des lois prohibant les représentations furent même appliquées dans le Massachusetts en 1750, en Pennsylvanie en 1759, et à Rhode Island en 1761.
Annexes
Articles connexes
Troupes de théâtre américaines
Pièces de théâtre américaines
Dramaturges américains
Théâtre yiddish
Metteurs en scène américains
Récompenses de théâtre aux États-Unis
Broadway (théâtre),
Catégorie : Theatre festivals in the United States,
Arts de la marionnette
Bread and Puppet Theatre
Liens externes
Notes et références
|
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"redpajama_set_name": "RedPajamaWikipedia"
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Q: Datafactory : forloop not running in parallel I am using data factory foreach activity . however the sub-sequence activity are running in sequence . Even issequence is not check . Let's have a look on the following screen.
A: What is that your sub activity doing ? I did tried a very simple test and I think it is working as expected . Please share more info and I will like to investigate further in this .
This is what i tried .
Created a pipeline with array as a variable , just add four elements to that and in the foreach loop printed the value using the Set Variable activity . I added a wait activity to make it more visible . I set the sequential as false and batch count as 2 .
This is what i see the , the set variable is grouped together and the value for the SV activity is not in order , very much on the expected lines .
Now when i set the seqiuential = true , i see that only one value is picked at a time .
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{"url":"https:\/\/biomedical-imaging-group.github.io\/GlobalBioIm\/abstract.html","text":"# Abstract Classes\u00b6\n\nThis section describes the abstract classes of the GlobalBioIm library. It provides general properties for every derived classes.\n\n## Map\u00b6\n\nclass Abstract.Map\n\nBases: matlab.mixin.Copyable\n\nAbstract class for Maps which maps elements from $$\\mathrm{X}$$ to $$\\mathrm{Y}$$ $$\\mathrm{H}: \\mathrm{X}\\rightarrow \\mathrm{Y}.$$ where $$\\mathrm{X}$$ and $$\\mathrm{Y}$$ are either $$\\mathbb{R}^N$$ or $$\\mathbb{C}^N$$.\n\nParameters:\n\u2022 name \u2013 name of the linear operator $$\\mathbf{H}$$\n\n\u2022 sizein \u2013 dimension of the left hand side vector space $$\\mathrm{X}$$\n\n\u2022 sizeout \u2013 dimension of the right hand side vector space $$\\mathrm{Y}$$\n\n\u2022 norm \u2013 norm of the operator $$\\|\\mathrm{H}\\|$$ (if known, otherwise -1)\n\n\u2022 isInvertible \u2013 true if the method applyInverse_() is implemented\n\n\u2022 isDifferentiable \u2013 true if the method applyJacobianT_() is implemented\n\n\u2022 memoizeOpts \u2013 structure of boolean (one field per method, see details below).\n\n\u2022 doPrecomputation \u2013 boolean true to allow doing precomputations to save time (will generally require more memory).\n\nNote on the memoize option This option allows to store the result of a method such that if an identical call to this method is done, calculations are avoided. Example: memoizeOpts.apply=true will store the result of H*x.\n\napply(this, x)\n\nComputes $$\\mathrm{y}=\\mathrm{H}(\\mathrm{x})$$ for the given $$\\mathrm{x} \\in \\mathrm{X}$$.\n\nCalls the method apply_()\n\napplyJacobianT(this, x, v)\n\nCompute $$\\mathrm{x}=[\\mathrm{J}_{\\mathrm{H}}(\\mathrm{v})]^{\\star}\\mathrm{y}$$ where\n\n\u2022 $$[\\mathrm{J}_{\\mathrm{H}}(\\mathrm{v})]$$ is the Jacobian matrix of the Map $$\\mathrm{H}$$ computed at $$\\mathrm{v} \\in \\mathrm{X}$$\n\n\u2022 $$\\mathrm{y} \\in \\mathrm{Y}$$\n\nCalls the method applyJacobianT_()\n\napplyInverse(this, x)\n\nComputes $$\\mathrm{x} = \\mathrm{H}^{-1} \\mathrm{y}$$ for the given $$\\mathrm{y} \\in \\mathrm{Y}$$. (if applicable)\n\nCalls the method applyInverse_()\n\nmakeComposition(this, G)\n\nCompose the Map $$\\mathrm{H}$$ with the given Map $$\\mathrm{G}$$. Returns a new map $$\\mathrm{M=HG}$$\n\nCalls the method makeComposition_()\n\nplus(this, G)\n\nOverload operator (+) for Map objects $$\\mathrm{M}(\\mathrm{x}) := \\mathrm{H}(\\mathrm{x}) + \\mathrm{G}(\\mathrm{x})$$\n\nCalls the method plus_()\n\nminus(this, G)\n\nOverload operator (-) for Map objects $$\\mathrm{M}(\\mathrm{x}) := \\mathrm{H}(\\mathrm{x}) - \\mathrm{G}(\\mathrm{x})$$\n\nCalls the method minus_()\n\nmpower(this, p)\n\nReturns a new Map which is the power p $$\\mathrm{H}^{p}$$ of the current $$\\mathrm{H}$$.\n\nCalls the method mpower_()\n\nmtimes(this, G)\n\nOverload operator (*) for Map objects $$\\mathrm{M}(\\mathrm{x}) := \\mathrm{H}(\\mathrm{G}(\\mathrm{x}))$$\n\n\u2022 If $$\\mathrm{G}$$ is numeric of size sizein, then apply() is called\n\n\u2022 If $$\\mathrm{G}$$ is a Map, then a MapComposition is intanciated\n\ntimes(this, G)\n\nReturns a new Map which is the element-wise multiplication of the current $$\\mathrm{H}$$ with $$\\mathrm{G}$$ $$\\mathrm{M}(\\mathrm{x}) := \\mathrm{H}(\\mathrm{x}) \\times \\mathrm{G}(\\mathrm{x})$$\n\nCalls the method times_()\n\napply_(this, x)\n\nNot implemented in this Abstract class\n\napplyJacobianT_(this, y, v)\n\nNot implemented in this Abstract class\n\napplyInverse_(this, y)\n\nNot implemented in this Abstract class\n\nplus_(this, G)\n\nConstructs a MapSummation object to sum the current Map $$\\mathrm{H}$$ with the given $$\\mathrm{G}$$.\n\nminus_(this, G)\n\nConstructs a MapSummation object to subtract to the current Map $$\\mathrm{H}$$, the given $$\\mathrm{G}$$.\n\nmpower_(this, p)\n\nWhen $$p=-1$$, constructs a MapInversion object which is the inverse Map of $$\\mathrm{H}$$. When $$p\\neq-1$$, this method is not implemented in this Abstract class\n\ntimes_(this, G)\n\nConstructs a MapMultiplication object to element-wise multiply the current Map $$\\mathrm{H}$$ with the given $$\\mathrm{G}$$.\n\nmakeComposition_(this, G)\n\nConstructs a MapComposition object to compose the current Map $$\\mathrm{H}$$ with the given $$\\mathrm{G}$$.\n\ncopyElement()\n\nPerform a deep copy of $$\\mathrm{H}$$\n\nCalled by the function copy()\n\n## LinOp\u00b6\n\nclass Abstract.LinOp\n\nBases: Abstract.Map\n\nAbstract class for linear operators $$\\mathrm{H}: \\mathrm{X}\\rightarrow \\mathrm{Y}.$$ where $$\\mathrm{X}$$ and $$\\mathrm{Y}$$ are either $$\\mathbb{R}^N$$ or $$\\mathbb{C}^N$$.\n\nAll attributes of parent class Map are inherited\n\nSee also Map\n\napplyAdjoint(this, y)\n\nComputes $$\\mathrm{y=H}^*\\mathrm{y}$$ for $$\\mathrm{y} \\in \\mathrm{Y}$$\n\nCalls the method applyAdjoint_()\n\napplyHtH(this, x)\n\nComputes $$\\mathrm{y=H}^*\\mathrm{Hx}$$ for $$\\mathrm{y} \\in \\mathrm{Y}$$\n\nCalls the method applyHHt_()\n\napplyHHt(this, y)\n\nComputes $$\\mathrm{y=HH}^*\\mathrm{y}$$ for $$\\mathrm{y} \\in \\mathrm{Y}$$\n\nCalls the method applyHHt_()\n\ntranspose(this)\n\nReturns a new LinOp which is the Adjoint $$\\mathrm{H}^{\\star}$$ of the current $$\\mathrm{H}$$.\n\nctranspose(this)\n\nDo the same as transpose()\n\napplyAdjointInverse(this, x)\n\nComputes $$\\mathrm{y} = \\mathrm{H}^{-\\star} \\mathrm{x}$$ for the given $$\\mathrm{x} \\in \\mathrm{X}$$. (if applicable)\n\nCalls the method applyAdjointInverse_()\n\nmakeHtH(this)\n\nCompose the Adjoint Map $$\\mathrm{H}^{\\star}$$ with $$\\mathrm{H}$$. Returns a new map $$\\mathrm{M=H}^{\\star} \\mathrm{H}$$\n\nCalls the method makeHtH_()\n\nmakeHHt(this)\n\nCompose the Map $$\\mathrm{H}$$ with its adjoint $$\\mathrm{H}^{\\star}$$. Returns a new map $$\\mathrm{M=H}\\mathrm{H}^{\\star}$$\n\nCalls the method makeHHt_()\n\napplyAdjoint_(~, ~)\n\nNot implemented in this Abstract class\n\napplyHtH_(this, x)\n\nThere is a default implementation in the abstract class LinOp which calls successively the apply() and applyAdjoint() methods. However, it can be reimplemented in derived classes if there exists a faster way to perform computation.\n\napplyHHt_(this, y)\n\nThere is a default implementation in the abstract class LinOp which calls successively the applyAdjoint() and apply() methods. However, it can be reimplemented in derived classes if there exists a faster way to perform computation.\n\napplyAdjointInverse_(~, ~)\n\nNot implemented in this Abstract class\n\nplus_(this, G)\n\nIf $$\\mathrm{G}$$ is a LinOp, constructs a LinOpSummation object to sum the current LinOp $$\\mathrm{H}$$ with the given $$\\mathrm{G}$$. Otherwise the summation will be a MapSummation.\n\nmakeAdjoint_(this)\n\nConstructs a LinOpAdjoint from the current current LinOp $$\\mathrm{H}$$\n\nmakeHtH_(this)\n\nConstructs a LinOpComposition corresponding to $$\\mathrm{H}^{\\star}\\mathrm{H}$$\n\nmakeHHt_(this)\n\nConstructs a LinOpComposition corresponding to $$\\mathrm{H}\\mathrm{H}^{\\star}$$\n\nmakeInversion_(this)\n\nConstructs a LinOpInversion corresponding to $$\\mathrm{H}^{-1}$$\n\nmakeComposition_(this, G)\n\nReimplemented from parent class Map. If $$\\mathrm{G}$$ is a LinOp, constructs a LinOpComposition object to compose the current LinOp (this) with the given LinOp$$\\mathrm{G}$$. Otherwise the composition will be a MapComposition.\n\napplyJacobianT_(this, y, ~)\n\nUses the method applyAdjoint (hence do not need to be reimplemented in derived classes)\n\n## Cost\u00b6\n\nclass Abstract.Cost(sz, y)\n\nBases: Abstract.Map\n\nAbstract class for Cost functions $$C : \\mathrm{X} \\longrightarrow \\mathbb{R}$$ with the following special structure $$C(\\mathrm{x}) := F( \\mathrm{x} , \\mathrm{y} )$$ where $$F$$ is a function takin two variables, both in $$\\mathrm{X}$$.\n\nParameters:\n\u2022 y \u2013 data vector (default 0)\n\n\u2022 name \u2013 name of the cost function\n\n\u2022 lip \u2013 Lipschitz constant of the gradient (when applicable and known, otherwise -1)\n\n\u2022 isConvex \u2013 true if the cost is convex\n\n\u2022 isSeparable \u2013 true if the cost is separable (R^n basis)\n\nAll attributes of parent class Map are inherited and norm is fixed to -1, sizeout is fixed to for all Cost\n\nSee also Map, LinOp.\n\napplyGrad(this, x)\n\nComputes the gradient of the cost function at $$\\mathrm{x} \\in \\mathrm{X}$$ (when applicable) $$\\mathrm{g} = \\nabla C(\\mathrm{x})$$\n\nCalls the method applyGrad_()\n\napplyProx(this, z, alpha)\n\nComputes the proximity operator of the cost at $$\\mathrm{z} \\in \\mathrm{X}$$ (when applicable) $$\\mathrm{prox}_{\\alpha C}(\\mathrm{z}) = \\mathrm{arg} \\, \\mathrm{min}_{\\mathrm{u} \\in \\mathrm{X}} \\; \\frac{1}{2\\alpha} \\| \\mathrm{u} - \\mathrm{z} \\|_2^2 + C(\\mathrm{u}).$$\n\nCalls the method applyProx_()\n\napplyProxFench(this, z, alpha)\n\nComputes the proximity operator of the Fenchel Transform $$C^*$$ at $$\\mathrm{z} \\in \\mathrm{Y}$$ (when applicable)\n\nCalls the method applyProxFench_()\n\napplyGrad_(this, x)\n\nNot implemented in this Abstract class\n\napplyProx_(this, z, alpha)\n\nBy default, if the cost $$C$$ isConvex, computes the proximity operator of $$C^*$$ at $$\\mathrm{z} \\in \\mathrm{X}$$ using the Moreau\u2019s identity which uses the applyProxFench() method $$\\mathrm{prox}_{\\alpha C}(\\mathrm{z}) = \\mathrm{z} - \\alpha \\,\\mathrm{prox}_{\\frac{1}{\\alpha}C^*}\\left(\\frac{\\mathrm{z}}{\\alpha}\\right).$$\n\napplyProxFench_(this, z, alpha)\n\nBy default, if the cost $$C$$ isConvex, computes the proximity operator of the Fenchel Transform $$C^*$$ at $$\\mathrm{z} \\in \\mathrm{Y}$$ using the Moreau\u2019s identity which uses the applyProx() method $$\\mathrm{prox}_{\\alpha C^*}(\\mathrm{z}) = \\mathrm{z} - \\alpha \\,\\mathrm{prox}_{\\frac{1}{\\alpha}C}\\left(\\frac{\\mathrm{z}}{\\alpha}\\right).$$\n\nplus_(this, G)\n\nIf $$\\mathrm{G}$$ is a Cost, constructs a CostSummation object to sum the current Cost $$C$$ with the given $$G$$. Otherwise the summation will be a MapSummation.\n\nminus_(this, G)\n\nIf $$\\mathrm{G}$$ is a Cost, constructs a CostSummation object to subtract to the current Cost $$C$$, the given $$G$$. Otherwise the summation will be a MapSummation.\n\nmakeComposition_(this, G)\n\nReimplemented from parent class Map. Constructs a CostComposition object to compose the current Cost (this) with the given Map$$\\mathrm{G}$$.\n\napplyJacobianT_(this, y, v)\n\nUses the method applyGrad (hence do not need to be reimplemented in derived classes)\n\nset_y(this, y)\n\nSet the attribute $$\\mathrm{y}$$\n\n\u2022 has to be conformable with the sizein of the cost\n\n\u2022 can be a scalar.\n\n## Opti\u00b6\n\nclass Abstract.Opti\n\nBases: matlab.mixin.SetGet\n\nAbstract class for optimization algorithms to minimize Cost objects\n\nParameters:\n\u2022 name \u2013 name of the algorithm\n\n\u2022 cost \u2013 minimized Cost\n\n\u2022 maxiter \u2013 maximal number of iterations (default 50)\n\n\u2022 verbose \u2013 bollean (default true) to activate verbose mode\n\n\u2022 OutOpOutputOpti object\n\n\u2022 ItUpOut \u2013 number of iterations between two calls to the update method of the OutputOpti object OutOp (default 0)\n\n\u2022 CvOpTestCvg object\n\n\u2022 time \u2013 execution time of the algorithm\n\n\u2022 niter \u2013 iteration counter\n\n\u2022 xopt \u2013 optimization variable\n\nSee also OutputOpti Cost\n\nrun(this, x0)\n\nRun the algorithm.\n\nParameters:\n\nx0 \u2013 initial point in $$\\in X$$, if no argument restarts from the current value xopt.\n\nnote: this method does not return anything, the result being stored in public attribute xopt.\n\ninitialize(this, x0)\n\nImplements initialization of the algorithm\n\nParameters:\n\nx0 \u2013 initial point\n\ndoIteration(this)\n\nImplements algorithm iteration\n\nReturns:\n\nflag with values\n\n\u2022 OPTI_NEXT_IT (= 0) to go to the next iteration\n\n\u2022 OPTI_REDO_IT (= 1) to redo the iteration\n\n\u2022 OPTI_STOP (= 2) to stop algorithm\n\nupdateParams(this)\n\nUpdates the parameters of the algorithm at each iteration (default: no update). This method can be overloaded to makes some parameters varying during iterations (e.g. descent step, lagrangian parameters\u2026)\n\nstarting_verb(this)\n\nGeneric method to display a starting message in verbose mode.\n\nending_verb(this)\n\nGeneric method to display a ending message in verbose mode.\n\n## OperationsOnMaps\u00b6\n\nThe following classes implement basic operations between Map (LinOp and Cost). They are not abstract but generally they do not need to be instanciated. They are mainly used inside the methods of the abstract classes Map, LinOp and Cost for the operator algebra machinery.\n\n### MapComposition\u00b6\n\nclass Abstract.OperationsOnMaps.MapComposition(H1, H2)\n\nBases: Abstract.Map\n\nMapComposition : Composition of Maps $$\\mathrm{H}(\\mathrm{x}) = \\mathrm{H}_1 \\left( \\mathrm{H}_2(\\mathrm{x}) \\right)$$\n\nParameters:\n\u2022 H1 \u2013 left hand side Map\n\n\u2022 H2 \u2013 right hand side Map\n\nExample H=MapComposition(H1,H2)\n\nSee also Map\n\napply_(this, x)\n\nReimplemented from Map\n\napplyJacobianT_(this, y, v)\n\nReimplemented from Map\n\napplyInverse_(this, y)\n\nReimplemented from Map\n\nmakeComposition_(this, G)\n\nReimplemented from Map\n\n### MapInversion\u00b6\n\nclass Abstract.OperationsOnMaps.MapInversion(M)\n\nBases: Abstract.Map\n\nMapInversion : Builds the inverse Map\n\nParameters:\n\nMMap object\n\nExample Minv=MapInversion(M)\n\nSee also Map\n\napply_(this, x)\n\nReimplemented from Map\n\napplyInverse_(this, x)\n\nReimplemented from Map\n\nmpower_(this, p)\n\nReimplemented from Map\n\nmakeComposition_(this, G)\n\nReimplemented from parent class Map.\n\n### MapSummation\u00b6\n\nclass Abstract.OperationsOnMaps.MapSummation(Maps, alpha)\n\nBases: Abstract.Map\n\nMapSummation: Sum of Maps $$\\mathrm{H}(\\mathrm{x}) = \\sum_i \\alpha_i \\mathrm{H}_i(\\mathrm{x})$$\n\nParameters:\n\u2022 Maps \u2013 cell of Map\n\n\u2022 alpha \u2013 array of coefficients\n\nExample H=MapSummation(Maps,alpha)\n\nSee also Map, LinOpSummation\n\napply_(this, x)\n\nReimplemented from Map\n\napplyJacobianT_(this, y, v)\n\nReimplemented from Map\n\nmakeComposition_(this, G)\n\nReimplemented from Map\n\n### MapMultiplication\u00b6\n\nclass Abstract.OperationsOnMaps.MapMultiplication(Map1, Map2)\n\nBases: Abstract.Map\n\nMapMultiplication: Multiplication of Maps $$\\mathrm{H}(\\mathrm{x}) = \\mathrm{H}_1(\\mathrm{x}) \\times \\mathrm{H}_2(\\mathrm{x})$$\n\nParameters:\n\u2022 Map1Map object\n\n\u2022 Map2Map object\n\nExample H=MapMultiplication(Map1,Map2)\n\nSee also Map\n\napply_(this, x)\n\nReimplemented from Map\n\napplyJacobianT_(this, y, v)\n\nReimplemented from Map\n\nmakeComposition_(this, G)\n\nReimplemented from Map\n\nclass Abstract.OperationsOnMaps.LinOpAdjoint(TLinOp)\n\nBases: Abstract.LinOp\n\nParameters:\n\nTLinOpLinOp object\n\nSee also Map, LinOp\n\napply_(this, x)\n\nReimplemented from LinOp\n\napplyAdjoint_(this, x)\n\nReimplemented from LinOp\n\napplyHtH_(this, x)\n\nReimplemented from LinOp\n\napplyHHt_(this, x)\n\nReimplemented from LinOp\n\napplyInverse_(this, x)\n\nReimplemented from LinOp\n\napplyAdjointInverse_(this, x)\n\nReimplemented from LinOp\n\nmakeAdjoint_(this)\n\nReimplemented from parent class LinOp.\n\nmakeHHt_(this)\n\nReimplemented from parent class LinOp.\n\nmakeHtH_(this)\n\nReimplemented from parent class LinOp.\n\n### LinOpComposition\u00b6\n\nclass Abstract.OperationsOnMaps.LinOpComposition(H1, H2)\n\nBases: Abstract.OperationsOnMaps.MapComposition, Abstract.LinOp\n\nLinOpComposition : Composition of LinOps $$\\mathrm{H}(\\mathrm{x}) = \\mathrm{H}_1 \\mathrm{H}_2\\mathrm{x}$$\n\nParameters:\n\u2022 H1 \u2013 left hand side LinOp (or a scalar)\n\n\u2022 H2 \u2013 right hand side LinOp\n\nExample H=LinOpComposition(H1,H2)\n\nSee also Map, LinOp, MapComposition\n\napply_(this, x)\n\nReimplemented from LinOp\n\napplyAdjoint_(this, x)\n\nReimplemented from LinOp\n\napplyHtH_(this, x)\n\nReimplemented from LinOp\n\napplyHHt_(this, x)\n\nReimplemented from LinOp\n\napplyAdjointInverse_(this, x)\n\nReimplemented from LinOp\n\nmakeAdjoint_(this)\n\nReimplemented from parent class LinOp.\n\nmakeHtH_(this)\n\nReimplemented from LinOp\n\nmakeHHt_(this)\n\nReimplemented from LinOp\n\nmakeComposition_(this, G)\n\nReimplemented from MapComposition\n\n### LinOpInversion\u00b6\n\nclass Abstract.OperationsOnMaps.LinOpInversion(M)\n\nBases: Abstract.OperationsOnMaps.MapInversion, Abstract.LinOp\n\nLinOpInversion : Builds the inverse LinOp\n\nParameters:\n\nMLinOp object\n\nExample Minv=LinOpInversion(M)\n\nSee also Map, LinOp, MapInversion\n\napplyAdjoint_(this, x)\n\nReimplemented from LinOp\n\napplyAdjointInverse_(this, x)\n\nReimplemented from LinOp\n\nmpower_(this, p)\n\nReimplemented from MapInversion\n\nmakeComposition_(this, G)\n\nReimplemented from MapInversion\n\n### LinOpSummation\u00b6\n\nclass Abstract.OperationsOnMaps.LinOpSummation(LinOps, alpha)\n\nBases: Abstract.OperationsOnMaps.MapSummation, Abstract.LinOp\n\nLinOpSummation: Sum of linear operators $$\\mathrm{H}(\\mathrm{x}) = \\sum_i \\alpha_i \\mathrm{H}_i(\\mathrm{x})$$\n\nParameters:\n\u2022 LinOps \u2013 cell of LinOp\n\n\u2022 alpha \u2013 array of coefficients\n\nExample L=LinOpSummation(LinOps,alpha)\n\nSee also Map, LinOp, MapOpSummation\n\napplyAdjoint_(this, y)\n\nReimplemented from LinOp\n\nmakeAdjoint_(this)\n\nReimplemented from LinOp\n\nplus_(this, G)\n\nReimplemented from LinOp\n\n### CostComposition\u00b6\n\nclass Abstract.OperationsOnMaps.CostComposition(H1, H2)\n\nBases: Abstract.OperationsOnMaps.MapComposition, Abstract.Cost\n\nCostComposition: Compose a Cost with a Map $$C(\\mathrm{x}) := F( \\mathrm{Hx})$$ where $$F$$ is a Cost and $$\\mathrm{H}$$ a Map\n\nParameters:\n\u2022 H1Cost\n\n\u2022 H2LinOp\n\nExample C = CostComposition(H1,H2)\n\nSee also Map, MapComposition, Cost\n\napplyGrad_(this, x)\n\nReimplemented from Cost\n\napplyProx_(this, z, alpha)\n\nReimplemented from Cost\n\nIf this.H2 is a LinOp and $$\\mathrm{H} \\mathrm{H}^{\\star}$$ is a LinOpScaledIdentity\n\nmakeComposition_(this, G)\n\nReimplemented from Cost\n\nset_y(this, y)\n\nSet the attribute $$\\mathrm{y}$$\n\n\u2022 has to be conformable with the sizeout of the Map$$\\mathrm{H}$$,\n\n\u2022 can be anything if $$\\mathrm{H}$$ is not yet set (empty),\n\n\u2022 can be a scalar.\n\n### CostMultiplication\u00b6\n\nclass Abstract.OperationsOnMaps.CostMultiplication(C1, C2)\n\nBases: Abstract.Cost\n\nCostMultiplication: Multiplication of Costs $$C(\\mathrm{x}) = C_1(\\mathrm{x}) \\times C_1(\\mathrm{x})$$\n\nParameters:\n\u2022 C1 \u2013 a Cost object or a scalar\n\n\u2022 C2 \u2013 a Cost object\n\nExample F = MulCost(Cost1,Cost2)\n\nSee also Map, Cost\n\napply_(this, x)\n\nReimplemented from Cost\n\napplyGrad_(this, x)\n\nReimplemented from Cost\n\napplyProx_(this, x, alpha)\n\nReimplemented from Cost\n\nmakeComposition_(this, G)\n\nReimplemented from Cost\n\n### CostSummation\u00b6\n\nclass Abstract.OperationsOnMaps.CostSummation(costs, alpha)\n\nBases: Abstract.OperationsOnMaps.MapSummation, Abstract.Cost\n\nCostSummation : Sum of Cost $$C(\\mathrm{x}) = \\sum_i \\alpha_i C_i(\\mathrm{x})$$\n\nParameters:\n\u2022 costs \u2013 cell of Cost\n\n\u2022 alpha \u2013 array of coefficients\n\nExample F = CostSummation(ACost,alpha)\n\nSee also Map, Cost, MapOpSummation\n\nmakePartialSummation(this, Lsub)\n\nInstanciation of CostPartialSummation.\n\nParameters:\n\nLsub \u2013 number of Cost used for computation\n\napplyGrad_(this, x)\n\nReimplemented from Cost\n\napplyProx_(this, z, alpha)\n\nReimplemented from Cost in the case of the sum between a CostRectangle $$i_C$$ and a Cost $$f$$ which is separable [1] $$\\mathrm{prox}_{\\alpha(i_C +f)}(z) = \\mathrm{prox}_{i_c} \\circ \\mathrm{prox}_{\\alpha f}(z)$$\n\nReference\n\n[1] \u201cA Douglas?Rachford splitting approach to nonsmooth convex variational signal recovery\u201d P. L. Combettes, and J.C. Pesquet, Journal of Selected Topics in Signal Processing, 1(4), 564-574, 2007\n\n### CostPartialSummation\u00b6\n\nclass Abstract.OperationsOnMaps.CostPartialSummation(costs, alpha, Lsub)\n\nBases: Abstract.OperationsOnMaps.CostSummation\n\nCostPartialSummation : Sum of Cost with apply, applyGrad,\u2026 computed from a subset of Cost $$C(\\mathrm{x}) = \\sum_i \\alpha_i C_i(\\mathrm{x})$$\n\nParameters:\n\u2022 costs \u2013 cell of Cost\n\n\u2022 alpha \u2013 array of coefficients\n\n\u2022 Lsub \u2013 number of Cost used for computation\n\n\u2022 partialGrad \u2013 parameter for subset selection (0: no partial gradient; 1: stochastic gradient descent; 2: equally spaced indices)\n\nExample F = CostPartialSummation(ACost,alpha,Lsub)\n\nSee also Map, Cost, MapOpSummation\n\nsetLsub(this, Lsub)\n\nSet Lsub parameter\n\napply_(this, x)\n\nReimplemented from Cost\n\napplyGrad_(this, x)\n\nReimplemented from Cost","date":"2022-08-10 05:02:43","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 2, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.4444888234138489, \"perplexity\": 11310.276579811236}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-33\/segments\/1659882571147.84\/warc\/CC-MAIN-20220810040253-20220810070253-00155.warc.gz\"}"}
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Q: How do I access variables from the main class from another class (java)? I'm making a cookie clicker clone in java to practice my java skills and I have a small problem, I have variables that are declared in the main method that I want to access from an ActionListener class. Here is some sample code from the ActionListener class. the the int variables (ex. clicks, grandamaCost) and the JTextFields (ex. display, cpsDisplay) are all in the main method. I was wondering how I could have access to variables in the main method so that this code could work in the other class. Thanks!
@Override
public void actionPerformed(ActionEvent e) {
JButton b = (JButton) e.getSource();
button(b.getText());
}
public void button(String input) {
switch (input) {
case "Cookie":
clicks++;
display.setText("Cookies: " + clicks + "");
cpsDisplay.setText("CPS: " + cps);
break;
case "Buy grandma":
if (clicks >= grandmaCost) {
grandmas++;
clicks = clicks - grandmaCost;
grandmaCost = (int) ((.15 * grandmaCost) + grandmaCost);
cps++;
}
display.setText("Cookies: " + clicks + "");
prices[0].setText("$" + grandmaCost);
cpsDisplay.setText("CPS: " + cps);
break;
case "Buy monkey":
if (clicks >= monkeyCost) {
monkeys++;
clicks = clicks - monkeyCost;
monkeyCost = (int) ((.15 * monkeyCost) + monkeyCost);
cps = cps + 2;
}
display.setText("Cookies: " + clicks + "");
prices[1].setText("$" + monkeyCost);
cpsDisplay.setText("CPS: " + cps);
break;
case "Buy Teemo":
if (clicks >= teemoCost) {
teemos++;
clicks = clicks - teemoCost;
teemoCost = (int) ((.15 * teemoCost) + teemoCost);
cps = cps + 3;
}
display.setText("Cookies: " + clicks + "");
prices[2].setText("$" + teemoCost);
cpsDisplay.setText("CPS: " + cps);
break;
}
}
}
A: Your variables should be fields.
Fields are declared outside of a class's methods and are usually found right below the class declaration. Fields can be accessed by all methods of a class.
They can also be accessed from other classes (unless they are private) using the dot operator.
*
*If a field is marked with static, its class name is used to reference it.
*If a field is not static, an object of its class is used to reference it.
Example
public class Man {
public String name; //this is a field
public static String gender = "Male"; //this is a static field
public Man(String newName) {
name = newName; //assigns the value of a field from within a method
}
}
and another class...
public class Main {
public static void main(String[] args) {
Man bob = new Man("Bob");
System.out.println(bob.name); //referenced from object, prints Bob
System.out.println(Man.gender); //referenced from class name, prints Male
}
}
And to have more control over the access of your fields, you can use getters and setters. Take a read!
A: public class ActionClass {
{
private static int clicks;
@Override
public void actionPerformed(ActionEvent e) {
clicks++;
}
public static void setClicks(int c){
clicks = c;
}
public static int getClicks(){
return clicks;
}
}
public class AnyClass {
{
// now you have access to your clicks count .
int clicks = ActionClass.getClicks();
// set value of clicks
ActionClass.setClicks(0);
}
A: Here, I will give you an example for exactly what you need. In this code you simply just need to set anything you would like to add to actionPerformed as static.
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import javax.swing.JButton;
public class testJava implements ActionListener {
protected static JButton b; // since this is static you can
// now access it in other classes
public static void main(String[] args) {
}
@Override
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
if(e.getSource() == b) {
// do stuff here
}
}
}
A: Using fields and their accessor methods. An example here.
A: You would have to make the variables public class variables instead of method variables, thereby increasing the scope and visiblity of the variables. Like so:
public class ActionClass {
{
public string MyPublicVariable = "woot";
@Override
public void actionPerformed(ActionEvent e) {
...
}
}
A more popular/recommended way to do this is to use a getter/setter instead of making the variable explicitly public. You would access a private variable through public methods like so:
public class ActionClass {
{
@Override
public void actionPerformed(ActionEvent e) {
private string MyPublicVariable = "woot";
public void setMyString(string newString){
MyPublicVariable = newString;
}
public string getMyString(){
return MyPublicVariable;
}
}
}
This way, you have more control over what your variables are set to.
A: You can pass main class instance reference to another class instance, or register callback.
For the first way
Class MainClass {
private int mValue;
public void init() {
AnotherClass cla = new AnotherClass(this);
}
public void setValue(int value) {mValue = value;}
public int getValue(){return mValue;}
}
Class AnotherClass {
private MainClass mMain;
public AnotherClass(MainClass ref) {
mMain = ref;
}
public void controlValue() {
if (mMain != null) {
mMain.setValue(1);
mMain.getValue();
}
}
}
For the second way
1. declare an interface
2. implement this interface in main class
3. register this implementation to another class.
4. get and set value in another class.
public interface ClassListener {
public void setValue(int value);
public int getValue();
}
public class MainClass implements ClassListener{
private int mValue;
public void registerListener() {
AnotherClass cla = new AnotherClass();
cla.registerListener(this);
}
@Override
public void setValue(int value) {
// TODO Auto-generated method stub
mValue = value;
}
@Override
public int getValue() {
// TODO Auto-generated method stub
return mValue;
}
}
public class AnotherClass{
private ClassListener mListener;
public void registerListener(ClassListener listener) {
mListener = listener;
}
public void controlValue() {
if (mListener != null) {
int value = mListener.getValue();
mListener.setValue(++value);
}
}
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,872
|
Kensington Car Show Makes MORE Noyes!
The Fifth Annual Kensington Car Show raised nearly $10,000 for the Noyes Children's Library Foundation's Make MORE Noyes Renovation Campaign for a universally accessible and creatively expanded Noyes. Mario Bruno, owner of Kensington Service Station, Lake Liberty Chevy Chase, and Bruno's Classic Muscle, donates all funds raised through Car Show sponsors to the Campaign. With this year's donation, the Kensington Car Show has topped $25,000 in contributions to the Make MORE Noyes Campaign.
Our thanks to the 2018 Kensington Car Show Sponsors:
Premium Sponsors: Trick Trucks • Reliable Tire Co. • Winners Coating & Tinting • Strohsniders Hardware
Gold Sponsors: Frankly... Pizza! • R&B Steel Fabrications • Eaglestone Tax & Accounting • Lafayette Federal Credit Union
Silver Sponsors: Sage Consignment • International Oil & Energy Corp. • RBR Hunter Parts & Service • Dish & Dram • Myers Cycle Engineering • Splaine Security Systems
Bronze Sponsors: Unity Thunder • JHLM, Inc.
Also thanks to Kristen Haddeland, Independent Consultant, Usborne Books & More for donating a portion of her proceeds to the Make MORE Noyes Campaign!
Make MORE Noyes Campaign Nears Halfway Point
Kensington, MD - What do muscle cars, teddy bears, chocolate owls and the Maryland State Assembly have on common? They are all examples of the wide range of community support that the Noyes Children's Library Foundation has received for its Make MORE Noyes Renovation Campaign.
With the announcement last month of a $100,000 2018 Bond Bill from the State of Maryland, and the success of April's Read! Dream! Fly! Make MORE Noyes Gala, the Foundation is poised to reach the halfway point of its $1.6M fundraising goal.
"We are pleased to announce that we have raised over $750,000 toward our commitment to Montgomery County for the Make MORE Noyes Renovation," said Diana Ditto, Co-president of the Foundation. "With the strong support of the State of Maryland, the Town of Kensington, local businesses, family foundations, and countless community donors, we are charging full speed ahead toward raising the remainder of our commitment to a fully accessible and creatively expanded Noyes. "
Ditto cited examples such as the Kensington Car Show, to be held Saturday, May 12, 2018 in Kensington, as an example of the groundswell of community support for the Campaign. Other unique events and contributors – such as the upcoming Teddy Bear Picnic on the grounds of Warner Circle Park, and the enterprising local high school student who has sold over $2,000 in handmade chocolate owls to benefit Noyes – have enabled the grassroots, all-volunteer organization to maximize its fundraising efforts.
"So many local businesses come to us with ideas for fundraising, because so many of their customers are patrons of the Noyes Children's Library, or benefited from the Library in their childhood," said Ditto. "Noyes has been welcoming children for generations, and the Make MORE Noyes Campaign will ensure that it welcomes them for generations to come."
The Noyes Library in Kensington, MD was built in 1893, joined the Montgomery County Library system in 1950, and became a children's library in 1972. Noyes is one of only a handful of public libraries in the U.S. dedicated to children, and is also the oldest public library in the Washington, DC area.
The Make MORE Noyes Renovation Campaign will help fund a major renovation of the historic Noyes Library, creating three stories within the original footprint and adding an innovative glass elevator. The Renovation expands both the Library's space and its early literacy mission, enabling it to reach more children, families, librarians and teachers with early literacy programs and training.
For more information go to www.noyeslibraryfoundation.org/make-more-noyes, or contact Sheila Dinn at info@noyeslibraryfoundation.org.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,595
|
\section{Introduction}
In this note
we show some numerical computations of representations of the fundamental group
in $\mathit{SL}(2;\mathbb{C})$
and Reidemeister torsion for the homology 3-sphere
obtained by $1/n$-Dehn surgeries along the figure-eight knot.
More precisely we enumerate all conjugacy classes of irreducible representations
in $\mathit{SL}(2;\mathbb{C})$
and compute Reidemeister torsion for these representations.
Reidemeister torsion was originally defined by Reidemeister, Franz and de Rham in the 1930's.
It can be defined in more general situation,
but in this paper,
we consider this invariant for a homology 3-sphere $M$
with an irreducible representation $\rho$ of the fundamental group $\pi_1(M)$ into $\mathit{SL}(2;\mathbb{C})$.
It is denoted by $\tau_\rho(M)\in\mathbb{C}$.
In the 1980's Johnson \cite{Johnson} developed a theory of Reidemeister torsion
for representations in $\mathit{SU}(2)$, or in $\mathit{SL}(2;\mathbb{C})$.
That was studied motivated by the relations to the Casson invariant.
Further he proposed a torsion polynomial of a $3$-manifold.
In this paper, we define the torsion polynomial as follows.
Let $M$ be a homology 3-sphere.
We denote
the set of conjugacy classes of representations from $\pi_1(M)$ in $\mathit{SL}(2;\mathbb{C})$
by $\mathcal{R}(M) $
and the subset of conjugacy classes with nontrivial value
of Reidemeister torsion
by $\mathcal{R}'(M)$.
Now assume that $\mathcal{R}'(M) $ is a finite set.
\begin{remark}
In general $\mathcal{R}(M)\ne \mathcal{R}'(M)$. For examples, please see \cite{Johnson, Kitano94-1}.
\end{remark}
\begin{definition}
A one variable polynomial
\[
{\sigma}_{M}(t)
=\prod_{[\rho]\in\mathcal{R}'(M)}(t-\tau_\rho(M))
\] is called the torsion polynomial
of $M_n$.
\end{definition}
\begin{remark}
If $\mathcal{R}'(M)=\emptyset$,
then we define
$\sigma_M(t)=1$.
\end{remark}
In \cite{Kitano15-2, Kitano16-1}
we gave explicit formulas of $\sigma_M(t)$
in the case of Brieskorn homology 3-spheres obtained by surgeries along torus knots.
\begin{remark}
The torsion polynomial was defined as
\[
\pm
\prod_{[\rho]\in\mathcal{R}'(M)}(t-1/\tau_\rho(M))
\]
under another normalization in \cite{Kitano15-2,Kitano16-1}.
\end{remark}
In this note,
we show numerical computations for Dehn-surgeries along the figure-eight knot
by using Mathematica.
\section{Setting}
First we explain geometric setting,
which is the same one in \cite{Kitano94-2, Kitano15-1}.
Please see them for details.
Let $K\subset S^3$ be the figure-eight knot and $E(K)$ the exterior.
It is well-known that
$\pi_1 E(K)$ has the following presentation;
\[
\pi_1 E(K)=\langle
x,y\ |\ wx=yw
\rangle
\]
where
$w=xy^{-1}x^{-1}y$.
One can take $x$ as a meridian element in $\pi_1 E(K)$.
As a longitude, one can do
\[
l=w^{-1}\widetilde{w}
\]
where
$\widetilde{w}=x^{-1} y x y^{-1}$.
Let $M_n$ be the homology 3-sphere obtained by $1/n$-Dehn surgery along $K$.
The fundamental group $\pi_1 M_n$ has the presentation as
\[
\pi_1 M_n=\langle
x,y\ |\ wx=yw,\ xl^n=1
\rangle.
\]
Let $\rho:\pi_1 M_n\rightarrow \mathit{SL}(2;\mathbb{C})$ be an irreducible representation.
Simply we write $X$ for $\rho(x)$, $Y$ for $\rho(y)$ and so on.
It is well known that we may assume that $X$ and $Y$ have the following forms;
\[
X=\begin{pmatrix}
s & 1 \\
0 & 1/s \\
\end{pmatrix},
Y=\begin{pmatrix}
s & 0 \\
-t & 1/s \\
\end{pmatrix}
\]
where
$s\in\mathbb{C}\setminus\{0\},t\in\mathbb{C}$, after taking conjugations.
Now define the matrix $R$ by
$R=W X-Y W$
where
$W=X Y^{-1} X^{-1} Y$.
The equation $R=\begin{pmatrix}0 & 0\\ 0& 0\end{pmatrix}$
induces a system of defining equations of the space of conjugacy classes
of $\mathit{SL}(2;\mathbb{C})$-representations of $\pi_1 E(K)$.
By direct computations, we have only one equation
\[
f(s,t)=3-\frac{1}{s^2}-s^2+3 t-\frac{t}{s^2}-s^2 t+t^2=0
\]
from $R=\begin{pmatrix} 0 & 0\\0 & 0\end{pmatrix}$.
Solve $f(s,t)=0$ in $t$,
\[
t=\frac{1-3 s^2+s^4 \pm \sqrt{1-2 s^2-s^4-2 s^6+s^8}}{2 s^2}.
\]
Hence the parameter $t$ can be eliminated
by substituting
\[
t_\pm=\frac{1-3 s^2+s^4 \pm \sqrt{1-2 s^2-s^4-2 s^6+s^8}}{2 s^2}.
\]
Under $f(s,t)=0$,
there are four choices on a pair of $(X,Y)$ as follows.
\begin{enumerate}
\item
$
X=
\begin{pmatrix} s & 1 \\ 0 & 1/s
\end{pmatrix},
Y=
\begin{pmatrix} s & 0 \\ -t_+ & 1/s
\end{pmatrix}$
\item
$
X=\begin{pmatrix} s & 1 \\ 0 & 1/s
\end{pmatrix},
Y=\begin{pmatrix} s & 0 \\ -t_- & 1/s
\end{pmatrix}$,
\item
$
X=\begin{pmatrix} 1/s & 1 \\ 0 & s
\end{pmatrix},
Y=\begin{pmatrix} 1/s & 0 \\ -t_+ & s
\end{pmatrix}$ ,
\item
$
X=\begin{pmatrix} 1/s & 1 \\ 0 & s
\end{pmatrix},
Y=\begin{pmatrix} 1/s & 0 \\ -t_- & s
\end{pmatrix}$.
\end{enumerate}
\begin{lemma}
Among the above 4 pairs of $(X,Y)$,
$(1)$ and $(3)$ give the same conjugacy class and $(2)$ and $(4)$ also gives the same one.
These two classes are not same.
\end{lemma}
\begin{proof}
Let us consider the trace of $XY^{-1}$:
$\mathrm{tr}(XY^{-1})=2-t$.
Then by elementary arguments in linear algebras,
one can see the above.
\end{proof}
Solve the inequality
\[
1-2 s^2-s^4-2 s^6+s^8>0
\]
under the condition $s\neq 0$
in the real numbers,
one has
\[
\begin{split}
& s\leq -\sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)},
-\sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)}\leq s<0,\\
&0<s\leq \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)},
\sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)}\leq s
\end{split}
\]
and
numerically
\[
s\leq -1.61803, -0.618034\leq s<0,0<s\leq 0.618034, 1.61803\leq s.
\]
If $s$ belongs to the above intervals,
then the corresponding $(s,t)$ gives an irreducible representation of $\pi_1 E(K)$ in $\mathit{SL}(2;\mathbb{R})$.
Here the matrix corresponding to a longitude is given by
\[
L=W^{-1}\widetilde{W}=X^{-1} Y X Y^{-1}X^{-1} Y X Y^{-1}.
\]
By direct computations, each entry $L_{ij}(1\leq i,j\leq 2)$ of $L$ is given as follows:
\[
\begin{split}
L_{11}
&=
-1+\frac{1}{2 s^4}-\frac{1}{2 s^2}-\frac{s^2}{2}+\frac{s^4}{2}\pm\frac{1}{2} \sqrt{1-2 s^2-s^4-2 s^6+s^8}\\
& +\frac{\sqrt{1-2 s^2-s^4-2 s^6+s^8}}{2s^4},
\\
L_{12}
&=\pm\frac{\sqrt{1-2 s^2-s^4-2 s^6+s^8}}{s^3}\pm\frac{\sqrt{1-2 s^2-s^4-2 s^6+s^8}}{s},
\\
L_{21}
&=0,
\\
L_{22}
&=-1+\frac{1}{2 s^4}-\frac{1}{2 s^2}-\frac{s^2}{2}+\frac{s^4}{2}\mp\frac{1}{2} \sqrt{1-2 s^2-s^4-2s^6+s^8}\\
& -\frac{\sqrt{1-2 s^2-s^4-2 s^6+s^8}}{2s^4}.
\end{split}
\]
Here the double-sign corresponds in the same order, which is depended on the choice of $t$.
\begin{remark}
Remark that $\mathrm{tr}(L)$ is not depended on the choice of $t$.
\end{remark}
We consider a $1/n$-Dehn-surgery along the figure-eight knot.
Because we do not consider the 3-sphere, then we assume that $n\neq 0$.
Note that $X$ is corresponding to a meridian.
Then we compute the relation as
\[
D=X-L^{-n}=(D_{ij}).
\]
\begin{remark}
It holds that $D_{21}=0$ identically,
because $X$ and $L$ are upper triangular matrices in $\mathit{SL}(2;\mathbb{C})$.
\end{remark}
Any solution of the systems of equations
$D_{11}=D_{12}=D_{22}=0$
gives a conjugacy class of $\mathit{SL}(2;\mathbb{C})$-representation.
Because $X$ and $L$ are upper triangular matrices,
then it holds that $D_{11}=0$ for some $s\in\mathbb{C}$ if and only if $D_{22}=0$ for the same $s\in\mathbb{C}$.
Hence we consider system of two equations $D_{11}=D_{12}=0$.
\begin{remark}
For any solution $s_0\in\mathbb{C}\setminus\{0\}$,
then the complex conjugate of $s_0$ is also a solution.
Because the complex conjugate $\bar{\rho}$ of $\rho$ is alway a representation for any $\rho$.
If $\rho$ given by $s_0$ is an conjugate to a representation in $\mathit{SU}(2)$,
then $\rho$ and $\bar{\rho}$ are conjugate each other.
Because $\bar{s_0}=1/s_0$.
\end{remark}
We consider Reidemeister torsion $\tau_\rho(M_n)$
for $M_n$ with
$\rho:\pi_1(M_n)\rightarrow \mathit{SL}(2;\mathbb{C})$.
For the precise definition of Reidemeister torsion $\tau_\rho(M)$
for an $\mathit{SL}(2;\mathbb{C})$-representation $\rho$,
please see \cite{Kitano94-1, Kitano94-2, Porti15-1}.
In the case of $1/n$-surgeries along the figure-eight knot,
we obtain the following formula
of Reidemeister torsion in terms of the trace of the meridian image.
\begin{proposition}[Kitano\cite{Kitano15-2}]
Assume $n\neq 0$.
If $\rho:\pi_1(M_n)\rightarrow \mathit{SL}(2;\mathbb{C})$ is an acyclic representation,
then one has
\[
\tau_\rho(M_n)
=\frac{2(u-1)}{u^2(u^2-5)}\in\mathbb{C}\setminus\{0\}
\]
where $u=\mathrm{tr}(X)=s+1/s$.
\end{proposition}
Here $\rho$ is called to be an acyclic representation
if the chain complex of $M$ with $\mathbb{C}^2_\rho$-coefficients is an acyclic chain complex.
By numerical computations we compute $\tau_\rho(M_n)$ and $\sigma_{M_n}(t)$
by using this formula.
Here we mention the Casson invariant and the $\mathit{SL}(;\mathbb{C})$-Casson invariant.
Please see \cite{Akbult-McCarthy} and \cite{Curtis, Boden-Curtis06, Boden-Curtis12}
for precise definitions and properties.
In 1980's
Casson defined the Casson invariant $\lambda(M)\in\mathbb{Z}$
for a homology 3-sphere $M$
as the half of algebraic count of conjugacy classes of irreducible $\mathit{SU}(2)$-representations.
In 2000's
Curtis~\cite{Curtis} defined
the $\mathit{SL}(2;\mathbb{C})$-Casson invariant $\lambda_{\mathit{SL}(2;\mathbb{C})}(M)\in\mathbb{Z}$
for M
by counting conjugacy classes of irreducible $\mathit{SL}(2;\mathbb{C})$-representations.
In the case of $1/n$-surgeries along the figure-eight knot,
one has the following by applying general formula by Casson, and the one by Boden and Curtis.
\begin{proposition}
\noindent
\begin{itemize}
\item
$\lambda(M_n)=-n.$
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_n)=4n-1.$
\end{itemize}
\end{proposition}
\begin{remark}
For any positive $n$,
the above proposition implies that the number of conjugacy classes of $\mathit{SU}(2)$-representations is algebraically
$2n$
and
the one of conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations is $4n-1$.
\end{remark}
\section{Computation}
Here we show computations from $n=1,\dots,10$ by using Mathematica.
We make a list of the values of $s$, $u=s+1/s$ and $\tau_{\rho}$.
We remark the followings.
\begin{itemize}
\item
We choice the value $s$ in $|s|\leq 1$.
Because the inverse $1/s$ can be done if $|s|>1$.
\item
For a representation which is conjugate to the one in $\mathit{SU}(2)$,
we choice only one value and omit its complex conjugate.
\end{itemize}
\subsection{Summary}
We compare computations with the values of Casson invariant $\lambda(M)$ and the $\mathit{SL}(2;\mathbb{C})$-Casson invariant
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M)$.
For any cases, we could find numerically $2|\lambda(M)|$ conjugacy classes in $\mathit{SU}(2)$
and $|\lambda_{\mathit{SL}(2;\mathbb{C})}(M)|$ conjugacy classes in $\mathit{SL}(2;\mathbb{C})$.
Further we could also do only one $\mathit{SL}(2;\mathbb{R})$-representation.
We also compute torsion polynomials $\sigma_{M_n}(t)$.
We simply write $\sigma_n(t)$ to $\sigma_{M_n}(t)$.
From the definition,
it is a polynomial over $\mathbb{Q}$,
because $\tau_\rho(M)$ is an algebraic number for $[\rho]\in\mathcal{R}'(M)$.
All previous examples in \cite{Kitano15-2,Kitano16-1} are polynomials over $\mathbb{Z}$.
To compute torsion polynomials,
values of imaginary parts which are sufficiently small
as compared with their real parts are regarded as 0.
Because theoretically we can see a torsion polynomial is a polynomial over $\mathbb{Q}$.
\begin{remark}
In the case of a torus knot,
there is a 3-term relation among $\sigma_{n+1}(t),\sigma_{n}(t),\sigma_{n-1}(t)$.
However we see that there is not such a relation in the figure-eight knot case,
as $\sigma_{-1}(t)=\sigma_1(t),\sigma_0(t)=1$.
\end{remark}
\subsection{The case of $n= 1$}
The first example $M_1$ is the Briskorn homology 3-sphere $\Sigma(2,3,7)$,
which is not a hyperbolic manifold.
Now one has
\begin{itemize}
\item
$\lambda(M_1)=-1$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_1)=3$ .
\end{itemize}
We find 2 conjugacy classes of $\mathit{SU}(2)$-representations
and totally 3 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
But the third one that does not come from $\mathit{SU}(2)$-representations is an $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|} \hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ & $-0.400969+0.916092 i$ & $-0.801938$ & $1.28621$ \\ \hline
$\circ$ &$ 0.277479\, +0.960732 i$ & $0.554958$ & $0.615957$\\ \hline
& $0.611406$ & $2.24698$ & $10.0978$\\ \hline
\end{tabular}
\end{table}
In this case the torsion polynomial is given as
\[
\sigma_1(t)=t^3-12 t^2+20 t-8.
\]
This computation coincides with the one in \cite{Kitano15-2}.
\subsection{The case of $n=2$}
The next $M_2$ is a hyperbolic homology 3-sphere.
One has
\begin{itemize}
\item
$\lambda(M_2)=-2$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_2)=7$ .
\end{itemize}
In this case we find 4 conjugacy classes of $\mathit{SU}(2)$-representations
and totally 7 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
The last representation is an $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|} \hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ & $-0.423608+ 0.905845 i$ & $-0.847217$ & $1.20196$ \\ \hline
$\circ$ & $\ -0.194046 + 0.980992 i$ & $-0.388092$ & $3.80096$ \\ \hline
$\circ$ & $\ \ \ 0.156335\, + 0.987704 i$ & $\ 0.31267$ & $2.86834$ \\ \hline
$\circ$ & $\ 0.476693\, + 0.87907 i$ & $\ 0.953386$ & $0.0250713$ \\ \hline
& $-0.69314\pm 0.0194149 i$ & $-2.13472 \mp 0.0209638 i$ & $2.98853 \pm 0.563052 i$ \\ \hline
& $0.61642$ & 2.23869 & 42.1266 \\ \hline
\end{tabular}
\end{table}
The torsion polynomial is given by
\[
\sigma_2(t)
=t^7-56 t^6 +660 t^5 -3384 t^4+ 8720 t^3 -11008 t^2 +5376 t -128 .
\]
\subsection{The case of $n=3$}
Next one has
\begin{itemize}
\item
$\lambda(M_3)=-3$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_3)=11$.
\end{itemize}
In this case we find 6 conjugacy classes of $\mathit{SU}(2)$-representations
and totally 11 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
The last representation is an $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|}\hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ & $-0.489756+ 0.87186 i$ & $-0.979511$ & $1.02124$ \\ \hline
$\circ$ & $-0.31143+ 0.950269 i$ & $-0.622859$ & $1.814$ \\ \hline
$\circ$ & $-0.125581+ 0.992083 i $ & $-0.251162$ & $8.03486$ \\ \hline
$\circ$ & $0.108419\, + 0.994105 i$ & $0.216838$ & $6.72584$ \\ \hline
$\circ$ & $0.352641\, + 0.935759 i$ & $0.705282$ & $0.263178$ \\ \hline
$\circ$ & $0.462835\, + 0.886444 i$ & $0.92567$ & $0.0418747$ \\ \hline
& $-0.649243\pm 0.009109 i$ & $-2.18919 - 0.0124968 i$ & $6.0064 + 1.53513 i$ \\ \hline
& $0.762562\, \pm 0.0145425 i$ & $2.07345 - 0.010457 i$ & $-0.709789 + 0.0436681 i$ \\ \hline
& $0.61732$ & $2.23723$ & $95.5058$ \\ \hline
\end{tabular}
\end{table}
The torsion polynomial is given by
\[
\begin{split}
\sigma_3(t)
=&t^{11} -124 t^{10}+3036 t^9 -31696 t^8 +161024 t^7 -364128 t^6\\
& +152640 t^5
+426752 t^4 -262144 t^3 -142336 t^2 +55296 t-2048.
\end{split}
\]
\subsection{The case of $n=4$}
Next one has
\begin{itemize}
\item
$\lambda(M_4)=-4$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_4)=15$.
\end{itemize}
In this case we find 8 conjugacy classes of $\mathit{SU}(2)$-representations
and totally 15 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
The last representation is an $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|}\hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ & $-0.478316+0.878188 i$ & $-0.956632$ & $1.04682$ \\ \hline
$\circ$ & $-0.413993+ 0.91028 i$ & $-0.827986$ & $1.23604 $ \\ \hline
$\circ$ &$-0.242737+0.970092 i$ & $-0.485475$ & $2.64583$ \\ \hline
$\circ$ & $-0.0926466+ 0.995699 i$ & $-0.185293$ & $13.9046$ \\ \hline
$\circ$ &$0.0829193\, +0.996556 i$& $0.165839$ & $12.1993$ \\ \hline
$\circ$ & $0.268854\, + 0.963181 i$ & $0.537707$ & $0.67882$ \\ \hline
$\circ$ & $0.382257\, + 0.924056 i$ & $0.764514$ & $0.182491$ \\ \hline
$\circ$ & $0.49426\, + 0.869314 i$ & $0.98852$ & $0.00584088$ \\ \hline
& $-0.808705\pm 0.0102842 i$ & $-2.04505 \mp 0.00543826 i$ & $1.77945 \pm 0.0421084 i$ \\ \hline
& $-0.635184\pm 0.00517794 i$ & $-2.20943 \mp 0.00765508 i$ & $10.2737 \pm 2.88241 i$ \\ \hline
& $0.693187\, \pm 0.00964411 i$ & $2.13552 \mp 0.0104227 i$ & $-1.12125 \pm 0.112867 i$ \\ \hline
& $0.617633$ & 2.23672 & 170.236 \\ \hline
\end{tabular}
\end{table}
\[
\begin{split}
\sigma_4(t)
&=
t^{15}-224 t^{14} + 10320 t^{13}- 211776 t^{12}+ 2.2964 \times 10^6 t^{11} - 1.35709 \times 10^7 t^{10}\\
&+ 4.11722 \times 10^7t^9- 4.96721 \times 10^7 t^8- 3.55295 \times 10^7 t^7 + 1.56351 \times 10^8 t^6 \\
&- 1.13653 \times 10^8 t^5 - 5.89578 \times 10^7 t^4 + 1.15933 \times 10^8 t^3 - 5.0004 \times 10^7 t^2\\
&+5.89824 \times 10^6 t -32768
\end{split}
\]
\subsection{The case of $n=5$}
Next one has
\begin{itemize}
\item
$\lambda(M_5)=-5$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_5)=19$.
\end{itemize}
In this case we find 10 conjugacy classes of $\mathit{SU}(2)$-representations
and totally 19 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
The last representation is an $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|} \hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ & $-0.496333+ 0.868132 i$ & $-0.992666$ & $1.00743$ \\ \hline
$\circ$ & $ -0.420075+0.907489 i $ & $-0.840151$ & $1.21421$ \\ \hline
$\circ$ & $-0.343785+ 0.939049 i$ & -0.687569 & $1.57697$ \\ \hline
$\circ$ & $-0.19818+0.980166 i$ & $-0.396359$ & $3.67066$ \\ \hline
$\circ$ & $-0.073359+ 0.997306 i$ & $-0.146718$ & $21.4005$ \\ \hline
$\circ$ & $ 0.0671124\, +0.997745 i $ & $0.13422$ & $19.2915$ \\ \hline
$\circ$ & $0.215697\, + 0.97646 i$ & $0.431395$ & $1.26939$ \\ \hline
$\circ$ & $ 0.319062\, +0.947734 i $ & $0.638124$ & $0.386993$ \\ \hline
$\circ$ & $0.44386\, + 0.896096 i$ & $0.88772$ & $0.0676542$ \\ \hline
$\circ$ & $ 0.485869\, +0.874031 i$ & $0.971739$ & $0.0147589$ \\ \hline
& $-0.628893\pm 0.00332617 i$ & $-2.21894 \mp 0.00508349 i$ & $15.7706 \pm 4.61071 i$ \\ \hline
& $-0.731102\pm 0.00855695 i$ & $-2.09871 \mp 0.00744981 i$ & $2.35697 \pm 0.11268 i$ \\ \hline
& $0.840595\, \pm 0.00745097 i$ & $2.03014 \mp 0.00309303 i$ & $-0.568873 \pm 0.00810636 i$\\ \hline
& $0.664373\, \pm 0.00643176 i$ & $2.16941 \mp 0.00813843i$ & $-1.66792 \pm 0.199595 i$ \\ \hline
& $0.617778$ & $2.23648$ & $266.318$ \\ \hline
\end{tabular}
\end{table}
The torsion polynomial is given by
\[
\begin{split}
\sigma_5(t)
&=
t^{19}- 348 t^{18}+ 24428 t^{17}- 756768 t^{16}+ 1.22252 \times 10^7 t^{15} - 1.05049 \times 10^8 t^{14}\\
&
+ 4.39482 \times 10^8 t^{13} - 6.05556 \times 10^8 t^{12}- 1.45911 \times 10^9 t^{11} + 5.73225 \times 10^9 t^{10}\\
&- 3.56966 \times 10^9 t^9 - 9.32096 \times 10^9 t^8
+ 1.48101 \times 10^{10} t^7 - 1.9304 \times 10^9 t^6 \\
& - 7.91541 \times 10^9 t^5 +3.44198 \times 10^9 t^4+ 1.05592 \times 10^9 t^3 - 6.28883 \times 10^8 t^2\\
&+ 4.45645 \times 10^7 t
-524288
\end{split}
\]
\subsection{The case of $n=6$}
One has
\begin{itemize}
\item
$\lambda(M_6)=-6$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_6)=23$.
\end{itemize}
In this case we find 12 conjugacy classes of $\mathit{SU}(2)$-representations
and totally 23 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
The last representation is an $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|}\hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ & $ -0.490087+0.871673 i$ & $-0.980174$ & $1.02053$\\ \hline
$\circ$ & $-0.460559+ 0.887629 i$ & $-0.921118$ & $1.0908$ \\ \hline
$\circ$ & $ -0.366341+0.93048 i $ & $-0.732683$ & 1.44635 \\ \hline
$\circ$ & $-0.290911+ 0.95675 i$ & -0.581822 & $2.00486$ \\ \hline
$\circ$ &$ -0.167221+0.985919 i$ & $-0.334442$ & $4.8814$ \\ \hline
$\circ$ & $-0.0607063+ 0.998156 i$ & $-0.121413$ & $30.5197$ \\ \hline
$\circ$ & $ 0.0563605\, +0.99841 i$ & $0.112721$ & $28.0037$ \\ \hline
$\circ$ & $0.179677\, + 0.983726 i$ & $0.359355$ & $2.03702$ \\ \hline
$\circ$ & $0.272216\, +0.962236 i$ & $0.544432$ & $0.653532$ \\ \hline
$\circ$ & $0.387688\, + 0.921791 i$ & $0.775376$ & $0.169874$ \\ \hline
$\circ$ & $ 0.442457\, +0.89679 i$ & $0.884914$ & $0.0697032$ \\ \hline
$\circ$ & $0.497456\, + 0.867489 i$ & $0.994912$ & $0.00256369$ \\ \hline
& $-0.625531\pm 0.00231384 i$ & $-2.22415 \mp 0.00359944 i$ & $22.4926 \pm 6.72158 i$ \\ \hline
& $-0.693197\, \pm 0.00642162 i$ & $-2.13566 \mp 0.00694108 i$ & $3.11879 \pm 0.197207 i$ \\ \hline
& $-0.863685\pm 0.00558142 i$ & $-2.02147 \mp 0.00190054 i$ & $1.61843 \pm 0.0115852 i$ \\ \hline
& $0.762163\, \pm 0.00726544 i$ & $2.0741 \mp 0.00524079 i$ & $-0.714588 \pm 0.0221273 i$ \\ \hline
& $0.64957\, \pm 0.00453793 i$ & $2.18897 \mp 0.00621643 i$ & $-2.34127 \pm 0.304635 i$ \\ \hline
& $0.617856$ & 2.23636 & 383.752 \\ \hline
\end{tabular}
\end{table}
The torsion polynomial is given by
\[
\begin{split}
\sigma_6(t)
&=
t^{23}- 504 t^{22}+ 52020 t^{21} - 2.40364 \times 10^6 t^{20} + 5.93684 \times 10^7 t^{19}\\
&- 8.20377 \times 10^8 t^{18}+6.23643 \times 10^9 t^{17}- 2.42844 \times 10^{10} t^{16} + 2.55758 \times 10^{10} t^{15}\\
& + 1.64156 \times 10^{11} t^{14} - 6.99939 \times 10^{11} t^{13}+ 8.04666 \times 10^{11} t^{12}+ 1.36418 \times 10^{12} t^{11}\\
&- 5.35777 \times 10^{12} t^{10}+ 6.06942 \times 10^{12} t^9 - 6.38688 \times 10^{11} t^8 - 6.38688 \times 10^{11} t^8\\
&- 4.81682 \times 10^{12} t^7 + 4.04757 \times 10^{12} t^6 - 2.98072 \times 10^{11} t^5 - 1.0991 \times 10^{12} t^4\\
&+ 5.29833 \times 10^{11}t^3 - 7.9675 \times 10^{10} t^2 + 3.47288 \times 10^9 t
-8.38861 \times 10^6 .
\end{split}
\]
\subsection{The case of $n=7$}
First one has $\lambda(M_7)=-7$, $\lambda_{\mathit{SL}(2;\mathbb{C})}(M_7)=27$.
In this case we find 14 conjugacy classes of $\mathit{SU}(2)$-representations like previous examples.
We do more 13 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
Among them we do only one $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|} \hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ & $-0.498132 + 0.867101 i$ & $-0.996264$ & $1.00376$ \\ \hline
$\circ$ & $-0.456717+ 0.889612 i$ & $-0.913434$ & $1.10105$ \\ \hline
$\circ$ & $-0.415538 + 0.909576 i$ & $-0.831076$ & 1.2304 \\ \hline
$\circ$ & $-0.322425+0.946595 i$ & $-0.644849$ & $1.725754883584166$ \\ \hline
$\circ$ & $-0.251181 + 0.96794 i$ & $-0.502362$ & $2.50781$ \\ \hline
$\circ$ & $-0.144537+ 0.989499 i$ & $-0.289074$ & $6.27537$ \\ \hline
$\circ$ & $-0.0517713 + 0.998659 i$ & $-0.103543$ & $41.2613$ \\ \hline
$\circ$ & $0.0485751\, + 0.99882 i$ & $0.0971502$ & $38.3362$ \\ \hline
$\circ$ & $0.153815\, + 0.9881 i$ & $0.30763$ & $2.98291$ \\ \hline
$\circ$ & $0.236804\, + 0.971558 i$ & $0.473608$ & $0.982802$ \\ \hline
$\circ$ & $0.340083\, + 0.940396 i$ & $0.680166$ & $0.304734$ \\ \hline
$\circ$ & $0.397537\, + 0.917586 i$ & $0.795074$ & $0.148438$ \\ \hline
$\circ$ & $0.470809\, + 0.882235 i$ & $0.941618$ & $0.0320157$ \\ \hline
$\circ$ & $0.492671\, + 0.870215 i$ & $0.985343$ & $0.00749368$ \\ \hline
& $-0.623523\pm 0.001701i$ & $-2.22730105 \mp 0.002675 i$ & $30.4388 \pm 9.21537 i$ \\ \hline
& $-0.671806\pm 0.00487214 i$ & $-2.16025 \mp 0.00592043 i$ & $4.04034 \pm 0.29538 i$ \\ \hline
& $-0.787517 \pm 0.00610802 i$ & $-2.05725 \mp 0.00374012 i$ & $1.88117 \pm 0.0331674 i$ \\ \hline
& $0.881081\, \pm 0.00431162 i$ & $2.01602 \mp 0.00124229 i$ & $-0.534338 \pm 0.00285536 i$ \\ \hline
& $0.719217\, \pm 0.00597955 i$ & $2.10952 \mp 0.00557931 i$ & $-0.905083 \pm 0.0385034 i$ \\ \hline
& $1.560218\pm 0.008182 i$ & $2.201129\pm 0.0048120i$ & $-3.13933 \mp 0.427727 I$ \\ \hline
& $0.617903$ & $2.23628$ & $521.407$ \\ \hline
\end{tabular}
\end{table}
Now we can see that
\[
\begin{split}
\sigma_7(t)
&=t^{27} -684 t^{26}+94916 t^{25}-5.8661\times 10^6 t^{24}+1.92341\times 10^8 t^{23}-3.348194\times 10^9 t^{22}\\
&+3.38294\times 10^{10} t^{21} -1.60441\times 10^{11} t^{20} +1.21022\times 10^{11} t^{19}+ 2.40426\times 10^{12} t^{18}\\
&-1.09427 \times 10^{13} t^{17}
+ 1.08363 \times 10^{13} t^{16} +5.40471\times 10^{13} t^{15}-1.97613\times 10^{14} \times 10^7 t^{14} \\
&+2.18273 \times 10^{14} t^{13}
+1.14878 \times 10^{14} t^{12} -4.92745 \times 10^{14} t^{11}+3.10448 \times 10^{14} t^{10}\\
&+2.31893 \times 10^{14} t^9
-3.52574\times 10^{14} t^8+4.61951\times 10^{13} t^7 +1.12401 \times 10^{14} t^6\\
& -4.44561 \times 10^{13} t^5-8.4299 \times 10^{12} t^4+6.16006\times 10^{12} t^3\\
& -7.89939\times 10^{11} t^2 +2.34881\times 10^{10} t ^1 -1.34218\times 10^8.
\end{split}
\]
\subsection{The case of $n=8$}
In the next case, one has
\begin{itemize}
\item
$\lambda(M_8)=-8$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_8)=31$.
\end{itemize}
In this case we find 16 conjugacy classes of $\mathit{SU}(2)$-representations
and totally 31 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
The last representation is an $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|}\hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ & $-0.494366+0.869254 i$ & $-0.988732$ & $1.011492842$\\ \hline
$\circ$ &$ -0.47754+ 0.87861 i$ & $-0.95508$ & $1.048631249$ \\ \hline
$\circ$ & $-0.419132+0.907925 i$ & $-0.838263$ & $1.217529795$\\ \hline
$\circ$ & $ -0.37393+ 0.927457 i$ & $-0.747861$ & $1.407484041$\\ \hline
$\circ$ & $ -0.28699+0.957934 i$ & $-0.573979$ & $2.045827739$\\ \hline
$\circ$ & $-0.220616+ 0.975361 i$ & $-0.441231$ & $3.081126266$ \\ \hline
$\circ$ & $ -0.127229+0.991873 i$ &$-0.254459$ & $7.85131636$\\ \hline
$\circ$ & $-0.0451269+ 0.998981 i $ & $-0.0902538$ & $53.6247182$\\ \hline
$\circ$ & $ 0.042678\, +0.999089 i$ & $0.0853561$ & $50.2893844$\\ \hline
$\circ$ &$0.134395\, + 0.990928 i$ & $0.268789$ & $4.107704808$ \\ \hline
$\circ$ & $0.209299\, +0.977852 i$ & $0.418599$ & $1.375414532$\\ \hline
$\circ$ &$0.301435\, + 0.953487 i$ & $0.60287$ & $0.471324314$ \\ \hline
$\circ$ & $0.357963\, +0.933736 i$ & $0.715926$ & $0.247016780$\\ \hline
$\circ$ &$0.434268\, + 0.900784 i $ & $0.868535$ & $0.0820955924$ \\ \hline
$\circ$ & $0.466323\, +0.884614 i$ & $0.932646$ & $0.0374963906$\\ \hline
$\circ$ &$0.49857\, + 0.866849 i $ & $0.997141$ & $0.0014358346$ \\ \hline
& $ -0.622227\pm 0.00130353 i$ & $-2.22935\mp 0.0020633 i$ & $39.60771 \pm 12.09283 i$\\ \hline
& $ -0.658499\pm 0.0037911 i$ & $-2.17705\mp 0.00495151 i$ & $5.1134622 \pm 0.4078733 i$\\ \hline
& $-0.742103\pm 0.00541306 i$ & $-2.08955\mp 0.00441555 i$ & $2.2312104 \pm 0.05871724 i$\\ \hline
& $-0.894615\pm 0.00341956 i$ & $-2.0124\mp 0.000853032 i$& $1.56555866 \pm 0.00477245 i$\\ \hline
& $0.808382\, \pm 0.0051452 i $ &$2.04537\, \mp 0.002728 i$ & $-0.61198103 \pm 0.008329177 i$\\ \hline
& $0.693201\, \pm 0.00481418 i$ & $2.13571\, \mp 0.00520388 i$& $-1.1321183 \pm 0.05702512 i$ \\ \hline
& $0.635422\, \pm 0.00258348 i$ & $2.20915\, \mp 0.00381495 i$ & $-4.060877 \pm 0.5708212 i$\\ \hline
& $0.617934$ & $2.23623$ & $682.674$\\ \hline
\end{tabular}
\end{table}
The torsion polynomial is given by the following.
\[
\begin{split}
\sigma_8(t)
&=
t^{31}-896 t^{30}+164160 t^{29}-1.34889\times
10^7 t^{28}+5.9497\times 10^8 t^{27}-1.48208\times
10^{10} t^{26}\\
&+2.0792\times 10^{11} t^{25}-1.5975\times
10^{12} t^{24}+5.113\times 10^{12} t^{23}+1.609\times
10^{13} t^{22}\\
&-2.199\times 10^{14} t^{21}+7.892\times
10^{14} t^{20}-2.13\times 10^{14} t^{19}-8.29\times
10^{15} t^{18}\\
&+3.230\times 10^{16} t^{17} -5.36\times
10^{16} t^{16}+1.31\times 10^{16} t^{15}+1.119\times
10^{17} t^{14}\\
&-1.979\times 10^{17} t^{13}+8.35\times
10^{16} t^{12}+1.488\times 10^{17} t^{11}-2.212\times
10^{17} t^{10}\\
&+7.22\times 10^{16} t^9+7.11\times
10^{16} t^8-7.047\times 10^{16} t^7+1.329\times
10^{16} t^6+9.403\times 10^{15} t^5\\
&-5.4486\times
10^{15} t^4+1.03031\times 10^{15} t^3-7.2572\times
10^{13} t^2\\
&+1.59773\times 10^{12} t-2.14748\times
10^9
\end{split}
\]
\subsection{The case of $n=9$}
First one has
\begin{itemize}
\item
$\lambda(M_9)=-9$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_9)=35$.
\end{itemize}
Here we find 18 conjugacy classes of $\mathit{SU}(2)$-representations
and totally 35 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
The last one is also a $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|} \hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ &$-0.498871+ 0.866676 i $ & $-0.9977413535$ & $1.002267596$\\ \hline
$\circ$ &$ -0.473084+0.881017 i$ & $-0.946168$ & $1.059217610 $ \\ \hline
$\circ$ &$-0.447444+ 0.894312 i$ & $-0.894888$ & $1.126969070$ \\ \hline
$\circ$ &$ -0.38399+0.923337 i$ & $-0.76798$ & $1.359403531$\\ \hline
$\circ$ &$-0.338051+ 0.941128 i$ & $-0.676101$ & $1.614264278$\\ \hline
$\circ$ &$ -0.258146+0.966106 i$ & $-0.516291$ & $2.403508579$ \\ \hline
$\circ$ &$-0.196502+ 0.980503 i$ & $-0.393003$ & $3.722604536 $\\ \hline
$\circ$ &$ -0.113601+0.993526 i$ & $-0.227202$ & $9.60858961$\\ \hline
$\circ$ &$-0.0399928+ 0.9992 i$ & $-0.0799857$ & $67.6097739$\\ \hline
$\circ$ &$ 0.038057\, +0.999276 i $ & $0.0761141$ & $63.8633816$\\ \hline
$\circ$ &$ 0.119296\, + 0.992859 i$ & $0.238591$ & $5.41179913$ \\ \hline
$\circ$ & $ 0.187397\, +0.982284 i$ & $0.374793 $ & $1.831789427$ \\ \hline
$\circ$ &$ 0.270042\, + 0.962849 i$ & $0.540083 $ & $0.669766151$\\ \hline
$\circ$ &$ 0.324317\, +0.945948 i$ & $0.648634$ & $0.364749065$ \\ \hline
$\circ$ &$0.398186\, + 0.917305 i$ & $0.796372$ & $0.1470861889$\\ \hline
$\circ$ &$ 0.434658\, +0.900595 i$ & $0.869317 $ & $0.0814873088$ \\ \hline
$\circ$ &$0.482193\, + 0.876065 i$ & $0.964385$ & $0.0188178181$\\ \hline
$\circ$ &$ 0.495536\, +0.868587 i$ & $0.991072$ & $0.0045248460$ \\ \hline
& $ -0.621342 \pm 0.00103034 i$ & $-2.230757 \mp 0.001638478 i$ & $49.99999 \pm 15.35354 i$ \\ \hline
& $-0.649630 \pm 0.00302325 i$ & $-2.18893 \mp 0.00414036 i$ & $6.335051 \pm 0.5346564 i$\\ \hline
& $-0.712996 + 0.00458062 i$ & $-2.115470751 \mp 0.004429539 i$ & $2.6499001 \pm 0.08729179 i$\\ \hline
& $ -0.825748\pm 0.00436242 i$ & $-2.03674 \mp 0.00203523 i$ & $1.71886598 \pm 0.01444799 i$ \\ \hline
& $ 0.905426\, \pm 0.00277285 i$ & $2.00987 \mp 0.000609481 i$ & $-0.52058388 \pm 0.001326408 i$\\ \hline
& $0.76209 \pm 0.00484289 i$ & $2.074218042 \mp 0.003495356 i$ & $-0.71548730 \pm 0.014788047 i$ \\ \hline
& $0.676202 \pm 0.00390501 i$ & $2.155000405 \mp 0.004634916 i$ & $-1.3928905 \pm 0.07775814 i$\\ \hline
& $0.631699 \pm 0.00204673 i$ & $2.214715401 \mp 0.003082315 i$ & $-5.106041 \pm 0.7321823 i$\\ \hline
&$ 0.6179549394855183$ & $2.236195908$ & $864.16$\\ \hline
\end{tabular}
\end{table}
The torsion polynomial is given by
\[
\begin{split}
\sigma_9(t)
&=
t^{35}-1132 t^{34}+260580t ^{33}-2.68147\times 10^7 t^{32}+1.47405\times 10^9 t^{31}\\
& -4.5380\times 10^{10} t^{30}+7.7496\times 10^{11} t^{29}-7.064\times 10^{12} t^{28}+2.387\times 10^{13} t^{27}\\
&+1.509\times 10^{14} t^{26}-1.968\times 10^{15} t^{25}+7.51\times 10^{15} t^{24}+4.22\times 10^{15} t^{23}\\
&-1.523\times 10^{17} t^{22}+6.24\times 10^{17} t^{21}-1.058\times 10^{18} t^{20}-2.9\times 10^{17} t^{19}\\
&+4.70\times 10^{18} t^{18}- 7.64\times 10^{18} t^{17}+ 5.\times 10^{17} t^{16}+1.39\times 10^{19} t^{15}\\
&-1.62\times 10^{19} t^{14}- 1.3\times 10^{18} t^{13} +1.61\times 10^{19} t^{12}- 9.5\times 10^{18} t^{11}\\
&-4.19\times 10^{18} t^{10}+ 6.28\times 10^{18} t^9-9.4\times 10^{17} t^8-1.364\times 10^{18} t^7\\
&+ 5.645\times 10^{17} t^6+4.88\times 10^{16} t^5- 6.157\times 10^{16} t^4+1.06336\times 10^{16} t^3\\
&-6.2396\times 10^{14} t^2+ 1.02048\times 10^{13} t-3.4360\times
10^{10}
\end{split}
\]
\subsection{The case of $n=10$}
One has
\begin{itemize}
\item
$\lambda(M_{10})=-10$,
\item
$\lambda_{\mathit{SL}(2;\mathbb{C})}(M_9)=39$.
\end{itemize}
In this case we find 20 conjugacy classes of $\mathit{SU}(2)$-representations.
We do totally 39 conjugacy classes of $\mathit{SL}(2;\mathbb{C})$-representations.
The last one is an $\mathit{SL}(2;\mathbb{R})$-representation.
\begin{table}[htb]
\begin{tabular}{|c|l|l|l|}\hline
$\mathit{SU}(2)$ & $s$ & $u=s+1/s$ & $\tau_\rho$ \\ \hline
$\circ$ &$ -0.496377+0.868107 i$ & $-0.992753$ & $1.007339322 $\\ \hline
$\circ$ &$ -0.48554+ 0.874215 i$ & $-0.971079$ & $1.030431375$\\ \hline
$\circ$ &$ -0.446171+0.894948 i$ & $-0.892342$ & $1.130661131$\\ \hline
$\circ$ & $-0.416138+ 0.909302 i$ & $-0.832276$ & $1.228229667$\\ \hline
$\circ$ &$ -0.352798+0.9357 i$ & $-0.705596$ & $1.521863518$\\ \hline
$\circ$ & $ -0.307592+ 0.951518 i $ &$-0.615185$ & $1.846942653$\\ \hline
$\circ$ &$ -0.234348+0.972153 i$ & $-0.468696$ & $2.797187148$\\ \hline
$\circ$ &$ -0.177045+ 0.984203 i$ & $-0.35409$ & $4.43107072$\\ \hline
$\circ$ &$ -0.102597+0.994723 i$ & $-0.205194$ & $11.54680784$\\ \hline
$\circ$ &$ -0.035907+ 0.999355 i$ & $-0.0718141$ & $83.2162915$\\ \hline
$\circ$ &$ 0.0343385\, +0.99941 i$ & $0.068677$ & $79.0582830$\\ \hline
$\circ$ &$ 0.107229\, + 0.994234 i$ & $0.214457$ & $6.89542715$ \\ \hline
$\circ$ &$ 0.169577\, +0.985517 i$ & $0.339153$ & $2.352207686$\\ \hline
$\circ$ & $0.244255\, + 0.969711 i$ & $0.488511$ & $0.900301681$\\ \hline
$\circ$ &$ 0.295848\, +0.955235 i$& $0.591697$ & $0.501615796$\\ \hline
$\circ$ &$0.365525\, + 0.930802 i$ &$0.73105$ & $0.225388067$ \\ \hline
$\circ$ &$ 0.403589\, +0.91494 i$ & $0.807179 $ & $0.1361160029$\\ \hline
$\circ$ &$0.457052\, + 0.88944 i$ &$0.914104 $ & $0.0493695567$ \\ \hline
$\circ$ &$ 0.478014\, +0.878352 i$ & $0.956027 $ & $0.0235490782$\\ \hline
$\circ$ &$0.499085\, + 0.866553 i$ & $0.998171$ & $0.0009170979$\\ \hline
& $ -0.62071\pm 0.000834778 i$ & $-2.23176 \mp 0.00133189 i$ & $61.6152 \pm 18.99773 i$ \\ \hline
&$ -0.643412\pm 0.00246298 i$ & $-2.1976 \mp 0.00348646 i$ & $7.703480 \pm 0.6759767 i$ \\ \hline
&$ -0.693202\pm 0.00385058 i$ & $-2.13574 \mp 0.00416238 i$ & $3.1296446 \pm 0.11879778 i$\\ \hline
&$ -0.779549\pm 0.00431503 i$ & $-2.0623 \mp 0.00278537 i$& $1.9275656 \pm 0.02619426 i$\\ \hline
&$ -0.914251\pm 0.00229081 i$ & $-2.00804 \mp 0.000449854 i$ & $1.54165231 \pm 0.00241772 i$\\ \hline
&$0.84037\, \pm 0.00372875 i$ & $2.0303 \mp 0.001551 i$& $-0.56939143 \pm 0.004072021 i$\\ \hline
&$ 0.731022\, \pm 0.00427557 i$ & $2.09892 \mp 0.00372495 i$ & $-0.83854998 + 0.02192047 i$\\ \hline
&$ 0.664454\, \pm 0.0032119 i$ & $2.16941 \mp 0.00406293 i$ & $-1.6862274 \pm 0.10076482 i$\\ \hline
& $0.62906 \pm 0.00166081 i$ & $2.218722556 \mp 0.002536113 i$ & $-6.274506 \pm 0.9124470 i$\\ \hline
&$0.61797$ & $2.23617$ & $1067.00$\\ \hline
\end{tabular}
\end{table}
In this case
we see that the torsion polynomial is given by
\[
\begin{split}
\sigma_{10}(t)
&=
t^{39}- 1400 t^{38}+400500 t^{37}-5.1428\times 10^7 t^{36}+3.5518\times 10^9 t^{35}\\
&-1.39149\times 10^{11} t^{34}+3.1030\times 10^{12} t^{33}-3.8975\times 10^{13} t^{32}+2.301\times 10^{14} t^{31}\\
&+4.36\times 10^{14} t^{30}-1.779\times 10^{16} t^{29}+1.244\times 10^{17} t^{28}-2.49\times 10^{17} t^{27}\\
&-1.99\times 10^{18} t^{26}+1.807\times 10^{19} t^{25}-6.86\times 10^{19}t^{24}+1.25\times 10^{20} t^{23}\\
&+2.6\times 10^{19} t^{22}-7.2\times 10^{20} t^{21}+1.64\times 10^{21} t^{20}-1.09\times 10^{21} t^{19}\\
&-2.45\times 10^{21} t^{18}+6.5\times 10^{21} t^{17}-5.1\times 10^{21} t^{16}-2.8\times 10^{21} t^{15}\\
&+8.8\times 10^{21} t^{14}-5.6\times 10^{21} t^{13}-2.00\times 10^{21} t^{12}+4.88\times 10^{21} t^{11}\\
&-2.09\times 10^{21} t^{10}-7.32\times 10^{20} t^9+9.85\times 10^{20} t^8-2.549\times 10^{20} t^7\\
&-6.95\times 10^{19} t^6+5.507\times 10^{19} t^5-1.2432\times 10^{19} t^4\\
&+1.21881\times 10^{18} t^3-4.8826\times 10^{16} t^2+6.4321\times 10^{14} t-5.4976\times 10^{11}
\end{split}
\]
\section{problem}
In \cite{Kitano15-2,Kitano16-1},
the torsion polynomial for a Brieskorn homology 3-sphere
obtained by surgeries along a torus knot,
which is not exactly same with the one given in this paper,
it can described by using Chebyshev polynomials of the first kind.
It seems that it is natural, because any value of $\tau_\rho$ is given by some special values of the cosine function.
In the case of the figure-eight knots, or in more general cases of hyperbolic knots,
how to treat the torsion polynomial, it is a problem.
\vskip 0.5cm
\textit{Acknowledgments}.\/
This study was starting while the author were visiting
Aix-Marseille University from September, 2015 to March, 2016.
The author would like to express their sincere thanks for Michel Boileau and Luisa Paoluzzi.
He also thanks Joan Porti for pointing some errors in the first one.
This research was supported in part by JSPS KAKENHI 25400101 and 16K05161.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,213
|
# PRAISE FOR KIM PAFFENROTH'S
" _Dying to Live_ is not just a thinking man's horror novel, it's a zombie book for philosophers. There's plenty of action—and we enter the story while it's already in gear—and we get inside the head and heart of a moral man trying to understand the cosmic implications of the apocalypse."
—Jonathan Maberry, author of _Ghost Road Blues_
"Kim Paffenroth put me on the edge of my seat from the opening scene of this apocalyptic thriller and never let up. His prose is gritty and tough, as horrific and as multilayered as a Brueghel painting, but it is always intelligent, always with something insightful to say about the human condition. _Dying to Live_ is a truly powerful achievement. Don't miss it."
—Joe McKinney, author of _Dead City_
"Kim Paffenroth writes with passion, bringing a human element to a world of the inhuman. His love of the zombie genre is matched only by his insight in posing philosophical questions of those surviving the apocalypse. Intelligent and never boring, _Dying to Live_ is as good as the zombie genre gets."
—Scott A. Johnson, author of _Deadlands_
"A grave new world... a startlingly original vision of the direction of the whole human race.... This is as bloody, violent and intense as it gets. An intelligent novel that will make you think and make you squirm with disgust in equal measure."
—David Moody, author of the _Autumn_ series
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Dedicated to St. Augustine
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of the dark side of human nature
_O death, where is thy victory?_
1 Corinthians 15:55
_To sue to live, I find I seek to die,
And seeking death, find life. Let it come on._
Shakespeare, _Measure for Measure_ 3.1.42–43
##
I AWOKE TO find a lone zombie underneath my little hideaway. The tree house I had spent the night in was poorly constructed—the bottom was just a square of plywood, reinforced with a couple boards, with plywood walls on three sides and the fourth one open. It had no roof, but the sky was clear, so no bother. All the pieces were irregular and unpainted, with big gaps between them in many spots, and the walls were only between two and three feet high. But it was higher up than most, a good twelve feet off the ground (the kid's mom must've been one of the ones we always called a "cool mom," to allow such a dangerous playhouse), so I was even more surprised to see my unwanted visitor.
I scanned the surrounding field and trees and saw that the zombie and I were alone; my heart slowed down. In a few moments, my situation had gone from peaceful morning reverie, to possible or near-certain death, to minor inconvenience. In that respect, this was a typical morning.
Tree houses, and any other little platform above the ground, were my favorite places to catch a couple hours of sleep at night as I made my way across the country. Going inside a building required a careful search, and later on, as you tried to sleep, you'd start to worry that maybe you had missed some hiding place, from which the real Boogie Man, who doesn't need sleep, would rise up during the night. And building the necessary barricades on the doors and windows often made so much noise you could end up with a growing crowd of the undead, whose moaning and clawing at the doors would probably keep you up, on top of the danger they would pose when you tried to leave your shelter in the morning. Unless you were in a group, a building was not a good choice for your little motel in hell.
Little platforms above ground, on the other hand, were ideal. Not comfortable, but ideal. You usually had to lash yourself to them so you wouldn't fall off in the night, and you almost always had to sleep sitting up, but that was nothing for a few blessed hours of relative peace of mind. The undead are by nature incurious and almost never look up, so the chances of being spotted once you were in your little eyrie were low. For exactly the same reason that hunters once used them, back when humans were the hunters rather than the hunted, your scent wouldn't usually carry down to the creatures below, either. The tree houses always made me a little sad, 'cause they reminded me of my kids, but what could you do? All in all, my little sky boxes were the best places I had found to spend the night, so long as the living dead were afoot. But best, of course, had never been the same as perfect, and that was infinitely more true now.
One reason the zombie and I were alone this morning was that it lacked the ability to make sound. Like so many of its kind, its throat was torn open, leaving its windpipe a ragged hole, and the front of its suit stained brown with blood.
It looked up at me with its listless, cloudy eyes that lacked all expression—not hatred, not evil, not even hunger, just blanks. It was chilling in its own way, like the stare of a snake or an insect. Its look would never change, whether you drove a spike through its head, or it sank its yellow teeth into your soft, warm flesh; it lacked all capacity to be afraid, or to be satisfied. Its mouth, however, had a great deal more bestial expression to it, for it was wide open, almost gnawing at the bark of the tree as it clawed upward.
I stood looking down at it for a few moments. It was times like this—and there had been several in the last few months—that I had always wished that I smoked. In a few seconds, I would fight this thing and one or both of us would cease to exist—"die" is obviously the wrong word here—and just to stand here and contemplate that inevitability cried out for some distraction, some mindless and sensual habit like smoking, to make it less horrible. I guess I could've chewed gum, but that would make the whole scene ridiculous, when it was really as serious, overwhelming, and sad as any that had ever occurred to a man.
With nothing to distract me, I just felt the full weight of a terrible and necessary task, and the tediousness and unfairness of it. I had just awakened from a relatively peaceful sleep, and I already felt a crushing weariness coming over me. Again, it was developing into a pretty typical morning.
People had come up with lots of names for the walking dead in the preceding months. While we weren't fighting them off or running like hell, we usually came up with humorous labels. "Meat puppets" was a popular one. Somebody came up with "Jacks and Janes," like they were just some annoying neighbors from the next circle of hell, or as a variation on "Jack-offs."
Sometimes, when they'd get especially noisy and rambunctious, but didn't pose any immediate threat, we'd call them "the natives," as in "the natives are restless." Maybe that was a little racist, I don't know. "Walking stiffs" was pretty accurate. But mostly we'd go for the tried and true—zombies. That's what they were, and we'd always be one breath away from becoming one—a mindless, shambling bag of flesh.
My zombie this morning looked to have been a middle-aged man in its human life, slightly graying, average build. Its suit was intact, and other than its throat wound, there were no signs of further fights with humans or other zombies. Decay had taken its toll, and it looked more desiccated than gooey, a brittle husk rather than the dripping bag of pus that some of them became.
At first, I looked it over to size up its threat and plan my attack, but that quickly turned into contemplating its human existence. Maybe his kids had built the tree house, and that's why he'd been hanging around here, almost as if he were protecting it, or waiting for them to come back. Or even worse, maybe his kids had been the ones to tear out his throat, when he had rushed home in the midst of the outbreak, hoping against hope they were still okay. Or, just as bad, maybe he'd been bitten at work or on the way home, only to break in to his own house and kill his kids.
My mind reeled, and I clutched the wall of the tree house. I'd heard of soldiers in other wars having a "thousand yard stare," a blank look that signaled they were giving in to the hopelessness and horror around them, soon to be dead or insane. As for me, I was suffering the thousand yard stare of the war with the undead: once you contemplated the zombies as human beings, once you thought of them as having kids and lives and loves and worries and hopes and fears, you might as well just put your gun in your mouth and be done with it right then, because you were losing it—fast. But, God knows, if you never looked at them that way, if they were just meat puppets whose heads exploded in your rifle's sights, then hopefully somebody would put a bullet in your brain, because you had become more monstrous than any zombie ever could be.
I shook myself free of my paralysis. I'm not exactly sure why, but I wasn't ready to give up yet. I tossed my backpack beyond where the zombie stood. It turned to see where it landed, then immediately looked back up at me. Its head lolled from side to side, and I was again glad that it couldn't vocalize, as it was clearly getting worked up and would've been making quite a racket if it could.
You never used a gun if you didn't have to, for its noise brought lots of unwanted attention, so I pulled out a knife, the one I carried with a long, thin blade, like a bayonet, as that would work best. I stood at the edge of the plywood platform. "I'm sorry," I said, looking right in the zombie's eyes. "Maybe somewhere, deep down, you still understand: I'm sorry."
I took a step forward and started to fall. I tried to hit it on the shoulder with my right foot, but its arms were flailing about, and my boot hit its left wrist, sliding along its arm. I sprawled to the right and then rolled away as the zombie was shoved into the tree.
As it turned to face me, I scrambled up, took a step forward, and drove the knife into its left eye. Its hands flailed about, either to attack me or to ward off the blow. The blade was long and thin enough that it went almost to the back of its skull. The whole attack was noiseless, without so much as the sound of a squish or a glitch as the blade slid through its eyeball and brain.
As I drew the blade out, I grabbed the zombie by the hair and shoved it downward to the side, where it fell to the ground and lay motionless.
And that was that. Like everyone, I always used to imagine deadly fights would be much more dramatic. But in my experience, there were seldom any Chuck Norris flying, spinning kicks, or any _Matrix_ -style running up the wall while firing two guns on full auto. If anyone's ever around to make movies about the wars against the undead, maybe there will be such moves in them, I don't know. But usually, like this morning, there were just a couple of savage, clumsy blows, and it was over.
I was barely breathing at all, let alone breathing hard, the way I felt someone should when they kill something that was somehow, in some small way, still human. A few months ago, I would've at least felt nauseous, but not anymore. Looking down at the creature from the tree house had been much more traumatic than delivering the killing blow.
I bent down over my would-be killer and cleaned the blade on his suit jacket. I then reached into his pocket. It was a little ritual I still followed when I could, though the horrible exigencies of a zombie-infested world usually made it impossible. I pulled out his wallet and got out his driver's license. Rather than look at the bloody horror at my feet, with its one undead eye and one bloody, vacant socket, I stared at his driver's license picture—smiling, happy, alive, years and decades of life ahead of him. I cleared my throat to speak clearly. "I have killed Daniel Gerard. I hope he's somewhere better now."
I cast the wallet and license on top of his motionless body, scooped up my backpack, and hurried away.
It had been close to a year since all the worst parts of the Bible started coming true. Armageddon. Apocalypse. The End of Days. God's righteous judgment on a sinful humanity. Whatever the self-righteous jerk who railed at you once a week from a pulpit used to call it. Well, he might have been self-righteous and a jerk, and now he was probably lurching around like most everyone else, drooling on himself with half his face torn off, but it sure seemed as though he had had some inside information that we all wish we'd gotten a little sooner.
For most people, I assume it started like every other day. Brush your teeth. Kiss your spouse without any feeling. Go to work. Grab whatever it is you grab to eat on your way to work. Eat it, not really noticing or enjoying it. But then at some point that blessed, kind, comforting routine goes horribly awry and someone—maybe your neighbor, or coworker, or worse, your kids or your spouse—staggers up to you with a blank look and tries to tear your throat out with his teeth. If he gets you, then you don't have to worry anymore, because you'll be dead, and then you'll get up and wander around like him, with no more thoughts or feelings, just shuffling around trying to bite people. If you get away from him, then you'd be one of the survivors, at least for a little while, and then you'd have lots of worries, and your only feeling would be fear. Either way—welcome to hell.
Theological assessments aside, the automatic assumption was that the dead were rising and killing because of some infection, and the infection was spread by their bites. The next logical assumption—since there was not much reliable evidence of zombie infestations before the 21st century (horror movies notwithstanding)—was that we had tinkered with viruses and DNA and had brought all this shit on ourselves.
Here again, a theological assessment was hard to avoid. We had created a hell on earth through our own arrogance and ignorance, and now we were reaping what we had sown—with a vengeance. Worse than any couple who ate an apple or any bozo who slapped some brick and mortar on the Tower of Babel, we'd messed with The Man's prerogatives, and either He'd given us the biggest damn smack down of all time, or we'd just set off something that only He could control. Shit, you didn't need to believe in the Bible to see how much sense it made. You remember that crazy Greek myth you read about in fifth grade—Pandora's box. Same damn thing. A box full of walking cannibal corpses who wouldn't let you close it once it got open.
Now, how that box got opened, that was a hot topic of debate among survivors, when we weren't fighting to prolong our miserable existence and could afford the luxury of conversation or discussion. Outright warfare or a terrorist attack was probably the least popular theory, though it had vigorous proponents. I don't know why more of us didn't subscribe to that hypothesis. I suppose it's funny to say, but I think we didn't because it was the least comforting of all our speculations. It was too horrible to imagine that even terrorists could unleash the hellish plague of undeath on the whole world, even their own people, including women, children, and the elderly. And at the same time, the theory made it too pat and simple, like it was just some crazies who did this, some tiny band of malcontents—horror of this magnitude seemed to require a more powerful, far-reaching source.
That's probably why more people bought into the paranoid, conspiratorial theory that the disease had been released by our government or someone else's as a horribly botched attempt to test it on a real population. Proponents of this theory almost delighted in its vindication of every real and imagined form of government-sponsored terror, from Andersonville to Tuskegee to Gitmo to putting fluoride in the tap water. Their tales could almost be the bedtime stories of the apocalypse, lulling us to sleep with some tiny and bizarre shred of hope that even now, the world made some weird kind of sense, that undeath was not a new and incomprehensible kind of evil, but just a continuation of this world's madness and brutality, like Jackie scrambling on the trunk of the Lincoln to grab the big chunk of her husband's head that was sliding around back there, or like bulldozers pushing mountains of emaciated bodies into pits in Dachau. Strange comfort, that, but it was often all we had.
But the most popular theory—the one I personally advocated, though without much conviction—was simply that there had been a horrible accident. Nothing malevolent or calculated, just plain old human error. Somebody dropped a test tube somewhere. A lab monkey bit through somebody's glove. The kind of thing that happens a thousand times a day for thousands of days with no fatal outcome. It was the most blackly humorous theory, I suppose, for it made the misery and violent deaths of billions of people just the result of a stupid mistake, but it had its own cold comfort. If all this was just some blunder, then maybe, if we could ever shoot every zombie in the head—the only way we had found to kill them permanently—or if they would just eventually rot and fall apart—what everyone had hoped for initially—then we could go back to life like it used to be. We weren't evil, just stupid and clumsy. Like poor Pandora.
That's how some of us theorized that it had begun. But whatever had happened—and I've left out the more exotic theories, like an extra-terrestrial source of infection—we ended up in the same place. Almost one year after the first corpse rose, the world was ruled by the undead, who wandered about with no discernible goal other than to kill and eat living people. The undead were everywhere, the new dominant species that took the place of the old, extinct one. Places where there had been large human populations were especially thick with the walking dead, though they never took any notice of one another.
The living, meanwhile, as was their wont, almost always congregated in little groups. The government or society or culture had imploded or disintegrated with terrifying speed as the infection spread. Within hours, there had been no telephone service, no police or rescue response to the terrified calls for help. Within days, there was no power or television. And within weeks, the last organized military and government resistance collapsed, at least in the U.S.
But groups of survivors quickly came together into little groups, little communities with a pecking order and rules and authority, but also some of the little perks of being around other people—companionship, conversation, sex, someone to hold your hand when you die, someone to put a bullet in your brain when you went to get back up as a zombie. (And if you've ever seen a zombie—and God love you, I hope you haven't, but if you're reading this, I suspect you have—then you know that last perk is by no means the least important one.) You didn't have to be a damned philosopher to know that we're social animals, and would be till the last zombie bit the last human and dragged us all down to hell, which, judging by the zombies, looked like it was going to be the most unsociable place imaginable.
Yes, humans always build their little communities in order to survive, and in order to make surviving a little more bearable. Except me. I was alone. And it sucked. It was dangerous and it sucked.
By midday, I was moving closer to what looked like a small-sized city. I had thrown my maps away a few days ago when I had failed to find my family. After that, I figured, I didn't have much need for maps: if I didn't have any place to go anymore—and I had decided that I didn't—what difference did it make where I was at the moment? Besides, the end of civilization had wreaked a lot of havoc with the things depicted on maps: I guess the rivers and mountains were still there, but cities were gone, roads were clogged with wrecked cars, bridges and tunnels and dams had been blown up to try to stop the rampaging hordes of the undead. So long as I was out of reach of those things, and had one bullet for myself if it came to that, I was in about the best location I could hope for.
It was a late spring day, bursting with a sunshine that didn't make it hot, but just made things seem better, brighter, more alive than they were on other days. I still had the instinct to call it beautiful as I looked around and forgot the obvious shortcomings of the day for a moment. One shortcoming I couldn't forget, however, was the gnawing hunger I felt.
Never one for breakfast, I had definitely been put off from eating anything this morning after killing Daniel Gerard, a man who, after all, had only been looking for something to eat, just as I was. I had some supplies in my backpack, but if I was near an area where I could forage for more and conserve what I had, that would be the much wiser course.
The undead weren't exactly afraid of sunlight—they weren't afraid of anything—but they did seem to avoid it unless aroused and provoked. Maybe it hurt their skin or eyes, or maybe they could sense that it was speeding their decay and that brought them some discomfort. Whatever it was, during bright daylight, you could walk through places where the walking dead were nearby without immediately attracting a crowd, so long as you were quiet and downwind. Still, I never went too far into an urban area. Right now, I just wanted to find some food and get back out to the sticks before nightfall.
From what I'd seen, many cities had burned more or less to the ground, once fire crews were no longer there to put out the inevitable fires. But here, for whatever reason of wind or rain or luck, many buildings were still standing. Some were gutted or damaged by fire, and all had the usual marks of looting, ransacking, and the final, desperate battles between the living and the dead. There were few unbroken windows.
In the street, wrecked or abandoned cars were everywhere. There were a few bodies and pieces of bodies in extremely advanced stages of decay, and paper and dead leaves rustled about on a light breeze.
The sight of the burnt-out remains of a city was almost as overwhelmingly depressing as the human wrecks that wandered everywhere as zombies: this place should be bustling and alive, and instead it was—quite literally—a graveyard.
I always wondered why there weren't more animals around now, since zombies didn't eat them, but everywhere I went, it always seemed like there were even fewer animals than when people had ruled the earth. I almost never heard a bird sing. I seldom saw pigeons or squirrels. It was almost as though even the animals fled from such horror, fled when the ruler of the animal kingdom died, and left the king's mausoleum in peace, until it could completely crumble away and they could reclaim it after a suitable mourning period. I know it seems almost delusional in its anthropomorphism, but sometimes you can't help thinking like that when you're alone in these dead places.
I checked the remains of a couple stores, barely venturing inside the darkened buildings, for fear of the dead hiding in ambush. The inventories of a clothing store and a jewelry store were barely touched: it was funny how quickly things had been re-prioritized in the final, chaotic days of the human race.
I looked at what appeared to be hundreds of thousands of dollars of diamonds, now mixed in with the smashed glass of the cases that had once displayed them: both sparkled in the sun, but their value had been radically and traumatically equalized a few months ago. I imagined that during last winter—the first winter of a world that would now remain more or less dead in every season—the snow too had sparkled just as brightly when it blew in and covered the diamonds that, in better times, would've adorned hundreds of brides.
A quick look into a liquor store revealed much less remaining stock—human nature and appetites being what they are—but there was a bottle of some bad bourbon just a few feet inside the door, so I reached in and grabbed it. I didn't know when I'd be able to drop my guard enough to partake, but since I wasn't carrying that much, it made sense to take it.
I knew I was getting too far into the dead city, but on the next street was a convenience store where there might be food. It was facing perpendicularly from the stores I had examined, so at least it would be brighter inside. The big front windows were still intact, but the glass of the front door had been smashed. Looking up and down the street and still seeing no movement, I went inside.
I was looking for snack cakes. When the final crisis of humanity had begun, people had instinctively stocked up on canned food: I guess Spam is forever etched in our collective consciousness as the foodstuff of the apocalypse. People at first had bought up everything canned, and then, within just a couple days, as cash became utterly worthless and stores weren't even open, the stronger smashed and grabbed from the weaker. I had never seen a can of food in a store since I had started foraging: you could only find cans in people's houses, and even then they were getting pretty rare at this point. So, for now, snack cakes were the way to go. What I would do when those finally went bad and the last few cans ran out—that was a question still a few months off, and therefore way beyond any reasonable contingency plans.
I don't know if all the old urban legends that Twinkies and those pink Snow Ball cakes could survive a nuclear explosion were true, but they and their kind definitely had a shelf life well over a year, if the box wasn't opened and you weren't fussy, which I clearly wasn't at this point.
There was a treasure trove of them in the second aisle into the store, and I smiled when I saw there were no chocolate ones: I guessed some priorities remained effective right up till the last gasp of humanity. I made my way quietly to them, tore open the boxes, shoveled a bunch of the wrapped ones into my backpack, and proceeded to gorge myself on what I couldn't carry. I was licking white créme filling off my fingers when I heard the crunch of a shoe stepping on broken glass.
##
THE ZOMBIE WAS about twelve feet away from me, at the end of the Twinkie aisle. It was staggering toward me with the usual slow, stiff motions of the undead. It had been a teenage girl, blonde and pretty, as far as I could tell now, wearing her boyfriend's high school letter jacket, way too big for her. Its mouth moved noiselessly, except for the clacking of her bloody, yellow teeth.
The jacket was open, and the lower half of her t-shirt was flayed and soaked with blood, which also had soaked her jeans down past her knees. Her abdomen was torn wide open in a wound about a foot wide. They'd ripped all her organs out when they killed her. She wasn't moaning now the way zombies usually did, because she didn't have lungs. You could see right through to her ribs and spine, not glistening and drippy the way a wound on a living person would be, but dark and dry and caked, like mummies I'd seen in museums.
It was coming closer, slowly but inexorably, but I couldn't look away from that horrible tribute to mortality and incarnation. You saw all kinds of wounds on the living dead, but some still commanded shock, almost a reverential awe at the miracle of life and the horrible mystery of death. Partly you were aghast at the perverse will to go on "living," despite terrible mutilation and decay: why couldn't it just lie down and die? Just rest, damn it, and stop struggling. Ashes to ashes, dust to dust—that was how it was supposed to be, and this was some hideous violation of nature.
But mostly, you couldn't help the pity that spasmed up from your own gut, putting a lump in your throat, at how awful and degrading and unfair the person's death must have been. People—even young and pretty ones—died in car crashes, or from diseases, or in war, or from horrible crimes, and their young, healthy bodies might even be mutilated and disfigured. Such deaths were hard enough to take without anger and despair. But no one was supposed to be gutted like a fish, butchered like an animal, and left to dry out like a damn piece of jerky. You might see shit go down most every day, but if you were going to go on living, you had to know, deep down, that some things were still just plain wrong, and you could still let out a primal scream against them as some kind of evil abomination. And what I was staring at in that convenience store, on a glorious spring day, licking sweet white créme off my fingers, was as wrong as anything ever could be.
There was that damn thousand yard stare again, closing me off, tunneling my vision and lulling me to just let go.
To my left, something roared, and I turned. Over the shelves, which were about at chin level, like they usually were in convenience stores, I could see what could only be described as a hairless bear, its arms out in front, Frankenstein-like, lurching toward me. I swear the thing looked like it had been a professional wrestler in its human life—probably 350 pounds, almost a head taller than me, covered in tattoos, though its flesh was now a mottled gray that obscured much of the artwork.
It crashed into the shelves, tipping them over onto me and the other zombie. I was pushed back and pinned against the opposite shelves as the monster scrabbled at my face with its foul nails; the shelving unit kept it from getting closer. The girl zombie wasn't pinned as tightly as I was, so she was still slowly working her way toward me, teeth clacking.
The top shelf was pressing into my upper chest and arms, making it hard to breathe, as well as almost impossible to get to a weapon, and even if I did, I wouldn't be able to bring it up to eye level to get a head shot at either of them. I had no leverage to push the shelving unit off me, and I wasn't sure I could do it anyway, as the zombie was so much bigger than me.
I struggled and drew my .357 magnum from the holster in the small of my back. I'd have to shoot from the hip. I fired, and my ears started ringing from the roar of the magnum. The plate glass window behind the big zombie shattered as the bullet went through his torso. A bullet anywhere other than through the brain won't put a zombie down, but this one made it stagger back just enough for me to push the shelving unit off of me.
The zombie lunged again—I stuck the barrel in its face and fired. Its arms shot up as it spun around and dropped on its face, the back of its head blown off.
I turned as the girl zombie grabbed my shoulder. This was it. Another second and she'd sink her teeth into me—then it wouldn't really matter if I shot her or not: the bite would kill me and turn me into a zombie in a matter of hours or days.
I grabbed her hair, wrenched myself free from her grip, and shoved the barrel of the gun under her chin. I yanked her head down and to the left, so she wasn't looking at me. "I'm sorry," I rasped as I pulled the trigger. Her brains were blasted out in a gray slop all over the ceiling surveillance camera and the cigarette display case above the counter.
I shoved her away from me, and she fell on her back with a cracking sound. I was panting and drenched in sweat. I grabbed my backpack and looked down at her one more time. Thankfully, her long hair covered her face. I pulled the flap of the too-large jacket across her belly. What a world, where that'd be considered an unusually kind gesture, covering up the magnificent corpse I had made out of what had been a ninety-five pound girl.
As I stood up, I heard the moaning underneath the ringing in my ears, and I suddenly felt icy cold. One zombie was already stepping through the broken window. At least ten were closing in on the shattered storefront, and I knew there were dozens more nearby, and hundreds more behind them.
It was getting seriously close to being time for me to eat a bullet. At least then I might get to see God and ask Him what all this shit had been about. Some days, like the days when I blew a teenager's brains all over the ceiling, I wouldn't even mind meeting the _other_ guy. At least with Old Nick, it seemed like you knew better where you stood.
I made my way to the back exit of the store, holstering the magnum, shouldering my backpack, and drawing my Glock. The magnum was wild overkill against zombies anyway, and with three of its six shots fired and no time to reload, the seventeen-round magazine of the 9mm would increase what little chance I had.
I was only a few steps ahead of the growing horde of zombies filling the store. If I couldn't get the back door open, or if there were more outside when I opened it, it was all over. You couldn't risk a close-in fight with them: they might grab your gun arm and make it impossible to shoot yourself.
The closest zombie was at the end of the hallway that led to the backdoor, maybe fifteen feet away from me. Several were right behind it, and more were shuffling in steadily—old women in housecoats, men in suits, young people in shorts, men and women in aprons or uniforms. Most were white, while several were black, Hispanic, or Asian. Normally, they would've staggered around without even noticing each other, but their hunger had united them in a way that would've been quite remarkable in life. The human species had finally overcome racism. Too bad we had to give up our intellect and turn into mindless cannibals to do so. The plausibility of the whole apocalypse/judgment thing occurred to me again as I turned away from them and grabbed the door handle.
The door was a big, heavy metal one. That was a huge bonus for me, as was the fact that it opened inward, though for the undead, these two facts slightly lessened their chance for lunch that day. Before the first zombie could figure out to push down on the thumb latch and pull the handle toward itself, the others would have pressed up against him and mashed him against the door in a writhing, moaning mass. Then the only way they were coming through the door would be when enough zombies in the back of the mob lost interest and wandered off, so that the pressure was taken off the front zombie and he could pull the door back. Given their monomania and their inability ever to get bored or distracted, that could take hours, if not days.
I squeezed the handle and yanked back on it. I couldn't afford to examine the alley behind the store before I went outside: so long as a bony hand didn't grab me immediately, I was going out that door.
No bony hand.
I stepped through the door and closed it.
With my left hand, I drew my knife—not the thin-bladed eye-poking one this time, but the big Crocodile Dundee-type one, the kind you could use to hack off a grasping hand, or bash-in the side of a zombie's head with the pommel. Within seconds, I heard the thumping of the dead assaulting the door from within, but as I had suspected, there was no sign that the door was opening.
At the end of the alley, several zombies were staggering toward me, and they let out a moan which would surely bring more. I had no choice but to go the other way, though this would probably take me farther into the city, which could be even worse. Again, there wasn't much choice. I ran that way till I reached the next cross street.
Zombies this time were everywhere, though there were definitely more to the left, closing in on where I had originally fired the shots. The farther I could get from that zombie magnet, the better my chances got, especially if I could do it without firing more shots.
I turned right and began running down that street. I dodged between the scattered undead, only once getting close enough to actually fight one off. It was an older woman, and it came around the front of a van that was up on the sidewalk as I ran between the vehicle and the building. The hair was matted to the left side of her head with blood from where her ear had been bitten off.
Her left arm reached out for me, clutching, even though much of the flesh of her forearm had been torn off, so much so that you could see the bones and tendons in her forearm moving back and forth. Her soulless moan sounded the alarm to any other zombies nearby.
"Die, bitch!" I growled as I brought my left hand up as hard as I could, driving the blade up under her chin until the tip of the blade shattered through the decayed top of her skull. I quickly drew the blade out and let her fall. For the first time that day, I felt exhilarated, and I almost wanted to spit on her body. I shivered at my reaction. Like the thousand yard stare, if you got the bloodlust, your chances of survival went down, because it made you careless and foolhardy. I wanted to get out of that town and to somewhere relatively safe before I descended further into that or some other species of madness.
I was making good and uneventful progress, not running too fast, conserving my strength, and not taking anymore shots that would draw more zombies. For almost a block, I was able to jump from the top of one wrecked car to another to avoid the grasping dead.
As I came over a rise, the street descended slightly to end in a cross street, beyond which was a park on the banks of a fairly large river. On the other side of the river looked to be a continuation of the park, and then lower buildings, not like the small downtown district I was in at the moment. The bridge across the river was one block to the left. All I had to do was run there, across the bridge, and I would be outside the city proper, on my way to the suburbs.
But as soon as I turned left, something moaned behind me. On the cross street that paralleled the river, at least a hundred zombies were heading my way.
I needed to get way ahead of them before nightfall, but that in itself was not a huge problem. Zombie top speed seemed to be about two miles per hour, so even a brisk walk—so long as you didn't get hung up with more obstacles—meant you could pull very far ahead of them in a short time, easily out of eyeshot. They weren't herd creatures by nature, which maybe says something about people—I don't know. They'd all follow the same goal, which was always the same: find someone to kill and eat. But they were never really a herd, much less a pack; they were just separate individuals who happened to be going in the same direction at the same time. And once the mob didn't see you, it would start to disperse. So as terrifying as a crowd of a hundred zombies looks, if you keep moving, it's not nearly as dangerous as a small crowd in an enclosed space, like I had just faced in the convenience store.
I kept running and made it to the bridge. It was a broad, low bridge, with four lanes plus a sidewalk on each side. At my end, a barricade had been built: two Humvees, parked perpendicularly across the roadway, supplemented with some cop cars, sand bags, concrete traffic barriers, and barbwire. It may well have held, for whatever good it had done, as the vehicles still effectively blocked the bridge. They did not appear to have been moved from their original spot, nor was there any sign of fire or explosion, common at such scenes.
The machine guns had been removed from the Humvees. It was probably just as well, in case I felt tempted to play Rambo and try mowing down the pursuing mob of zombies with a full auto weapon, a tactic that clearly hadn't worked for whoever had built the barricade, guys who probably had a lot more training than I did. For me, it would've been briefly satisfying on one level, but far more dangerous than simply retreating, as the noise would attract more and more of them from both sides of the river.
As usual when you came across a battle site, there weren't many bodies lying around, as most had gotten up and walked away, but there were a few scattered before the barricade, most in civilian garb, with a couple in military and police uniforms. There was no smell of decay, beyond the usual in a city of the dead, as the bodies had—unlike zombies—almost completely rotted away.
As was also usual at a battle site—I suppose from any war, but the war with the undead was the only one I knew—it was impossible to guess the details of what had gone on here: how many had fought, or died, or even whether the barricade had been intended to keep the undead on this side of the river, or to keep them from coming over from the other side.
Well, it all seemed pretty moot now. It was just a few vehicles and lifeless bodies, with weeds growing up through the cracks around them. It wasn't like there were going to be any people to make a monument here, like it was some kind of Gettysburg or Normandy. Just one of probably ten thousand places where the human race had just puttered out. In a few years, it'd be like finding the campsite and spearheads of some Neanderthals, the odd and poorly designed remnants of some species that didn't have what it takes to survive.
I looked back as I climbed over the barricade. Although, for the long haul, the dead seemed well-suited to survival, at the moment, they were falling behind me. The roadblock would probably slow them down enough that, by the time some of them made it over, I'd be way out of sight and they'd sit down on the bridge and forget all about me.
I ran across the bridge. The wrecked vehicles made it impossible to see all the way to the other side, but there were no signs of zombies anywhere, and I almost started to relax. I looked down at the water, crystal clear and fast moving from the center of the channel to the far bank, shallower nearer the side I had left. I dodged past a few more vehicles and I was to the opposite side of the bridge. I could no longer see the barricade, but I was sure the dead had not surmounted that yet.
To my right was the park I had seen from the other side of the river, but to my left was a parking lot, beyond which was a high, brick wall, brightly painted. It ran from the river, along the parking lot, to where it connected to a large, irregularly-shaped brick building, maybe four stories tall. In the wall facing the parking lot, there was a large metal gate, while along the wall was spray painted "R U DYING 2 LIVE?" I wished I had time to ponder that, as it had been some time since I had someone other than myself to pose abstract questions, but there was an obvious impediment to such philosophizing—the crowd of zombies, probably almost two hundred, that was crowded in front of the wall, pressing against it.
They hadn't seen me yet. They were pretty intent on the wall, and they must've been for some time, as they weren't moaning or agitated, but just kind of milling around.
The street on this side was not as clogged with wrecked cars, so I couldn't dodge between them and hope to remain unseen. I would be running along an empty street, less than fifty yards from them. Still, if I just started running, I'd be in no worse situation than I was with the previous mob: I just had to keep running for long enough that I was out of their sight, then keep going till I was in a safer area. It was either that, or jump off the bridge into the river. Although the fall looked survivable, the chance of spraining an ankle, losing all my supplies and equipment, and coming up somewhere downstream that was just as bad made me think that it was not the better option.
I set off at a good sprint, trying to get as far down the street as I could before they started pursuit. Sure enough, after just a few yards, the moan started, and the chase was on. They turned, almost as a group, and begin staggering toward me. I kept running. But then another group emerged from a grove of trees and from behind a building in the park. It was just a dozen or so, nowhere near as big a mob as the zombies at the wall, but with just a few lurching steps, they had effectively cut off the street ahead.
I stopped. Now I either had to turn right, into the park, and hope the trees held no more surprises, or turn back and jump into the river. I didn't like either.
"You there!" an amplified voice called. The zombies stopped their march toward me, and I looked around. Over the top of the high brick wall, two platforms appeared on either side of the gate, the kinds of platforms on scissor-type lifts that people used to paint tall buildings or clean their windows. On each platform, two men stood, together with the .50 caliber machine guns from the Humvees. On the platform to my right was the guy with the bullhorn. I hadn't seen people in weeks, and these were, obviously, an especially welcome sight.
The zombies were temporarily frozen. It was one of the many disadvantages of almost completely lacking a working intellect—they couldn't handle multiple threats at all, or change from one target to another easily. They looked at their enemies above the wall, then back at me, swaying uncertainly. I, too, was frozen, as I wasn't sure at all what I was supposed to do. There were still about two hundred zombies, fanned out now in a more or less crescent-shaped wall of rotting, grasping flesh, between me and the people.
"Start moving toward the gate," the guy with the bullhorn said. "We're going to get you."
He sounded confident, and their set-up indicated a good deal of planning and equipment, like they had done this before, but I still wasn't too enthusiastic about moving toward a mob of mindless cannibals. I took a few slow steps, and again, the zombies moved toward me. But then we both heard the gate rattle as it slid to one side.
Again, the zombies were confused, and many at the back turned toward the gate. I took a few more steps, and then a crowd of about twenty people came rushing out from the gate. Like the guy on the cherry picker, they seemed pretty disciplined and organized, letting out a loud "Arrrrrr!" as they charged the zombies. They looked like the crazy, post-apocalyptic bikers and villagers in _The Road Warrior_ movie, all decked out with various kinds of impromptu armor—football pads, paintball and fencing masks, pieces of tires cut up and bound to their arms and legs as armor, hubcaps and garbage can lids for shields. They crashed into the zombies, wielding bats, clubs, machetes, axes, shovels—any hand-to-hand weapon that could deal a fatal blow to the head.
The zombies were now completely confused, and they began to fall back before the assault. I was impressed and grateful for the people's bravery, but I didn't see how they stood a chance of clearing a path.
Up on the cherry pickers, the two people who were not on the machine guns were swinging things at the end of a rope, the way you would a sling, but the objects were bigger, so they were using both arms, like in a hammer throw. "Set!" the guy on the bullhorn commanded, and they let go their projectiles, which flew over the crowd and crashed down slightly in front of me, one to either side. When they hit the ground, I heard loud popping, and then splashing sounds. I wasn't sure, but I started to catch on, so I stopped and took a couple steps back.
The people who had thrown the objects were now wielding bows with flaming arrows, and from where I was standing, it looked like they were aimed right at me. I also got a whiff of something I hadn't smelled in years, that smell you always associated with summer evenings, when Dad went out and lit the Kingsford in the backyard. I kept backing up as the zombies again advanced on me.
"Fire!" came the command from the guy on the cherry picker, and the arrows shot into the zombie crowd right in front of me on either side. I ducked down, brought my right arm across my face, and hoped these people knew what they were doing.
##
WHEN THE ARROWS hit and ignited the lighter fluid, the hair on the back of my hand singed and curled in the heat and blast of the expanding fire ball. Unlike the zombies, I needed to breathe, and I staggered back a step to catch my breath as the flames receded slightly after the initial flare up. The people in the cherry pickers pressed the attack, throwing another pair of fuel bombs, redoubling the flames and driving me another step back.
Just a few feet closer to the centers of the two conflagrations, the zombies were faring much worse than I. With their dried-up flesh and hair, most of them were burning briskly, and their moaning now turned to screams as they flailed about in whatever it was they experienced as pain. It smelled like a cross between a barbecue and the seventh circle of hell.
Though horribly burned, many of them were still capable of motion, with their limbs still moving, even though scorched bones could now be seen through their burned clothes and flesh. But even the more hardy ones were losing their struggle to carry on the fight, as their eyelids had shriveled up in the first blast of flame, and their eyeballs looked like singed marshmallows, with sizzling goo running down their dried, cracked cheeks. They would walk into each other, or collapse to their knees, their burning hands clutching their faces in a slow agony that looked appallingly like a final supplication to the God who had made them, punished them, and was now punishing them again.
Between the edges of the two puddles of burning fuel, there were only a few zombies who had completely escaped the flames. I started walking toward them, as this was the gap in the midst of the two burning mobs that led to the gates. The first zombie to get close to me I shot in the face, then kicked him in the stomach and sent him crashing into the burning zombies to my right. Unfortunately, another burning one grabbed my gun arm and lunged for it with its mouth. I twisted away as I drove my knife into its mouth. It flailed around, still burning, with the tip of my knife stuck in the back of its throat. I wrestled my right arm out of its grip and stuck the barrel in its left eye. I fired as I pulled my knife out, and the zombie fell back into the burning crowd.
This altercation had slowed me down, and two more were closing in, one from my left and the other right in front of me. The one on my left was horribly cadaverous, even by zombie standards. It had been a very old woman before its death, and from the look of its torso, it had been run over and crushed by some large vehicle since then. It couldn't move its arms, all its bones were so crushed, so its two limbs just hung at its sides, swaying randomly as it walked. Its dress was torn, revealing the shriveled, dried flesh underneath, crisscrossed with feathery lines of dried blood and caked with dirt. Its insatiable maw kept coming nevertheless, and would keep on doing so no matter what.
The one in front of me, on the other hand, was a fairly robust male, with just the typical neck wound and blood stain down his shirt. I leveled the Glock at him and fired, sending him falling back into another zombie behind him. At almost the same time, I slashed the old zombie's throat as hard as I could with the serrated back edge of my knife. The blow spun her around and dropped her, with her neck severed almost all the way to the spine. She landed on her face, but her head bounced up and twisted around, so she was looking completely backward, up at me, before the head flopped back down on its side.
Even then, she started to pull her knees up under herself and struggle to rise. She'd be able to get up, doubtless, but not before I got out of there.
There were just a few more zombies between me and the people who had come out from the gates. I kept moving, but the zombie that the robust male had fallen on was getting up, just as another was coming at me from the right. I kicked the rising one in the head as I shot the standing one in the face. I was just a few feet now from rescue, when something grabbed my left wrist.
I turned and raised the Glock, but saw that I was aiming too high. I was held by something less than four feet tall, what had been a little boy of six or seven. Its jugular was torn open on the left side, but there were no other marks on it. It was slowly bending its mouth toward my wrist, ignoring any danger I might pose in its obsession for human flesh, its only remaining goal or desire. I raised my left arm, lifting the child zombie off the ground even as it continued craning its neck, its bared teeth yearning for my arm.
Oddly enough, the color of this zombie's flesh was like that of milk, like all his blood had drained out when he died, but had not been replaced with the horrible putrefaction and discoloration that inevitably accompanied undeath, instead leaving him pristine and undefiled. Here was flesh without blood, but also flesh without decay. It was animal existence at its purest—deadly, unholy, and unstoppable.
I holstered the Glock and grabbed the horrible, beautiful thing by the throat as I wrenched my left arm free of its grip. I sheathed the knife and held the thing with both hands around its neck. It wouldn't have been so bad if I could've throttled it to end its eternally pitiable existence, letting it slip slowly into a merciful death, but zombie physiology wouldn't allow this. It didn't help that this thing in my hands was the same age as my youngest son last year. The only minuscule consolation was that he didn't look at me, but up at the sky, unblinking even though he stared right at the sun, his jaw still working in his hellish, animal hunger.
"Sorry" fell so far short of what was going on here and what I was feeling that I wasn't going to bother with it this time. "Damn you," I whispered instead, and I flung the little thing away from me and back into the flames. Damn who? The zombie? Me? God? The asshole who invented the disease that caused the dead to rise? What the hell, it looked like there was plenty of damnation to go around, so why not just damn us all together, Lord, in one big mass of suffering, with you as the King of it all. Unlike earlier that day, this time I really did feel nauseous.
Two of the people from the gate had reached me by this point. "Come on," one shouted, grabbing me by the shoulder, "let's get inside." I followed them dumbly through the gate as it rattled closed behind us.
Before me was a grassy area with several dozen people on it, as well as a few trees and some odd-shaped sculptures of metal and stone. It was enclosed by the brick wall behind us, which ran down to the river on the one side. To the right, meeting up with the wall, was the large building I had seen from outside. And about two hundred feet in front of me was another wall like the one behind us, again running from the building down to the river over there. The river was the fourth side of the enclosure.
The people who had brought me in were smiling and patting me, encouraged and pleased with their own work. But I was immediately met by a woman a little younger than I was, one who had not gone outside in the attack. Like everyone there, her garb was a hodgepodge of different outfits, but it definitely conveyed the sense of a scientist or doctor rather than a soldier—smiley-face hospital scrubs as pants, canvas loafers without socks, a stained man's dress shirt, and some kind of blue vest with pockets, sort of like the ones that greeters at Wal-Mart wear.
She looked me over. "Your arms, show me your arms," she said, not exactly gruffly, but definitely not friendly, either.
I rolled up my sleeves and showed my arms, turning them over, feeling very awkward and embarrassed. Nobody's hygiene had been what it should be since the dead rose, but you were still made aware of it at odd moments like this.
She had already moved on to my neck and torso, raising her eyebrows and looking down the front of my shirt, as well as tilting her head to see both sides of my neck. "No bites? You're sure you haven't been bitten?"
I knew she had to ask. After the initial outbreak, most people had been killed when someone in their group was bitten and they tried to take care of him, only to have him then rise as a zombie and attack the others. Almost all the hospitals were taken out in the first few hours because of that. And then once they realized how quickly it spread, people were faced with the awful burden of having to execute anyone who had been bitten. "No," I said, shaking my head, "nothing, I swear."
She kept looking me over, though she still didn't feel bold enough to touch me in order to turn me around or lift my clothing. "No fever, chills, burning thirst, loss of appetite, vomiting?"
I kept shaking my head. "No, really."
"Open your mouth." I did. She winced. "Not the prettiest sight, but whose mouth is these days with no running water or toothpaste?"
She tried a different tactic. "Don't be afraid to tell us if you're sick, we won't send you back outside. We're not barbarians. We'll treat you humanely. We've quarantined people before. Some even pulled through."
"No they didn't," a voice behind me said. The guy who had been giving the orders on the bullhorn was walking up to us. He was big, not body-builder big, but a little taller than me, and he had obviously been in shape before months of tight rations and fighting off those things had whittled him down a bit. Now he just looked sort of gaunt and tenacious. He was about my age, late thirties, and dressed in the remnants of a military uniform, though I don't think the various parts matched or were from the same branch of the service.
Again, I didn't know much about the military or wars before the one we were in right then, and in this one we couldn't stand much on conventions. It was amazing and welcomed just to meet people who had a pulse. If they had themselves a little compound with a wall and some weapons and supplies, they could dress like the archbishop of Canterbury or flaming drag queens—or, what the hell, even both—it made absolutely no difference. Which meant, of course, that it never really did.
Oddly, the presence of a never-sleeping army of the undead just outside the gates really didn't change the dynamics of petty squabbles and insults and power plays, so the woman, whose pride was somehow hurt by the remark, shot back, "Yes, that one guy did!"
Military guy smiled. "Okay, Jones did, because he was just barely nicked on the hand and you chopped his damn arm off at the shoulder about two minutes after he'd been bitten. It was so tiny he might've pulled through with his arm if you hadn't. Don't sugarcoat it for this guy."
The woman flushed. "I did the best I could! I always do the best I can, you macho asshole! I was a damn dental hygienist before... before... Oh, why do you have to be so _mean_?!" Her voice was starting to crack and she turned and stomped off.
There was an awkward silence. I was glad that their military precision was better honed than their interpersonal skills, or I probably would've been incinerated, or eaten, or both. "I'll go talk to her later," he said, a little surprising and almost comical in his sheepish contrition. "We call her Doc, even though she's... well, you heard that she's not completely, all the way, exactly... a doctor. But we try to be nice to her and show her some respect, 'cause she's helped a lot of people. Damn it, I shouldn't have said anything." Another awkward silence. "You'll have to excuse us, but it's just hard with new people. It's so embarrassing, admitting how little we have here and how many things we can't do, and how scared we all are. But we have to be careful."
"No, no, I understand. What you did was amazing, and I'm grateful you got me in time."
He picked himself up a little, forgetting his faux pas and reasserting some authority. "But she's right—if you're bitten or hurt or sick, we'll do the best we can for you, even if we have to quarantine you." He took a step toward me. "But if you lie about it and we find out, it'll be different."
I wasn't trying to be a bad ass, but I sensed this guy was again jockeying for position, and I still understood enough about people to know that I had to have some credibility and respect in this new group, so I met his gaze and didn't back down. "I said I understood."
We'd gotten our male posturing out of the way early, and to everyone's satisfaction, which seemed to suit him fine. He stuck out his hand. "Sorry. I'm Jack Lawson."
We shook hands. "Jonah Caine," I said.
We walked over to a table near the gate where a young man sat with a clipboard. On one side of him, one of those big Rubbermaid storage sheds shelved dozens of weapons, and on the other side there was a big plastic garbage can. The people who had attacked the zombies at the gates were handing in their weapons to the man at the table, and he was marking them down and putting them in the locker. They tossed their armor and shields and larger items into the garbage can.
Jack put his own pistol, an old Colt .45, on the table. He turned to me. "Rules. You'll have to hand over your weapons. We'll keep them and give them back if you ever want to leave."
This was a little more than just looking me over for bites, more than them protecting themselves; this was them asking for a lot of my trust so they could enforce some crazy rules they'd thought of for their group. And it was taking away the only things that had kept me alive for weeks. "I'm not handing over my weapons. What if they break in? I don't know how safe this place is."
"It's plenty safe, trust me. We can't always go outside whenever we want, but they can't get in. And we're rigged up pretty good if they do. There are weapons lockers all over the compound, and they're all guarded, but you'll never be far from a gun or a club if you need to defend yourself. You saw how many of us could get armed and go get you, as soon as our lookouts spotted you. It's just part of our community that when we're in here, we don't want to have weapons to remind us of how we have to live, or to ever tempt us to use them on each other. We don't have much here, but we don't have to live like people used to." He paused, and we were once again back to posturing, unfortunately. "It's not a request."
"I know." I put the Glock, the magnum, and four knives on the table.
"Anything else in the bag?"
"Just clothes and food and stuff, no weapons."
"Okay." He made a gesture, and a boy of about thirteen came running up. "Is it all right if he takes the bag inside?" he asked me. He looked at the boy and made his tone stern. "He knows not to open it."
"Sure," I agreed. The boy took my backpack and ran off toward the building.
Jack eased up a little and visibly relaxed. "Well, welcome to our little place. Let's take a walk around."
We headed toward the river. "Jonah... Jonah... Wasn't he in the Bible? Didn't he—no, Noah was the guy in the ark, with the animals. What did Jonah do?"
"Jonah was swallowed by a whale."
"I thought that was Pinocchio."
"Him too."
"Oh, yeah, right." He laughed a little. So did I, for the first time in weeks.
"What was this place?" I asked, looking around.
Jack stopped and turned, pointing back at the big building. Near the top was a large sign that read, "MUSEUM OF SCIENCE AND HISTORY." Jack laughed again. "It was kind of a miscellaneous museum for the whole area, since they didn't have too much of either science or history that was unique to here." He gestured to the sculptures we were walking by. "I'm not sure how these qualify as either science or history, but they eventually added the sculpture garden, too. No big discoveries or battlefields or artists around here, but they had some nice displays. Kids liked it. We all used to come here when we were kids, and some of the older people in our group used to bring their kids here. And being by the river was always nice. They'd have concerts in the summer, and you'd come down to see the fireworks over the river on the Fourth of July. Now it's all we have. I guess it's not where any of us would've picked to be when the world ended, but it's held up pretty good."
We had reached a low fence and hedge that bordered the river. I looked over and saw that we were at the top of a concrete wall that dropped about six feet down to the water. It was an unusually good setup for defense against attackers who didn't have weapons or vehicles or machines. It almost seemed fated, or even providential. But I couldn't quite make myself believe it, even though I partly wanted to.
Over the hedge and fence, there was a string, with various bells and wind chimes every couple feet. Jack held one to show me. "Just in case. We keep a close eye on this point, since it's not perfectly secure. Sometimes, one of them will manage to climb around the edge of the wall. Usually they just fall in the water and the current takes them downstream, but a couple have gotten a foot up on the wall here and have grabbed on to the hedge before somebody whacked them and killed them."
"How'd you all get set up here?" I asked. "We heard on the radio and TV to go to military outposts, forts, those kinds of things."
Jack looked down at the water, then across the river at the city. "Yeah, for the first few days, that made sense. If there are corpses walking around and the only way to stop them is to shoot them in the head, then it made sense to go somewhere where there are lots of guys with guns."
He sighed. "But as the tide turned, those became death traps. All those people in one place, shooting guns and making noise, and driving and flying vehicles in and out: the dead converged on them and overwhelmed them. Whoever could get out and make a stand somewhere else, did. No, an army camp in the states hasn't been set up for a siege since the days of cowboys and Indians.
"Nowadays, they just have barbed wire fences and gates with some barricades you have to drive around. Hell, they were set up for stopping suicide truck bombers, not armies of walking corpses. What the hell does a zombie care about a waist high concrete barrier or spikes that shred your tires? He doesn't know what tires are anymore."
He fell silent for a second, but then he laughed again and gestured at the walls, and I definitely got the feeling that he thought of them as _his_ walls. "No, if you needed a place that was made for keeping people out, and you were lucky and it hadn't already been blown open or burned down, the place to look was where they used to charge _admission_! And preferably a place where not too many people would think to hole up. At least there you would have a shot. We got lucky."
He looked at me quizzically for a moment. "Or God was looking out for us? You believe in God, Jonah?"
I looked down at the rushing water. I didn't know the answer to his question. If I were being practical, I might have thought just to tell him what he wanted to hear, whether it was true or not, but I wasn't sure yet from his comments what he thought the right answer to his question was. "I don't know. I'm sorry."
He took it in stride. I was beginning to get the impression that he took a lot in stride, and I liked that. "No need to be sorry. I'd think you were crazy or retarded if you could answer that with two thumbs up or two thumbs down at this point." He looked so thoughtful, gazing across the river at the city of the dead. "No, I think this has definitely taken all the certainty out of it, if you ever had any. You think of all the good people, the honest and kind people, who've died since this began, and you just can't help thinking the Big Guy is on vacation and he left the other guy in charge. What do they call that?"
I knew the Bible a little, but I wasn't up on all the end of the world stuff. "The Apocalypse?"
"No, no, I had an aunt who was into all that junk.... It was a special part of the apocalypse.... Oh, yeah, the 'Tribulation' she called it, when the Beast would rule the world for seven years. Anyway, you see this and you think either there isn't a God, or if there is, then He must be taking a big, divine siesta, and we're catching hell—literally—until he wakes up.
"But then you look at something like this," again, the gesture to his walls, his domain, and he clearly wasn't just proud of it, he was grateful for it, and would even pray for it, if he could, "and you know we'd all be dead if it weren't for a million little coincidences and lucky breaks, and you can't help thinking maybe, just maybe we'll make it, and we were supposed to make it, and it'll all work out."
Jack was getting his own thousand yard stare, I could see. In his position, he probably had to give a lot of pep talks and bolster people's courage in tough times, so he hadn't had a chance to speak his real, more ambiguous and pained mind in days.
"Well, Jonah, to go back to your question: everything here kind of came together haphazardly, little by little. For the first couple days, there were just a few museum employees holed up here. They grabbed all the food from the cafeteria that they could, and they barricaded themselves on the top floor. As they looked out the windows, though, they got a little braver. They saw that nobody—living or dead—was paying any attention to them, and so long as they stayed away from the windows and doors in the main lobby that face the outside street—don't worry, we walled those up later—they could move all over the museum and the grounds.
"They got brave enough to open the gates—there's another gate to the employee's parking garage in the back—to let in survivors sometimes. Fortunately, they only did that after they'd heard on the radio and TV about the bites, so they knew to be careful. They were brave people. They even went up to the roof."
He pointed to a little glass enclosure on top of the building. "See that? The local TV station had a camera up there. It'd show you a view of downtown all the time on their website, and they'd show the view from it at the end of the 11 o'clock news. Before the electricity went off, the people got brave enough—or lonely enough—to stand in front of it and try to get out a message that there were people here and it was fairly safe and they'd try to let you in. Fortunately, not too many people saw it, or tried to get here, or it probably would've been mobbed by the living, and then by the dead, and there'd be nothing here now.
"A few days into it, we showed up—the military. We also didn't think anything of one more shut-up building, and we drove right by here without a thought. We rounded up some police and firemen who were still listening to their radios, and we built the barricade you probably saw on the bridge. We were sent to try and keep the dead in the city.
"At first, it was easy. Shoot one every couple minutes, wait, shoot another. But little by little, you were shooting them constantly, and they were still coming. It was obvious we couldn't stay there. We were the only ones still trying to hold on to one of the bridges, so they could just get around us on another bridge and come at us from behind. I know, I know, they couldn't figure that out as a conscious strategy, but the point is, our position was exposed, and it wasn't even keeping the dead in the city, since they had other ways out.
"We knew we had to get out of there, but at that point, we'd stopped hearing anything from our base. The crowds of dead were getting bigger, and we had no idea where to fall back to. It was then that somebody saw people on the shore here, pretty much right where we're standing now. You know, from a distance, you can't tell if it's a real person or one of them, but through binoculars you could, and we could see they were real people, alive and waving. We were lucky: we were able to pull back and only lost a few men. We even retrieved the .50 calibers from the Humvees and a lot of ammo. God knows, we've needed it since.
"We waited with the people who were already here, listening to the radio and watching the TV. But in a few days, there was nothing. You didn't hear shots in the distance or explosions anymore, either. It was just still, just the sound of the river, with the reeking of the dead rolling across you when the wind shifted. And at that point, supplies became an issue.
"We were a lot luckier than some people, because we had the river for water. But all we had to eat was what had been raided from the museum snack bar and whatever people had brought with them when they'd been let in. So we launched raids for supplies.
"We got better organized as time went on. Like I said, we walled up the lobby completely, so we had no worries there. There was construction going on at the museum when the outbreak began, so there were some construction supplies to do the work, as well as the cherry pickers you saw, and we could see how to use those to defend the gates when we needed to open them. We got especially good at distracting the stiffs—setting up a ruckus out by the employee parking lot till they all congregated there, then launching a raid out the main entrance, or vice versa. We can go through the sewers to a couple spots on this side of the river.
"Oh, and to the other side of the river?" He got a mischievous grin. "See that?" He pointed overhead to a wire that ran from the roof of the museum to the city on the other side. "That was my idea: a zip line. I swam across during the night and waited till dawn, then climbed out and tied the thing down, so we'd always have a way to get to the other side quickly. It's another way we set up distractions for the stiffs. Cool?"
"Very cool, Jack." I was truly impressed with everything, and also didn't want to let him down, as he was obviously getting wound up.
"For a while, we kept picking up survivors. At first, there were quite a few. We'd spot smoke somewhere and go check it out and bring a few people back. That's how we got Popcorn and Tanya."
"Popcorn?"
"Long story. Somebody will explain it to you when you meet him. Sometimes, someone would get close, and we'd let them in, like we did you. That's how we got Milton. You'll meet him later. He's kind of important."
"Important? Important how? Like a leader?"
"No, not exactly. It's another long story. I'm sorry, it seems like everything here is a long story, but that's just how everything's come together in a little less than a year, and it's hard to explain in a few minutes. So anyway, we've been scrounging for supplies, but this year might be better. We've caught some fish from the river, and now we're going to try growing some vegetables here, so we might eventually not be so reliant on Spam and Twinkies."
"I was getting tired of them, too."
"I know, a hazard of the apocalypse."
"Who put the sign up outside?"
He laughed again. "Our little motto, or half of it? Milton thought of it, to let people know there's someone still in here, and to say we're dying to really live, not just living to die."
"I hope that's not his most profound idea."
Jack laughed harder. "No, not at all. He's got some odd ones, I'll admit, but they all kind of work out and make sense. I don't know how to explain him, either, but you'll see. I'm kind of the joint chiefs of staff or the secretary of state, and he's more like the pope or the dalai lama."
I thought about that for a second. "Who's the president in that scenario?"
"That's one we haven't had so much use for in our new setup."
This time we both laughed, and more heartily.
##
WE WALKED BACK toward the museum. The side of the building that faced the river and sculpture garden was all windows, with glass doors near the front wall. Jack led me through these into the lobby, a large circular room that extended the height of the building, with a huge Calder-inspired mobile hanging in the midst of it.
I could see the construction Jack had spoken of—roughly made brickwork had been erected on the inside, shutting off the large windows and doors that fronted on the street outside. At irregular intervals, some bricks had been left out at eye level, to function as peepholes, I assumed, and possibly gun-slits, if it came to that. Walling it in had made the room dark and cavernous, though the sun was now shining in from the western side of the building, over the dead city and across the river and sculpture garden. As Jack had promised, there was another weapons locker here, with a young man seated next to it.
From the lobby, a spiral staircase ascended to the second, third, and fourth floors, and a large archway led into a huge room on the first floor. A sign above the archway read, "MAIN EXHIBIT HALL." Jack and I went up the staircase, all the way to the fourth floor. After the third, the sign above the stairs read, "EMPLOYEE OFFICES."
"We kept their original idea of barricading ourselves in on the top floor," Jack said. You could tell that he loved arranging and organizing things, and then describing the arrangements and all the thought that had gone into them. "If the stiffs ever got this far, I don't suppose it would make much difference, but I think it helps if people think there's a plan. Without electricity, the staircase is the only way up, except for the fire stairs at the end of the hall. Those are sealed up on the lower floors, so you can only get to them from up here, and they lead up to the roof, so we could fall back to there if we ever had to."
"All you need is a helicopter."
Jack looked around and gestured toward the ceiling. "I don't think the roof would take a big one, one big enough for everyone, but maybe a small one. Hey—are you making fun of me?"
I smiled. "Only a little. I actually think a helicopter would be good, as a last resort, and possibly for getting supplies, too."
"Well, I'll put that on my to-do list." He went back to his tour. "This is where most everybody sleeps, unless they're on duty. We run three shifts, same as people used to on bases or factories or wherever. If someone petitions me and Milton, sometimes we let people bunk on the lower floors. Real estate here isn't at too much of a premium. People still need some privacy, especially if we're going to start having babies."
I knew how unlikely but inevitable that seemed. "Has that happened yet?"
He smiled, part lascivious, part humorous, part just happy. "Yes, just a few, and more on the way." But the smile faded. "But not nearly as many as we've planted in the ground out by the sculptures, or burned up out back in the parking lot. We bury our dead, even if they've turned, but we don't bury any zombies we kill attacking the place. I guess it's not fair, but it doesn't seem right, treating them like people, when you didn't know them when they were people."
"I suppose not." I began to realize how many of the finer points of living and undeath I had missed, surviving on my own.
There were people milling about near the other end of the hall. They stayed back, though some looked our way. Jack opened the door to a very small office, in which the desk and furnishings had all been pushed to one corner. The window was open and what looked like a very old Native American blanket lay on the floor, with my backpack next to it. "This is kind of our guest room, for new people till they're more used to us and we're used to them. We took the blanket from a display, obviously. We've ended up using a lot of the displays for one thing or another. I guess we should be trying to preserve them better."
"I'm sure you're doing fine." I picked up my backpack and unzipped it. "I assume you have some rule about sharing food?"
"Yes, that was one of the first, even before the weapons rule. I guess someday our Bill of Rights will be in one of the displays."
I held the bag open, showing a couple dozen snack cakes on top. "Well, I don't know how you're going to carry these to wherever you'd put them."
He laughed. Looking around, he reached over to the desk and grabbed one of those baskets that people use to hold papers and mail. I loaded it up, eventually uncovering the bottle of bourbon underneath.
"Whoa," Jack whispered when he saw it, looking around, partially closing the door, and setting down his basket of snack cakes. "Now, now, let's not get too carried away with all this stuff about everyone sharing everything. You said _food_ , not _booze_. Giving everyone a thimbleful, versus giving two or three people a chance to forget their problems for one evening—well, I think that's an easy matter of weighing the greater good for the greatest number, don't you?"
"I leave that up to your leadership, Jack. I don't want to cause trouble. I'll pour it out on the ground if you want."
He looked at me and raised an eyebrow. "Now I know you didn't just say that. I'd write a new law myself telling what happens to people who are crazy enough to waste booze." He pulled the bottle out of the bag. When he saw the brand, he wrinkled his nose. "I hope you didn't risk your life for _this_?"
I laughed. "No, no, it was lying right out in the open."
"I can see why." Then he shrugged. "Well, we've made due with a lot worse inconveniences than bad booze." He slipped it into an inside pocket of his camouflage jacket. "Let's just not mention this to anyone and see if we can maybe enjoy it a little later on with some young ladies I know."
"I wouldn't want to intrude, Jack. You already said how important privacy is in those situations."
He laughed harder than he had all afternoon. "Oh, God, don't worry about that. I'm going to go talk to Doc, just to make sure she'll even talk to me anymore, and Tanya's made it clear that the only men she likes less than soldiers are cops. If anybody's going to be asking the other guy to make himself scarce so he can get lucky, it'll be you."
"Then why are you inviting them?" I asked, laughing along with him.
He turned suddenly serious, and since we were alone, I was pretty sure it wasn't going to be bluster or posturing. "Because I mostly talk to the people here my age, the ones who've seen things and lost things and been hurt and who probably aren't ever going to come through to the other side of this and become the new Adam or Eve. Those two gals are like that. And I don't mean to put you down, but I'm thinking that describes you pretty closely, too."
"I understand, Jack. It's not a put down; it's just the way it is."
"Good," he said, brightening up again. "So I guess all I'm saying is, I wouldn't mind talking to you and them and sharing your booze, and I'm not going to fight you for any of the gals here, though you'll have to excuse me if I still try to talk a big game—a man's got to have some pride left. If the youngsters want to go off and repopulate the earth, I'm perfectly happy to be guarding the gate from those things while they do it." He leaned a little closer. "All the more reason not to begrudge me a few drops of bad booze."
He picked up his office basket of snack cakes and started to leave. "Meet me back in the lobby a little after sundown."
"Thanks, Jack."
"See you in a bit."
After Jack left, I circulated some on the fourth floor. There were many larger offices with several people living in each, and two really big rooms in which they had rearranged the cubicle partitions to make them into living space. Everyone was friendly, though they clearly had some boundaries and some set rituals and restraints about getting to know newcomers. Given Jack's description of the growth of their community, I seemed to be the first new addition for some time, so that probably made it more difficult. On the other hand, it also made me quite the object of curiosity, so everyone was eager, if not to get to know me, then at least to see me and say hello.
Before it got too awkward talking to people in the living quarters, I made my way back to the sculpture garden and down to the river's edge. The sun was lowering in the sky, just sinking beneath the tops of the buildings across the river. City skylines were perhaps one of the most remarkable and disorienting sights in the world of the dead. During the daylight hours, the line of tall buildings was probably not that different from how it had appeared in the city of the living, but at night, it became one big, black, silent outline of rectangular shapes against the stars, sort of the way mountains looked at night, except mountains were never so angular, and one didn't expect them to twinkle cheerfully with thousands of lights. Such weird, artificial shapes as the outlines of big buildings demanded the softening, gladdening glow of artificial light to make them bearable. Without it, they became oppressive, monstrous, nightmarish. And in the gloom of twilight, as now, the nightmare was taking hold of the city.
The river, on the other hand, had retained its beauty, or the darkness had even enhanced it. In the city of the living, the river would have been a cold, dark void under the banks and the bridge. But now, it was the city that was dark and threatening, while the river's constant murmuring was a comforting, lulling blanket, taking my mind off of the dead, who were blessedly silent now after our battle, however many of them were still left out there.
As I gazed at the water, I thought of how it would reflect the stars and the moon later this evening, the only visible sign that perhaps the hellish, cannibal earth would ever again reflect the quiet peace of heaven. Certainly none of the edifices of man on the other side of the river had ever done that: now they were just part of a giant, broken corpse, madly swarming with thousands of little, insane corpses. I shook off that thought and focused back on the river. After a moment I smiled, having forgotten how much water always calmed me.
It was getting dark. I turned back to the museum, seeing a few lights. I walked back to the lobby and found Jack, who led me into the main exhibit hall. With its ceiling four stories high, the side of the hall that faced the river was all windows. The room was lit with many candles and a few large torches, and there were many round tables laid out with folding chairs.
In the gloom above, I could just make out a biplane and the skeleton of a mosasaur, hanging from the ceiling. To see an ancient fighter plane and the skeleton of a prehistoric reptile by flickering torchlight made the place feel like a cross between Valhalla and a ride at Epcot—not creepy so much as surreal, almost funny. The reality, doubtless, was less grandiose and less comical.
"Big fundraiser tonight, Jack?"
"What? Oh, the tables. Yeah, darndest thing, isn't it? We found all these tables in a big storage area. Now we take our common meals here."
I laughed a little. "I volunteered at a museum years ago, and they all have their most impressive, breathtaking hall set up to hold fundraising banquets for rich donors."
"Yeah, the employees here told us, but it's still funny. They'll never have another fundraiser or rich person in here, and they've just got to trust us to be careful when we're sleeping on the ancient Indian blankets and cooking in some antique copper pot."
"I forgot to ask: is this where I'll meet Milton?"
"No, I was hoping you might, but he's sick again. I talked to him and he said tomorrow definitely."
"Sick?"
"It's another of those things that's really hard to explain around here. He's not going to turn, if that's what you mean, Jonah. I keep telling you—a million little accidents that keep us alive, and Milton's like the biggest one of them all. But you just wait till tomorrow for him."
Jack led me to a serving line where we got the usual post-apocalyptic fare: Spam, canned vegetables, and canned fruit. The one surprise was some flat, slightly burnt biscuits. I hadn't had anything even pretending to be bread in months, and I would gladly have taken more than the two I was given. It was funny, but before all this, you wouldn't have thought bread could ever be an interesting food, but as with the river, its simple attractiveness had been enhanced.
After the serving line, Jack led me to a table where I recognized Doc, though she had shed the blue vest and had changed into a sweatshirt from the museum's gift shop. I learned her real name, Sarah, and it fit her perfectly: ordinary, capable, solid. She was a lanky woman, and the shirt looked slightly too small for her because she had rolled up the sleeves. So that they wouldn't come to just below her elbows, I suspected. Her hair wasn't pulled back into a ponytail as before, and you could see she might have been pretty before all this. She was still self-conscious, but Jack clearly had made amends with her, and she was much more at ease in this setting.
With her was an African-American woman who remained, even in her disheveled state, quite striking. This was the Tanya that Jack had mentioned. She was only slightly shorter than I was, about my age, with her hair closely shorn, and she had the toned but not bulky body of a swimmer or gymnast. She was slightly butch, to be honest, and reminded me of my gym teacher in middle school.
She seemed somehow to be less haggard than the rest of us, without the clear weight loss, sunken cheeks, and the dulled gaze. Tanya's eyes were deep and brown and infinitely soothing, but full of life. To be honest, she reminded me of the Homeric tag-line "ox-eyed Hera," like she was meant to look down on mere mortals. More than anyone I'd seen for months, more than I had thought possible anymore, she exuded vitality.
But when Jack introduced me, there was something pained and forced in her smile, and I could tell immediately she never laughed. Here was the pain and loss of which Jack had spoken, the misery that put us all in the same category and allowed us to communicate and sympathize. I could see the truth of Milton's little riddle or motto: about a year ago, we all had died, and now we were just living to die a final time.
Morbid thoughts notwithstanding, we were all in the mood to have as good a time as we could at our table. I don't think it would be possible to _savor_ such fare as we had, but we did linger on it, not so much to enhance any flavor, but just because the prevailing wisdom held that it made you feel fuller for longer if you took your time, while wolfing it down was supposed to make you hungry again sooner. As with all such folk wisdom, though, there were proponents of both sides.
Jack was all business at dinner, but in a jovial way. "I was especially glad with how fast we got out to you," he said. "We practice it a lot, but I was afraid, since we hadn't done it in so long, that you'd be running around out there for a lot longer before we got enough people to the gate. And the fire—they did really good this time. We have to practice that with watered-down paint, but we don't do it so much, 'cause it gets the neighbors all riled up, obviously. But they really got the two kill zones nicely spaced—just enough room in between for you to get in."
Slightly less morose and a little more playful than I'd seen her, Sarah said, "Enough, Jack, it's not a post-game show." .
"I know, I know. But, hey—Jonah here brought us something special, and I was wondering if you ladies would join us over in Frontierland to see it?"
"It better not be anything disgusting," Tanya said, "or I'll put a hurt on you." She was also a little more playful. "And you too, new guy," she added, glaring at me, "even if G. I. Joe here put you up to it."
"Nothing of the kind," I reassured them.
"Meet us there in five," Jack said, and he and I left to take our dishes back.
On our way to the rendezvous, Jack turned on a flashlight and led me through some much smaller exhibit halls. "We try to conserve batteries as much as possible," he whispered as we walked past cases of Native American artifacts, "but walking with a candle is just too damned hard."
After two more rooms of artifacts, we entered the recreated interior of a frontier cabin. Jack lit a couple candles and placed the bottle on the roughhewn table, then arranged four mismatched mugs of pewter, glass, and porcelain around it. We sat down, and a couple minutes later, Sarah and Tanya entered from the cabin's other entrance.
Both women gave the requisite "Oooh," at the sight of booze, like teens at a party about to do something wrong and forbidden and dangerous. As I watched them, I finally figured out what I had been going over in my mind at dinner—how the two of them complemented each other. Sarah was by far the more nervous of the two—even now, she was glancing about, just like a kid worried about being caught at his illicit activity—while Tanya was utterly unflappable. If anybody complained to her about her drinking, she'd break the bottle over his head.
Sarah, on the other hand, could be saddened more easily, but she could also be more easily cheered, while inside Tanya there was clearly such a well of pain and anger that it was almost uncomfortable to be around her, as bold and vivifying as she was.
But as nervous or sad as they were, there was something infinitely comforting—in an emotional, if not sexual, way—about having them around, as Jack had tried to describe before. Their attractiveness, too, had been enhanced by our horrible situation.
Jack played host, pouring the bourbon. "To new friends." We all toasted and took our first drink in ages.
Somewhat predictably, the women's drinking was opposite as well: Sarah barely sipped, Tanya slammed the booze down. Till it kicked in, we didn't talk much. I knew when Jack suggested the get-together that that was part of the value of the booze—to get people to talk.
About halfway into the bottle, Sarah spoke up. "So, Jonah, what's your story? How'd you get here? People say you came from a secret underground bunker, or that you fought your way out of an overrun military base, or God knows what else."
"I was on a ship when it started," I said. The bourbon had done the trick, as I would otherwise have shrugged it off with something vague.
"You were like in the Navy or something?" Sarah asked. "What's that group? The SEALS? That's what someone was guessing."
Jack guffawed. "No, he was most definitely _not_ a SEAL. No offense, Jonah."
"I know.... No, nothing military. I'm... well, I _was_ a college English professor, just working at a community college, and I'd work on cargo ships during the summer for extra money."
"Family?" Tanya asked, between gulps.
"Yes, a wife and two kids. The ships paid better than teaching summer school, so I was doing it until I could get a better position. The disease must've broken out right after we got on board. Just the day after we left L.A., the television and radio were talking about it. We just thought it was local and we'd go ahead to Honolulu and everything would be normal. But then it was obviously not. The captain just stopped the ship, and we watched and listened to the reports. Until they stopped. We'd still get short wave stuff, but we didn't know what was going on, all just garbled cries for help.
"We sailed back to within sight of the docks. There were fires everywhere, and we got our first look at... them. You'd think they were people at first, of course, but then we'd see them through the binoculars, and we'd understand what the TV had been talking about. We sailed up the coast, still hoping the reports were wrong and it was just in the city, but it was everywhere—fires, and those things staring at us from the shore. You didn't see anything human on the shore, but a few survivors in boats pulled up alongside us. We'd heard about the bites, so we didn't let any infected on board, fortunately.
"Little by little, we had a flotilla or colony floating around out there offshore. But after a couple weeks, no more showed up, like Jack said happened with you guys here. We'd share supplies, fish for food, and some of the guys rigged up a distillery to desalinate the water. I guess it was as safe as we could hope for, but some of us with families weren't satisfied. We had to know if they were still alive, if they made it."
"I know," Jack sighed. "It's a pull, a pull too strong for some people, stronger than the will to survive. We had people leave to look for family. None of them ever came back. I hope they found them and are off on an island somewhere."
"Me too, but we all know the chances of that." I took a gulp. "Those of us who wanted to leave took some supplies, guns, and one of the smaller boats. We found a place to land where we didn't see any of them, and we went our separate ways to look. I don't know what happened to any of the others, but I stayed alive and kept looking. I found my town, found my house. Nothing there. No bodies, no blood, the car was gone, but what does that prove? Just that they didn't die in the house. After that, I didn't know what to do, so I just wandered. And that's pretty much how I ended up on your doorstep. Sorry, it's nothing too dramatic, I guess."
We were silent for a while. "I was at work," Sarah said very quietly. "Me, and the doctor, and the receptionist."
"I thought you said you were a dental hygienist?" I asked.
"The dentists usually prefer to be called 'doctor.' It was a crowded little strip mall, and it all went to hell in a couple hours, with people outside being attacked in the parking lot, cars crashing, explosions, sirens, gunshots. We just locked the doors and drew the blinds and watched the TV. I'm sorry, it seems so cowardly now, like I should've tried to help or do something."
Jack and Tanya both knew to put their hands on her shoulders. "It's okay, baby," Tanya whispered. "You did plenty here. Everyone knows. You just hold on to that."
Sarah resumed. "But when the TV started saying that we were supposed to get to a rescue station, they wanted to make a run for it. His car was parked close, and it really looked like maybe we could make it. This was just the first day, so we didn't know about... we didn't know that they... you know . . ."
"That they rip your guts out and eat them while you're still watching," Jack muttered, looking into his cup. Tanya glared at him.
"Yes, I think they wouldn't have tried it if they knew that. The people we had seen attacked in the parking lot, it just looked like they were fighting with them, and then they'd fall behind a car or something. We didn't know. And the TV hadn't said anything about that. They just said to go to a rescue station. So they wanted to try it. I just couldn't. I told them that I couldn't, that I'd just freeze and scream. So they said okay, they'd go and send back help later.
"They tried it, and I just closed and locked the door behind them. There were so many of those things in the parking lot that we hadn't seen, and they got them. And I saw it all. It was quick, at least. I don't know what I would've done if it hadn't been quick."
She stopped and put her head on Tanya's shoulder. Then she picked back up. As I had thought, she was quick to break down, but more resilient in coming back. She took a sip of the bourbon. "I'm okay, thanks. I was in the office for a while. It seemed like forever. I didn't make a sound, after I saw what had happened to them. I barely moved or breathed.
"When I saw there were none of them right outside, I wrote 'HELP' on the window with lipstick. I filled a bunch of containers with water before the water stopped running, but I didn't have food. I tore the place apart and just found the usual stuff that you throw in a drawer and forget about—little bags of saltines or oyster crackers, ketchup and sugar and soy sauce packets, the after dinner mints you get from restaurants. I ate the damn fern we had growing on the counter. I was pretty weak when the people from here picked me up."
The silence was a little longer this time, and then, inevitably, Tanya spoke. "I was at home. In the kitchen. No TV on. It was summer. I didn't let the kids hardly watch TV. But then I heard them scream. First one, then the other. They came running in, all bloody. They were babbling something crazy about the neighbor bit them and they ran away from him. They both had what looked like teeth marks on their arms. I was looking at them when the neighbor came through the door, all weird and crazy-looking and covered with blood.
"I pushed my babies behind me and told them to go upstairs and lock themselves in the bathroom. He was coming toward us, mouth open, all drooling and shit. He was bit too, on the neck, I could see, but I didn't care about him. I grabbed the frying pan and told him to just stop, I was going to call the police. But he was between me and the phone anyway, and I already thought that this was way past what the police could do anything about.
"He came at me, and I kicked him in the nuts, but he didn't even flinch. Nothing. Like I'd kicked the damned wall next to him and not him. He kept coming at me, and I gave him the frying pan across the head. Hard. And two more times. The first two times I heard a crunch, and the last time it sounded wet and squishy. He went down. The blood started seeping out from under him, all thick and dark. The kitchen looked like a damned horror movie, with all the blood from my kids and him. I dragged him outside quick and locked all the doors.
"I thought to get my babies to a hospital, but then I looked outside and there were more of those things in the street. Just like Sarah said—cars crashing, houses on fire, shooting, all hell breaking loose. I turned on the TV and started to see what was going on. Then I went upstairs to my babies, bound their wounds, kept them warm and safe." She snickered, but bitterly. "Gave them chicken soup. My babies' last meal was damned chicken soup from a can." Even she was breaking down at this memory, but she inhaled through gritted teeth and kept going.
"I kept them in there for three days, as quiet as I could make them. Nobody else bothered us. Like Sarah, I thought I should have some kind of sign, in case people came by to help us, so I put a sheet out the one window, with 'HELP' written on it in marker. When my babies were asleep, I watched the TV, and I knew what was going to happen to them. We only had a small house. They shared the same room. I couldn't leave one in there when the other... turned. So I watched them closely.
"When I thought my daughter was nearly gone, I made her a little bed on the floor of the closet in their room and tucked her in it. When I was sure she was gone, I closed the closet door and shoved the back of the chair up under the door handle to hold it. It was only a couple minutes and that horrible banging started on the closet door. Steady. Like a damn drum. She was tiny, though, and I was pretty sure she couldn't get through. I whispered her name, in case it was really her and she could still hear me, but there was never any voice, just the steady banging.
"My son passed a few hours later, and I closed the door on their room and jammed it with another chair the same way. He was bigger, though, and I worried it wouldn't hold, so I risked making a lot of noise by nailing the door shut. I tried to time the hammering with the banging he was making on the door. I was lucky and none of them came to check it out. I went downstairs and I never went upstairs again.
"I ate, I guess, and stayed alive, but I didn't think anything until Jack and them came to get me later. All I remember was that damn banging."
We'd said what we needed to say, that bizarre mixture of way too much and not nearly enough, but just what was necessary and what the alcohol made possible. I stood up, drunker than I thought, but then I hadn't had anything to drink in months. "You all saved my life," I started. "I couldn't find my family, but I'm glad I found you, and I'll try to never let you down."
It was getting to be one of those sloppy drunk moments, and Jack—either holding up better himself, or wishing to avoid any messy disclosures of his own—was the one to shut it down. He raised his glass and toasted, "To the future."
We finished off what we had in our cups.
"Now let's go sleep the sleep of the just," Jack said. The ladies left the way they had come, arm in arm, propping each other up, and Jack took me back to my little cubicle. It was the kind of evening that you wouldn't ever call fun, but you'd look back on it for the rest of your life and know that it was one of the most important times you'd ever have.
##
I AWOKE THE next morning with only the slightest of hangovers, and an overall feeling of well-being, of safety and belonging, that I hadn't had since all this had begun. Not even on the ship.
Although last night's food was about the same as I could've scrounged up on my own, breakfast turned out to be slightly better than I had grown used to. My heart leapt when I saw there were more of the burnt biscuits. I think they were leftover, as they seemed harder and drier than before, but the cooks offset this with their post-apocalyptic version of biscuit gravy: since sausage, bacon, or butter were all out of the question, I had to assume this was some mixture of cooking oil and flour, but it tasted like someone had poured the whole pepper shaker in it, making it a pretty lively addition. The thing they served as coffee was the first hot beverage I'd had since leaving the ship, but it was definitely not Starbucks.
"How you doing?" Jack asked as he came up beside me. "Ready for training this morning?" He shadow-boxed a little, like a great big kid. "Show us your moves? See some of ours?"
"I don't know, Jack," I said, finishing off the coffee, but I went along anyway. I followed him to the second floor, into a small auditorium. There were windows on the east side, with the blinds open, so it was fairly bright. Several people were onstage. Jack gestured for me to sit in the audience section first as he went up. I looked around and saw Tanya and sat next to her. She didn't seem unhappy to see me.
"Thanks for the social hour last night," she said, leaning over. "You need that once in a while."
"I know I did. How's training go here?"
"Jack will start with a big group, going over some basics with them, what everyone needs to know, even those who aren't really designated as fighters. It's like karate, only you really work on hitting the other guy in the head, and breaking holds."
"Did you do karate before this?"
"Dance instructor. Taught little girls to dance ballet, tap, and jazz. I don't like karate moves, they seem unnatural to me, but I guess dancing helped me pick them up better than some people do."
Jack was on the stage in front of ten people, taking them through what looked like a karate class. They obviously had learned to do a series of set moves like a kata at the beginning of class, then Jack had paired people up and improvised a sort of zombie sparring: one person played the zombie assailant, while the other went through set counterattacks, mostly blows to the head. The class went on for some time, then Jack dismissed them and came down to us.
"Hey, you guys, we're going to take a break, then it's time for my advanced students. That means you, big gal," he gave Tanya a couple playful punches on the shoulder, then turned to give me a couple. "I don't know about you, tough guy, but we'll see if we can get you to where you don't need us saving your ass anymore. Meet me up there in five, Tanya."
"Don't worry," she said, looking over her shoulder at me as she walked up to the stage, with the hint of a smile. "Me and Popcorn will have him pretty softened up before you get up here."
Some of the beginning class had sat in the auditorium to watch, but I looked around to see who this mysterious Popcorn was. At the far left side of the auditorium was a boy of about nine or ten, and I knew it was him. There was no word to describe him except "feral"—he was wiry, tanned like he never came inside during the day, and his hair looked like it had never been cut; it just flowed off his head in a shaggy mess past his shoulders. He looked like Mowgli's evil twin. Between the very in-shape Tanya and the Wild Child over there, I could well imagine that Jack was going to be thoroughly worked over. I was glad, as I had no formal combat training, and I wasn't looking forward to getting up in front of these people who clearly had been working on it pretty hard for some time.
A kid pushed three homemade dummies out onto the stage: one in the back, and one on each side. They were crudely made, like scarecrows. Tanya stood in the middle. Jack came out from the right side, next to the one dummy. He was padded all over, with gloves and a Kevlar helmet, so much that he looked like the Michelin Man. Even though it wasn't a very good place to hit a zombie, as Tanya's story proved, Jack's crotch was understandably the most cushioned part. This was clearly going to be full contact.
"Batons," he said, and the kid gave Tanya two police batons. She twirled them, as she shook out her legs. "Ready?"
"Ready!"
"Begin!" Jack walked toward her, trying to imitate a zombie's slow, lurching gait.
I still think that most fights don't look like anything special, but Tanya definitely made it look more like Chuck Norris than I ever could. She took two steps toward the dummy in the back as she brought her left hand across to her right hip. With a snarl, she unleashed a backhanded blow with her left baton that sent the dummy's head bouncing across the stage. Just for effect, she shoved the torso to the ground and kicked it.
With a flourish, she brought both batons up to either side as she took four steps across the stage. She gave a shout as she slammed both batons together on either side of the second dummy's head, then she kicked it in the midsection and sent it sprawling.
She turned to face Jack, who had only taken a couple steps. Tanya strode up to him, and raising the baton in her left hand as a feint, she swung with her right, trying to hit him in the side of the head. Jack, moving a bit more dexterously than a normal zombie did, caught her right arm.
Enraged, she brought the left baton down on his forearm to break the hold. He groaned and let go. Then she gave him a backhanded blow with the baton in her left hand, followed by the blow from her right that she had originally intended. He staggered back, and she reversed it—backhanded blow with her right, forehand with the left. She raised both batons, and Jack raised his hands. "That's a takedown," he said, I thought a little weakly. "Finish your last opponent."
The batons clattered on the stage as Tanya threw them aside. She walked up to the last dummy, giving it a backhanded fist across the face with her right hand, then grabbed its head and twisted it 180 degrees. She jerked it up, just for effect, like she was going to tear its head off, then shoved the whole dummy to the floor.
"Good job." Jack said, and Tanya returned to sit next to me. "Popcorn, you're up in five."
"So what's the kid's story?" I asked Tanya.
"He was one of the people trapped in the city at the rescue center downtown. They didn't have a chance. The city was the last place you wanted to be, once it started. It was just a few cops and firemen downtown, so once they saw they were surrounded, they didn't try to make a break for it. They just fought and fought and kept losing ground, with the building full of women and kids. I saw it on the TV; they were broadcasting it from the news helicopter, but there was no one to help them.
"Popcorn and his mom were some of the ones who tried to scatter and get away when it was all over. They ran from the center, down the street, and climbed up on a dumpster. His mom lifted him up so he could reach the bottom of a fire escape, but before she could pull herself up, they got her. He had to watch her get eaten, hear her screams. I guess it's like Sarah said—at least it was quick. You don't know how much more messed up he'd be if it hadn't been quick. He looks so tough now, but you can hear him sometimes at night, crying for her."
"Poor kid," I said, watching him ascend to the stage. You could never get used to hearing these stories of horror and sacrifice as civilization collapsed and individuals tried to save those closest to them.
"He climbed up the fire escape and broke a window to get inside the building. Lucky for him, it led to the balcony of the old downtown theater. The balcony had been closed off for years, condemned by the fire department, so the stairway leading up to it was locked. There was a stairway up to the roof that was blocked off from the main theater, too, and he could go anywhere there without them seeing him or being able to get to him. He lived there for two months before we picked him up."
"I take it he lived on popcorn?"
Tanya laughed a little. "Actually, somebody called him Jujube first, but he didn't like that, and we found that pissing him off isn't the smartest thing you can do. Yeah, he got lucky again in the theater: the stairway to the balcony had been used to store stuff, and there was candy, soda, and popcorn, and he lived on that. He'd climb up to the roof and build a little fire to pop it. He even found a huge box of drink cups: he dragged those up to the roof and put them all over to catch rain water."
"How'd he start a fire?"
"More luck. The theater had some kids' meal deal where you got a toy surprise, and one of them was a little plastic magnifying glass. He'd take the wrappers and cardboard boxes and use the sun to light them with that. That's how we spotted him—the smoke from the fire. God was looking out for that kid."
"You think?" It was just a little surprising, coming from her, after the story she had told last night.
"What, some shit happens and I can't believe in God anymore? I don't know why He took my babies, and I don't know why Popcorn's still here, but that doesn't mean I can't hear his story and know, deep down, that he's here for some reason, and I'm supposed to love him like my own."
I looked at her intently. Coming from someone else, such a profession would've struck me as trite and ignorant. But from her, I knew it was the deepest wisdom. I even hoped that maybe someday I could share it.
"Give him the stilettos," Jack was saying as they prepared for another round. "He likes those." It was hard to see from this far back, but it looked like the assistant gave Popcorn two thin-bladed knives. "Ready?"
Tanya lowered her voice. "When he was in the theater, one of those things managed to climb up on the dumpster and got in."
"Begin!"
Popcorn's arm came down, there was a _thwack_ , and the dummy right next to Jack had a knife sticking out of its face. It looked like it was right about where its left eye would be.
"Poor zombie," I said quietly.
Jack didn't like the knife-throwing. "Hey! You may not have killed it, and now you're down to one knife!"
"All I need is one," the kid hissed, turning toward the dummy in the back.
"Yeah," Tanya said, continuing her story of the zombie intruder in the theater, "Popcorn whacked it with a broom, but the broom handle broke without bashing its head in."
Popcorn leaped on the dummy at the back of the stage, planting his feet where its hips would be; he grabbed it by the neck and plunged the knife into its eye. As the dummy fell to one side, he jumped off, turned to the left side of the stage, and crouched.
"So he took the broom handle—it was all pointy where it had broken off—and shoved it in its eye. Before he dumped the body, he lowered down a string and pulled the top of the dumpster open, so they couldn't do that again."
"Smart kid."
If Tanya had made knocking the dummies' heads in look like a scene from a Chuck Norris movie, then Popcorn definitely made it look more like the _Matrix_. With a shriek, he ran at the back wall of the stage, took two steps up it, then launched himself at the dummy on the left side of the stage. He stabbed the knife into the side of its head and knocked it over, then kept stabbing it when it was on the ground.
He got up and faced Jack. Popcorn ran right at him, then at the last moment, he threw his feet forward and slid past Jack, like a runner in baseball. Before Jack could turn around, Popcorn had leapt onto Jack's back, where he was pounding both sides of his head, one with the pommel of the knife, the other with his empty left fist.
"Okay! Okay!" Jack conceded defeat, and the kid jumped off him. Popcorn left the stage and walked out of the auditorium. "Okay, Jonah, you're next," Jack said.
I really didn't want to do this. I had no training. All I had done for the last few weeks was hit, shoot, or stab dead people in the head, with no grace or accuracy. Many of them hadn't even gone down permanently dead, but had just kind of flopped around, twitching, and I was lucky to have gotten away from them. My survival in spite of my ineptitude was as sure a proof of God looking out for me as Popcorn's little plastic magnifying glass and soda cups catching rain water. Oddly, as I walked up to the stage, I realized I had never looked at my situation that way until now, though I also realized how little it would save me from embarrassment in front of all these people.
I was in the middle of the stage. "Okay, Jonah," Jack said, "we'll make it easy for you. Give him a bat."
The kid tossed me an aluminum bat. It felt good in my hands, and it was kind of the ideal weapon for whacking dummies in the head.
"Jack," I said, "I don't know about this. Maybe I should just sit this out."
He laughed at me. It was mostly good-natured, I knew at this point, but there still might have been just a little more posturing. From some time in middle school, you learned that guys couldn't completely get past that. "Come on, Jonah, anybody should be able to kill four zombies with a bat, especially when three of them don't even move. Ready?"
"Ready," I said, with no conviction.
"Begin!"
I had seen from the first two combatants that Tanya's was the only logical sequence: take out the one in the back, the one on the left, then Jack, and finally the one on the right. I turned toward the back, took two steps as I raised the bat, then swung it horizontally and took the dummy's head off. I turned to the left, took three steps, and brought the bat down on the other dummy's head. It was all perfunctory and graceless, but at least I hadn't slipped and fallen, or anything else embarrassing.
I turned to face Jack. "I still don't know about this," I said.
"You're halfway there, just finish it."
I closed the distance with him and brought the bat down. Again acting a little more dexterous than a zombie, Jack raised his arms to defend himself from the blow and grabbed the bat with his left hand. This is what I was afraid would happen, and it was starting to annoy me.
I wrestled with him for the bat, then jerked it out of his grip, throwing him off balance. Then I shoved the end of the bat into his face. I think it stunned him more than anything, and he took a step back.
Holding the bat with my left had, I reached across with my right to grab his wrist, so he couldn't block or grab. I jerked him forward again to put him off balance, as I raised the bat and brought it down on his right shoulder blade. It was a savage blow, but it only made me want to hit him more, so I gave it to him on top of the helmet next. I was raising the bat for a third time when he yelled, "Okay, enough there, killer! This dummy's down, go get the other."
I let go of him and walked to the last dummy. I brought the bat down on its head.
Jack walked up to me, his helmet off, and put his hand on the bat. "Easy, easy there," he said soothingly.
"I don't... I never trained... it's just embarrassing. I can do it for real, but not like this."
Jack laughed a little, rubbing his shoulder. "That's kind of the problem; you do it a little _too_ real."
"I'm sorry," I said. "I didn't mean to."
"It's okay, really. I should've known, wandering around out there alone, you wouldn't be ready to fight just for practice."
Tanya had walked up to us. "Guys, not in front of the kids," she whispered. "Make it look good."
Jack knew what she meant and said in a louder voice, "Great job, really got me there." He slapped me on the back and looked out at the audience.
I realized then some of the difficulties of joining this new community. Like Popcorn, I'd picked up killing on my own, but it was a private, emotional, and, most of all, shameful ordeal each time. Now I'd have to do it in front of others, and play at it, and joke about it. It would definitely take some getting used to.
Jack caught up with me later that day. I had gone back to the edge of the river. It was the most enlivening, calming place in our very circumscribed little world. "You okay?" he said calmly.
"Yeah, I don't know what happened."
"Don't worry, I should've guessed before it happened. There are so many things you have to adjust to here, sometimes I forget and do things out of order. Usually it goes the other way. We had so many people here at first who had never hit anyone before in anger. What were we supposed to do with them? There weren't enough of us with training and weapons to protect the place, let alone go out and try to forage for supplies.
"And it's not like when you're a kid, and some jerk gets tired of you being afraid of water, so he throws you into the deep end of the pool and yells, 'Sink or swim!' We couldn't open the gates and shove people out there and yell, 'Kill them before they eat you!' So we trained them. Katas, sparring, dummies—we got them to pretend to hit things. We got them to get good at pretending to be good at hitting things.
"But it was just pretend. And the first time they went out to defend the gates, or on a raid to scrounge up supplies, they usually froze. And crapped their pants. And then, if they managed to pull it together and bash some stiff's head in, and they had brains splattered all over themselves, they usually stopped to puke. And hopefully they didn't get themselves killed. But sometimes they did. And sometimes the guy next to them, too. A lot of times they came back bitten and sick. And guess who had to sit up with them? And when they turned, guess who had to split their skull with an axe, so we could save a bullet?" He stopped and shook his head.
"But you're different," he continued. "You've gotten okay at killing. Not great, but you're definitely not going to freeze if one of those things comes up out of the water right now. But you don't like practicing it. It's not pretend for you: it's _too_ real for you, it's _too_ personal, and you can't pretend. I'm sorry. I should've guessed, but Popcorn was one of the few people we picked up that was like that. Tanya was a little, too. But they were both eager to practice, so they could get better and kill as many of those things as possible. They have a lot of rage that you don't. You're just resigned to it. And that's good. I respect that. And it'll work good here. I think I have enough hot heads and enough people who are too scared. So don't worry, Jonah."
"Thanks, Jack."
"Anyway, since we had so much trouble training people right for combat, that was one of the things Milton thought to change when he arrived. He wanted an initiation rite, I guess you'd call it."
"Sounds a little bizarre."
"Well, don't be too quick. People need rituals, they need some kind of structure or plan, and most of the old ones are gone. We don't have to go to church, or school, or work. We can't vote, or pay taxes, or take tests to get driver's licenses or black belts or whatever. It puts people too much up in the air, and then they get in trouble, either in here with one another, or out there when they're fighting."
"I guess so." Again, the subtler points of the new society were constantly eluding me, but I was trying to adapt.
"So we started making rules, and getting them accepted by the group. Some were straightforward, like you've seen: no weapons, no hoarding food, no stealing, hands off anybody else's man or woman. The usual Ten Commandments stuff. Rules for sanitation and for dealing with infected people. Rules for settling disputes. But once we had those, Milton still thought there had to be something more, something other than prohibitions. Responsibilities, something that made us a community and not just a bunch of people who still had a pulse and had ended up at the same place. So we started having two levels of citizenship. First, if you're part of the community, then you work. You do whatever you're good at it, and we all take turns with the jobs nobody wants. If you don't work, you don't eat. We'd already pretty much been doing that, but we made sure it was a rule, and everybody agreed to it."
"Makes sense."
"Then we had the trickier one, the one that would involve an initiation rite. We decided that anybody could stay here and be protected, so long as they did some kind of work, but they couldn't be full citizens and participate in making the group's decisions unless they fought. If you don't fight, you don't vote."
"Seems kind of harsh."
"I thought so too, but I was surprised how little fuss there was over it. Everyone seemed to like it, even if it meant they wouldn't vote. Like Sarah—she'll never try to get citizenship, and she thinks it's fair that way. I guess the idea of having less responsibility but less rights, or more rights and more responsibility, just made sense to most people. So we've been doing it that way since."
"And this initiation rite?"
"We didn't know what to do exactly. I mean, in the regular world, I guess these things just developed over time; you didn't sit down and make them up. We didn't know whether to make it just symbolic, you know, like being knighted with a sword. People actually didn't seem to like that. They wanted it to be some kind of real, first act of fighting that would initiate you as a warrior and a citizen. But we had to be practical: there wasn't any sense in fighting and risking your life unless there was something more than a symbolic payoff. We wanted it to yield some real benefit for the group, as well as initiating the person.
"So for our 'citizenship test,' we've been sending people out in little groups, without guns, to raid the city for 'special,' non-essential supplies. We go out in force to get the food and fuel, but everybody looks forward to initiation days, because they'll bring back a few little things that make us more human and remind us of what we've got to fight our way back to—soap, CDs, pens and paper. The remnants of civilization, I guess you'd call them. When a group of three or four new people has trained enough that they think they're ready, then we send them out."
"How many come back?"
"All of them. They have something to work for, they're not rushed into anything, and they're ready. And they have a walkie-talkie: if they call for help, we go get them. They just don't get citizenship then. I've lost people on regular raids, but not on initiations."
"And you want me to do this, I take it?"
"Maybe someday. I was just telling you the kinds of things Milton has thought of, and how we live here, because I guess it must be strange, just being thrust into it. I think you should go meet him now. It's always an interesting conversation."
"Yes, that sounds good. A lot better than the practice fighting."
He smiled. "You like the water, don't you?"
"Yeah, I do. Sometimes I forget how much. I guess that's why I always liked working on the ships. I missed my family, but it was a good time to recharge and regroup."
"Yeah, I used to like to go fishing in a rowboat with my dad when I was little. Didn't do it so much when I grew up. Maybe that's why I was never regrouped enough." He looked thoughtful and smiled. "Well, Jonah, I hope someday you and I can get on a boat and do some fishing. But until then, let's take you to see Milton. I think you'll find he kind of recharges you in a way, too. Definitely always gets me thinking."
##
I FOLLOWED JACK to the third floor of the museum, to the end of the corridor where a glass door was labeled "LIBRARY." The room was on a corner of the library, with windows on two sides, and therefore very brightly lit. Wherever there weren't windows, bookcases were built into the walls, and these were full. There were several tables and chairs for studying in the room. Milton was sitting in a comfortable chair, reading, and he rose to greet me.
He was in late middle age, I'd guess in his mid fifties. He was as tall as Jack, but looked like he had always been slim, even before a life of privation. He was dressed in some sort of baggy pants, and a covering like a poncho or smock on top of them: it was really just a big piece of fabric with a hole cut out for his head, drawn tight around his waist with a piece of rope. His hair and beard were completely white, but not as unkempt as most people's.
Overall, he was perhaps the only person I'd met since the end of the world who I would say looked dignified, but he was not especially imposing or mysterious-looking. I guess I had expected Yoda and had gotten Obi Wan.
Milton extended his hand in greeting. He wore wire-rimmed granny glasses, and I had to keep myself from smiling; the only celebrities I could remember wearing those were John Lennon and Heinrich Himmler: though I suppose both had some claim to charisma, neither was a particularly striking or intimidating looking man, and neither was Milton.
In a way, he too was somehow appropriate to our particular apocalypse. The world had ended in such a mundane way, with your utterly ordinary neighbors attacking you and turning you into yet another member of a mindless, anonymous mob. So maybe it made a sort of sense that the new leader or prophet of the apocalypse would be an entirely regular-looking man.
But then I thought of Tanya and her steadfast love of God and Popcorn, deeper than any theologian's, and her Stoical acceptance of her children's horrible deaths, as strong as any Greek philosopher's or Roman statesman's, and I knew there needn't be anything aristocratic or exotic about wisdom. Milton was what he was, and if he had brought some guidance to this community, then I had better respect that.
"Jonah Caine," Milton said happily. He paused a moment, then added, "We have all killed many of our brothers, haven't we?"
"Yes, unfortunately, we have," I said, not knowing where else to go with his reference to my name.
Jack spoke up. "Milton, I'll leave you two alone for a bit. I have some things to do."
"Yes, thank you, Jack," Milton said amiably. "Amiable" was a good word for him. He just seemed easy-going, more than guru-like.
As we sat there, I caught a whiff of something foul. It was the odor we had all grown used to in the last year—the smell of decay and rot, of gangrenous infection and lingering death. The windows to the room were all open, and a breeze was blowing through, but I didn't think it was from outside. We were too high up, and the direction of the wind wasn't right for it to be blowing around from the front, where we had fought and incinerated so many zombies the day before, and it wasn't a burnt smell, either. In fact, I thought the windows were left open to air the place out and get rid of the smell.
Milton noticed my discomfiture. "I'm sorry, Mr. Caine, it's part of my... condition. I'll explain more later, if that's all right."
"Oh, of course," I said, embarrassed, "I didn't mean anything. And call me Jonah, if you want." I suddenly wondered whether everyone was on a first- or last-name basis with Milton, but this wasn't the time to ask.
He smiled. "Thank you. Tell me how you came to us."
I repeated my story more or less as I had told it the night before, though more briefly and matter-of-factly in this setting. Milton watched and listened as though it was imperative to note every nuance and detail and word choice. When I finished speaking, he leaned back and looked off dreamily, seeming to digest everything I'd said, when, of course, the whole story was quite straightforward and uninteresting to me.
"You're so lucky," he finally said. "You must've read so many books as an English professor."
I was taken aback, to say the least. Jack would've asked me about the details of killing zombies or how I survived, Tanya would've asked about my family, and Sarah, well, she probably would've just cried. But Milton was the only person I could imagine who would've been interested in what it was like to be a damned English professor at a community college. It was hard for me to understand, but again, it seemed to fit him.
"Well, yes," I said, "I suppose I did. I always liked reading. I mean, I used to. Not so much anymore."
"I hope you reacquire the habit someday. Let's go over by the window. It should make the smell more tolerable for you as well." He got up and walked past the bookshelves. "I never read books before coming here. But once I was here, it was all I wanted to do."
He looked at me intently again. "You see, there were all these people here, and they seemed automatically to look up to me, like I was a messiah sent to save them. But I'm not, Jonah, and I keep telling them that, but it doesn't seem to sink in. I'm glad I can talk to you because I doubt you'll have that misconception. But they did, right away, once they saw my... condition."
He paused just a moment, looking at his hands. I now noticed they were mottled, like with liver spots on old people, but these were light-gray patches. I still wasn't sure why the people here longed for a smelly messiah with weird eczema, but I was eager to find out.
Milton continued, more calmly again. "Anyway, I was with all these people here, and they seemed to need me to lead them, guide them, help them. And I didn't have a clue what to do for them. So I wanted to read books, maybe get some ideas." He gestured to the shelves. "Most everything here was about science, or nineteenth century local history. That's not what I needed, though a few books on psychology were useful, as were some on the hardships faced by settlers here, and on the wisdom of the Native Americans.
"Fortunately, the traveling exhibition downstairs was on Ancient Rome, so they had a few books on history, literature, politics, philosophy, mythology, religion: that's what I needed.
"And one time when Jack went out foraging for food, I begged him to take me along, so I could get more books. I didn't want to risk anyone else hunting around for books, so I would do it myself, but I had to have more books. Jack looked at me like I was crazy, but even he thinks I have some gift, so he humored me. I needed books on what makes people tick, on what they value, on how they get along with each other. So I read, and I learned. And with Jack's help, we built up the community."
The story was more preposterous than I ever could've imagined. If the guy could perform telekinesis or levitate, it would've made more sense to me. If he was just some glib, charismatic, fast talker who had conned everyone, that would really make sense. I'd hate him for it, but it would make sense at least. But a reluctant messiah who educated himself with books looted from the local Barnes and Noble—it really seemed comical.
Still, I had to admit that, unlike all the religious and political hucksters who had run a con before the dead rose, at least Milton had never taken any money from anyone; he hadn't gotten some revelation that he should be surrounded by a harem and fed grapes, and he seemed quite eager to debunk himself.
So, again, if it worked for this community that I wanted to be a part of, it seemed silly to look down on or ridicule them. But I did want a little more proof of profundity before I embarked on an initiation rite that involved running around, unarmed, among the hungry dead, looking for Air Wicks and silk flowers, or whatever it was I was going to be doing.
"What, exactly, did you learn?" I asked, and it came out a little more accusatorily than intended.
"Jonah, I learned about human nature, which I have to say, I had never once considered before coming here. I don't think many people do. Reading all those wonderful books, I am sure you thought about it before, didn't you?"
He was so simple that it was wonderfully disarming. I still didn't quite understand what the people here thought was so miraculous about him, but I did see how they could come to him with their problems and respect his answers. It was impossible to imagine him being arrogant or selfish or forcing his opinion on someone.
"Yes, I did," I replied. "I'm not sure I came to any conclusions, though."
He laughed for the first time, and it was even more disarming than his little, ingenuous questions. "Good! I couldn't very well talk to you if you had, because I know I don't have answers, only questions, and you'd be quite bored with how stupid I must seem. But if you have questions, too, then maybe we can ask some together. Or does Jack have some eminently practical, necessary thing for you to do? Dig a ditch or build a wall or kill something? He loves that kind of thing so much, it's funny."
"No, no, he didn't mention anything."
"Wonderful!" He leaned back and looked off dreamily again, his own thousand yard stare. I could see that Milton was pretty much _all_ thousand yard stares, and it was nice to see someone enjoying them for a change.
"I think the thing that surprised and interested me the most was how so many people agree that people's souls have several parts. They differ on what to call them, or how many there are, but they agree that there are parts. Had you always known that there were several parts? I found it so amazing!"
Again, his innocence was quite captivating, and also, as Jack had said, energizing, either to someone who hadn't considered the things he was talking about, or to someone like me, who had thought of them too much, in the desiccated, constipated way that academics or experts do. "Well, Milton, I don't think I had been raised to believe that, but I do remember learning it in college, and I remember it was exciting to me then."
"Yes, that's what I mean! What a marvelous life you had, to learn such an important thing so young. It made it so much easier for me to see how people would interact here, or what they would need. Look at Jack: can you think of someone who's more the embodiment of the rational part of the soul? All his plans and schemes and calculations!"
I smiled. "Yes, I think that's his part of the soul, Milton."
"So I know when I talk to Jack, it has to be all logic and business. But he knows now that sometimes he leaves people's emotions out of his calculations, and now we work much better together. For instance, did he tell you about burying the bodies?"
"Yes, he said you did that for the people here, but not for those you killed."
"Yes, but even before I got here, people had argued with him about what to do with the bodies. He wanted to dump them all over the back wall and be done with it. He said we couldn't afford the calories we'd spend digging holes for them, and we didn't have the fuel to burn them. He would compromise, grudgingly, but once I could understand what was going on, how he was looking at it differently, it made more sense to me, and we could discuss it."
He paused just a moment, and I could guess he was switching gears. "And while we're discussing that—tell me, if it's all right, why did you go looking for your family?"
That question took me by surprise again. I was trying to tie it in with his observations on human nature, but then I thought that I should just answer it as straightforwardly as possible. "Because I loved them and missed them?"
"Yes, I know, but I wonder if there was something more going on. You don't think, for example, that you loved them more than all the thousands of people who didn't go looking for their loved ones, who just assumed, once it all started, that they were dead and there was no point?"
"No, no, I don't remember ever thinking that I loved them more than other people loved their families."
"And I believe you were right not to think that. And please don't think I mean that you didn't love them, but it just seems to me that it also has to do with your will being the predominant part of your soul. Once you made a decision, a commitment, then the logic or the emotion involved were really beside the point. You went alone across hundreds of miles with literally millions of walking corpses all around, because you thought it was your responsibility. It was quite remarkable, too. I don't think Jack could've done it, even with all his training. Or Tanya—I don't think she could carry on like that for so long, though think of how much explosive power she has at any given moment. You've seen her fight?"
"Yes."
"It's amazing, isn't it? And a little frightening, to have so much rage. If you were being attacked, hand to hand, who would you want fighting alongside you?"
"Tanya—pure, raw emotion."
"Exactly. Each part has its uses, has a place to fit, and has its own people devoted especially to it."
"And which one is the most important?"
His eyes really sparkled, and he smiled. "Oh, I am so glad you came here! No one here would think to ask me that! They just let me ramble on, and they're so convinced I've been touched by God or some such thing that we never just _talk_! I have to say, I haven't yet read anything that I agree with about that question. I really don't see how you could say one part is better, or more important."
"And which one are you?"
He laughed again. "Oh, my, now a question like that would really not be asked by anyone else here! But so much more fun! I have to be honest and say I'm not sure. It's not a cop-out: I'm really not sure. I definitely have no will at all. Look at how I just went along with being their leader, even though it's the thing I least want! I have some reason, surely, but not like Jack. And nothing like Tanya's emotions. Again, it would surprise the people here, but I'm afraid I'm pretty average."
He had me so intrigued and enthralled by such an abstract conversation. I'd say I hadn't had one like that in months, but really, it had been long before the dead rose since I could talk to someone like this, probably all the way back to my days as a student. "And what about the flesh, our bodies? A lot of religions and philosophies make a lot of that, or say we've made too much of it."
He shook his head. "That's another one where I'm not sure anyone I've read has it completely right. But there is something wrong with the flesh, isn't there? I mean, isn't that what this is all about?" He gestured toward the window. "Look at them out there. Have you ever really looked at them—closely?"
I remembered the girl in the convenience store, and Daniel Gerard, and the little boy outside the gates, and I shuddered. And that wasn't even one day's worth of carnage at my hands. "Yes, I have. And I can't stand it."
Milton nodded. "I thought you had, and there again you see how we're each different. Jack can slice them open or blow them up or set them on fire, all without blinking an eye, because it just makes sense to do so, like cutting off a wart. Tanya can do it because she's so angry at them. Either one could be knee-deep in gore, and it'd be like rain drops off a goose.
"But if you don't have either of those two reactions, the walking dead are most disturbing, aren't they? They're flesh all by itself—without reason, or emotion, or even will—simple, unguided, unadorned. Just flesh that won't lie down and die and go back into the earth. And, I have to admit, it is the flesh that causes me a good deal of pain and inconvenience, and gives the people here all their overblown opinions of me."
I paused, but the suspense was getting to me. "I'm afraid I don't understand, and maybe it's none of my business, but why do they think all these things about you and your... condition?"
He sighed. "Yes, I have to tell you, before you get some crazy, embellished version from someone else here. Please don't expect anything too exciting. It's just another story of coincidences and luck that brought me here, signifying nothing."
I smiled. "Well, I think now that at least it's not being told by an idiot."
He laughed harder than before. "Oh my, that is good! I should've known not to drop Shakespeare references with an English professor, but that one really slipped out without my remembering where it was from! I am so going to like having you here!"
Milton finally began his story. He had been a scientist before the dead rose. It was completely illogical, but I think, like anyone, I had to ask, "You weren't, you know, involved with... what happened?"
"Oh my, no! I worked in biotech, but nothing that was related to what happened. It was just cancer research."
I smiled grimly. "Funny to call it 'just' cancer research, like it was some minor thing you were trying to fix."
He nodded and smiled back, less grimly than I. "Yes, I suppose it is." He shook his head as he told me how he had survived when the crisis had first unfolded. "I was a terrible coward, Jonah. I've told everyone that, too. I just cowered in my townhouse, with everything barricaded as best I could, and I didn't make a sound."
His wife had been on a business trip at the time and was almost certainly dead. For his daughter he had a little more hope. "She had been on a camping trip. Maybe she was far enough out in the wilderness that she might have survived. I like to think so, but I wasn't like you. I couldn't just walk across the country, looking for her. No, I just hid."
Like thousands of people in the same situation—exactly like Sarah, Tanya, and I had done at the beginning—Milton had sat and watched things unfold on the television till the power went off. Then he just sat in the dark—until his cell phone rang.
"It was some government agency," he said. "It was almost as scary as what was going on out in the street, the fact there was a list somewhere of anyone who worked in biotech. But there must've been one, and they'd worked their way down to me. They must've been desperate to be calling someone whose work was totally unrelated to weapons research or epidemiology, but I guess by the time they thought to research what was happening, most everyone with any applicable knowledge was already dead, or walking around without a functioning brain. So at that point they were just trying to find anyone with a Ph.D. or an M.D."
Milton told whoever it was on the phone that he was in his townhouse. The caller informed him that they would be there any second and that he should get ready by the front door. A few minutes later, Milton saw a helicopter hovering over his front yard, with two soldiers sliding down ropes.
They took him to a military airbase, but the undead had already overrun it. They were all over the runway, and the pilot nearly gave up at that point. "But the commander of the group was more calm and in command of the situation," Milton said. "He pointed at a big transport jet that some soldiers were defending, and he told the pilot just to land next to it for a few seconds, then he'd be on his own if he wanted."
The soldiers on the ground formed a ring around the ramp in the back of the plane, shooting, trying to hold back the dead till the group from the helicopter could get on board and take off. The plane was loaded with soldiers, some civilians, and piles of crates and supplies. They flew for a long time, and finally set down at an airstrip in the Rocky Mountains.
Milton smiled, recalling his own naiveté about the area. "I was so surprised to see snow on the highest peaks. I'd always heard of such places in the West, but I'd never bothered to travel. I guess I picked the worst possible time to start.
"When we got off the plane, I could see we were in a place where there weren't any zombies around. But when I looked at the cyclone fence in the distance, I could see shapes crowded there, and I knew what they were. They were everywhere, not just in the cities or on the east coast. We were so isolated there weren't enough at the fence to threaten us, but I knew then there really wasn't any escape from them."
Their group consisted of five scientists and a dozen soldiers. Since they'd been assembled so haphazardly and with no real organization, there wasn't much they could do. "We didn't have any plan, or any way to direct our work, especially since we didn't even work in the same fields. And the supplies were just a random assortment of lab equipment. Every time someone had any kind of an idea that might go somewhere, we had to abandon it because we didn't have the right equipment or supplies."
With such meager resources, they did little other than confirm what everyone already knew—that it was a virus spread by bites. With nothing useful to do, they puttered through the summer and early fall.
"All we really wanted to do was just sit back and enjoy what time we had," Milton said with that dreamy, wistful look of his. "The base was in a breathtaking caldera, one of the most beautiful places I'd ever seen. A spring bubbled up in the midst of the camp. I'd never seen water just bubble up from the ground before; it looked like something in a storybook. There were flowers everywhere. It made me remember that, when I was little, my mother and I would pick wild strawberries in fields sometimes. I guess we were at the caldera too late in the summer because I didn't see any, but I bet there are some there now.
"Every day, I'd wish that we could just plant some crops and stay there. And I'd wish that there were some women in the group, if not for me, then at least to give us some hope of survival for the group in that paradise."
Slowly, it became clear that they couldn't stay there. Their food would eventually run out. The jet barely had any fuel, and they didn't know where to fly anyway. But they did have two old jeeps on the base, and they knew of other facilities like theirs nearby. Initially, they'd been in communication with them, though by the autumn, they hadn't heard from anyone in weeks.
The nearest base was about thirty miles away, and apparently, it was a much larger one. They thought maybe it would have more supplies, so they decided to check it out. Milton went with two soldiers in a jeep.
"There were no zombies at their fence," Milton said, much quieter now, not as animated. "The gate was still locked, but we'd brought bolt cutters for just such a situation.
"Everything was deserted, silent, still. I'm sure you know, Jonah, being out there alone for so long, sometimes the loneliness and desperation make you get so optimistic that it just seems crazy. As I talk about it now, I know we should've run the minute we got there. A completely deserted compound, where we knew there had been dozens of armed men? How could we possibly hope to find something _good_?
"But we kept thinking maybe they had left the place, and maybe there were still some supplies there, and we could just grab them and go. Maybe it really was a good thing that they weren't still there, or they might not want to share with us. We'd tell each other such nonsense as we went from empty room to empty room."
He paused again, and this time he bowed his head and rubbed his eyes. "I'm sorry," he said quietly. "I told you, I'm such a little coward. I should be able to tell a simple story without getting all emotional. You'll have to give me a minute." He shuddered. He was shaking. "It was most definitely _not_ a good thing that there were no people left there. No, no, _not_ a good thing at all."
##
THEY FOUND THE labs in the compound, which were more damaged than the other rooms, with splotches on the floor of what looked like dried blood. The labs contained cages of various sizes, all empty. In one of the larger labs, they found all the equipment smashed and scattered all over the floor, and more empty cages all over. Some of the cages looked like the bars had been ripped apart. Puddles and smears of blood had dried everywhere.
They finally decided they'd seen enough and it was time to retreat. "We turned back to the door to the smaller lab, and then we saw them to our right," Milton said, very quiet at this point of the story. "I guess our eyes had adjusted enough to see in the gloom, or they had stepped forward from the deeper shadows in the back of the lab. Three big dogs. Their fur and skin were torn off in spots, matted with blood. And they had those eyes, you know, like the dead have, all cloudy."
"How could they?" I asked. "The infection has never affected animals."
"I know, that's what we said, but it didn't make any difference. The scientists must've been working on a different strain of the virus at that facility. Or they created it by mistake. I don't know. But the dogs were definitely standing there, and they definitely weren't alive.
"The ranking soldier whispered to back slowly toward the door. As we did, he shoved his .45 in my hand. He told me, 'Do what you can, Doc. I'm sorry we got you into this.' I told him I wouldn't be any better off still cowering in my townhouse, and he just nodded."
The three of them edged toward the door. The soldier who gave Milton his .45 was the same one who had made the pilot land next to the jet back at the airbase. I smiled at how kind and grateful Milton sounded when he described such bravery. I guess, like Jack or Tanya, things had hardened and calloused me more, and I haven't appreciated such little things as another person's bravery. That's where Milton seems to have survived better than many of the rest of us: he can still see the moments of grace and virtue amidst the horror.
But the horror in the lab had continued. The other soldier couldn't take the pressure and ran for the door. Apparently, zombie dogs—if, God forbid, there are more such abominations out there—aren't as slow as zombie humans, and they were all over him. Milton and the commanding soldier started shooting, but in the dark, they weren't too accurate. They probably hit the soldier more than once before they finally put down the dogs.
They went to drag the wounded man out. "But then we heard scraping noises in the main lab, and we saw them." Again, his voice dropped lower. "Monkeys, dogs, cats, even smaller animals like rabbits and rats—all their lab animals, undead, edging toward us. And in the smaller lab, some of the undead scientists and soldiers were staggering toward us. I thought for sure we were finished.
"The commander took the wounded man's submachine gun, so he had one in each hand. He whispered to me, 'Doc, try to take out the human zombies in there. They're bigger targets, you might get lucky. These killer bunnies and shit are going to be the real problem. Damn things have brains about the size of a damn walnut.'
"We started shooting again. It seemed like it went on for hours, though it must've been only a few seconds. We shot until we were out of bullets, then we fought them hand to hand. I'm sure you know—the .45 is a big, ugly piece of iron in your hand, and I did the best I could with it, bashing them in the head. In the end, we put them all down, but we were bitten up by the animals, so it was only a matter of time before it was all over. The soldier who'd first been attacked by the dogs was already dead. The commander took the .45 from me, reloaded it, and finished him when he got back up. He had the decency to say, 'God forgive me,' as he did it. People's decency has also constantly amazed me in all this, Jonah. He told me to do the same for him when it came time, if he went first, and to be sure to save a bullet for myself."
But the attack hadn't killed them, and even though they thought that death was inevitable, the will to survive almost never listens to rational thought or logical analysis—it just fights and scraps. In that sense, the zombies are just exaggerations of what we always do—fight to live, whether we should or not. So the two survivors searched and found bandages, and Milton dressed their wounds as best he could. They took shelter in a different building and found some food.
"We even boiled some water with a Bunsen burner and had tea." Milton laughed and shook his head. "Sipping tea right next to a damn slaughterhouse, with poison coursing through our veins—what were we thinking? But on the other hand—why not?
"It hurt like crazy, of course. Burning, wracking pain, with fever and shaking. We took turns, only one of us sleeping at a time, so the other could shoot him in case he turned. We didn't know what to do if the guy who was awake turned while the other one slept, but we didn't have a better plan.
"After a couple days, we actually seemed to be improving. You know how most people are gone in hours, unless it's just a scratch, and we were both bitten all over. After a week, it seemed as though, so long as you didn't die from loss of blood, the infection from the animal strain wouldn't kill you. I don't know whether the virus had just mutated again, or if this strain the animals had contracted had always been that way, and the people originally on the base had all bled to death, and that's how they became zombies. I still have no idea why we survived.
"But a definite downside was that the pain didn't go away. It was like it had become a low-grade infection, and it would always hurt—forever. It makes it difficult and painful to breathe or move. At least it has for me, ever since. And sleeping—you can just about forget about sleeping. And then we noticed the smell. I mean, we noticed it at first, but we thought it was because we were dying. But it never went away. It's like the rot and death are inside us, never healing, but never quite finishing us off."
The two of them had gathered up some food and headed back to their own base. When they got there, Milton had taken the .45 and the commander had gotten out a shovel, so they could kill the zombies at the gate and get inside. "But that's when we noticed their reaction to us.
"As soon as we got out of the jeep, when we must've still been about fifteen or twenty feet away from them, they all started to cower, with their hands up to their faces, like we were really bright lights that were blinding them. The commander motioned for me not to shoot. We walked toward them, and they started to back away from us. We could just shoo them away from the gate. It was like in the cartoons, when an elephant is stomping everything in sight, and then it sees a mouse and runs away. We were thrilled. Now we could gather food or anything else and bring it back to the camp.
"But when we went in, we found that everyone else had left. The other jeep was gone. They must have waited a couple days, and when we didn't come back, they gave up and went their own way. We waited a couple days to see if they'd come back, and to enjoy our mountain paradise a little longer, then we split up and went our separate ways, to look for them, or any other human survivors that we could help with our new gift."
He'd wandered for weeks before he found the museum. He was lucky, as it was starting to get cold by then, and Milton freely admitted he lacked any kind of outdoors skills to survive in the wilderness on his own, even if he was safe from the undead.
"When the people here saw my gift," he said, smiling, "they thought I was their salvation, and I could do anything. I tried to explain that it was just some wild, unforeseeable side effect of a mutant virus, but I heard them whisper—that I'm sent from God, and I can kill zombies by looking at them, and if you touch me then a zombie can't bite you for twenty-four hours. I can't do all that crazy stuff. I really can't do that much of anything. Sometimes the pain is too intense for me to go out and gather supplies or help people. And even when I do, I can't carry everything by myself, so we still have to risk other people on our raids. And I can't protect them. It's not like I have mind control over the zombies—they just avoid me.
"I assume it's because of some smell, maybe pheromones. They think I'm one of them, maybe even some kind of alpha or king zombie, though God forbid I could be something so horrible as that. But unless someone is pressed up against me, I can't keep the dead from attacking them. I do what I can, and we've lost very few people since I came here, I'm proud to say. My pain is a little price to pay to help build up this community."
He sighed and leaned back. "That's my story, Jonah. I'm sorry it took so long to tell, and I'm sorry I got so worked up, when others have been through so much worse."
"No," I said, "it's amazing. I know you don't like to hear it, but it really does seem like your coming here was a miracle for these people. And since they saved me, now it's a miracle for me, too."
He smiled. "Oh no, not you, too! Aren't there real miracles all around us? Isn't that enough? Why do people look at one thing rather than another and put all the pressure and stigma of 'miracle' on it? It's a miracle the people here survived for months before I got here, or there wouldn't have been anything for me to 'save,' and I probably would've frozen to death during the winter. No one is really a savior, I don't think. I think we just help each other. Or, we're supposed to."
We sat, just staring out the window. It was another beautiful spring day, and I was feeling quite content, even pleased, with Milton's way of looking at things.
Milton broke the silence again after a moment. "Isn't it wonderful that the place they could seal off and defend was a museum? I just assumed, as I wandered around, that I'd find survivors at an army base, or storage facility, or some other place with nothing but concrete walls and barbed wire and fluorescent lighting. But instead, we're here, surrounded by so many beautiful and amazing things, things we can use to teach the children about how life was, and to remind us that life isn't just canned food to eat and a wall to keep you from being eaten. It keeps us from becoming merely animals, I think."
"Yes, I've been thinking that as I looked around here. It was very fortunate."
"Another of your miracles, I suppose?" He smiled. "Maybe so. Maybe, somehow, events conspired to make sure we had just a little more than the barest of necessities. And now I think we need to work on getting more such things, just as much as we need food and other supplies."
"Jack told me about your initiation rite, that you use it as an opportunity to get such non-essentials."
"And what do you think of such a task?"
"I was skeptical at first, to be honest."
He nodded. "I can imagine. I think you have to be a part of the group for it to make sense. I'm sorry you weren't here when we started it."
"Yes, I think that's what I'm realizing. But I think I can see its value. And if you have Jack sold on the idea, then it can't be totally unreasonable, can it?"
He smiled. "Quite right, though he had to be convinced to see its value, just like with burying the bodies. But anyone can see, I think, that people need more than just food and shelter. Allow not nature more than nature needs . . ."
"Man's life is cheap as beast's," I finished the quotation.
Milton laughed. "Now that time I was being just a little naughty and trying to catch you. Do you know I hadn't read Shakespeare since high school, before I came here? And the only play I remembered from high school was _Macbeth_. Witches and ghosts and bloodstained hands—all quite appropriate to how we live now, I suppose. But now I'm amazed at what else I've read in his plays."
He looked up again with his dreamy look. "Isn't that strange—we had all his plays, just sitting around, and I never bothered to read them? And now we have to fight and kill to get some copies of his books and others, books that are blowing around at the smashed-up local bookstore, quickly turning into dust. Maybe that was what was wrong with the way we used to live—so many luxuries sitting around that we didn't appreciate them."
"Too many, and the wrong kind."
"Exactly. I think we should have to work for luxuries and not just be handed them." He gave a mischievous little grin. "Well, there was one time when I thought the people here just needed to be handed some luxuries and shouldn't have to fight for them."
"Oh, when was that?"
"I hardly think it'd be considered too indulgent. We had a long, wet fall last year. By December, everything was all muddy and gross, but not really cold."
"Global warming? I wish that was all we had to worry about anymore."
"Yes. My point is, everyone was feeling down. So I snuck out one night. I'm sure Jack would've blamed me for wasting batteries and risking getting hurt, but I had to try. I didn't know my way around that well because I had been too sick to go on many of the raids, but I did find what was left of one big store. The dead seemed especially docile that night, parting before me as I invaded their little castle or tomb. They didn't seem to cringe from me so much as bow and scrape. I know it's crazy, but maybe even they can be made to feel peaceful, under the right circumstances. I filled a huge sack with various things and dragged it back here. We didn't have a bad Christmas, all things considered."
I smiled. Milton had again lapsed into painting a rather absurd image, a Santa trudging through walking corpses to bring back a bag of goodies from the remnants of Wal-Mart. But again, in a world of unrelenting ugliness and brutality, it had a certain charm, a kind of humble beauty and value that you couldn't ignore.
"But other than that one night, I think perhaps we got too spoiled in our old world, and I wanted a world where we'd have beauty, but we'd appreciate it better, not take it for granted. So I wanted to work it into the initiation, the thing that made people ready to enter the community, by having them show that they could fight and struggle and present everyone with something that wasn't just useful, but beautiful or uplifting, even if it was just in some tiny little way. Maybe _especially_ if it was only a tiny little thing! God, do you remember how pretty women's toes used to look with nail polish on them, wearing sandals in the summer? Now _that_ was something!"
I laughed and shook my head. So did Milton, and for a long time. "Oh my God, Milton, now how did you go from Shakespeare to toenail polish as the essence of what makes us human and separates us from the beasts or zombies!"
"I'm sorry, I'm sorry. I told you I haven't had anyone to talk to, and it all just comes spilling out now. But really," he turned serious again, "I hope that some day we have the place built up enough that we can just take the food and safety for granted, and people can start to work on their own for better things, whatever it is that excites them, whether it's cosmetics or Shakespeare or disco. Wouldn't that be something? To have the luxury again of people being poets, or musicians, or fashion designers, or athletes? Don't you want that for your children? And is it so much to ask?" He got a little quieter. "Did we mess up so bad that our children don't even deserve _that_?"
Talking to him was like careening down a twisting road, and he'd made the next curve and was on to his next phase, or his next role. Now he wasn't the reluctant messiah, or the guy infected with a mutant virus that allowed him to defeat the living dead, or a post-apocalyptic Santa with a toe fetish: now I could see a glimpse of how he could motivate people to organize politically, to have an agenda and make sacrifices to accomplish something and become part of something greater than themselves. I don't think Milton was all the way to an "Ask not what your country can do for you" speech, or an "I have a dream" speech, but he was miles ahead of the "No child left behind," or "I feel your pain" kind of rhetoric to which we'd become so debased and accustomed.
"I think you'll be able to pull that off here, Milton, maybe sooner rather than later."
"I hope you're right, Jonah. And I hope you'll help us."
"Oh, I'm sure I will. You and Jack just let me stay here a little longer before you send me back out there. I've been among the dead too long. Now I'd like to enjoy life here a little. I've been dying to live, I guess you could say."
Milton smiled and rose. "Not my best pronouncement, I hope, but it seemed right as a slogan for people to come in and rest."
"It worked for me," I said as we shook hands.
"I'm so glad it did," Milton said. He led me to the door, and we parted for the day.
##
A FEW WEEKS passed. I moved from my guest room to the main living quarters. I got used to fighting practice with Jack and Tanya in the mornings, and I got to where I could fight without getting all worked up, finally moving with some grace and ease. When I was done there, I'd help out around the compound—hauling water from the river, planting and watering the crops we hoped could help feed us, shoring up defenses where they were needed.
Afternoons I usually spent talking with Milton. He never tired of it. I smiled that it had taken the zombie apocalypse for a scientist and an English professor to appreciate and discuss the finer, more humane points of civilization. On balance, perhaps it was not the most surprising irony one could imagine, but it was still funny.
Some evenings I spent with Tanya, far too few for my taste. She was devoted to Popcorn and spent most of her time with him, reading with him, helping him with his mathematics lessons. It was nice to see them together, as they both obviously needed it very much.
On the other hand, I was constantly reminded of how much I disliked other people's children, and the fact that this one was the most extreme charity case imaginable only slightly offset that. I was honest enough to admit—only to myself, of course, never to Tanya—how much I resented him for threatening or diminishing what little opportunities I had to work on a sexual or romantic relationship with her. But a couple evenings with Tanya were better than I had expected ever to have again, and much better than I deserved.
On balance, like Milton's meager little Christmas among the undead, everything was a lot better than I had dared hope for, under the circumstances, and it made me quite optimistic for the future.
I learned that, spats and posturing notwithstanding, Jack did have an infrequent and more or less completely physical relationship with Sarah. I usually smiled at it, seeing how right Milton was that categorizing people made it so much easier to understand them and work around their idiosyncrasies. Whereas some objective standard of morality might have made the relationship of Jack and Sarah seem wrong, seeing it from Jack's rational point of view made it much easier to appreciate and weigh its merits.
As Jack explained it—in completely rational, cost-benefit terms—the physical relationship was satisfying to both, and the relationship brought Sarah some increased status among the women in the community, by attaching her to the strongest male leader and removing the stigma of not having a mate or children. Though Jack didn't include it in his list of benefits, I could've added that it also removed him from the inconveniences of sexual advances from either males or females in the group, thereby cutting down on jealousy and competition, as well as eliminating any doubts about his potency or preference, making it easier for him to command.
But, of course, to understand a relationship was not the same as to want to imitate it: I wanted something more with Tanya, but was also willing to wait until the opportune time.
Further complicating this was the initiation rite. I learned that Tanya and Popcorn would be the next ones going out. Milton explained they could've added an age requirement to exclude Popcorn until later, but it hardly made sense in his case. If anything, they were afraid he'd grow more sullen and withdrawn if he didn't have this to work for. It made sense for me to go with them, though there was precedence in the community for two people to go out with someone who had already been initiated as a full citizen, if a suitable, new, third person was not ready at the time of their initiation. But I told Jack and Milton that I would be going with them, and everyone seemed happy with that.
The day before our big outing, I was sitting up with Tanya, back in the recreated frontier cabin. Jack had been kind enough to sneak us a bottle of wine on his last raid to the supermarket, and Tanya was considerate enough to me that she tucked Popcorn in early. But the evening was still mostly business. Using various utensils and crockery, Tanya was recreating a map of the city on the table and going over the plans for tomorrow.
"If we head west, we can see what's in this part of the town. The hospital's there." She set a cup down on the table to represent the hospital. "The only other things I remember there were offices and some restaurants, so we might not find anything. By that time, it should be getting sunny and hot, so we should be alone on the streets. Then if we turn south, there's a big grassy hill, with the library in the middle of it. Depending on how far south we go, we can make it across the bridge right in front of the museum, or across the next one down. If we do that, we'll be in the park, but it's just a few hundred yards before we're back here and we're done."
I looked at her. I'd thought she was stunning from the first, but candlelight and familiarity and respect only compounded it; I hadn't been so smitten since I met my wife back in college. "You know, whatever happens, I think it'd be nice if the kid made it back." It wasn't quite what I'd expected to say, but I'd turned into such a lightweight when it came to drinking that it didn't surprise me when I blurted out things I might have rethought, if given the chance. Living on a diet of mostly burnt biscuits and canned peaches, I guess it made sense that a half bottle of wine would go to my head and drop my guard this way.
Tanya's eyes sparkled and she smiled. "Jonah, coming on to me by acting like you're interested in Popcorn when I know damn well you're not? Oh my, you are slick. Jack would just say something crude and flex his biceps, the big, macho doofus."
I put my hand on hers and said, "I love you, Tanya." I really hadn't been planning on that, either.
She raised an eyebrow and leaned close. She was trying to hide it, but she was as tipsy as I was. We weren't the first couple to ease the awkwardness with alcohol. "I think I knew that, Jonah, but thanks for saying it. Most men take way too long getting around to the saying part. And I don't think now's the time to be taking too long saying things."
She put her other hand on mine. Then she leaned across the table, and we kissed, just lightly at first. Between kisses, she breathed, "I love you, too." Then she leaned back. "You're not going to ruin everything by saying, 'Let's make love,' are you?" She smiled as she asked it.
I'm sure I had the usual crushed look that men always have in that situation. "Why not?"
She walked around the table, and I stood up. She started kissing my neck, working up to my ear. "Because it sounds goofy as hell when a man says it." She leaned back and put her forefinger on my lips. "But don't say it the other way, either. Just don't talk." Then she really kissed me.
I'd never been a prizefighter, and I'd never stormed the beach on D-Day or gone over the top of the trench at the Battle of Verdun. But if I remembered the film conventions appropriate to each type of combat, then I knew it was imperative—indeed, a matter of life and death—that one never have sex before the big fight. Yet it was equally imperative to lose one's virginity before going into battle in the war to end all wars. As I kissed Tanya more and more passionately, as my hands found her breasts and then her buttocks, I decided that, even though I wasn't a virgin, what _was_ the war to end all wars if not the one we were in?
That night, I found that Tanya's passionate nature extended to something other than fighting and raising kids. And between her extremely muscular thighs, on some musty, 150-year-old quilt, I got much, much more optimistic about the future.
##
WE WERE UP before dawn, on the museum's roof with Jack, Milton, Popcorn, and some others. Popcorn, Tanya, and I were all wearing denim jackets and jeans for the little protection they'd afford us from bites. In the summer heat, though, it would wear us down quicker, so we each had to carry a canteen to offset dehydration.
Jack handed each of us a walkie-talkie. "The bridge to the north is pretty clear, so we'll send a vehicle across if you call. They should be able to pick you up pretty quick wherever you are. We'll watch all the bridges. When we see you coming across, we'll distract the stiffs away from either the front or the back gate to let you in."
On a table, Jack had set the weapons he had chosen for us. To Popcorn he gave four spikes—huge nails about ten inches long. I remembered having aluminum ones almost that long for sticking in baked potatoes, but these were thicker and looked a little rusty, although the couple inches closest to the point were all silvery, like someone had been sharpening them.
"Remember what we went over," Jack said. "Stab them in the eyes, ears, or temples. And no throwing!"
To Tanya he gave the particularly brutal weapon of a machete. "Careful with that thing: I've been sharpening it every night."
"Oh, Jack, now I know you care," Tanya teased. "Or did Sarah finally come to her senses, and you've been lonely at night, playing with your big, sharp 'machete'?"
Jack cocked an eyebrow at her, but he was smiling, too. "Easy, gal, the stiffs aren't known for their sense of humor, and we need yours back here."
He handed me an aluminum baseball bat and smiled. "I think this has always been your weapon. Be careful."
He took us over to his favorite little addition to the museum, his insane zip line across the river. The sun was just coming up behind us, and I could begin to make out the gloomy streets and buildings of the dead city. Looking down the zip line, I had a clear view of where I'd touch down, and one man was watching the landing zone through the scope of a rifle. Never mind the dead on the other side; the main fear I had was how long the zip line was: it must've been well over five hundred feet across the museum grounds, the river, and into the city. "Jack," I said, "how fast are we going to hit the other side?"
"Not a problem. The slope is really shallow. Some people have actually gotten stuck in the middle, but if that happens, you can just drop in the river, and we can pick you right up. If you're to the other side, you'd be close enough to the ground that you should be able to drop off. But with you three, I'd only worry about Popcorn not having enough momentum, since he's so small, so he should go second. Tanya could knock him along if she had to, and hopefully not lose all her momentum."
Jack handed us each a leather belt. "They're all greased up in the middle, so don't touch them there. Put it over the cable, wrap the ends around your wrists, and give yourself a good push off the edge here. All of you be careful."
"Good luck," said Milton. "Don't take any unnecessary risks, and hurry back to us."
Part of me still thought the whole thing was an unnecessary risk, but I wanted to prove myself to the group. I'd lived for weeks on my own out there; a few hours with two people who were fairly brutal and efficient killers should be easy. I was pretty confident neither of them would get careless, and I knew I wouldn't.
I got up on the edge of the roof and did as Jack instructed. I'd never been afraid of heights, but this was definitely daunting, looking down the four stories to the ground, then scanning the hundreds of feet across the river. There was nothing to do about it but just go. I launched myself forward and started sliding along the cable. As Jack had said, it was a fairly slow and steady ride. I let go as my feet were about to hit the ground, then rolled forward and got up without too much difficulty.
Popcorn was right behind me. I caught his legs as he came sliding across the cable, so I could stop his descent before he let go. I did the same for Tanya. Then all three of us started forward, down the middle of the street to avoid anything hiding in doorways.
As Tanya had planned, we would move straight ahead, up a slight hill to the hospital. The stores and restaurants in this part of town were all thoroughly ransacked, and we saw no sign of the dead as we worked our way west among the abandoned cars.
When we got to the third large cross street, we saw the hospital on the northwest corner of the intersection, definitely a scene of much greater carnage and destruction than mere looting and ransacking. Most of the hospital's windows were smashed out, and from some, the blinds were flopping in the wind. Above others, blackened marks reached up the side of the building, as though many rooms had been on fire, with the flames licking upward. Since the building was still standing, most of this must've happened in the first few days of the crisis, when there was still water-pressure in the sprinkler system to put out the fires.
The wrecked cars in front of the building were especially dense, essentially shutting off the entrance, and many of them were also badly burned, as though they had kept coming there and crashing, even when the wrecks were already two deep and up on the sidewalk. I couldn't help but shake my head at the sign above the entrance: "MERCY HOSPITAL."
I had thought perhaps of checking out the building when Tanya had mentioned it the night before, but it didn't look too inviting now. With such a traffic jam in front of the doors, I couldn't even see any way in. "Come on," she whispered, "let's go. I hated hospitals before this, and now they must be crawling with them. Let's go."
Just then we heard a wail. At a smashed-out window on the third floor, one of the dead had spotted us. It wore a nurse's uniform, covered in blood, and was partly burned all over. It was pointing at us.
The room behind her must've been crowded with other zombies, for at her signal of new prey, they pressed forward and toppled her out the window. She flipped in the air and landed on her back with a horrible, dull thud. The impact was hard enough that her upper torso and arms bounced up, then flopped back down. The fall must've broken her back, as her arms kept writhing and her head kept lolling around, yet she made no move to get up.
As my gaze moved back to the window, another zombie crawled onto the sill. It was burned so badly I couldn't tell what it had been in life. The others pushed it out. Unlike the other, it landed face first, a more fatal landing, as the head snapped back, then flopped forward, and the whole body stayed still. Mercy Hospital was continuing to claim victims.
More zombies stood at the window, flailing and clamoring and killing themselves to get at us. I turned to see both Tanya and Popcorn just as mesmerized by the grotesque display of undead, human lemmings. And I saw, almost at the same time as I smelled its hideous stench, a zombie's rotted hand reaching for the back of Tanya's neck.
As I shoved her to the side, I raised the baseball bat up with my left hand.
The zombie wore coveralls. He was hunched over, his right arm out in front, his head lolled to the left. The right side of his face had been torn off, revealing blackened teeth through flayed flesh and caked blood. The right eye had been gouged out as well, leaving a black, sightless hole. And as I pushed Tanya out of the way, I could see he still clutched a hammer.
Though it happened frequently—either when zombies obsessively clutched things they had been holding when they died, or when their random groping latched onto something—there was still an extra element of terror in seeing a zombie wielding a weapon. It wasn't that it made the creature more dangerous: their insatiable, plague-contaminated teeth would always be their most terrifying weapon.
No, seeing them clutch anything made by human artifice—a hammer, a pistol, or, in one of my most horrible visions, a little girl's doll—was a terrible reminder of what we all tried desperately to overlook every minute of the day: that the undead were not some alien invader that had descended upon us. They were what we, the temporarily living, would inevitably become, each and every one of us—a rotting, tottering, mindless parody of ourselves.
I gave a backhanded blow with the bat. It smashed him across the mouth, sending teeth flying like a shower of bloody raisins, snapping his head around, and making him stagger back. I grabbed the bat again with both hands and raised it high to finish him, but he raised his one good eye to look at me. As usual, there was no emotion or feeling in his eye—not rage, not fear—but, like the hammer in his hand, they held just enough residual humanity to make him both horrible and pitiable.
He reared up, growling, wheezing, and raising both his arms. Whether it was to attack with the hammer or to ward of my blow, it was completely impossible to tell.
With a sudden flash to my right, Tanya's machete went through his left wrist and neck. The hammer, with the bloodless hand still clutching it, clanked at my feet, while his head fell to the left and rolled under a car. The remaining limbs and trunk stayed where they were, the right leg and arm twitching slightly.
Tanya took a step forward and shoved it to the ground. "Son of a bitch! Try to touch me, you son of a bitch!" She wiped the machete blade on the leg of his coveralls, then stood up and spat on him. I remembered my little ritual over the dead, and knew she would think I was either stupid or crazy for doing it, just as I was feeling increasingly uncomfortable with her savagery. But I also knew there was no sense judging her. As bestial as our lives had become, the only question was how to maintain some respect and humanity among the living: whether you did it by granting some tiny shred of respect to the dead, or by completely dishonoring them, that was a choice you had to make for yourself, based just on what drove you the least crazy.
I looked around. Fortunately, Headless had not brought any of his undead friends along.
I looked Tanya over. She was panting, teeth gritted, veins bulging out of her neck, sweat on her brow, and she was holding the blood-smeared machete down with her left hand. Like Milton said, such an unbelievable and frightening amount of rage. I'd never seen a lover decapitate someone trying to kill us; it was as disconcerting as I would've imagined, though strangely arousing in some savage, primal way. I guess you'd know she'd always have your back, and if anyone could raise and defend your kids in this insane, charnel-house of a world, it had to be her. But you really didn't want to piss her off.
I grabbed her by the arm, and we all ran toward the grassy plaza across the street, trying to escape for just a minute the endless violence of both the living and the dead.
The large grassy hill of the plaza occupied six city blocks: three north-south by two east-west. On the corner near where we entered, we saw a playground, with trees, fountains, statues, and benches throughout the park. The dead seemed to have deserted the place in the heat of the late morning sun. As we ran along one of the walkways, the wails from the hospital receded, though I swore I could hear a different, higher-pitched sound from somewhere else nearby. It faded as well as we ran.
We stopped under a tree to catch our breath and saw no dead in pursuit from the hospital, so we calmed down a little. At the westernmost edge of the plaza, at the top of the hill, there was a large, modern building of glass and concrete, four stories high. Popcorn pointed to it. "We going to the library?"
Tanya looked at me. "You want to? I don't see any of them. It looks pretty good, and we can see all around if any start coming from the buildings at the edge of the square."
"Yeah, I think we should. You know how Milton loves books." I looked at Popcorn and smiled, though I knew pleasantries or kidding around were wasted on him. "Besides, you probably need some more books to help kids with their lessons, don't you?"
Popcorn glared at me. "Don't make fun of me, old man, just 'cause you're sleeping with her." I knew I could always count on him to remind me why I didn't like other people's children, tragedies notwithstanding.
Tanya had said they avoided pissing Popcorn off, but like I just said, you really didn't want to piss her off, either. She grabbed him by the ear and yanked him around. "'Her'? Who's she—the cat's mother?" she hissed. "Now, I know you didn't just call me 'her.' And I know you didn't say anything disrespectful."
The kid flushed and squirmed. Even in our mad world, I was glad there were people like Tanya, not just to chop off zombies' heads, but more importantly, to make sure that human relations stayed on the right footing. "No, Ms. Wright."
She yanked his ear so he was facing me. "Now, say you're sorry to Mr. Caine, too."
He glared at me twice as sourly as he had before, with his eyes narrowed to slits and his teeth bared, but his back was to Tanya. "I'm sorry, Mr. Caine."
"Thank you, Popcorn, but I'm not sure this is the time. Let's get going."
We made it up the hill to the library, still with no sign of the dead. All the windows on the ground floor were smashed out, but on the upper floors, almost all were intact. There were no signs of a fire or other damage. With so many windows, and with nothing of immediate, practical value inside, I doubted anyone had tried to hole up in there, so it hadn't been the site of a siege or a pitched battle.
We stepped through the broken glass doors and into the main reference room. It had been cursorily ransacked, with most of the books and magazines still there, just scattered around on the floor, but with the windows gone, everything had been ruined by the weather. With the whole east side of the building being windows, the room was well lit. We still seemed to be alone. "Let's go upstairs," I said.
"You sure?" Tanya never sounded nervous, but she definitely sounded like she didn't like or approve of the idea.
I kept looking around. "There's nothing worth taking here. There aren't any of them on this floor, so they probably aren't upstairs. And if there are, we can always just run out here. And we can still see if any are coming from other buildings, like you said."
I should've remembered Milton's analysis: I was giving logical arguments, and that would've worked with Jack, but I could see that Tanya just had a bad vibe from the place. On the other hand, I also suspected there weren't many buildings outside the museum where you wouldn't get a bad vibe. Most of all, I knew I had to have books—I _had_ to—and Milton had observed the power that my will had over me.
I led the way to the back of the reference room, to the stairwell door. It was a fire door and had a small window, the kind made out of glass with wire mesh embedded in it. The stairwell beyond was pretty dark, but I couldn't see any immediate threats. I motioned to Tanya to get ahead of Popcorn and be ready. I opened the door.
The light from the room made it much easier to see in the stairwell, which still looked empty. There were no sounds of anything moving on the stairs, either. I went in, and we started going up as quietly as possible.
We made it to the second floor landing, and I looked through the window on the fire door there, into a big room of books. It looked relatively undisturbed, with still no sign of the dead anywhere. We entered. At the southern end of the room, to our right, another fire door gave access to a second set of stairs. Some bookshelves lined the walls, and a row of them ran down the one side of the room.
On the other side, by the windows, there were tables, study carrels, chairs, and the dried-up remains of several large potted plants. Only one window was broken, and the plant by the broken window had gotten enough rain to survive and was looking quite hale, while its fellows were just dried sticks. Funny how the same rules apply in any kingdom—plant, animal, or undead. Sometimes you're in the right place at the right time and you survive, while the guy next to you is killed and eaten. And sometimes the guy next to you does fine, and you're dinner. Funny.
Although some of the furniture was tossed about, again there didn't seem to be any signs of a battle or siege—no blood, bullet holes, bodies, or burn marks. With only one broken window, there was no weather damage to the books. I thought of giving Milton directions, so he could come here whenever he liked, since it was a lot less risky for him. I walked over to the row of book shelves, while Popcorn climbed onto a desk by the door and Tanya looked at the shelves along the wall.
On this floor, we were in the fiction and poetry section, which was just what I wanted, of course, either for myself or Milton. We had brought a duffle bag for our finds, and I laid it on the floor, unzipped it, and started tossing books into it. I wanted them all, but books were about the heaviest thing we could choose to bring back. On the other hand, we had to hurry, so I couldn't pore over every choice. My earlier kidding aside, I also knew that Tanya really did like to teach Popcorn and the other kids, so I asked her to pick some as well, while I walked over to the windows.
I looked down on the plaza, and saw no motion whatsoever. It was a hot and sunny day, so maybe we would get lucky and all the dead would stay indoors. From this position on the second floor, plus the elevation of the hill, I could see the roof of the hospital—and the EMS helicopter parked there. I would have to tell Jack about that, though I doubted we could get to it.
Looking back down at the floor of the library, I was again surprised, this time by a woman's purse on the floor by one of the carrels. I gave it a nudge with my foot, and some stuff spilled out—a wallet, Kleenex, lipstick, a bottle of Tylenol, and a bottle of nail polish. Pink with sparkles, to be exact. I grinned as I picked up the Tylenol and the nail polish. I walked over to Tanya, tossed the Tylenol into the bag, and handed her the other little bottle. She wrinkled her nose at me. "I don't think I've worn nail polish since before I was married, Jonah."
"Milton asked for it."
She cocked an eyebrow at me. "Are you saying that... you know... that he's . . ." She shot a glance over to Popcorn and lowered her voice. "You mean that he plays for the other team? Not that there's anything wrong with that . . ."
I smiled. "No, I didn't mean it was _for_ him, I just meant he asked _for_ it. I mean, he mentioned it. I mean... just slip it in your pocket. Books ready to go?"
"Sure," she said as she zipped up the bag.
I tested the duffle to see how much it weighed. It was heavy, but not impossible to carry. We could even pick up something else small, if we saw anything on the way back to the museum.
As I turned toward the door we had come in, I heard Popcorn hiss, and saw him crouching on top of the desk he had been sitting on.
The handle on the door next to him was moving.
Like the zombies on the third floor of the hospital, the ones in the stairwell must've been pushing on the one at the front of their group, because the door suddenly opened and the first zombie staggered in, almost falling on its face.
Before Tanya and I could run to help Popcorn, he'd launched himself off the desk, just as he'd leapt the first time I watched him train. He hit the zombie from the side, driving a spike through its right temple as he brought the other spike down on top of its head. He spun in midair, twisting the spikes and pulling them free.
As Popcorn landed on his feet, the zombie—what had been a middle aged woman, still wearing her glasses—swayed for a moment, eyes rolling back in her head, tongue lolling. Then she slumped to the side.
Tanya and I rushed over to Popcorn as the second zombie made it through the door. Pushed by its fellows, it tripped over the first zombie and fell on its face. I brought the bat down, smashing its skull and spattering its rotten, reeking brains on the front of my jacket.
Popcorn, meanwhile, had thrown himself at the door and was trying to force it shut, but one undead hand was clutching the edge of the door and stopping him from closing it. I got next to him, and we pushed as hard as we could. With a popping and crunching sound, the door severed the four dried-up fingers and closed all the way. The fingers fell to the floor in a pile of desiccated flesh that I knew I'd remember the next time we ate Vienna sausages back at the museum.
As Popcorn and I struggled to hold the door shut, Tanya slid the biggest table over and told us to get out of the way. Popcorn went first, and then I slid aside just as they slammed the table against the door. Eventually, the zombies could push past this barricade, but we only needed time to get to the other stairwell.
Unfortunately, as I looked on in surprise and alarm, the other fire door opened, and in lurched another big, fat, lethal pile of undead flesh.
##
I DIDN'T KNOW where the hell they were all coming from, but it was definitely becoming a problem. It was as if every floor _except_ the first was crawling with the book-loving undead. I raced to the other side of the room, and with a snarl, I shoved the end of the bat into the forehead of the first zombie. It had been a large man, and it staggered into the ones behind it.
Still wielding the bat more like a spear, I hit the zombie in the forehead again. I shoved it all the way back into the stairwell, throwing the other zombies off balance. I shut the door, then threw myself against it. Immediately, the undead began to beat on it.
"Popcorn, leave Tanya to hold that table, and slide one over to me!"
Popcorn sprang to do so, and we secured the door. But now, if we stepped away from the tables on either side of the room, the dead would start to push past our barricades. Plus, both entrances were now blocked, and we had no idea how many were in the two stairwells. I was struggling to make a plan, and half considering just calling on the walkie-talkie to finish it. "The window!" Tanya said.
"Yeah," I agreed. "Go look out the window, Popcorn, make sure the lawn isn't crawling with them!"
He ran to the window. "No," he answered, shaking his head, "nothing out there. It's still deserted."
"Good," I said. "What's under the windows? Concrete? Grass?"
He leaned a little out the broken window. "Bushes. It doesn't look too bad."
"Okay. Toss the bag of books out, then go out the window. Tanya, go out right after him, then I'll go."
She nodded. It wasn't much of a plan, but it would have to do. The windows went from the floor to the ceiling in this room, so it was just a matter of stepping through the broken one and hoping the bushes broke our fall.
Popcorn dragged the bag across the floor and heaved it out the window. He watched it land, then jumped right after it. As soon as he had gone out the window, Tanya was across the room and had stepped through the opening after him. I heard the bushes rustle both times, and no screams, so those were good signs. But the door Tanya had been securing immediately started to open; undead hands wriggled around the edge, their eager, greedy fingers grasping. I could hear their moaning now, rising in pitch, almost as though they sensed victory, and it even seemed as if the zombies pushing against my door redoubled their efforts in response.
I left my table and ran to the window. Better to get out before they could see where I went, so they might not dive out the window after us. If the fall didn't kill us, it definitely wouldn't hurt the undead, and we'd have a mob of them chasing us down the lawn, plus however many their moaning attracted.
Outside, Tanya and Popcorn waited for me several yards away. I took the step out the window. The bushes were a good break for the fall.
The three of us ran down the hill, heading for some trees near the southeast corner of the plaza. I kept looking over my shoulder to see if any of the dead were tumbling out the window after us, but none were.
At the trees, we stopped and looked around. We were all pretty hot and worn out, so we drank from our canteens and tried to calm down a little. No question that it had been tense in the library, but now we were out and halfway done with our work. We moved quietly along the street, heading east toward the river.
As Tanya had said, most all the storefronts were restaurants or office buildings, useless to explore, but one undisturbed window caught all our eyes. It was a toy store, of all things. The window wasn't smashed, and everything looked as though it hadn't been disturbed in a year. Like most non-chain toy stores, it was extremely high-end stuff—Brio, Playmobil, Steiff—all stuff that I'd never been able to afford. I tried the door, but it was locked.
"Too bad," I whispered. "Jack seemed excited about people raising kids."
"Forget it!" Tanya whispered.
"Yeah, I guess so." Just then, a human form emerged from the gloom in the store. He had been an older man. There were no big wounds on him, but his right forearm, mouth, and chin were covered with dried blood. He must've crawled in there to die, locked himself in, and been trapped since.
Something about us really set him off. Maybe it had been his store, and some part of him still regarded us as vandals and thieves. With a gurgling roar, he raised his two bony fists above his head and charged at the door. His head and fists all hit the glass at the same time, and it was enough to smash through. And all that glass shattering and crashing to the ground was loud—really loud.
Loud enough to wake the dead, you might say.
The zombie storekeeper staggered onto the sidewalk with us. Tanya dropped the bag so she could wield the machete better. But she didn't need to. Popcorn was behind the guy, and that's the only opening we needed. The boy sprang onto his back and plunged both spikes into the old guy's temples.
The zombie's hands flew up, his eyes rolled back, and he took one step forward before falling on his face. As soon as he did, Popcorn was back on his feet, but he was pointing into the store and gasping, "Look out!"
An old lady zombie was coming at us through the shattered door. She was hunched over almost double. The left shoulder of her dress was shredded and soaked in blood from two massive wounds on her neck and shoulder. I guessed she had been the old guy's wife. He must've eaten her after he turned, and they'd gotten to spend nearly a year getting cozy in the store together. In a different situation, I would've found their story touching and sad, but right now all I could think was that "Till death do us part" made a lot more sense.
She was nearly on Popcorn, so I swung the baseball bat upward into her face. The blow straightened her up to a standing position, and she staggered back a couple steps into the store. I went in after her and brought the bat down hard. It crushed her skull, splattering her brains onto the wall next to her.
As she fell backward, I turned to go, but I knew that I had to have something for all this trouble. These two territorial bastards had just alerted half the town; we'd be lucky to get out alive now. I grabbed a stack of Playmobil sets with my free hand and stepped out the door.
"What are you, nuts?!" Tanya yelled as I tossed them into the duffle bag and shouldered it. All over, the dead were coming out of doors. Fortunately, there seemed to be a lot more of them back toward the plaza, while the way to the river still looked passable.
"Out in the street! Between the cars!" I yelled. "Popcorn, get up on the cars, you can move faster! Go! Go!"
Tanya was ahead of me, with Popcorn jumping from car to car next to us. The dead were mostly staggering around on the sidewalks, bumping into each other and into the wrecked vehicles, so it wasn't as bad as it had seemed.
"Up ahead, on the right!" Popcorn warned us. A big, dead guy had navigated between the cars and was moving to intercept us. Tanya didn't hesitate. The machete flashed, and the headless trunk swayed a second before collapsing to the pavement.
I heard Popcorn give a yelp. A zombie from the sidewalk had grabbed his left ankle and had tripped him up. Tanya shrieked and ran back toward us as I jumped onto the car bumper.
The zombie was pulling Popcorn by the ankle as it pulled itself onto the car hood to sink its teeth into his leg. I couldn't get a good swing at its head; Popcorn was in the way.
He wriggled around and, shoving the zombie's face back with his foot, he plunged a spike into its left ear. Popcorn twisted the spike around in its brain before pulling it out.
The zombie's head snapped back, its eyes wide, like it had just heard something really interesting. Then it twitched, lost its grip, rolled onto its left side, and slid off the hood, leaving a long trail of thick, black blood across the metal. Popcorn rolled over and got up.
"Get between us!" I said to him as he jumped off the car.
We reached the street that ran beside the river and looked back. The zombies were bouncing around between the cars like marbles in an old pachinko game, working their way toward us, but they were seriously slowed down. We could still make it.
The bridge that led directly to the museum entrance was to our left, but there were a few dozen of them coming from that direction. Considering the slothful habits of zombies, they could have been the remnants of the mob that had pursued me a few weeks before. The other bridge to our right would take us into the park, and we made for that.
When we got to the far side of the bridge, we looked back again. The mob moving parallel to the river had been joined by those zombies that had managed to navigate the maze of wrecked cars down the perpendicular street, and now the growing horde was following us to the bridge. They moved so slowly, but they never tired or got distracted, and they'd never relent, so we couldn't slow down.
The park was foreboding, with too many trees for my liking right now. We started down one of the walkways, and all I could think of was Dorothy and her friends in the _Wizard of Oz_ and "Lions, and tigers, and bears! Oh my!" Unfortunately, we had much worse things to fear in there.
We worked our way along slowly and quietly, looking at every tree as though it were a threat. I kept looking back at our pursuers: they had reached the end of the bridge on the other side of the river. We were doing all right, if we could just keep moving like this—but then I heard a growl. A zombie had come out from behind a tree and was coming for us.
Tanya stepped toward it, raising the machete. She buried the blade in its forehead, all the way down to the middle of its face. She had to put her foot in its chest to pull it out.
Then something hit me in the shoulder, and Popcorn yelled, "Look out!"
I turned and stepped back to see a huge, lumbering figure right at my side. It must've been a motorcycle or equestrian cop because it wore that kind of helmet, with the visor down. Its left arm had been torn off at the shoulder, leaving a dangling mass of ragged flesh with one thick bone sticking out and a bloodstain running the length of its body. In its right hand, it still held its police baton, which is what it had hit me with.
I swung the bat and connected solidly. Its head jerked to one side, but came right back to an upright position. The helmet was enough to protect it from my blows.
Quickly, Popcorn dodged under the cop's raised arm and drove a spike up under its chin. It probably hadn't gone that far into its brain because the zombie started to twitch, and its head slumped forward, looking down at Popcorn as it raised the baton again.
Before it could strike, I shoved it backward. It toppled over and lay on the grass, writhing, perhaps unable to get up.
We started moving forward again. But then, as I looked at a big tree ahead and to the right, an indistinct, dark shadow at its base started to move and resolve into separate, distinct figures.
I pulled Tanya by the elbow to the left, but she pointed to a tree there, where a similarly sized group was rising slowly to their feet. After a few seconds, we saw another group even farther to the left, and another one directly in front of us.
As they ended their siesta and started their pursuit, they set up their low moaning, and my skin crawled. They slowly rose and walked, forming an undead wall between us and the museum. And the pursuing mob behind us was halfway across the bridge. We stopped, then started inching back.
"Uh, guys, we may have to think about getting citizenship some other day," I whispered. "As embarrassing as it might be, I think it's time we asked the cavalry to come get us."
"I don't think they'd get here in time," Tanya replied. "There are too damned many of them, and they're closing too quick. We need a place to make a stand until help gets here."
"There!" Popcorn pointed to a large clearing by the river, with a bandstand in the middle.
"All right," I agreed, "go, go!"
We ran to the bandstand and went inside. Past it, a wall dropped about six feet to the river. We'd be able to hold them for a while, even once they got to us. Like every bandstand, it had only one entrance, with a low fence around the rest of the platform. The platform itself was raised two feet off the ground, and a hedge surrounded it as well, making it harder for them to get at us quickly from anywhere except the one entrance.
"Jack?" I said into the walkie-talkie.
"We're here, Jonah. We saw you come across the bridge. Where are you?"
"On a bandstand in the park. We need help."
"That's okay. Milton went out looking for you. Just sit there one more minute."
From under the trees, the first of them came out into the light of the clearing. As in the convenience store, they were all races and ages and sizes. And they all had only one thought left in their rotted brains—to bite into our warm flesh.
About a dozen had entered the clearing when we heard the moaning trail off. Milton ran toward us, bashing several of them in the head with a large staff. He got to the step up to the bandstand. "Stay close to me!" he said. "Walk along right on the edge of the wall by the river. I'll clear a path through them. You've only got a little ways more to go."
We did as he said. He walked in front and slightly to our right, arms out, trying to keep them away. Fortunately the mob from the bridge still hadn't reached us, so we only had to make it past the ones from under the trees. They would approach, cringe from Milton, but then reach for us as the hunger overcame whatever fear it was that they had of him.
Popcorn was in the very middle of the three of us. I was in front, and if one got too close, I'd have to give it the bat across the head, or simply shove it into the river, though it was hard for me to strike or grab over Milton's outstretched left arm. Tanya was bringing up the rear, and more of them were getting around Milton's right arm to clutch at her. She lashed out with the machete, careful of Milton when she struck. We inched along, with Tanya leaving a trail of heads and arms along the river wall behind us.
After just a couple minutes, we were out of the park and crossing the street to the museum. The cherry pickers were raised again, and as we neared, the gates opened and people came out to help us the way they had the day I arrived. This time, there was no mob of undead; we easily entered, and the gates were secured behind us.
Milton embraced all of us, bloody and reeking though we were. He raised our hands with his, and the crowd gave us a cheer. At least we were home.
Milton walked into the sculpture garden and climbed onto the base of one sculpture. "Friends, we must decide now on the status of our brave warriors, whether their partial victory is enough to grant them citizenship. I can only counsel you that many times we have decided the spirit and not the letter of the law should prevail in our community. Jack, could you please present the case and examine them?"
I hadn't heard about this part of it. Maybe there was no set protocol for deciding disputable cases and Milton was just making it up. It would be like him. I knew that he liked the theatrics of the whole thing, and I couldn't blame him for his showmanship.
Jack came forward and stood under Milton. "Citizens, these three have gone out, fought, and brought back prizes for the community. They did all this with one of their group, as brave as he is, being only a boy. Within sight of our gates, Milton went out to help them—before they could ask for our help. However, they did, in fact, radio me for help before Milton found them. This is the evidence before you today."
Milton called for questions from the group. "How many did they kill?" someone asked.
I paused to count: the one by the hospital, two in the library, two at the toy store, two running down the street, one in the park. But it was hard to calculate. I wasn't sure I'd killed the one on the stairway in the library. And did the two who fell out the hospital window count? And did it count if we only incapacitated them, as I had the motorcycle cop? And I had no idea how many Tanya had killed as we followed Milton. I assumed I should give us the benefit of the doubt, but I was still faltering and trying to think of what to say.
"More than twenty," Tanya said loudly. For a little more effect, she raised her bloody machete and added, "Their rotten heads lie scattered all over the city!" There was a loud rumble of approval. It wasn't an impossible estimate, surely. (She told me later that it would put us over the number killed by any other group, and that she had long suspected that other groups exaggerated, too. I could see that the new community was only slightly less prone to the excesses and deficiencies of the old world.)
"And what did they bring?" someone else asked. A little greed and bribery were worked into the system too, I saw.
Jack unzipped the duffle bag. He could see the contents before the crowd did, and I could see he was considering which item to present first; to create the best effect for us, I assumed.
He held up the Tylenol. There was only a very slight rumble of approval that dissolved into whispers of, "Could come in handy... We don't have much left."
He held up the books, handfuls of three or four at a time. I think he showed some more than once. The approval was less than for the Tylenol.
Finally, he held up the Playmobil sets. There was a very loud, "Awww! We don't have any of those!" Okay, so they were greedy, but for their kids. As human nature went, it wasn't too bad.
After that, I was pretty sure we'd win on a decision, and we did. There was another cheer, and the crowd dispersed. Popcorn took the Playmobil sets and went off to present them to the other kids. Tanya and I stood with Jack and Milton.
"Thanks," I said. "I hope that doesn't compromise the justice here too much."
Jack was his usual jovial self. "What? You feel bad about it? The kids got some stuff for a change. Milton got his books. And I have a feeling Tanya didn't exaggerate the body count too much, hmm?" He looked sideways at her. "People used to inflate body counts when it was real people they were killing, so they could get the budget increase they wanted. No, I don't think we're doing too badly, by human standards."
"And those are the ones we must live by, Jonah," Milton said. "We will be judged—if, God willing, there are ever again historians to judge us—by how we fared compared to other human societies. Don't be so hard on yourself.".
"You all worry about the details," Jack said as he got up to leave. "I have to check the back gate, see how they're doing with the stiffs piling up back there."
"Besides, we didn't even show them everything we got," Tanya added after Jack left. She got out the bottle of nail polish and showed it to Milton. "I'm not sure why, but Jonah said you wanted this."
Milton actually blushed. "No, no, my dear, I'm sure you misunderstood him. I think Jonah meant that he thought you might like it. It's so hard to find and present a gift to someone in our wretched world, isn't it?"
"Yes, I suppose it is," Tanya said, still eyeing the two of us as she slipped the bottle back into her pocket.
Later that evening, we were feted at another of our meager feasts, though someone was thoughtful enough to serve canned hams instead of the less appetizing varieties of canned meat to which we were condemned. Some wag even put together that glop of green beans and cream of mushroom soup that everyone has for Easter. My guilty conscience and nit-picking about rules notwithstanding, I felt almost as optimistic about the future as I had the night before.
##
THE FIRST THING the next morning, I told Jack about the helicopter on the hospital roof. In hindsight, I guess it should have been counted as one of our accomplishments at our judgment the day before, but I just hadn't remembered it at the time. Jack listened with rapt attention and fascination to the details of the hospital, obviously hatching a plan.
"The buildings next to it aren't taller than the hospital, are they?"
I really hadn't remembered to note such details when I was looking at it from the library. "I don't think so."
"I don't think so either, from what I can remember. And the main entrance looks impassable?"
"Definitely. You'd need a bulldozer to push the wrecks out of the way, and even once you did, you'd be fighting off all the zombies trapped inside."
"And if they were lined up to take a dive off the third floor, it sounds like the whole inside of the place is pretty full of them."
"It sure seemed that way."
He paused to think. "Okay. I think we can send a vehicle over the bridge to the north, and have it circle around to get close to the hospital. When the zombies come out of the hospital to investigate, it'll start to drive away. They'll follow it, then you, me, and Franny—the only person here who knows how to fly a helicopter—will sneak up there and fly it home." Jack smiled, obviously pleased and satisfied with his plan. "I even made sure I had a 24 volt battery on hand, so just in case we ever did find a chopper, we'd be able to jump start it."
"But how will the zombies be able to get out of the hospital to follow the vehicle? That was why they were falling out the window: the entrance was so blocked they were all jammed in there with no way to get out."
"Oh, now that's going to be a little bit of a fun part, if you like that sort of thing. Come with me."
Jack took me to the main exhibit hall. On the one side was an archway labeled "GEMS AND MINERALS." I'd seen it every day, but had never been inside; a big metal gate was closed across it and locked. It seemed to be part of the original design of the museum because it was the one room where they kept things valuable enough to warrant such security. I hadn't even thought that the survivors used it for anything special, but today Jack unlocked it and we went in.
The room only had the one door, with no windows, so it was dark and cool inside, like a cave, even during the day. The beam from Jack's flashlight fell on various crystals and gems, which sparkled with an unearthly magnificence. On top of several of the glass cases were laid rows of firearms and ammunition.
"This is where we keep some of the bigger or more unusual weapons, the things that we don't have in the lockers for people to grab just for everyday use." He chuckled a little. "People used to have towels that were only for special company—we have guns and bombs that are only for special company!"
His light fell on a pile of about a dozen wooden crates, each about four feet long and ten inches square at the ends. "We brought these with us when we tried to defend the bridge, but we never got a chance to use them. I've been wondering when they might come in handy."
We stepped close enough to be able to read the lettering stenciled on them. Jack shined the light on one, and it said "M136 AT4 HEAT." He moved the light to another box that was labeled "M136 AT4 HEDP."
"What are they?" I asked, being almost completely unfamiliar with any weaponry beyond civilian handguns.
"Shoulder fired, light anti-tank weapons. If you saw it, you'd probably call it a bazooka. These are much more modern, one-shot weapons. They fire a rocket with a shaped charge, very effective against tanks and other vehicles."
"So why don't you fire them into a crowd of zombies, to blow them up?"
He smiled at my naiveté. "You got to pay attention, professor: I said they were effective— _against tanks_. The projectile carries a shaped charge, for penetrating armor, so it doesn't make a big explosion outward, or send out lots of shrapnel. That's what you'd need to take down a group of stiffs. No, these are pretty useless against the undead, I'm afraid, so here they've sat."
"They'd only be of use against other people, since only people drive vehicles."
Jack looked sideways at me, the light from the flashlight illuminating our faces from below, the way you used to hold a flashlight to scare your sister when you were little. "You know, it's a little reassuring, _and_ a little sad, to meet someone as cynical as I am. But yes, that's just what I've been thinking. If some bad guys came sniffing around here, trying to hurt us or take our stuff, these might be just the thing to make them go away."
We walked back out of the minerals gallery, and Jack locked it back up.
"Okay," I said, "so why show me now, when we're talking about getting the helicopter off the hospital roof?"
"You wanted a way to let all those zombies out of the hospital, so they could follow our vehicle away from it, and we could get in without much trouble. So why not blow a hole in the wall and let them out?"
"I thought you said it didn't make a big explosion?"
"That's why it's good we have both kinds of ammunition. It comes in two main types: 'HEAT' stands for 'high-explosive anti-tank.' That would just make a small diameter hole if we shot it at a wall. But 'HEDP' is 'high-explosive dual purpose.' They made it for taking out walls and bunkers and buildings. Still not much good against personnel, but it would make a pretty big hole in the side of the hospital. And when the dust settled, out would come the zombies. And in we'd go."
"How do you know they're all going to come marching out?"
"Oh, I don't. The blast should skrag any that are unlucky enough to be standing right by that wall. A lot of the ones who are nearby won't be able to hear for hours, so that should help us sneak in. The rest should go to investigate what's going on, like they usually do."
"What if the chopper isn't gassed up?"
"Might not be. Maybe that's why they left it up there. But even if it's down to fumes, we should be able to make it the couple of blocks to get over here."
"Sounds pretty risky. I'm not sure I'm glad I brought it up."
Jack smiled. "Hey, you risked your life for Playmobil. If this works, we're going to have a _real_ helicopter on our roof. That'd make life easier, and safer."
And that was that. The plan was logical, and it had a practical, physical benefit. With Jack, once that was determined, there wasn't much discussion. So we were going the next day to try and get a helicopter.
By mid-morning, I was at the rear parking lot. We didn't leave at dawn, as we didn't expect this to take too long, and we wanted to take advantage of the relative scarcity of zombies in the midday sun.
Jack was inspecting the vehicle and its crew. It was a small dump truck, the kind a landscaping company would have. Apparently, it had been the truck assigned to help maintain the grounds of the museum and the park across the street, so it had been in the museum parking lot when the crisis began; they were lucky enough that there were keys to it in the museum. One man would drive the truck, with three in the back to fire the missile and to fight off the zombies that would try to climb up.
Jack made sure they had plenty of weapons, besides a couple of the AT4s. He was usually very conservative with ammo, as he had been the night I arrived, but to get his helicopter, Jack had armed them heavy, not just with the usual hand-to-hand weapons, but with plenty of guns and Molotov cocktails. This was to be his prize, his legacy to the community, leaving it better protected and provided for.
Once the truck crew was settled, I went with Jack to the roof of the museum. Franny was there waiting for us. She was a tall, blonde woman with big blue eyes. She wasn't gorgeous, and just a tad butch, but with her height, blonde hair, and athletic build, she was quite striking in her own way, especially in fatigues and a flight jacket, as she was dressed now. She had been part of Jack's group and was very competent at work in general and combat in particular. If I had picked anyone for Jack to hook up with, it would've been her rather than Sarah, but I was almost always wrong about such things.
Jack reached under his jacket and pulled out my Glock and my magnum. "A man should have his own hardware," he said, smiling and handing them to me.
The plan was for the truck to go do its thing, then we'd go across on the zip line as the men in the truck led the zombies away from our goal. Jack was carrying the battery; I'm sure he was the only one of us who could, as it was heavy. We watched the truck pull out of the parking lot, the gate closing behind it. It slowly circled around, across the bridge to the north, and into the city.
In a couple minutes, they were outside the hospital. They reported over the radio that a few undead were already investigating them, attracted by the sound of the engine. Then we heard the explosion. They reported that they were driving away with a large, undead crowd in slow pursuit. "If you're going to go," the speaker said, "you should go now."
Two men looking through the scopes of rifles reported that the landing area for the zip line was clear, and we went. When we got to the hospital, there was a hole about four feet wide and five feet high in the side. Up the street, about a quarter mile away, the truck led the mob away from us. We could hear their occasional gunfire, and there was no sign of the dead in our immediate vicinity.
"The first floor should be clear," Jack whispered, "but they're probably still all over the upper floors, so make for the stairwell and up to the roof as fast as possible. And don't make a sound."
We ran across the street and through the hole, into the hospital. As Jack predicted, the remains of several zombies lay right near the opening, either completely dead with glass and masonry stuck in their heads, or just immobilized by the flying debris shredding their legs and torsos.
We moved into the hall, which was thankfully deserted, though it was one of the more horrific building interiors I'd seen. Equipment and furniture was everywhere, with no way of telling whether it had been used as barricades, or randomly flung about by people fleeing, or shoved around by zombies in the intervening months. What seemed like millions of pieces of paper were scattered all over. Blood was everywhere, of course, some of it diluted by the sprinklers to a grimy, pink sheen, and some of it in darker splotches on the floors, or smeared across the walls in huge swatches, sometimes with handprints still visible in them. I could not imagine that the killing floor of a slaughterhouse would look any worse. But, of course, that is just what the hospital had been a few months ago.
The stairwell wasn't that far from where we came in, fortunately. As we moved toward it, a human upper torso slithered toward us in the hall. It was wearing a nurse's uniform, with bloody shreds of cloth, flesh, and intestines trailing off behind it like dried tentacles or tendrils.
I tried to step around it, while Franny brought her boot down on its head with a wet, crunching sound. Its arms, which had been reaching for Franny's foot, flew straight out, spasmed once, then went limp. Franny wiped her boot on the back of its uniform, then stepped away. Unlike Tanya, there was no anger or disgust in her movement: she would've shown more emotion stepping on a roach. I envied her, in a way.
We entered the stairwell and began to climb. It wasn't a huge hospital, only six floors, so we didn't have that far to go. The stairwell was clear, but we did duck at each landing as we passed the little windows in the fire doors, avoiding any undead eyes.
When we got to the sixth floor, we saw that the stairwell didn't go all the way to the roof. We would have to go through the building and find the one that did. Thankfully, this floor looked deserted.
However, the door had been secured. A thick chain ran from the handle inside the hall, between the door and the sill, and then was wrapped around a water pipe in the stairwell. It must've been padlocked or otherwise secured on the other side, after someone had run it around the water pipe and closed the door on it. The door couldn't close all the way, but the chain had been drawn tight enough that it couldn't be opened more than an inch, either.
Fortunately, Franny was carrying a pack with some tools that Jack thought we'd need. She got out bolt cutters, took care of the chain, and we opened the door.
The hall of the sixth floor was not nearly as ghastly as the first. People must've abandoned it sooner in the crisis, so it had not been the scene of such carnage. As we moved along the hall, we did see occasional evidence of violence, with spatters of dried blood at eye level.
When we got to the main nurses' station at the middle of the hall, boxes of baby formula, diapers, and other supplies were neatly stacked on it. We all looked at each other and shook our heads, unable to guess what had happened here.
The stairway to the roof was past the nurses' station, next to the elevators, and it didn't seem to be locked in any way. Before we went up, though, we saw a pair of double doors farther down the hall. They were chained shut and locked from our side.
The sign above the double doors said "NICU." That partly explained why this floor would be less damaged: they surely would've evacuated babies and mothers as soon as possible, thereby leaving the floor abandoned. On the doors themselves, faded red and white "BIOHAZARD" posters had been taped over the two windows, so one couldn't see inside. A crude skull and crossbones had been drawn on the door on the left. It looked like it had been done with a thick, black Sharpie. On the right hand door, "RIP" was written in big letters using the same kind of marker.
We had no reason to lift up the little posters and look inside. No reason other than curiosity, but that reason was too overpowering, even for a supremely rational person like Jack. It was yet another Pandora's box, where it was just human nature to look when you shouldn't.
As we peeled back the yellowing paper posters, I think we all wished we hadn't. We'd all seen horrible things, but I am sure none of us had seen anything like what those doors were meant to hide. That should've stayed hidden till the final trumpet blast.
The room was partially lit with a pale, jaundiced sunlight. Whoever had sealed it up had closed the blinds first, though here and there they'd been torn down. All over the room—lying, sitting, writhing, crawling over each other—was a myriad of the dead, in all the shapes and sizes the human body comes in, and with all the marks of death and decay that those bodies could bear.
Most were restrained in some way, with plastic police handcuffs binding their hands, or with straitjackets, or with bags over their heads. Some had gags on, made of belts or cloths. A few restraints had been torn off over the months, but the dead were too uncoordinated or disinterested to free themselves in their makeshift prison.
Like the nurse on the first floor, many were missing limbs or parts of their faces or torsos, with viscera and other organs spilling out. Their various open wounds had leaked every bodily fluid, and all this mortal slurry had now dried and decayed into a shiny slime all over them. They rolled around in their own insides just as cheerfully or obliviously as they would in a bubble bath, flesh reveling in flesh, with no respect or shame, and with all its hidden ugliness bursting to the surface. "Happy as pigs in shit," was all I could think.
The only thing that could be interpreted as fortunate was that the doors seemed to seal all the smell inside the room.
And like anything revolting, we could not turn away from it, as desperately as we wanted to.
"Shit," Franny whispered, "they tried to store them or contain them."
"I'll have to ask Milton what circle of hell this is," Jack whispered in disgust, pushing the corner of the poster back into place before turning away.
I'd taught Dante enough to know that this chamber of horrors resembled his description of several different circles of hell. But it was most like the last two parts of the eighth circle, where the deceitful are punished. Maybe Dante was right: we'd lied to ourselves for so long about who we are and what we want that now we'd be punished by having our faces shoved into what was basest and ugliest about ourselves—forever. "It's the eighth circle—for liars," I whispered as I turned away too.
"Really?" Jack said as we went back to the door of the stairwell up to the roof. "Remind me never to lie."
The stairwell past the nurses' station only led up to the roof, not down to the other floors, so we weren't too worried about any more nasty surprises. "Hey, grab a box of the formula," Jack said before we headed up. "We can't feed the moms enough to make much milk as it is."
I got up the stairs. The door to the roof didn't appear locked in any way. I put my hand on the handle. "Hey," Jack whispered. "Let's watch it. Get your boomstick out."
I set down the box of formula and got out the Glock. I pushed the handle. The sunlight was bright on the roof, enough to blind us for a second when the door swung open.
Framed in the light was a human figure. And it was sticking the barrel of a shotgun in my face.
Still blinded by the sun, I heard a hoarse voice pronounce, "Say something."
Even though seeing me holding a gun was probably enough to indicate I was alive, it was still a reasonable request when you saw three figures coming out of a building that had been full of the undead for months. "Ummm... hello? Don't shoot?" That's all I could think of.
The barrel lowered. "Wow, you're alive. How'd you get up here?" We emerged onto the roof and stood with the man with the shotgun. He was probably in his late twenties and incredibly thin, with wild brown hair and beard. "Wait—you didn't break my lock into the top floor, did you? Hey—what are you doing with my baby's formula?"
"Easy," Jack said. "We did break your lock, but we can secure it again, if you want. And we didn't know the formula was yours." He set his box down. Franny followed suit, while I had left mine on the step. "We were wondering, though, if maybe you'd want to come with us?"
"I'm not going down there, through all those things! My baby and I are fine up here!"
Though he seemed more nervous than deranged, we all looked around, fearing the guy was a little off. "Where _is_ your baby?" Franny asked. I was pretty sure she was trying to get to the guy's side, flank him, in case he got too worked up and started waving the gun around again. I hoped she was planning on grabbing the gun, and not on shooting him, but I was also sure that she wouldn't hesitate. I gripped the Glock tighter.
The guy pointed to a long ladder that sloped down, bridging the gap between the hospital's roof and the roof of the building next to it. It looked pretty secure, as it had been placed over the top of a metal post on the hospital side, and it seemed to be tied down on both ends. On the other hand, you definitely would need a compelling reason to take a walk on a creaking aluminum ladder six stories up, over a crowd of hungry, walking corpses.
"In the other building," the man answered. "I come over here for formula for her."
"And you've been living like this since the outbreak? Just the two of you?" Jack asked. I could see he was also sizing up any threat this guy might pose.
"Yes, after my wife... died. I didn't know what else to do, once they were everywhere and we were trapped inside our building."
"You didn't see any of our people, or make a signal?" Jack asked, still trying to figure out whether this guy was fully rational.
"I'd hear gunfire sometimes, or a vehicle's engine, but I never saw anyone. This morning I heard a huge explosion, and then when I was over here, I heard something coming up the stairs."
"Wow, that's amazing, that you made it all this time," Jack said. "But why didn't you just take all the supplies over to your building?"
"I thought it'd be good to leave some supplies in both buildings. In case they ever got into our apartment building, we could run over here and pull up the ladder. But I wouldn't want to take my baby downstairs, on that floor with that room full of those things." He shuddered and then we could see a little tent he had set up on the roof next to the helicopter. "So I set up a tent here. I know it's not much, but I didn't know how else to plan for us."
"We understand," Jack said, being both genuinely sympathetic and still trying to calm the guy down. "We know it's been hard. I'm Jack, by the way. This is Jonah and Franny."
"Frank," the guy introduced himself.
"Well, Frank, I wouldn't ask you to carry your baby through a building and a city full of those things, but Franny here can fly a helicopter. Any idea if this thing works?"
Understandably, this news did seem to brighten up Frank considerably. "No, I don't know. I mean, I've opened it up and got inside. I put some supplies in there, too. But I don't know anything about flying one. You mean we can get out of here? Where would we go?"
"We have a safe place, just on the other side of the river," Jack answered.
"May I?" Franny asked, reaching for the helicopter's door handle, and seeming less intent on killing the guy if he acted strange.
"Oh, yes, of course," Frank stammered. "I didn't mean everything here belonged to me, it's just... the formula, it's for my baby . . ."
"We understand," Jack said again, as Franny got into the helicopter.
Frank had turned from enthusiasm to just wonder. "Wow, getting out of here. I had no idea. I thought we'd just stay here until everything ran out, and then... I didn't know what'd happen then. I didn't have any plan. I didn't want to think about it."
"Eighth of a tank of fuel, everything looks fine," Franny said as she climbed out of the chopper. "The battery's dead, of course, but we expected that."
Jack put down his backpack containing the battery and jumper cables next to the chopper. "Frank, can you go over and get your baby? Franny can help you while I get the battery hooked up."
"I've never carried her across the ladder," Frank said, setting his shotgun down on the roof and looking worried.
Jack thought a minute. "There's rope in the pack of tools, right?"
"Sure," Franny replied.
"You got a baby carrier or car seat, something you can strap her into?" he asked Frank.
"Yeah. We bought all that before she was born."
"Okay, then carry her across in that. Run the rope through the handle or strap of it, while Franny and I stand on the two buildings and hold the two ends of the rope."
It still sounded terrifying to me, but it would have to do. Fortunately, crossing the ladder only took three long strides to the other side. Frank and Franny went over to the other building, while Jack and I loaded the formula and gear into the chopper and hooked up the battery we had brought. The helicopter was big enough inside that we even went back for two more trips, to get the supplies Frank had piled up at the nurses' station.
On the second trip, we heard the door to the stairwell at the end of the hall open, and watched as the door was pushed inward. A partially decayed, burnt face peeked through the window in the door, and we could hear the moaning from others in the stairwell.
"Get to the roof," Jack whispered as he set down the box he was carrying and picked up a mop off the floor. "I think it's time we got going."
##
ONCE WE WERE in the stairwell up to the roof, Jack wedged the mop between the door and the horizontal handle, making it harder to open it from the hall. Then we went upstairs. On the roof of the other building, Frank and Franny were returning to the ladder. She was carrying two suitcases, as incongruous as it looked, and he was carrying a baby in a car seat, the kind that snaps out of the base so you can carry it by a handle. We used to have one like it, for our kids. I didn't know whether to envy him for still having his child with him, or to feel sorry for him for having to raise her in this slaughterhouse we live in now. The two of them looked like "normal" people getting ready to go on a trip with their baby. Franny scampered across the ladder with the two suitcases she was carrying and walked over to us.
"Don't say anything to him," Jack said, quiet enough so that Frank wouldn't hear. "They're in the hall downstairs. Just put the stuff in the chopper and get them in and get it started, Franny." Turning his back so Frank couldn't see, he handed me a grenade. "Wait till they're in the stairwell down there, then pull the pin, throw this down, and close the door."
Franny tossed the suitcases in the back of the chopper and went back across the ladder to Frank, carrying one end of the rope while Jack held the other. Frank threaded it through the handle of the car seat and prepared to make his run across.
Holding the door open to the stairwell, I could hear them moaning down there; the mop handle banged around on the doorsill as they tried to wrench it open. I looked down at the grenade in my hand. It was so much smaller than I had imagined grenades were, but I really didn't like the idea of holding it, let alone arming and throwing it.
I looked back to the others. Frank was across, carrying the baby. I heard a cracking sound in the stairwell, and the moaning suddenly got louder.
As Franny hustled Frank and his baby over to the helicopter, he could finally see the danger we were in. "Oh, shit! You didn't say they were already in downstairs!"
"Just get in!" Jack said. "Franny, start her up! Jonah?"
The lead zombie was just coming around to the landing. In a hospital gown, horribly burned, it was using its right arm to pull itself up with the handrail; its left arm hung limp, and its legs moved stiffly.
"Okay," I said. I held the door half open with my foot and pulled the pin on the grenade.
"Frag 'em!" Jack shouted, as the helicopter's engine roared to life.
I tossed the grenade down the stairs. It bounced past the two zombies that were now up to the landing, and I quickly shut the door. The explosion wasn't as loud as I expected. Bits of shrapnel pinged off the fire door, and the dead shrieked as the jagged metal bits tore into their flesh.
I looked over at the chopper, its rotors spinning slowly. Jack held his hand up in a "stop" gesture. "Hold the door another second!" he shouted. "Until she gets going!"
I stayed at the door, pressing against it. As the rotors picked up speed, I could hear renewed motion in the stairwell. A grenade designed to maim and kill the living wouldn't significantly deter a crowd of determined, walking dead. If anything, it would slow them down because they'd be tripping over the shattered, immobile bodies of their fallen fellows. But they would keep coming.
In a few seconds, with the rotors' noise rising to a whine, and with paper and other things blowing around on the roof, I heard fingernails scratching on the metal door—it pushed against me. I pushed back, scrambling to plant my feet, and shouted for Jack's help.
He ran over and threw his weight against the door as well. "Franny!" he shouted.
She was flipping switches and looking all over the instrument panel. "Another minute, Jack! You know I'm supposed to warm this thing up for fifteen minutes before we try to take off! And that's if it's been sitting around overnight, not almost a year!"
"It's going to have to be a little quicker today, Franny!"
Jack fumbled around under my jacket and drew my magnum. "Duck your head, Jonah!"
I hunched down, keeping my shoulder against the door. Jack pressed the barrel of the magnum against it at eye level, angled down a bit, and fired. My ears started ringing from the blast. He moved it over six inches and fired a second bullet, then moved it another six inches and fired again. I wasn't sure he hit anything, but the pressing on the door let up.
"Okay! Come on!" Franny shouted over the engine, and Jack and I ran for the chopper. As we climbed inside, we could hear Frank's baby crying for the first time. As much as I disliked other people's kids, I couldn't blame her.
Slowly, the chopper started to lift off the roof. We were only about a foot or so high when the stairwell door swung open, and the dead staggered straight for us, seeming to find new energy at the prospect of fresh meat. Jack and I shot the two lead zombies in the face before sliding the helicopter door shut. Frank's baby redoubled its wails at the blasts from our guns.
"Franny, come on!" Jack shouted, as two zombies pressed their mangled, flayed hands up against the plexiglass.
"Just give it a minute!" she shouted back.
We were still inching upward, but now there were several zombies pressing on the side of the chopper, making it drift more and more to the side. I didn't understand the aerodynamics of it, but I had visions of us toppling off the roof, to die in a fiery explosion on the pavement.
I looked at Frank's baby. I guess calling a baby beautiful is kind of pointless: I mean, I think it's really the _idea_ of a baby that's beautiful—the thing itself is usually pretty unattractive, unless it's your own. Regardless, I looked at her little crying mouth, her face so wildly pink, her eyes scrunched shut, and her little fists quivering as she held them up, and I knew the only thing I wanted in the world right then was for her to live.
I hadn't prayed during all the various horrors of the previous months, but sitting next to a baby girl on a helicopter that was tottering six stories above the street, with a growing crowd of the hungry dead banging to get in—it seemed like a good time to start. I didn't have the words, which wasn't surprising, as I'd never been much for praying before this, so I just went with the direct approach: "God, let the kid live, please." I tried to stop myself from finishing the thought with other inconvenient details, like, "Unless you haven't killed enough kids already." No sense getting off on the wrong foot when you're praying for the first time in years, and you think you're about to die a horrible death.
I looked back out the window. We were about two feet above the roof and right at its edge. The walking dead were not only moving the chopper to the side, they were causing it to spin slightly to the right. The zombie pushing right on the nose of the chopper—oblivious as ever to the dangers around him—tumbled off the edge of the roof.
In the rear, the tail rotor was swinging around toward the zombies behind us. There was a shriek and a slight jolt, and a forearm flew up through the air. A second later, a huge shower of red shot along the side of the chopper as it took the face off another one.
Franny moved the stick, spinning us the other way, and we finally started to lift faster. But then there was another, bigger jolt, and we tilted down to the left. The main rotor tore into the crowd of zombies, throwing four of them back with their heads nearly torn off. Franny fought the stick and finally righted us, getting us moving straight up and out of their reach.
"Get them off! Get them off!" she shouted. "They're throwing us off balance! The others are getting a hold of them and pulling us back down!"
I looked down and saw what had tilted us so dangerously at the end of our lift off: two zombies were hanging off the left landing skid, and the others on the roof were clutching their feet.
"Everybody grab hold of something!" Jack shouted as he slid the door back. He himself had his right arm wrapped in a seatbelt and was reaching back to me with his left. "Grab my arm! You've got to lean out to get a shot at them!"
Suspended about twenty feet over a crowd of mindless cannibals has to be high on anyone's list of nightmares, way ahead of being in a spelling bee naked. But, if I was going to ask God to help out, I guess it was only fair to pitch in.
I grabbed Jack's forearm and he grabbed mine, then I leaned out the open door with the Glock in my left hand. I aimed at the zombie in front, which looked like it had been a doctor, with a bloodstained white coat. It was hard to aim because the zombies were rocking the helicopter and Franny was fighting with the stick. My shot hit it in the shoulder. It let go with that hand, lost its grip with the other, and tumbled back into the crowd below.
I turned to the other zombie, a young woman in a ragged dress, the left side of her face and neck chewed off. Her head lolled around, making it difficult to get a shot. I fired and hit her in the chest, but she wouldn't let go.
To hell with the head shot.
Her hands, with their death grip on the skid, were definitely not moving, so I aimed at her left wrist and fired again. Her arm ripped away from her hand, which refused to let go. And as the dozens of zombies pulled on her from the roof, her right wrist tore apart too. She fell, leaving her two hands still tenaciously gripping the skid.
The helicopter jerked to the other side as we were freed from the mob's grip, but Franny quickly righted us and pulled us up. We slowly angled off to the north, as Jack slid the door shut. We were all panting from the final crisis of taking off, and the baby was still crying uncontrollably. Frank was trying to calm her down.
"What's her name?" I asked.
"Zoey."
I almost always thought it sounded stupid when someone said that a name sounded pretty, but I couldn't help smiling and saying, "That's a really good name."
We were flying north over the city when I saw, far in the distance to the left, a thin line of smoke. I pointed it out to Jack.
"Wow, we go months without finding anyone, and now two sets of survivors in one day! Too bad we can't check it out." He leaned over Franny's shoulder. "Low on fuel, right?"
She nodded. "Very. I was just looking for the truck, and then we got to get back to the museum."
"There they are," Jack said, pointing off to the right. "They're on the bridge."
Down on the bridge that went to the north of the museum, we could see the truck. The crowd of undead from the hospital was closing on them, but was still forty or fifty yards away. I couldn't believe they hadn't moved farther while we were in the hospital, but I could see now that our ordeal had only lasted a few minutes.
Jack got them on the radio. "Guys, I don't want them following you back to the museum. See that bus in the street, with all the stiffs milling around it?"
"Sure do," came the reply.
"Put a HEAT round in it and then get the hell out of there."
"You got it."
A man stood up in the back of the truck, and a fireball blasted out behind him. There was a huge explosion as the projectile hit a bus in the street. The gas tank of the bus exploded, and a few seconds later, so did another in a vehicle nearby. The dead were trapped among the vehicles, staggering about, wounded, burning, and collapsing to the pavement.
The truck pulled away, across the bridge and back to safety. We followed them, landing on the roof of the museum to the cheers of the community. Jack was basking in his accomplishment, and even had the unexpected prize of two new survivors, one of whom was an adorable baby. I was glad Jack didn't have to run for election, or the whole scene would have crossed over into obnoxious grandstanding.
Though the whole thing had been a complete success, I did not feel as optimistic as I had on the two previous nights. I suppose I should have, since we had saved Frank and Zoey from certain death by starvation, or worse. But there was something in the things we'd seen that tinged the whole day with a weird, diseased melancholy, a glimpse of just how bad things could get, of what was lurking right at the edge of our consciousness, of how much had to be sacrificed and stomped into a bloody mash just to give one child a chance to live on the slightly scorched outskirts of hell, rather than right in the middle of the lake of fire. If we had seen the eighth circle, it only served to remind me that we were, at best, living in the first circle, what Dante would've called "limbo," a shadowy land where those who weren't damned, but who could not be saved, spent eternity in hopelessness and sadness.
As Frank, Jack, and I sat down with a purloined bottle in the frontier cabin that night, my foreboding grew, because I suspected that hearing Frank's story of how he had survived those months would be much worse than I could imagine.
##
WE DIDN'T PUSH him. I'm sure even Jack, as optimistic and cheerful as he was by nature, could feel the dread that hung over that man. I'm sure it was why we didn't invite any others, besides the fact that we needed Sarah to baby-sit Zoey. At first, we talked about the mechanics of survival. It was always a safe topic, as everyone had some story of how they'd found food in the most unlikely ways or places, and everyone was always proud of their own resourcefulness, thrilled and amused by their own good luck.
Frank had barricaded himself in their apartment, throwing all their furniture down the stairway that led to the street. None of his neighbors were home when the attack started, so he kicked down their doors and piled their furniture on to his barricade too, till he had a jumble of wood and metal, all the way from the street entrance up to the second floor. I could imagine that it must've been quite formidable when he was done, thousands of pounds of furniture and other household items filling the stairwell. Apparently, it was the only entrance, and he felt secure that at least the undead couldn't get in.
But he could easily see that his fortress would also be a trap, especially with a baby on board. "I was proud of what I'd accomplished, of course, to build a place that was safe for Zoey and me, but I was really worried about food and water. I put together the food supplies from all the apartments, and it was a pretty good supply, but there was nothing for Zoey. We'd had a little bit of formula on hand—they give you these goodie bags full of different stuff at the hospital when you take the birthing classes—but no one else in the building had an infant, so I had to find something for her fast."
Apparently, after the initial outbreak and complete victory of the dead, the area around the hospital had been pretty quiet, so Frank got the rather dangerous idea to cross over to the hospital roof—like all of us, he had just done what he had to; it only seemed dangerous or heroic afterwards. On the roof of his apartment building, he found some scaffolding and other equipment for cleaning windows and painting tall buildings, including the ladder that he used as a bridge between the two buildings. Like anyone would, he got excited at this point in his story, because it was the kind of plan that would've scared any of us into paralysis before all this happened, but which had saved his baby.
"I knew the sixth floor of the hospital was pediatrics," he said, "since we'd taken the tour before Zoey was born. I'd seen the things, the dead people, on television and down on the street, so I knew they were slow. All I could do was hope there weren't too many of them up on the sixth floor, and I could just grab some formula and run back to our apartment. I was just hoping to get enough for a few days, until help arrived. I got across to the hospital and didn't see or hear anything at all. The whole town smelled like hell—you remember it was summer when it happened."
"Yeah," Jack grunted in agreement, "like Satan's asshole." I couldn't help smiling at him: he probably didn't know the religious concept of the _axis mundi_ —the center or navel of the world—and how some writers had inverted it to the _anus mundi_ in describing places like Auschwitz. But, as usual with Jack's deep reserve of common sense, he had rightly intuited it as the aptest label for our situation.
"That's about right," Frank nodded. "But oh my God, the hospital was worse than anywhere else. You could barely stand it. So many different smells of decay and sickness and death, but with disinfectant and chemicals mixed in. You'd never have been able to imagine it before. It made you gag constantly. I was wearing a bandana over my face, to try to help with the smell, and I was carrying an aluminum baseball bat. I must've looked like some kid playing a game, dressed up like that.
"On the sixth floor of the hospital, I saw one of them there, but I ran right at it and hit it with the bat, and it went down. Another one came out of a room, and I smashed that one, too. And then it was all quiet and still again. There was a bunch of cans of formula right out where I could see them, and I grabbed those and ran. I felt so good. I'd saved Zoey. I thought now we were okay. I could just wait until help arrived."
He paused and drank before continuing.
In the days after his first trip to the hospital, Frank realized he was in this for the long haul. It was then that he wisely secured the top floor of the hospital, as we had found it, so he wouldn't have to fight zombies every time he needed supplies, and so he could retreat if his building were overrun. He had killed the few zombies on the top floor and locked the doors into the stairwells with padlocks and chains, then heaved the dead bodies out a window. In the process of clearing the floor, he had found the same locked room of horrors that we had, but couldn't do anything with it, other than leave it alone.
As with his barricade, Frank was understandably proud of his arrangements, even now, months later. He had everything he needed for his baby girl, months and months worth of diapers and formula, and he had a fallback plan. But now he could see that food for himself was going to be a problem. With ten apartments in his building, he gathered quite a bit of non-perishable food. But it wouldn't last forever, and now he didn't have any plan or expectation of being rescued, and he had effectively trapped himself in the building.
Water would've been an even more immediate problem, but Frank had found lots of bottles of that electrolyte stuff that they give babies, and there were also quite a few five-gallon water cooler jugs over there, though God only knows how he nimbly skipped across that ladder, carrying a five gallon jug. I would've passed out from fear. He had found a couple soda machines in a nurses' lounge and had taken all the pots and pans from every apartment up to the roof to collect rainwater. All told, he was set for liquid for months, but much less so for food.
"I remembered the first floor of our building was a fancy restaurant," Frank continued. "But how could I get down to it? And if I did, I was pretty sure from watching the dead people come and go out front that they had broken in down there. I had one crazy plan, and if it didn't work, I didn't know what I was going to do. I thought if I got in the very back of the apartment right above the back of the restaurant, _maybe_ I could pull up enough floorboards, smash through their ceiling, and make a big enough hole to drop through. Of course, I didn't know the layout of the restaurant, so it'd take a few tries to find their storage room. And all that assumed that when I found it, the storage room wasn't also full of dead people."
"I keep telling Jonah," Jack put in, "the only reason any of us survived is a million little coincidences and lucky breaks. You had your share."
Frank looked thoughtful. And still depressed, but maybe just a little less so. "Yes, I suppose I did. Or it's more like I think that Zoey had hers. I think they were all for Zoey. I didn't want them or ask for them, except for her."
So while his infant daughter slept, Frank got to work on his next insane plan. The dead in the restaurant obviously heard him, but couldn't devise their own plan to kill and eat him, so they just kind of milled around beneath him. "I could hear them as I worked, and when it sounded like there were a lot under me, I'd move to another part of the floor. I was working on a spot, and I hadn't heard any under me. I thought that was good, so I kept trying to get through there.
"Finally, I kicked through the restaurant's ceiling, and one of them grabbed my foot. He was the only one in the room, thank God, and I shook him off. It was a part of the kitchen where all the shelving had fallen across the doorway and kind of trapped him in there. He looked like he worked there—white coat, white pants, like a chef, but all covered in blood. Hell, he's probably still there now."
Before covering up the hole he'd made, Frank had looked through to estimate how far the next wall was, so he could figure where to break through the ceiling again. And when he did, he finally broke into the big storage room with the restaurant's non-perishable food. And it was untouched by the undead. Frank worked quickly and quietly and got everything upstairs without the other restaurant guests ever bothering him. It had been tough on him, but he'd made it and found a way to survive with his daughter.
It was an uplifting, almost inspirational story, but we all knew we couldn't leave it at that. It was another Pandora's box, only this one was now a part of our community, so he had to open up. Maybe only once, and then Jack and I could cover for him and tell people to leave him alone, that he'd been through a lot; but he'd have to tell us, at least. And I'm sure he needed to tell his story, no matter how horrible it was. I was pretty sure Jack was going to make me be the one to ask, so I began. "Frank, was Zoey born right before the outbreak? She looks the right age."
"No, she was born right after. In the apartment." He knew where it was going.
"So your wife was with you in the apartment. You didn't mention her."
He sighed. It was the equivalent of the thousand yard stare, containing all the resignation and regret a human soul could bear, and then that little bit more that it couldn't. Maybe that signal of pain was enough. Maybe I could've told him to forget about it. But we knew that wasn't how it would go. "Yes," he said, "she was, but only for a day after the outbreak."
"What was her name?" I asked, almost in a whisper.
"Mary," he said. It was almost a sob.
We both knew it had to continue. "Frank, what happened to Mary?"
"I don't know if I can tell you. I don't know if I should. Some things are too horrible, even in this world. You'll think I'm crazy. You might even think there's something wrong with Zoey. I don't want that. Blame me. It was all my fault."
"I don't know there is any blame, Frank. We've all done horrible things. I'm sure you know that."
Another sigh. Then he began in earnest. "We were coming back from lunch and looking at baby stuff. We were so happy, and Mary looked so pretty. God, she was so big then, so unbelievably huge, but so pretty. We were almost to the apartment. And this guy in a hospital gown was coming toward us. He had bandages all over him, and he was all bloody. I thought he was a crazy person. He was shoving people out of the way, and it looked like he was snarling at them and lunging at them, but they were just running away and screaming.
"I looked around for a cop or something, but I didn't see anyone. He came closer. I could see then that his eyes weren't right—you know how they look, milky and dead. He grabbed Mary. It was summer, you remember, and she had on a short-sleeved dress. And he sank his teeth into her arm. Oh my God, her scream, it was horrible. And I grabbed him by the throat, to pull him off her, but as I pushed him back, I could see the big chunk of her arm in his teeth, blood gushing down his chin, and I could feel the spray of blood across my face as she spun away from us.
"I held on to him, and he and I fell on the ground. I don't know how, he was a lot bigger than me, but I kept slamming his head into the pavement. I didn't know to do that, of course—I hadn't heard anything about the outbreak yet, or that you had to bash them in the head, but I just did it for some reason."
"Another lucky break for Zoey," Jack said softly. "You got to hold on to that."
Frank nodded. "It was, or he probably would've bitten me too, and then we all would be dead, wandering around out there. When he finally stopped struggling, I went to Mary. I took my shirt off and wrapped her arm, then I took my belt and put on a tourniquet. I had no idea if it was the right way to do it. I hadn't had first aid since we learned it in eighth grade, and then they always told us never to apply a tourniquet unless a limb had been amputated; but the blood was gushing out so badly, with an extra surge with each heartbeat, that I just did it anyway.
"Mary was woozy, but we started toward the hospital. When we got near, I could see more people like the man who'd attacked her. Crazy, snarling people, with hospital personnel and cops fighting with them. And blood everywhere—on them, on the people fighting them, in big puddles on the sidewalk. I'd never seen anything like it. I thought maybe it was some kind of riot or civil war or mass outbreak of insanity, and we should just go inside and wait for things to calm down, and then I could get Mary to the hospital or doctor. So we went into our apartment building."
Being next to the hospital, Frank and his wife were practically at the epicenter of the expanding pandemonium and despair. They watched the television and learned about the plague and the bites. And, of course, Frank's poor wife saw what it meant for her, and what it meant for their baby.
Frank's voice kept getting quieter as he told his story. "She looked at me, pleading, but still in complete control of herself, unlike me. 'Please, whatever you do, don't take off the tourniquet!' I remembered you were supposed to loosen it every few minutes, and never leave it on for hours at a time, or the limb would die and have to be amputated, but I didn't see a choice at that point. I didn't have any idea what to do."
That's when he had started building the barricade on the stairs. It gave him something to do, to feel useful, like he was accomplishing something in the face of all this madness and pain. He still thought he might be able to get his wife out the fire escape and take her to a hospital somewhere, even if the one next to their building was overrun. Frank shook his head sadly. "I didn't want to have to fight anymore of them. I had no guns or weapons in the apartment. I only found the shotgun later in one of the other apartments. I fell asleep next to Mary on the couch. I hope that I kissed her then, that last night, but I don't remember if I did or not, I was so exhausted.
"When I woke up, I didn't see her. I went to the kitchen. She was sitting at the little table there. She'd made coffee." He looked up, fighting back the tears even as he smiled grimly. "Isn't that funny? I guess she'd gone all those months of being pregnant without it, she really wanted some. And on the table, next to her coffee, was the biggest, sharpest knife we had, with the handle pointed away from her. She was really pale, sweaty, and she was having trouble breathing. 'You're going to have to deliver our baby,' she whispered.
"I still hadn't caught on to what she meant. She wasn't due for two weeks. She pushed the knife toward me, and she talked really slow, over-pronouncing each word. 'Frank, you need to deliver it now, before it turns into... before it dies.' I finally got what she was asking. I told her I couldn't just cut her open while she was still alive. I couldn't do that. God, it was too horrible. A person can't just do that."
He stopped a minute before continuing. "Oh God, she was always the decisive one, the logical one. She was so weak from the loss of blood, she was swaying slightly, but she still seemed to be considering what I'd said, and coming up with a solution. 'You're right,' she whispered. 'I understand. But, so we're clear—you promise me that if I die, you'll do anything to save our baby?' I told her of course I would. She nodded, really slowly. Then she snatched up the knife, and jerked it across her throat.
"A stream of blood shot up from her neck, a fountain, so much blood, all the way up the wall; then, as she slumped forward, it gushed out onto the table. I screamed and tried to stop it, but it was obvious she was going to be gone in just a few seconds.
"I held her as she went limp. We were both covered in blood, all hot and sticky, with that metallic smell everywhere. I tried to focus, though, so I picked up the knife. If I didn't know how to apply a tourniquet, how was I going to perform a C-section with a kitchen knife? I was shaking wildly, but I tried cutting her across the belly. I was so nervous and scared I barely broke the skin.
"I screamed like a girl when she grabbed my wrist and lunged for my neck."
##
FRANK WAS SHAKING as he continued. "We were both covered in her blood, so my hand slipped out of her grasp and I staggered back. She tried to come at me again, but the floor was covered with blood, and she slipped and fell. She was growling and moaning. I didn't know what to do. She got to her feet and took a step toward me. But then she stopped.
"She kind of looked up, sort of swaying where she was standing. And sniffing. She was sniffing at the air. Then she looked down at her belly. She put her hands on it, kind of rubbing or patting it, and she gave a low growl, kind of like purring. She'd lost interest in trying to eat me. I had no idea why, at first."
As if what we'd heard already wasn't bad enough, I could unfortunately now guess what was coming. "Oh, no, she didn't," I whispered. But there was no more closing the box. We were going to see it through to the end.
"Yes. She did. She realized there was other prey in the room, much closer than I was. Her fingers went into the gash I'd made in her belly, and she started ripping it open. She peeled the skin back, but then she really had to strain and claw to pull the muscles apart. Blood was gushing up around her hands in thick streams. She was reaching around inside, and then she started pulling out stuff in gobs and flinging it on the floor.
"Finally, her head went back as she got both her hands in there and dug around till she got a hold of our baby. She gave this horrible, animal howl as she ripped it out of herself. Then she was holding it, looking at it, with her mouth hanging open. Its tiny arms and legs were moving as she growled and raised it up to her mouth to eat it.
"I was frozen. I admit it. It wasn't just that I didn't know what to do anymore. I just couldn't think at all. All I could do was watch. But then I heard it cry. Not loud. Kind of just a little squeal, sort of like a kitten would make. And that jolted me out of it.
"I lunged at Mary. I grabbed her hair and jerked her head back, to keep her from biting the baby. I was going to stab her under the chin with the knife, but we both slipped in the blood again. As we fell, her head smashed into the corner of the stove. And it was over. She didn't move anymore.
"I looked around. I'd never seen anything like it. I don't think anyone ever had. The kitchen looked like one of those posters some anti-abortion person would hold up in front of a clinic—blood and fluids and stuff splattered all over. I cut the cord and got Zoey out of there. We were so lucky the water was still running at that point, so I could clean her off. I looked at her, to make sure she wasn't—you know, one of them. She looked fine. But what's going to happen to her? The first thing she saw in the world was her father killing her mother. That isn't right. That can't be good for you."
By this point, both Jack and I were sitting next to him, rubbing his shoulders. "It's okay," Jack said. "You did what you had to. You saved your daughter. Your wife understands. She's happy with what you did for Zoey. It's going to be okay."
Frank looked at me. I think he'd figured out Jack was the optimistic one, and now he wanted a more sober opinion. I never liked being the bad cop, and especially not in this situation. "What do you think?" he asked me. "I can see that you know there's something wrong with what happened. Tell me the truth."
"There's something wrong with all of us," I said softly, "with everything around us. You can't put the guilt for all that on yourself. I don't even think it's a matter of guilt and innocence anymore. It's a matter of just trying to keep beautiful things alive in an ugly world. And you did that with Zoey. I don't know what else anyone could ask of you at this point."
He shook himself out of it a little, but only to feel guilty about the present rather than the past. "I should be with her. I shouldn't be here drunk. Take me to her."
Jack and I took him to the little room, where we'd left Sarah with little Zoey. Sarah cocked an eyebrow and shook her head at us, with her mock, motherly chastising that she often heaped on us men. We tucked Frank in next to his baby. And although I felt something much more like resignation than optimism, I did feel happy that we had Zoey and Frank here with us. Even though the events Frank had described were ultimately unexplainable and nearly unendurable, the beauty of his daughter and his love for her were as real and as powerful as any of the horrors around us. We would help him to see that, if nothing else.
The following weeks were for Frank what the preceding had been for me. I knew that Jack and Milton had agreed not to mention the initiation rite to him right away, so he'd have time to adjust to the community. He made friends fairly easily and quickly. It probably helped that Jack and I never told anyone what he had told us that night. We just let him ease into a spot in museum lore: the mysterious man who'd lived the longest in the city of the dead, all the while caring for an infant child.
I don't suppose you ever get over something like Frank had been through, but he definitely seemed to pick up around other people, especially the others who had small children. It's a little embarrassing, but unlike me, Frank actually seemed to like other people's children. He also seemed very much to enjoy working for the community, and whenever he wasn't with Zoey, he was always doing something to help others.
Zoey took to everyone, and of course, everyone loved her. I had never seen such a happy infant, as though her disposition were designed to be the opposite of the horrific events that had surrounded her entrance into this veil of tears. She took her first steps as I watched. She played with other babies for the first time in her life. I think Tanya knew that Popcorn would not tolerate any perceived threat to the attention she showed him, so she didn't dote on the baby particularly, but Sarah had no such qualms and was with Zoey constantly. It seemed to brighten her up immensely, and I was glad for her. Jack was maybe a little less so, but he was more than good-natured enough to make it into a joke: he said if he kept doing Sarah like he had been, she'd soon have one of her own.
The Fourth of July came, and Milton again snuck out, as he had told me he had at Christmas, to try to make things a little special for us. Just some noisemakers and sparklers from a party store, but it certainly brightened up the sultry, summer night. And our garden had yielded some small watermelons for us to share that evening. Even the dead seemed just lazy and not so belligerent under the hot and dry skies. Most days, Milton would just go out and shoo them away from our gates and into the cool shadows of the park across the street, rather than sending out people to try and kill them. It was shaping up to be a good summer, all things considered.
But we knew there would soon be another mission outside the compound to check out the smoke I had seen when we got the helicopter. Evidence of more survivors immediately raised Jack's feelings of responsibility—he had to help them as he had helped all the people he'd already brought into the community. And, as with so many other things, there was the overwhelming power of curiosity; once you knew there were survivors just a few miles away, you had to see who they were.
I understood all that, but I began to wonder if maybe we had enough people here, and we needed to focus on helping them, and building up what we had, rather than trying to find others. We couldn't save or rebuild the whole world, after all. But in the end, this was all neither here nor there, for I knew I would back whatever plan Jack came up with. It wasn't his grandstanding or charisma or even his logic; it was just the gratitude most everyone in the community showed him for saving all of our lives at various times.
Based on the situations they'd rescued people from previously, Jack wanted to take just a small group: big enough to break through a few besieging zombies if we found survivors barricaded in some building, but not enough to use up extra fuel with more than one vehicle, or to weaken the defenses at the museum. Tanya wanted to go, mostly to kill more zombies, and I wanted to go, both to help Jack, and, I had to admit, to be by Tanya. As kind of an old-school soldier, Jack certainly frowned on the latter motive, but he also understood it was going to be nearly unavoidable in the very small, domestic army he'd assembled here.
Popcorn wanted to go for a combination of the motives of Tanya and myself—to kill more zombies, and to be near Tanya. If it had seemed like a more dangerous mission, Jack would surely have objected to him coming, but by that time, Popcorn had been on a couple of their raids for food, as well as the raid to the local airport to get fuel for the helicopter.
As the only pilot, Franny was ineligible for missions until she could train someone else.
Finally, Frank wanted to go, and Jack welcomed the opportunity to make him feel more a part of the community, even if technically he had not been through the initiation rite.
We took a smaller vehicle, a jeep that belonged to another person in the museum community. This would be farther than anyone had ever ventured from the museum, so we left early. The plan was to circle way to the north before turning west, avoiding the city proper entirely. We weren't optimistic about finding anything on the first try. We didn't know exactly where we should look, so we'd need to see the smoke again to zero in on them. In the heat of summer, it seemed unlikely that they'd be burning fires all the time, so we hoped that, if we left early enough, their fire from the previous evening's meal would still be smoldering in the early morning light.
Sure enough, that is what we saw as we weaved between abandoned cars on one of the roads north of town—a barely visible line of white trailing up to a long, faint smudge where a breeze had spread it during the night and early morning.
I watched Jack's face as he drove us closer to it. He was so laughably easy to read. It was obvious that he thought something was not quite right about the location. When we were close, he turned off the main road, down a side road, and across a field, till we were way out in the middle of the field, next to some scraggly saplings at the base of an electrical tower that rose high above us.
From here we had a good view for a long ways around, and saw no movement of any kind. I noticed the saplings and the bushes would hide the jeep from view, and that there was a hill between us and the source of the smoke. "Let's park it here and scout around before we go driving up," Jack said, trying to act nonchalant about it; I could tell there was a change of plans by what he said. "They ought to be just over that rise, whoever they are."
"You don't want them to know that we have a vehicle?" I asked as we got out of the jeep.
Jack looked at me sideways. "We don't know the situation. I don't want to advertise that we're here until we do. Normal tactical decision."
I remembered Jack's comments about the anti-tank missiles being better suited for use against other people than against the undead, and I sensed that this was a strategic, and not a tactical decision. But I let it drop in front of the others. We were still pretty far away from whatever it was. It couldn't do any harm to walk over the hill and check it out. There was a stand of trees a hundred feet to our left, and another way off to the right, but otherwise we were in the clear and could see any trouble with plenty of time to react or flee.
As we walked through the tall grass, I looked back and saw that Jack's parking job had in fact hidden the jeep, as I had thought it would. We came up and over the brow of the next hill and could finally see, maybe a little less than a mile ahead of us, at the top of another, lower hill, the source of the smoke. It was coming from a large building behind a high, gray wall, several stories tall. The wall also enclosed a water tower and some smaller buildings. The wall was surrounded by two enormous rectangles of cyclone fencing, both topped with razor wire.
"The regional correctional facility," Jack said. "I saw the signs on the main road as we got close. Our new neighbors."
We walked a few more yards downhill toward the prison to get a better view. "The gate must be on the other side," Jack said as he scanned with the binoculars. "I don't see any movement."
"Nice people being eaten alive, and some bunch of rapists and perverts gets to ride it out in style in a fortress I helped pay for," Tanya said with disgust, absentmindedly chopping at the grass with her machete. I'd never thought of discussing politics with her, as it seemed pretty moot by the time I met her, but I could see she was going to be a little to the right of most of my college professor friends. That didn't seem all that bad to me at this point.
I heard a whistling sound, then a thwack, and then Jack cried out in pain—an arrow stuck in his left shoulder. Another thwack, and he staggered with a second arrow in his left thigh.
I raised my pistol, but I could see nothing among the shadows in the trees, and then I felt a shooting pain in my right chest as an arrow hit me.
"Drop your damn guns," a voice came to us from the shadows.
"Jack?" I asked, not wanting to be the one to surrender, still scanning the trees for a target.
"You got to," Jack rasped. "We can't see them, we can't shoot back, you got to."
Jack dropped his pistol and raised his right arm, and his left arm part of the way, in surrender. I dropped my pistol and raised my arms, and the others did as well.
Nine men emerged from the trees. All carried various hand-to-hand weapons like clubs and knives, as well as homemade bows. All were stripped to the waist. Most were covered with tattoos. What seemed to be their leader, a tall black man, picked up our firearms. The others hung back and kept their arrows trained on us.
"Who are you?" the black man asked.
"We were just out foraging for supplies, and we saw the smoke and followed it here," Jack wheezed.
"Where are you from?"
"There's a bunch of us holed up in a compound to the west," Jack said, lying about our location.
"Well, you're not going back there today," the leader said. "You four, get going. That way." He pointed off to the right and toward the prison.
"What about him?" asked one of the other men, pointing at Jack.
"He can't move fast enough. Leave him." The black man bent down and twisted the arrow that was still in Jack's thigh. Jack winced and gritted his teeth through the pain. "If you do hobble your sorry ass back home, tell them not to mess with us." He got up and turned to the others in his group. "Get the deer and stuff," he said, and four of the others went back to the trees. When they emerged, they were carrying two deer, suspended by their feet from two long poles. Another pair of men went back and returned with large bundles of firewood.
They made us walk along ahead of them. I managed to pull the arrow out of my chest, and I could apply pressure to the wound as I walked, so I wouldn't lose too much blood. The arrow was rudimentary—no real arrowhead, just a sharpened wooden shaft, so it didn't have any barb that would tear a bigger hole when I pulled it out.
We walked along for twenty minutes, circling to the right, and they made us stop by a big drainage pipe that stuck out of the hillside under the prison. It was concrete, about four feet in diameter, and it had a metal grate across it. A stream of dirty water ran out of it and down the side of the hill.
The leader approached the grate and unlocked it. He swung it up and held it open. "Get in there," he commanded.
I got in first and started crawling along. It was foul, and after the first minute, it was completely dark. I had no idea how they managed to pull the deer along behind us, but I assumed they had done this before and were good at it.
After a while, I saw some light up ahead. When I got to it, I could see that someone had dug through and smashed into the pipe from above. I stood up through the opening, so that my eyes were at about ground level, even with the edge of the hole that they had dug down into the pipe. I was surrounded by men like those who had captured us, and I immediately heard their catcalls: "Good hunting! Fresh meat!"
They pulled me out of the hole, followed by Tanya, Popcorn, and Frank. The catcalls got really loud when the second two emerged. I shivered. Ironically, months of dealing with only non-human monsters had made us soft and naïve. We had forgotten about the ugliness and brutality that humans could so gleefully perpetrate on anyone weaker than themselves, imagining that ours was the worst possible hell, when we should've remembered that it wasn't. Not even close.
When our captors climbed out with their other prizes, they led us across the field into which we had emerged. We were behind the walls of the prison proper, but not in the building itself, to which we were now being led. Most of the field here had been planted with corn, which seemed to be doing well. Not being a farmer, I forgot if, by the Fourth of July, corn was supposed to be as high as your eye, or knee-high. This looked somewhere in between.
I guessed that the drainage pipe had really been an old stream that had been buried when the prison was built. It carried water down from the hills above, and supplemented it with the prison's filth before dumping it out where we came in. If the inmates had tapped into it, they'd have some water to stay alive and to cultivate their new crops.
As we walked toward the large gray building, we stopped at a basketball court under the baking sun. The macadam of several other courts had been smashed up and hauled away to make room for more corn; the hoops still stuck out, incongruously, above the crops. But this court remained blacktop, though it had plenty of cracks with weeds growing out of them. The men hoisted the two deer up by their hind legs, suspending them from the backboards.
"Better get started on these quick in this heat," the leader shouted to some other men who were just lolling about, staring at us.
"I could get a good price for you all out here, but I guess I got to take you to meet the man," the leader of the group said as we continued walking.
"Who's that?" Tanya risked asking.
"Coppertop," the leader replied. "That's what the brothers call him. He calls himself Copperhead—some big, damn snake from the south, where his big, dumb, redneck ass is from. Thinks he's all bad and shit."
"And he's not?" I asked.
He shoved me on my right side, and pain shot through me from the arrow wound. "Oh, I think we're all plenty bad enough for you, little man. Now keep moving and shut up. 'Nuff of your stupid questions."
We entered the main building and went through some areas that had obviously been checkpoints and entrances when the place had functioned as a prison; there were doors of bulletproof glass operated by electrical switches, where people had to be buzzed through, watched over by guardrooms with speakers and control panels. Now everything was just a smashed-up mess, and you stepped through the doors where the glass would've been.
Where there was enough light from the windows, weeds grew in the cracks in the floor. The prisoners hadn't really bothered to sweep up the bits and pieces of glass from whenever this attack had taken place—which I suspected was soon after the zombie crisis had begun—but had just left them everywhere; over the months, they had been pulverized into a fine, sparkly dust where people walked regularly.
Past these, we were into the main cell block. It looked like an old prison, with four floors of cells facing a central, open area. The roof had a row of large skylights down the middle. Not as organized or tidy as the people in the museum, the prisoners had nonetheless taken similar precautions against an undead break-in: they had demolished the concrete stairs where they led up from the ground floor to the second tier, so the zombies couldn't get up. It looked crude, as though they had smashed the concrete with a sledgehammer, then cut through the rebar somehow and curled it back like some weird, big plant or hairdo.
The catcalls here increased, though it was obvious as I looked up at the leering faces that there weren't that many men living here, probably only a few dozen, in a place designed to hold hundreds. We climbed a rope ladder to the second tier, then walked from there up to the fourth and topmost one.
We were led to a cell. The one wall had been knocked out to connect it to the adjacent chamber, and part of the roof and outside wall had been smashed out as well, all with the same careless demolition as the stairs and the pipe and the basketball courts, leaving dusty, crumbling holes with rusting rebar curling back like the eyelashes of giant Cyclopes. The hole let a breeze through in the heat of summer, and the room was much cooler and more comfortable than those on the lower tiers.
All over the room sat large, one-gallon cans labeled "PEACHES" and "PEARS," and the room reeked of rotted fruit and yeasty fermentation, like a brewery or an ethanol plant. The walls were covered with pretty typical pornographic pictures, though there was a definite preference for blondes. A few firearms leaned on the walls and sat on the floor. To these, the black man who had captured us now added our guns. The entire floor was covered with deer skins. These guys must've lived on nothing but venison. For some reason, I suspected that it was barely cooked, too.
In the midst of all these bizarre and hyper-masculine furnishings sat what I took to be Copperhead, a large, bald white man covered with tattoos, all of which were some combination of the Confederate flag, naked women, snakes, and flames. I remembered that Queequeg's tattoos were supposed to reveal the secret meaning of the universe. I felt sure that this guy's revealed the rather straightforward meaning of his own base urges and desires. And I was very sure that it would've been much better for everyone—himself included—if he had kept them secret.
He saw me looking at the fruit cans and grinned. "Good old-fashioned pruno!" he crowed. "Best we can do—so far! I like to keep it close, or the boys'll be stealing me blind. Still is a might hard to take, but we're working on the recipe. Got some boys out back, working on a still—that ought to smooth it out real good, 'cause it's all the rot from the fruit that makes it nasty. And when that corn comes in—could be some corn whiskey, I'm hoping! That, plus the big tractor-trailer full of cigarettes we found turned over on the interstate—we'll be set!"
Great. Our final days of being sodomized and beaten to death were going to be overseen by Boss Hogg's insane, inbred nephew. I'd been hoping for someone a little more Mephistophelean, but if the zombies had taught us one thing, it was that you don't get to pick your apocalypses or your devils—they pick you.
"Couldn't you go out and find some regular booze?" I asked, without even thinking of the danger of seeming disrespectful and incurring some physical punishment for it.
Copperhead apparently didn't think the question was impertinent, but just that it was kind of silly. "No, 'cause then we'd need cars and shit."
"You could get them from the interstate. You said you were by the interstate."
"Yeah, but then we'd have to have somebody guard the gate when they go out, and open it up when they come back, and fight all those dead assholes, and find gas! It's the same reason we only have a couple guns here, the ones we took from the guards—you can't just go out and get more!"
He announced it in triumph, as though he'd found some fatal flaw in a plan that required only the barest of forethought and effort to make it work. He—and, I take it, the others, since they went along with it—would rather drink something that smelled like a dumpster than make the effort to get something else. It would be hilarious if I didn't know this guy and his colleagues planned to rape and kill all of us. I let the matter drop. No sense hastening the process.
The tall black man who'd captured us explained how they had found us. Copperhead got up to inspect us, nodding approvingly. "Now that's nice, that's real nice, having company. We done just about used up all the old guards, hadn't we? Now we got us a whole fresh batch for the Pit! Oh, that's where you'll be staying when you're with us—it's what we call the first floor of our little home."
His attention and nodding lingered especially over Tanya. "Now, isn't this a fine-looking gal we got here?" He put his hand on her hip and slipped it around to caress his buttocks. "Yeah, I might have to breed you, you look so fine, and not get out the old coat hanger like we did when we knocked up those guard bitches."
"Killed three of them that way," the tall black man muttered.
"Yeah," Copperhead said ruefully. "We should've been more careful, but we didn't want any brats running around here, messing things up." He went back to fondling Tanya, this time moving up to her breasts. "You know, my daddy wouldn't have approved, you being a nigger and all." He leaned close to lick her ear, whispering now. "That's not why I killed him with that hammer, of course. But still, I never believed all that racist bullshit, did you?"
Tanya knew enough not to say anything belligerent, if for no other reason than that she didn't want Popcorn to see her raped and killed. But she also couldn't bring herself to say anything that was appropriately friendly.
Copperhead slid down her body, blowing kisses at her breasts and crotch as his flunkies cheered him on. Slowly, he brought his right hand across to his left hip, then he uncoiled and viciously backhanded her across the face. Popcorn, Frank, and I couldn't help but take a step forward, but there were hands all over us, and knives stuck in our faces almost before we'd moved. Copperhead showed he wasn't at all lazy when it came to meting out pain and degradation.
As Tanya staggered back from the blow, he grabbed her by the throat and slammed her into the wall with his right hand, as his left hand grabbed her crotch. "You're gonna learn to be a nice nigger, little missy," he hissed, "for me and any other ugly-ass hump that has a pack of cigarettes to pay the Pit crew. The only choice you got is whether you learn easy, or hard."
He turned to his helpers. "You tell them to keep the boys away from this one for a while. I'll catch up with her later on. I don't think she's had a real man for a while." He sneered at me and Frank.
Copperhead let Tanya down and returned to his jovial, good ole boy routine. "But not tonight," he said expansively. "I say that tonight's the Fourth of July! Must be some time about now! Y'all have a calendar where you come from?" he asked, turning to me.
I shook my head and shrugged. No sense ruining a twenty-four hour reprieve.
"Yep—that settles it! It's tonight! And it'd be disrespectful to the good ole U. S. of A. to be carrying on like cats in heat when we should be showing our gratitude to this great country, what built us this fine, zombie-proof castle to live in and drink our fine hooch, and smoke all that fine tobacco, and thank God for our _freedom_!" A little cheer went up from his flunkies.
He turned to Popcorn. "Except we really can't have _no_ cavorting tonight. I'm sorry, son," he said with an icy, horrifying mock sadness and crocodile tears, shaking his head slowly. "No, we've never had a young 'un in here. And, you know, some of the boys here—well, we don't know exactly why," his eyes went heavenward and he really did seem to get dreamy and thoughtful, though I'd already seen that he was equal parts sadism and playacting, "but the good Lord gave them this powerful hunger for a special, little friend. And some of them been living with that hunger for years and years in here, with no way to satisfy it."
He patted Popcorn on the cheek, and I could see the fear and anger in the boy's eyes that he'd never shown, even when he was surrounded by ghouls who would kill and eat him alive.
"And, son," Copperhead said, "you can help them with that—isn't that nice? Well, you can help the ones of them that would pay dearly for it." He turned back to his followers and went back to the jovial routine. "Because let's not forget, boys, the business of America—is business!" Another little cheer went up, this time with a chuckle.
We were led away to our cells in the Pit, to await the festivities and horrors that their Fourth of July celebration would bring us. I had met the self-styled ruler of this hell, and he was a gruesome, swollen, little clown who thought he could dictate orders to time itself. God knows how much damage he could do if he weren't so damned lazy and stupid. But the damage he could do to the four of us would be more than satisfying enough to his stunted, twisted mind, and more than our exhausted, ill-fed bodies could endure.
Like Sarah in her dentist's office, I found myself only hoping it would be quick, but doubting this time that it would be. The dead were capable of such a meager mercy, but I was sure that such living monsters were not.
##
THEY STUCK TANYA, Frank, and me in three adjacent cells, near the end of the block farthest from the entrance. Across from us were Popcorn and the last two guards they hadn't worked or raped to death. They were a man and a woman, and from what I could see, they both looked pretty listless. They probably welcomed our arrival, as we would now absorb some of the physical abuse, but they looked too worn out to register anything.
The doors to the cells couldn't be closed, I assume because the power was off. Actually, I don't know if they really were stuck open, or if the inmates were just too lazy to bother closing them manually. Regardless, it meant there was a rather sizeable number of men—what Copperhead had somewhat predictably referred to as the "Pit crew"—to guard us constantly. They were armed with pieces of rebar, knives, and clubs, but I saw that no bows were allowed below the second tier, and no one had firearms outside of Copperhead's cell. I assumed they were imitating the rules that had been in force when the place was a regular prison—guards were not allowed guns when among the prisoners, lest one of the prisoners get a hold of a gun.
I also suspected that the Pit crew were of low social standing, for they seemed slightly more depraved even than the rest of the inmates—scrawny, cowering little creatures, more interested in the financial gain that could be gotten from physical cruelty, rather than the actual inflicting of it. Pimps and panderers, in the old-fashioned meanings of those words. They were probably next in line for rape and abuse, should the bottommost rung of their society ever run out.
Still, there were probably more than enough of them to beat us to death, should we ever try to fight back.
I sat in my cell with such thoughts, sullen and glowering. I thought of improvising a weapon, but didn't have the right kind of imagination for such handiwork. They had left us with nothing but our clothes, and the cell was utterly bare, beyond a filthy mattress and a non-functioning metal toilet and sink built into the wall. I also had no idea how to come up with any kind of plan for escape.
I thought that it might be possible that Jack might have made it back to the museum. But even if he had, it would take him some time to drive a stick shift back with his left leg hurt. And I also couldn't estimate how long it would take him to coordinate an attack on the prison, or how they would even be able to go about it. The people at the museum were set up for defense, not for mounting massive assaults on fixed positions. And they were used to fighting zombies, not this band of crazed sadists, armed with bows and guns.
And how much would Jack risk to save the four of us, who, for all he knew, were already dead? I knew him well, and we were good friends, I thought, but I also knew how logical he was, and how much he valued the community over any individual.
After a few hours, the odor of roasting flesh filled the prison. I have to admit, it was the one aspect of the prisoners' communal life that I found far preferable to that of our people.
We were taken outside by the basketball court, where the two deer were suspended on spits over a fire pit. Their heads were obscenely displayed on stakes stuck in the ground nearby—wide-eyed and tongues lolling out.
Copperhead emerged from the building and moved through the crowd to great acclimation for his magnanimity—not that any of them would've known to call it that. Two of his flunkies walked behind him, carrying a huge pot of the hideous fruit liquor, Copperhead's generous offering to his faithful subjects.
He cut the tongue out of one deer head and roasted it himself on the end of a knife, making a big show out of suspending it lewdly above his mouth and licking it before devouring it, bloody grease dripping down his chin. A cheer went up, and he raised his hands to speak.
"That's how I'm gonna do this fine sister tomorrow night!"
A bigger cheer went up.
"But don't you worry—every man who can afford it will have his turn, once I break that fine ass in! It ain't like the old days—race don't make no difference here!"
There was another big cheer.
"But tonight, boys, enjoy this feast! It's the Fourth of July! God bless America!" He gave a mocking salute to the barely recognizable, tattered flag that still flew on the flagpole outside. The biggest cheer of all rose up, from a bunch of goons who I felt sure had never celebrated the Fourth in any normal way since childhood, and who were now free to indulge their own sadistic, hedonistic version of freedom to their sick hearts' content. The whole scene made _The Lord of the Flies_ look like _Little Women_.
After Copperhead kicked off the festivities, we were treated to the spectacle of men devouring as much bloody flesh as they could. Like animals in the wild, they ate in descending rank. Copperhead ate first, like the leading male lion of the pride—even though, exactly like the chief male lion, he had done none of the work of procuring the feast.
Then it was the turn of the other lions of the pride: Copperhead's immediate henchmen and those from the hunting party, who lived on the prison's topmost tier, called "Park Avenue." Those others who lived on the second and third tiers—which I learned were called "Uptown" and "Downtown," respectively—came next, like the hyenas that descend on the lions' kill. Then the Pit crew was allowed to eat, like jackals, not wanting to offend or anger the more dangerous carnivores.
Finally, when there was no danger and all had torn off their share, the six of us who were the prisoners of the Pit were allowed to feed, like vultures, from the most unsavory scraps.
Starved as I was, in the presence of the first cooked meat I'd smelled in nearly a year, the barbaric feast tasted like the best thing I'd ever eaten, as I'm sure it did to the others as well. Gnawing on a bone as I looked at the inmates lolling about in a blood- and meat-gorged stupor, I again thought of how frighteningly little separated us from the other carnivores, staggering about outside the prison, with stupefied looks on their still-human faces.
Sitting on the ground in the twilight, once our hunger was sated, all we could feel was complete dread and helplessness at what was to come. And what could we say, especially to Popcorn? "I'm sorry," was far too meager and vague, while, "It'll be okay," was a lie. "Don't worry—it'll be our turn tomorrow night," was probably the most grimly honest, but wouldn't offer much consolation to him or us. Assuring him that we would fight would be true, up to a point. But we all knew eventually we'd have to stop and let it happen, or we'd be beaten to death, and then it would happen anyway.
Oddly, it was Frank who spoke up; he'd been silent almost since we left the museum that morning, which seemed a lifetime ago. "I think you guys are going to make it," he said. "And when you do, take care of Zoey for me. Tell her how much her mom and I loved her."
I think right then, we might have thought it was a little callous of him—making the situation about him, when it was obviously Popcorn who was going to suffer the most, at least that night. But Frank rubbed Popcorn's shoulder, and we just took his words to be an awkward expression of hopefulness for us. And for one of the few times I'd ever seen, Popcorn let someone other than Tanya express tenderness for him, so perhaps he knew what Frank meant, even if those of us who were older and supposedly wiser did not.
They rounded us up at that point and took us back inside. The men who lived in the tiers above ascended their ladders, and the Pit crew was now more vigilant in guarding us in our separate cells. Torches were burning, and the light from a nearly-full moon shined through the skylights, making it possible to see a little in the gloom of the Pit.
They tried to conduct things as quietly as possible, maybe out of some slight fear of unnecessarily provoking us to violence, maybe out of some tiny shred of vestigial humanity and shame at what they were about to do to an innocent child. I suppose they would've said they were being civilized or merciful about it, but words lose all meaning when stretched to such grotesque extremes.
A man came down the rope ladder to be Popcorn's first visitor. He went in the cell, with two guards watching from outside. One guard was standing right outside the door of each of our cells, with more in reserve. I stood up, and I'm sure Frank and Tanya did the same.
The guard at my cell door half raised a piece of rebar and growled. "Sit down before I bust you all up. I don't want to ruin that pretty face before I make you my bitch."
There were any number of witty repartees I could make at that point, but now was not the time. The only one I allowed myself was to think that, after a year of privation, I most definitely was _not_ pretty, no matter in what direction one's tastes ran.
If the three of us rushed our guards at the same time, we could probably get past them. And if we got a weapon away from each of them, we could maybe take out a couple more. Then the rest would beat us to death. Popcorn's inhuman degradation would be postponed maybe ten minutes.
Still, you had to make sacrifices for the payoff that was offered. We all can't die on Omaha beach, winning back freedom for millions of people. Some of us die on a filthy prison floor to defend a little boy, even though it won't make the slightest difference to what happens to him.
As I clenched my fists at my side and took a step forward, I knew how lucky I was. Some people died for nothing at all. I was going to die, smashing this ugly bastard's head into the floor and taking that piece of rebar from him, so I could smash a couple more ugly bastards' heads into pulp. That counted for a lot, in my book; and unless God was a much bigger asshole than I thought, part of me felt sure it counted for something with Him.
##
BUT THE GUARD and I were both stopped, before either could attack, by a hideous scream coming from Popcorn's cell. Some small, wet object landed with a splat on the floor outside the cell, and the man emerged, clutching at his face. "By doze! Son of a bitch bit by doze!"
There was a guffaw from the crowd assembled on the second tier, who had come out to see what the commotion was. The two guards went into Popcorn's cell; one pushed the bitten man out of the way. "You dumb bitch! Can't even do a kid without help?!"
There was more scuffling and yelling in the cell, and more guards went in. I never heard Popcorn's voice, only those of the men he viciously fought. But after a minute, it was still except for the dull thuds of fists hitting a body that wasn't moving or fighting back, but just being methodically pummeled into a bloody, submissive lump. Frank shouted, "Stop! You can't do that!" He charged his guard and tackled him. It distracted my guard for an instant, and I jumped on him, knocking him down.
I got his right arm—the one holding the rebar—pinned to the floor as I punched him in the face with my other hand. His grip slackened on the rebar, and I grabbed it. I raised it up to smash him in the face, but two other guys grabbed me from behind and pulled me to my feet.
I was still struggling, but it was useless at this point. Somewhere in the partial darkness above us, Copperhead shouted, "Go help those assholes in the Pit!"
He sent more men down the rope ladders from the second tier to help the Pit crew beat us into submission. I could see to my left that they had Tanya restrained similarly to how they had me.
The guard I had originally knocked down was back up. He kicked me in the testicles, then punched me in the face. I could taste blood, my ears were ringing, and spots were exploding before my eyes.
I roared, tore my left hand free, and tried to hit the guy holding my right arm, but my guard pinned my free arm behind me while another guy punched me in the stomach, then two more times in the face.
Now I could barely hear or see anything; my mouth hung slack to let out a steady stream of blood, and I wasn't able to draw in breath with the wind knocked out of me. I stopped struggling. It was a reflex. You couldn't will to go on with that kind of pain overwhelming you.
Well, I couldn't. Some of us are made of sterner stuff, under the right circumstances. And that night, it was Frank, for some reason. He threw off one of the Pit crew and grabbed the guy's knife. He slashed the man across the face, causing him to scream and fall back. Frank kept slashing as he yelled, "You can't do that! You can't! Leave him alone!"
The Pit crew hung back, afraid of getting cut.
I could see now why he had told us to take care of Zoey. It was because he'd had enough, and he knew he was going to die defending Popcorn. I should've seen earlier that he'd reached his own breaking point and was having his own thousand yard stare. Fighting zombies, killing his own wife, living on starvation rations for ten months—he'd somehow managed to survive all that for Zoey's sake.
But ironically, being safe with us had made him less able to carry on in the face of absurd and dehumanizing cruelty; he knew Zoey would be taken care of, so why should he turn his back on Popcorn's suffering and try to survive himself? I tried to yell, "No, Frank!" but I don't know what came out. Probably just a gurgling sound from a mouth full of blood.
Then there was that whistling and thwack sound as an arrow sank into Frank's back. He groaned and staggered. One of them tried to grab him, but he slashed again, and blood flew off of the guy's hand. There was another thwack, and Frank was hit with an arrow from the front. He staggered and finally fell.
The Pit crew descended on him like the cowardly, herd beasts that they were, brutally kicking and beating him. As in Popcorn's cell, after the first couple of blows, there was no sound or sign of any struggle, but just the terrible thuds of fists and feet hitting over and over.
They finally pulled him up, and he was covered from the waist up in blood. The spots around the broken shafts of the arrows were now no redder than the rest of him. Both his eyes were swollen shut, and his mouth hung open, dribbling blood. He could barely cough to clear his throat and draw in a wheezing breath through all the fluid.
One of the Pit crew yelled up into the darkness, "Copperhead, you sure we need both of these new bitch-boys? This one's a pain in the ass!"
"Can't you assholes do anything right?" Copperhead replied. "I bring you new toys, and you just screw it up!" There was a pause, then finally he said, "No, I guess we don't need both."
The jackals in the Pit seemed to like that. Now they could inflict pain not just for profit, and not even out of fear and rage. Now they could just be cruel for its own sake, as they had probably seen done so often by this hellhole's elite, either before or after the inmates took over.
Holding out Frank's arms, they tied his wrists to the bars of the cell door. Tanya and I were, of course, made to watch.
The two men whom Frank had cut were allowed to visit some special indignity or pain upon him for their troubles. The first took back the knife and put it beside Frank's head. "Son of a bitch cut my face!" he yelled. "I'll take something off your face!"
His arm moved back and forth in a sawing motion, and a cascade of blood fell at Frank's feet. He let out a gurgling scream.
Finished, the guy held up Frank's severed right ear. He slashed downward across Frank's face to punctuate his point. "Gonna wear it around my neck on a string when I do that brat kid you cut me over! Gonna do him twice as hard when I think of you, you crazy, dumb bastard!" The crowds on the second tier cheered.
The second guy took the knife and yelled, "This crazy asshole cut my hand!" And he stabbed Frank's right hand, driving the blade all the way through it. Frank screamed again, writhing against his bonds. The guy then walked over and did the same to his left hand. Again the crowd roared its approval.
We waited a moment, I guess to let Frank suffer more. It was deathly silent—unnaturally and painfully silent. You could just hear the animal panting of all of us, the throb of life, the life that inevitably craves another's death and suffering. The throbbing seemed to fill my head, seeping up through the floor into me, but then it became audible as the crowd on the second tier started chanting, "Kill! Kill! Kill!"
Given how Frank looked, I almost welcomed the chant of bloodlust, to hasten an end to his pain. One of the Pit crew walked up with a spear and stabbed Frank in the side with it. More blood gushed out, running down to puddle on the floor. I could never quite visualize until that night just how much blood is in the human body. The undead were usually a dry and crusty lot, or they exploded with puss and putrefaction: only a living body could spill out the incredible affluence of thick, rich, beautiful and horrifying blood. Frank was almost too weak to move at this point, but only flinched slightly at this final blow. The crowd gave another cheer.
"Will you numb nuts make sure he doesn't get back up again, please?" Copperhead asked the Pit crew.
Another member came up with a wooden baseball bat with enormous nails driven through it. I looked away, but I could hear the sickening glitch and crunch as he slammed it into Frank's head. It was finished. One of the most courageous men I had ever known—a man who had surely suffered enough already—had died, utterly forsaken and in agony.
"Let them take him down and bury him," Copperhead directed from above. "Oh—and if they're going to bury him, I guess they're going to use shovels?"
"Well, yeah, sure," one of the Pit crew replied.
"And what are shovels good for?" Copperhead asked.
The fellow hadn't caught Copperhead's sarcasm. "Uh, digging holes?"
"Yes—and hitting dumb asses in the head! So make sure they're guarded better this time!"
Speaking for myself, Copperhead needn't have worried too much. I could barely walk, breathe, see, or hear. I doubt I was much of a threat at that point.
Tanya and I got Frank's body down and dragged it outside with difficulty. We were accompanied by four guards, who gave us the shovels after we had set Frank's body down. We worked slowly, digging, sometimes almost hacking, through the dense, hard, red clay. The guards didn't seem to care if it took us awhile, as they were well supplied with the vile fruit liquor.
If I hadn't been busted up so bad, we might have entertained thoughts of attacking them, they were obviously getting so inebriated. But as it was, they sat around, laughing and not paying much attention to us. I also saw, lurking in the background, the large black man who had captured us that morning. I presumed he was there to make sure that Tanya was kept intact for Copperhead.
When we were down to a good depth of about four feet, Tanya and I stopped. It wasn't the official six feet, but we were exhausted.
We climbed out and heaved Frank's body into the hole. They hadn't given us anything with which to wrap or cover him, and he landed in a particularly grotesque and awkward pose, with his arms outstretched above him, his legs bent up under him, and the hideous nail hole in his forehead clearly visible, blood obscuring his whole face.
"Poor son of a bitch," Tanya said. "Can't leave him like that." She jumped back into the hole. Again, it simply was not in her just to ignore—as I would've been inclined to—something as emotionally weighty but physically inconsequential as a person's final posture in the grave.
She straightened his legs out and folded his hands across his chest. Then she raked some of his hair down across his bloody face and turned his head to the right, covering the bloody hole where his ear had been and hiding the nail hole in his forehead. While it was definitely an improvement, the extent of Frank's stigmata made it impossible to do too much. I shook my head at Tanya's kind and loving attempts, knowing that back in the normal world, this would most definitely be a closed casket affair.
I helped her climb out, and we stood there a moment. "They led him like a lamb to the slaughter," I half-sobbed, half-gurgled through my own blood and rising tears. I had no idea why, as I looked down at Frank's broken, humiliated body, I would remember that imagery from a biblical verse. I didn't even remember where it was from in the Bible. I guess it was supposed to mean Jesus, but I wasn't sure.
"They sure did," she agreed. "Poor guy toughs it out for months against the living dead, and these assholes kill him in less than a day. It's not right." She looked at me and asked, "You know the next verse?"
"What? After the one I said? No, I have no idea."
"What, you just sort of remembered it from _Silence of the Lambs_?" she sneered. Then she lifted her eyes to the warm, dark sky, her hands at her sides, palms forward. "O Lord of hosts, that judgest righteously, let me see thy vengeance on them."
Leave it to the spiritually profound Tanya to remember a verse that I could pray without hesitation. I repeated it over and over in my mind as we picked up our shovels and proceeded to fling the damp, dead clods onto Frank's body.
##
WHEN WE WENT back inside, Popcorn's cell was pretty much walled off from our view by guards, I assume to prevent further trouble from us. Tanya and I were led to our respective cells. I sat there, looking across at the shadows moving on the other side of the cell block, straining to hear anything, but I could detect nothing. We all knew what was going on. But there was nothing we could do at this point.
I slept fitfully sitting up. Close to dawn, I could hear one of the guards whisper to Tanya, "Hey, bitch, go over and sit with the kid. He needs you." I saw her walk over, then I dozed back off.
In the early morning light, I looked over again. The guards had dispersed at some point in the night, and I could see Tanya and Popcorn clearly. They both were asleep. Tanya was sitting up, leaning against the back wall of the cell, like I was. He was lying across her lap, on his side, turned slightly toward me.
As the light grew brighter, I could see them more distinctly. The way they were sitting, the morning light actually shined brightly across them. Popcorn was in as bad a shape as I had imagined he would be—bruised and bloodied from head to toe, lips cut and swollen, one eye swollen shut. The psychological or spiritual wounds that I couldn't see were probably much worse. Bathed in the morning light, his brutalized body graceful now and still, her beautiful and loving face bent toward him, both of them suffused with the peace of sleep and the vivifying glow of the sun—they could not have looked more like a pietà if they had deliberately staged it.
After a while, the prison began to stir with the more mundane and profane forms of life that dwelt in it. Eventually, we were led outside to stretch and spend some time in the light and fresh air.
While passing through the shattered remains of the guard room and entrance in order to go outside, Popcorn tripped on the doorframe and fell sideways, onto the floor of the control room. I reached down for him, but he batted away my hand. He scuttled a little ways with his left hand clutching at his side. His right hand was stretched out in front of him, and then he swept it around and out to his side, all the while making noises as if he were in pain.
I was so overcome with pity for him, I could almost have summoned up the strength and courage to fight those monsters again right then and there. One of them almost provoked me to it when he came up to see what was going on, and looked as though he was going to hit Popcorn. But the boy finally got up, bowing submissively to the guard, and we proceeded outside.
Out there, I paced back and forth; the two former prison guards sat off by themselves, and Popcorn refused any compassion or intimacy from Tanya now in the daylight hours, in front of others. She sat by herself, and he retreated to a corner by a wall and sat with his back to us the whole time.
When we went inside, Popcorn sat on the floor of his cell the rest of the day with his back to the outside. One certainly couldn't blame him for spurning all human contact, when so-called humans had so successfully broken him and dehumanized him more thoroughly than the undead ever could. Ill-fed and depressed, with my whole body aching and one eye swollen shut, I myself could do nothing all day but pass in and out of sleep, sitting up in the cell, and repeating my new favorite prayer from the night before: "O Lord of hosts, that judgest righteously, let me see thy vengeance on them."
Toward evening, with deer again roasting on the fire outside, it seemed to darken early in the prison, and thunder could be heard faintly in the distance. I smirked and grunted, the only right way to register enjoyment of a dark and deadly irony. The kind of sudden, violent summer storm that swept through this time of year seemed perfect for what I thought was coming that night.
Looking over at Popcorn's battered and bruised back, I had gotten my own thousand yard stare. And it was not out of pity for the undead, as I had felt it so many times before. It was out of rage and disgust for the living. And I felt some of the raw, primal energy of outrage and revulsion that Frank had tapped into the night before.
Tonight, I would help see God's vengeance extend even here, to the deepest pit of this manmade hell. God and I had let this place be stained with innocent blood, and I blamed both of us for it. Now it was time for these walls to be painted with the blood of the guilty, the way hell was supposed to be, with righteous judgment and richly deserved, never-ending punishment.
I looked heavenward. "Give me the strength, God," I said quietly. Thunder boomed closer. I clenched my fists and could extend them without as much pain as before. I looked outside my cell and could focus a little better with my one good eye, enough to have a little bit of depth perception.
I nodded. "Thanks," I said. "That'll do."
We were again treated to the barbaric venison feast outside, though there was none of the fruit liquor this time. I was grateful not to smell its nauseating odor this night, but it meant the inmates would be sober and better able to fight us off. I guess I didn't care at this point. We all ate ravenously as storm clouds swirled above us, though as yet no rain had fallen.
When I looked at Tanya, I felt sure that I saw a look of defiance and determination. I hoped it was the same look she saw on my face. And I hoped that it would end somewhat differently for us than it had for poor Frank.
With Frank dead, and me and Popcorn beaten into bloody pulps the night before, the Pit crew didn't seem to worry about our ability to fight them off. They had two guards on Popcorn, but only one on me and Tanya, as before, with several more hanging back, ready to jump in if necessary.
With the prison now in semi-darkness, punctuated by flashes of lightning, Copperhead descended the ladder and approached Tanya's cell. He too seemed satisfied that there would be no uprising, as no bodyguard accompanied him. He even felt optimistic enough to stop and taunt me before going in to Tanya's cell.
"Big storm tonight," he said with his mock cheerfulness—though, of course, I'm sure the prospect of sadism and degradation really did make him feel cheerful. "But I'm sure you'll still hear my new black bitch screaming my name when I show her how a real man gives her some hard lovin'. Ain't nothing gonna be loud enough to drown _that_ out once I get all up in her shit." He guffawed. I prayed it would be his last.
As Copperhead baited me, another group of pedophiles entered Popcorn's cell. I made no move toward the door. Better not to raise the alarm prematurely; I felt sure that Popcorn or Tanya would attack the monsters at any moment, and that would be the signal for me to jump in and do whatever I could before they beat me to death. I still assumed it would end with my death, though I hoped to take more of these ugly bastards with me than poor Frank had.
The lightning flashed, and I only counted to five before the sound of the thunder rolled through. The storm was getting close.
I stared intently at Popcorn's cell. Both guards had gone in with the visitor this time, I assume to administer another beating if necessary. Popcorn must've timed his attack just right, though, as I heard one of them yell, "Shit! Look out! Little bastard's got a . . ."
This switched abruptly to a gurgling scream as a huge arc of red shot out between the bars to splatter on the floor outside the cell.
"Get him off me!" another man yelled. "Get him off me!" This also trailed into another horrible scream.
Neither the inmates nor I had thought Popcorn would improvise a weapon, though, in hindsight, it was hard to believe we'd overlooked the possibility. Prisoners had been turning practically anything into a weapon ever since there were prisons, and usually with much less motive than Popcorn. If a man could spend weeks making something into a knife to kill another man for a pack of cigarettes, then certainly someone fighting against torture and humiliation could be counted on to fashion something sharp and deadly.
It suddenly hit me that when he'd stumbled that morning, it was all a ploy so he could hunt around on the floor for a piece of glass big enough to do the job. And judging by the screaming, it was just the right size.
I came out the door of my cell and went for the guard. It was the same guy as I had fought the night before—an ugly, bald, squat little bastard. He came at me with the rebar and a long, rusty knife.
We both snarled as we collided. I grabbed both his hands, and we wrestled for the weapons. He tried to kick me in the groin again, but I turned slightly to the side, and it did nothing; I tried to headbutt him, but he pulled back, and I grazed his nose, to no effect.
The adrenalin and the outrage pushed me on, but he was better fed and stronger, with a lower center of gravity. Neither of us could gain the upper hand.
On the tier above us, more men came out to watch. If some of them came down to help, as they had the night before, then it would all be over just as quickly as it had been then.
But as we struggled there, Copperhead come staggering out of Tanya's cell, with her hanging onto his back and screaming like an avenging fury. I couldn't exactly see, but she was strangling him from behind, with something wrapped around his neck. I guessed it was her shoelace, something else they'd overlooked in their laziness and stupidity, and which my naiveté had been unable to identify as a potential weapon.
He couldn't grab her, and he was staggering about now, looking for someone else to hit her for him, but it wasn't working. The guard in front of her cell, who carried the baseball bat with nails that had killed Frank, couldn't get a good shot at her, and he couldn't decide whether he should help the guy who was fighting me.
So Copperhead threw himself back against the bars of the cell, slamming Tanya into them with all his weight. I didn't think it was going to work, judging by how determined she looked. It also made it impossible for anyone else to take a swing at her.
When the crowds above saw Copperhead's predicament, they did not rush down the ladders to his aid. Instead, the same cheer as the night before rose up—"Kill! Kill! Kill!" This time it was punctuated by thunderclaps that were louder and closer each time. Clearly, the inmates were not only lacking in intelligence or a work ethic, but also in loyalty. It was hardly surprising—a place fueled exclusively on testosterone, barely-cooked red meat, sodomy, and fear would surely be lacking in those other qualities.
I suppose if Copperhead somehow came out on top, they could always claim later that they were cheering him on, so it made double sense not to get involved, but instead to enjoy the show. They regarded Copperhead fighting for his life as just an unusual and therefore very enjoyable entertainment—which, to be fair to them, was exactly how he would've regarded them in a similar situation.
This unexpected cheer also made the Pit crew hesitate. Several who had rushed to Popcorn's cell were now backing away and looking up at the crowd. Without a leader, and with its loyalties divided, the animalistic mob was much less frightening, and much less effective at either inflicting pain, or even at defending itself.
Perhaps our fight would last a bit longer than the previous night. I still assumed we would all die, but it now looked as though we had a real chance to kill Copperhead and several of the Pit crew. I could easily—no, _gladly_ —accept that outcome.
##
BUT AT THE MOMENT, I was still locked in a struggle with the guard. This ended abruptly when Popcorn flew in from my right and grabbed the guy's left arm. Popcorn was snarling like a beast and was already covered in fresh, hot blood from the men he had stabbed. He climbed up on the guy I was fighting, holding onto him and biting his forearm, as he plunged a shard of glass into the guy's neck. I was showered with blood as it shot from his neck and came flying off the shard as it repeatedly slashed up and down.
The guard screamed and staggered backward. I grabbed the rebar away from him as he collapsed. He fell to his knees, with his left hand clutching at his neck, blood pouring from between his fingers. The crowd's chant of, "Kill! Kill! Kill!" crescendoed, but I hardly needed any encouragement. There could be no mercy, both for what he had done, and for what he would become if I let him bleed to death. The last thing we needed was a zombie in here.
I brought the rebar down on his head once, then again when he fell onto his face. The crowd above us let out a cheer, just as they had when Frank was being murdered last night. As one might have expected, their cheering did not indicate approval of the winner, but merely excitement and near orgasmic joy at the maiming and killing they were witnessing.
Popcorn stood up beside me. Now his face and especially his mouth were covered with blood. It was even streaked throughout his long, wild hair. He was panting and licking his lips like a wild, rabid beast, which was not far from what he was at that moment. I couldn't say I blamed him, or even that I found the behavior all that disturbing, under the circumstances. I think anything short of drinking the blood or consuming the flesh of his tormentors would have been defensible, even decent, behavior.
I looked over, and Copperhead was still throwing himself backward against the bars of the cell, smashing Tanya into them. It didn't look fun for either of them, but she clearly seemed to be holding her own, and he seemed to be weakening.
The guy with the baseball bat finally decided to make a move toward me and Popcorn. I think at this point it was mostly an attempt to fight past us and just climb out of the Pit altogether. Good. We were no longer on the defensive, and we even had the crowd's support, if not their sympathy, for I doubt they had any. Maybe we wouldn't die that night.
The guard swung the bat at Popcorn, who nimbly jumped out of the way. He swung the bat at me, and I swung the rebar to counter it. The rebar stuck between some of the nails, so that we were then wrestling over the weapon. Popcorn dove for the guy's throat, but this time he let go of the bat to defend himself. They wrestled, and Popcorn continually slashed at his arms and throat. I disentangled the rebar from the bat and smashed the guy across the head with it once, then again, then one last time after he'd fallen. The crowd cheered wildly.
I handed the bloody rebar to Popcorn and took up the bat myself. With no more Pit crew near us, we finally ran over to help Tanya. She was wheezing and sweating from being slammed into the metal bars, but it was obvious now that she could feel the life ebbing from her tormentor. She looked at me, her teeth gritted, lips pulled back in a snarl, her eyes filled with rage, her mouth right next to his ear as his swollen, grotesque face turned blue.
He too was looking at me with his bugged-out eyes, and I imagined they were pleading, but I couldn't be sure. Perhaps worse, I'm not sure I would've cared whether or not they were. Worse still, the thought flashed through my adrenalin-soaked brain that if they were definitely pleading for mercy—something from which Frank and Popcorn had so bravely refrained—it might make what we all knew was coming next even more delectable. And I cringed, for the prospect of wreaking vengeance and punishment on this piece of filth was already terrifyingly sweet.
"You know, Jonah," Tanya hissed, "you probably don't know this, since you're not some inbred, redneck asshole who crawled out of some swamp—but you got to hit a snake in the head really hard if you want to kill its stupid, sorry ass."
I swung the bat back to deliver the blow. It was the cruel, up close and personal type of execution that a sadist like Copperhead would've found especially enjoyable, so I tried not to revel in it too much. But after the suffering of Frank and Popcorn, it was just plain impossible not to. You had to allow human nature some visceral, fleshly enjoyment from curing such a disease as Copperhead, like lancing a ripe boil, or even picking at a scab. I would've been much more inclined to show mercy to one of the undead.
Above us, the chant of, "Kill! Kill! Kill!" rose to an orgiastic crescendo.
"Die, you stupid son of a bitch." I slammed the bat into his forehead. The glitch and crunch was much louder this time than it had been with Frank, close as I was. I pulled the bat back, wrenching the nail loose from his skull, then Tanya shoved him off with a shriek of disgust as the crowd above us went wild. He fell onto his face with a thud that was barely audible above the cheers.
Tanya and I were panting, and our satisfaction was so intoxicating that we paused along with Popcorn to watch the puddle of thick, dark blood spread out from under his face. I looked at Tanya, and the bliss was almost of post-coital quality.
At that point, I really didn't care if the other inmates put my head on a stick. I'd sent the ruler of this pathetic little hell to the real thing. If anything else good ever happened to me now, or even if I just kept breathing for a few more minutes to enjoy this victory, then that was just gravy, and I'd put it on my list of things that hinted at a God interested in the guilty being punished. He had, at least, answered the prayer I had made when I buried Frank the night before.
The three of us stood there a moment, panting and covered with the warm and sticky blood, before two more screams tore through the prison, accompanied by lightning flashes and nearly immediate thunderclaps. The cheering above us stopped suddenly. The screams were long, piercing, as though from people who were being torn apart, and at the exact same moment that I heard them, I inhaled the strongest odor—even over the nearly overpowering metallic smell from all the blood—of rotting flesh. And then I could hear the other sound—a low and persistent moaning.
I really didn't want to, but I slowly turned around, away from Copperhead's body, and I saw that about forty feet away from where we stood, extending all the way back to the entrance to the prison, the ground floor was packed with swaying, shuffling human shapes. It must've finally started raining, as steam was rising off of them, as if they were soaking wet.
At the next lightning flash, I could see their rotten, undead visages—their blackened teeth, bloody mouths, foggy eyes, mottled flesh, and matted manes of straw-like hair. And though some were at present occupied with devouring two of the Pit crew, those in the vanguard were staggering toward us with their usual lack of coordination, and complete superabundance of determination and focus.
Defeating sadists and rapists only to be confronted by an army of the drooling undead—this place was about as close to hell as I hoped I would ever get. Now it seemed that we were most definitely going to die that night. It seemed it would be a lot quicker than I had previously imagined, but every bit as horrible, too.
I made sure to tack on a little extra prayer right then—that my guts were torn out and eaten before Popcorn's and Tanya's, so I wouldn't have to see that happen to them. No, wait, that would be selfish and unfair. But it didn't seem right to pray for them to die first. What the hell, I guess we could leave that part up to the Lord, as He always seemed to have the part down where innocent people died horribly, so I stopped praying and started to back up slowly.
##
IN THE FLASHES of lightning, we watched with a mixture of satisfaction and revulsion as the army of the undead took care of the two Pit crew members they had caught off guard. With a rending and a popping sound, one arm was torn from its torso, and a geyser of blood shot up after it. The two pieces were born in opposite directions in the writhing tangle of groping, eager hands. The other guy had already fallen into the crowd, and similar rending sounds could be heard as he was dispatched. Once their screams subsided, there was only a grotesque chorus of tearing and slurping.
To stem the tide of the undead, the men on the second tier pulled up the one rope ladder. Another guy had nearly reached the top of the other ladder; they pushed him off into the hungry horde, where the horrible screams and rending sounds started up again, as the inmates cut that ladder and threw it down.
As we slowly backed up, the undead reached the cell where Popcorn had been. The guy he had originally slashed lay outside on the floor. He must've still been alive, as they grabbed at him, their fingernails digging into the huge wound on his neck and tearing it wider.
He was too weak to scream, but a lightning flash made the overwhelming fear in his eyes quite apparent. Good. To be torn apart and eaten alive was indeed frightening, as I was starting to realize myself, for it seemed all too likely that it would be my fate as well. But I could still smirk at him because he had the added fear of being dragged before an angry God—for what other kind of God could possibly have created something as obscene and violent as the hungry undead?—right after beating a child nearly to death.
As more hands joined in the bloody rending, his eyes were covered, and they tore his head and his torso in two different directions, a river of blood spilling onto the floor when they finally tore the head loose, leaving behind a stump of ragged flesh.
The other Pit crew member Popcorn had slashed had staggered out, clutching at the mortal wound on his neck, blood still pouring out between his fingers and further inciting the undead's unholy hunger. Those zombies not already feeding on the headless corpse grabbed both his arms and pulled in opposite directions.
At first, it was a comical tug of war: the dead rocked him back and forth as he whimpered, too weak to muster a real scream. But when both sides finally pulled at the same time, the effect was less comical, at least for their victim, though I could still manage a cruel smile. No longer capable of fear, but just registering unspeakable pain, his eyes bugged out before both his arms tore off.
He hung there a moment, swaying slightly, mouth open, eyes rolled back in his head, blood spurting out both stumps at his shoulders with a flow that steadily weakened. Good—it seemed a more horrible version of Frank's suffering the night before.
Finally, he slumped forward and more zombies fed on his body.
I thought the undead were finished with Popcorn's tormentors at that point, but I had forgotten there was a third inmate in his cell, probably the evening's potential customer. As the undead's horrible feeding frenzy proceeded outside the cell, a weak voice came from inside, "Help! Damn kid stabbed my eyes! I can't see! Who's there? What? What? No!"
Again, the voice trailed off into screams as the undead found their blind and helpless prey. For some reason, his screams seemed to last especially long. Blinded, with claws and teeth tearing into his flesh from all sides, reducing him in seconds from a human being to a pile of meat—I hoped he spent eternity in hell like that, for what he'd done, or even just intended. I looked down at Popcorn, and he was smiling and grunting.
I guessed it was going to be the high point of our evening, for with the Pit crew devoured or retreated, and the means of getting to the second tier cut off, we were the only thing left on the menu. I clutched the bat tighter, and we kept backing up.
##
I LOOKED OVER my shoulder and saw that we'd been joined by the two former prison guards, who cowered behind us. "Back into the one cell," I hissed at them. We were already almost there.
"And then what?" they blubbered.
We were backing into the cell. "And then we'll take turns at the door to the cell, killing them," I said. "They can only come at us one at a time there."
They both scuttled to the back corner of the cell. "So what? There must be hundreds of them! And you know that none of those assholes are going to come down from upstairs to help!"
I handed the bat to Tanya, and I leaned down over the two guards, my fists and face still covered in blood. "Then we'll pile up their rotted bodies ten deep till they can't get at us! And then Tanya will take the rebar and smash our heads in, so we don't become one of them! If you can't help, then just stay the hell out of our damn way! How's that for a plan?"
I went back to Tanya. "Actually, I think it'll take two to cover the door," she whispered. "These nails will get stuck—somebody else better be bashing them with the rebar at the same time."
The dead were closing slowly. "Okay," I said. "That's what we'll do." I put my hands on Tanya and Popcorn's shoulders. "I'm sorry guys. I wish it would've turned out different."
They nodded.
Suddenly, the dead stopped, swaying and letting out a rumble of discontent or alarm. The lightning flashed again, and a ripple went through the crowd; a path opened up in it. The crowd parted, and a tall, lean figure emerged, carrying a staff.
As the thunder crashed in the darkness, we could just barely see the figure stride across the remaining yards between us and the army of the undead, and at the next lightning flash, it was right in front of us. It was Milton.
He embraced Tanya and Popcorn at the door, then pushed them farther into the cell, so he could take up a position guarding it. Now there was no way for the dead to get at us, past the leader whom they feared so much, for whatever reason.
"What's the old guy going to do?" the cowering guards bleated.
I glared at them. "Will you two just shut the hell up? Just trust us, okay?"
I patted Milton on the back, shaking my head; I couldn't believe he had the audacity to attack the prison with an army of the undead. "Thanks, Milton."
He looked over his shoulder and smiled. "You're most welcome. Where's Frank?"
I shook my head. "He didn't make it. They killed him."
Milton looked shocked and suddenly began shaking. "What—the dead I brought in here? They killed him? Oh my God!"
"No, no, not them," I said quickly, trying to calm him down. "The guys who run this place. They killed Frank last night."
"Oh, I'm so sorry. But I couldn't have lived with myself if it was because of me." He calmed down just a little bit in the pause. "But why would they do such a thing?"
With walking corpses shuffling around in front of him, sniffing at him and eager to tear the flesh from our bones, it was really quite extraordinary to see that the regular, human evil we had all lived with our whole lives could still so shock and astonish Milton. "Frank was trying to protect Popcorn," I said. "They wanted to... you know... they wanted to hurt him... like that."
Milton's eyes went wide, and I could see he was fighting back tears, trying not to look weak in front of Popcorn, let alone show him pity to his face. "Good God... But he's only a child. I'm sorry, I had no idea there was still such evil in the world. I thought we'd been through enough."
His eyes turned to rage, for the only time I'd ever seen, and he leaned farther out the door. "I brought these maggot-ridden corpses in here, you bastards! Hundreds of them! And they haven't eaten in months! And now they're going to tear you all apart and send you to hell, you sons of bitches!"
I patted his back. "Easy, Milton. We'd all like to see that, but what exactly are we going to do now?"
"I'm not sure," he admitted. "I think Jack has a plan."
Just then we could see that the inmates were finally mounting some kind of counterattack. Arrows started raining down from the second tier onto the undead. But arrows, as effective as they were against living deer and humans, were one of the least effective weapons against zombies. There were various roars of protest as the arrows lodged into torsos, limbs, and necks, but you could see that almost none of those they hit were falling down. I worried, however, that one of the arrows—either stray or intended—would hit Milton.
I grabbed the mattress off the floor and shoved it in front of him. "Here, hold this in front of you in case one of those arrows comes your way!"
He turned his face away from it. "Good Lord," he choked. "It smells worse than me!"
I smiled. "That's why you're here to rescue us."
He looked over his shoulder at me, one arm stretched across the doorway, the other hugging the stinking mattress to his chest. "On my belt," he said, "there's a radio. Get it. Call Jack."
I got the walkie-talkie. "Jack?" I said into it.
"Great to hear you!" came the reply. "Sorry I got you into this. Everybody okay?"
"The prisoners killed Frank last night," I replied.
There was a pause. "That's too bad. He'd been through so much." There was another pause, and then he was all business again. "Where are you all in the building? We need to get you out."
"We're on the bottom floor, at the end farthest from the entrance. Milton is the only thing keeping us from getting eaten right now."
"Milton's got them held back for the time being?"
"Yes. But we're taking fire from the upper floors."
"That I can help with. Franny?"
"Almost there, Jack," I heard her reply.
"Our guys are on the first floor, so aim for the second," Jack told her.
"Roger that. Second floor's the target."
Over the sound of thunder, I could hear the thumping of the helicopter. It got louder and louder, then held steady. The skylights exploded with a flash, and glass and metal cascaded onto the dead outside the cell. A moment later, I could see another flash at the smashed skylight, and with a whoosh, one of the cells on the second floor exploded amidst screams.
It must've been more of Jack's AT4s, being fired by someone now on the roof. As he had predicted, their real value would be proved should we ever have to fight the evil living, rather than the mindless dead.
With a flash and a whoosh, another cell exploded in flames and flying debris. Most of the upper cell block was now shrouded in a pall of dust and smoke as the groans of the injured drifted down to us.
No more arrows were raining down, so Milton lowered the mattress. "Glad I don't have to hold that anymore." He raised his hands up and shooed away some of the closer undead.
"Jack, we're not under fire anymore," I said into the radio, "but we're still trapped in here with no way out."
"Okay," he said. "I'm outside. You've got to describe the interior layout of the building to me, as best you can."
I tried to give him enough information for him to visualize the inside of the building, and where we were in it. Finally, he seemed satisfied. "Okay, there's a big wall in front of you, to your left?"
"That's right," I said.
"Then shield yourself from it, if you can, 'cause there's going to be a big hole in it in a few seconds."
"Okay, Jack." I lowered the radio. "Milton, cover yourself with the mattress again, as much as possible. Jack's going to blow some more things up."
"Well, all right," he winced as he raised the mattress again and turned his face away from it. "He does so like to do that, doesn't he?"
A second later, the wall just beyond Milton exploded with a roar that seemed ten times louder than when the rockets had hit on the second level. My ears were ringing like crazy this time. The zombies closest to the hole were thoroughly shredded by the blast, while those behind them were thrown back into the crowd, mangled and torn from the flying debris. There was now a path from the door of the cell to the hole in the wall, and we needed to go through it—fast.
I was right at Milton's back. "You okay?" I asked. He coughed slightly and nodded.
Jack and one of his men came through the hole, firing pistols. Headlights were shining through the hole as well. Jack spotted us. "Come on!" he shouted, as the dead regrouped and began pressing toward him, tripping over the body parts and corpses of their fallen comrades.
I shoved Milton forward and told him to hold back the zombies. Then I hustled the others out of the cell. "Run toward Jack!" I shouted.
They made it through the hole in the outside wall, and I was following them when a hand grabbed my ankle and tripped me up.
"No!" Milton shouted, and he slammed his staff down on the undead wrist. Its hold wouldn't break, and it was pulling its maw up to my ankle. But with a second blow, Milton severed the arm at the rotted wrist. I got up and dashed through the hole with the bony appendage still attached to my ankle. Jack and his partner were right behind me, leaving Milton to prevent his former army from coming through the hole after us.
Once outside, I kicked at the undead hand until I finally got it off of me. The small dump truck was parked thirty feet from the hole in the wall, and we all climbed into the back of it, while Jack got in the driver's seat. He pulled the truck right up next to Milton and opened the door. Milton turned and jumped in as Jack tore off, with dead hands grasping at the side of the vehicle.
As we drove off, the helicopter rose from the roof of the prison and headed toward the museum.
A few raindrops hit us in the back of the dump truck, but the storm was passing to the south and east. The stars were coming out above us now, and the moon was half hidden by the retreating clouds. The rain had cooled everything, and the air had the freshness that it has after a cleansing storm. It was especially pronounced after the stale and rancid air of the prison, and the reek of the undead.
I looked up and breathed in deeply, and almost in spite of myself muttered, "Thanks, God. Really, this time."
Four of us had gone into hell, and three had come out. We were hardly unscathed, but we had survived. In the world of the undead, this was as close to victory as one dared hope for.
As we drove through the gate in the razor wire fence, I could feel the truck stop. I jumped down to see what was going on.
Jack and Milton got out of the truck. Milton walked over to the gate and closed it. Then he threaded the chain that had locked it back through the gate and the fence. "Jack, do you have something to hold this with?" he asked.
Jack slapped his pockets, the way people do when they're trying to see if they have the exact change or something. "You mean people-proof, or zombie-proof?" he asked.
"Just enough to hold it for a while, if some zombies come sniffing around and press up against it." Jack fumbled around in the cab of the truck for a minute. He came back with the key ring from off the truck's ignition key.
"Will this do?"
"Perfect." Milton put it through two links of the chain holding the gate closed.
"What are you doing, Milton?" I finally asked.
"I think he wants to make sure they're stuck in there with his zombie army for a while," Jack replied. "Until they're all eaten." Almost on cue, screams and gunfire could be heard, coming from the prison.
"More than that," Milton said in his dreamy sort of way. "When I was herding the zombies into here, I began to think, why couldn't I just herd all of them in here? It would be perfect, a place to keep them, so they couldn't bother the living anymore, and we wouldn't have to kill them."
I shook my head. "Milton, there are several billion of those things wandering around on the earth now. The most you could push around in front of yourself with a staff would be a dozen or so. I assume you only got so many into the prison tonight because they were all bunched up at the gate."
Milton smiled back and shook his head. "Now don't jump so far ahead, Jonah. Just because there are too many in the world is no excuse for me not to round up the ones I can, to protect you all living at the museum, and make your lives easier and safer. I realized as I led this army in there, against those evil men, that it had been wrong of me to fight against the dead, once I learned they couldn't hurt me. I should've been trying to help them."
Jack laughed. " _Help_ them? Now you really are just talking crazy, Milton. Get back in the truck and let's get back to the museum."
"No, it's a beautiful night now," Milton said wistfully. "I'm going to round up a few of our dead brothers and sisters, and put them in their new home, away from you all, and where they can do some good, punishing and eliminating the evil living, and turning them into docile, calm dead. You'll see how well it works out."
Jack knew that arguing with Milton was usually pointless. He knew Milton couldn't be hurt by the undead, so letting him wander around outside the museum with a new project wasn't that bad of an idea. "Well, Milton, okay," he said. "You know where to find us. Stop by when you need food or supplies. Give him back the radio," Jack said to me, and I handed it to Milton.
"You know I will, Jack," Milton said. "I'll come back to see you all, and to see my beloved books, that helped me so much. And you'll excuse me if I don't round up the dead too much in the winter. But I have a few months before that, to help you out." Milton turned toward me. "I am so sorry that you all suffered so much in there, and that Frank is gone. Be good to one another and heal your pains. And I'm sorry I misinterpreted your name, way back when we first meant."
"I don't understand," I said.
"Cain slew his brother, as we all have now in this undead world, and that was all that you and I remembered about him. But I was rereading the story again, last night. I was looking for some guidance as I prepared to come into battle to save you all, and for some reason I turned there. And I saw I had forgotten that Cain also built the first city. Help build our city, Jonah, the right way. You've been in the belly of the beast—thank God for less than three days, but I think long enough to see how bad the city of man can become." He gestured back toward the prison.
I nodded. "Yes, Milton, I have. And yes, I want to help build our city."
He smiled and leaned close. "And one day, when I'm too old and tired to be rounding up the dead, I hope you'll let me take some people to a meadow I know, where there's plenty of fresh water and green grass, and we can live there and pick strawberries, and grow old in peace." I hugged him like the friend and mentor he had become to me.
He stepped back and turned to address everyone. "You will all see me again," he said. "This is not goodbye. Our friends' suffering, and the destruction of this terrible, evil place have made me see a new way! Together we will grow and live, like we never dreamed we ever could again! Finally, there will be more living, and less dying!" He waved to us, then turned and wandered off the road, into the moonlit fields, whistling a little tune.
The rest of us returned to the museum, where we wept for poor Frank, and for all the horrors we had seen, until we all dozed off finally, shortly before dawn.
##
MILTON'S IDEA TURNED out to be his best ever, by far. Within days, we saw the zombie population near the museum drop to where we could come and go much more safely, barely having to set up distractions. Within weeks, they were seldom seen anywhere near the museum.
Milton got so good at rounding them up that we had to clear the barricade from the other end of the bridge to allow him freer access to herd them across the river and out to the prison. It was good to see the barricade go, for it made us feel we were less under siege and could live a more normal—if still difficult and dangerous—existence.
I was so gladdened by the barricade being moved that I talked Jack into taking one of the small sculptures from the museum grounds and setting it up on the site, with a makeshift plaque, to commemorate the battle there. The war with the undead would have its own Gettysburgs and Normandys, and we would honor those who fought. We could even now afford to honor and respect the dead, rather than just cower in fear behind walls from them. Most of all, we could feel that this war would have the one thing that all wars were supposed to have, but which we had long since lost all hope of seeing—an end.
Jack began plans on extending the museum's walls to include the park across the street and some other nearby buildings. He had found a building site with plenty of supplies and began hauling them back to the museum. The park would be a special prize, as it would dramatically increase the amount of farmland for next year. And the farming this year was successful enough. By the fall, we were eating fresh fruits and vegetables every day, though we still didn't have enough to set aside for winter, nor did we have the equipment necessary for preserving them.
My experience in the prison had convinced me to talk the museum people into doing some hunting in the nearby countryside; we deserved to have some benefit from our brief and violent association with those carnivorous monsters. On one of these raids into the countryside, someone thought to grab some small farm animals, and we began to look more actively for them, acquiring a small number of chickens and goats, with plans for more once we had the space.
With material prosperity improving in the compound, Jack and I would gaze across the river and dream of the day we'd reclaim the city completely. "Looks like we're getting close to taking that rowboat out on the river, Jonah," he would kid me, "and do some fishing, like we planned."
The emotional and personal lives within the compound had kept pace with the material improvements as well. Sarah and Jack had, for all intents and purposes, adopted Zoey. Among the luggage Frank had brought, they found some few pictures of Frank and his wife, and they kept these to show Zoey when she grew up. She was the child of two very brave people, and she should know the pride and responsibility that came with such parentage.
One day, shortly after our ordeal in the prison, I found Tanya sitting on the grass, painting her toenails. "You and Milton seemed to get such a freak on about this stuff," she said coyly to me, "that I thought I should finally humor you all, before the damn bottle dries out." She grinned up at me. She almost giggled, and once you met Tanya—at least as I had known her in a world of the living dead—you knew giggling was not part of her behavior.
I sat next to her. "They look good," I said. "I knew they would. What are you so happy about?"
She put her arm around me and leaned close. "Oh, it's pretty normal, I think: happiest times in a gal's life are usually when she learns she's pregnant." And, allowing for all the irregularities of our situation, it was one of the happiest times of my life, too.
Soon after, we learned that Jack and Sarah were also expecting a baby. I had always thought the idea of pregnant women "glowing" was more or less concocted to distract them from the weight gain and nausea, but I had to admit that in Sarah's case, it was completely accurate. She was as radiant and joyful as a person could be.
Jack's reaction was harder to gauge. I could only suppose he had either seen the logic of procreating; either that, or the logical, foreseeable, and inevitable outcome of sex had simply caught up with him, and he just accepted it. It was impossible for me to tell with him, given his ever-joking manner about personal matters and feelings. But either way, they were both happy with having a future, and things to worry about other than shooting zombies in the head and consuming enough calories from canned food to stay alive.
And if we saw less of "our dead brothers and sisters," as Milton now liked to call them, we hardly saw any less of him. If anything, his constant sojourns to round up the dead and put them in their new home seemed to cure him mostly of his illness, so he had more energy and was in less pain.
Those of us without his affliction whispered darkly that it seemed like a terrible price to pay—having to walk among the reeking, rotting dead in order to feel more alive—and it seemed to bespeak a frightening, shameful kinship between Milton and his dead brothers and sisters, but there was no denying that it benefited all of us immensely. So even though Milton slept every other night out in the countryside somewhere between the museum and the prison, we saw him nearly as often as we used to, and he was always in the highest of spirits.
I remember one time, in the early fall, when the weather was about like it had been when I had first arrived at the museum—a glorious, clear, warm, almost painfully bright day. But in the fall, such a day would always hold the threat of cold and death to come, rather than the springtime's promise of rebirth and life and growth. But this day, such thoughts did not really make us sad, so much as they made us merely thoughtful and introspective, like Milton always was, regardless of the weather. Jack, Tanya, Sarah, and I were out on the roof of the museum, watching Milton move a particularly large mob of the dead. He had with him two dogs that he had found on his trips, and which had become instantly devoted and loving, while he had likewise become instantly enamored of their simple loyalty and good sense. He now used the dogs to help him herd the dead, so that he could now handle hordes over a hundred.
"It's so damned weird, watching him herd them," Jack said.
"I think it's cute," Sarah said. Unlike Tanya, giggling came quite naturally to her.
"It's still weird," Jack continued. "He looks like he's their shepherd. It's as if they like him or something. I don't know if that's right."
"He looks like a damn zombie Jesus," Tanya added. We all turned to stare at her. She was right, of course, but it still sounded strange, almost taboo, even if you weren't religious. "Well, he _does_ ," she insisted. "I mean, I don't remember Jesus having any dogs, but you got to admit—it's what he looks like. But maybe it's how it's supposed to be. Jesus was part man and part God, so he could save man. Milton's part man and part _them_ —no offense to him, you understand—so he can save them. And us."
Tanya was, as usual, quite right. We had our messiah, and so did the dead. We had our little community. We had our love and our children. And as hard as the past year had been, there was nothing more that we could legitimately ask for, and nothing more for which to blame God. We had been tried in a way more horrible than any of us could've imagined, but we had survived when none of us would've guessed it possible. Jack's million little coincidences and lucky breaks had all come together in such a way that the gates of hell—quite literally—had not prevailed against us.
##
THE PERSON WHO has read and commented the most on the manuscript is Robert P. Kennedy, with great enthusiasm and encouragement for my new project. At the same time, W. Scott Field—with his extensive knowledge of things military—has provided frequent and enormous assistance about the fine points of modern weapons systems, explosives, and tactics. William Lebeda's comments have been of a more aesthetic and less technical nature, but he too has been a constant inspiration to me to pursue my writing in this genre. Marylu Hill, as the token non-horror person among my initial readers, has often provided the outsider's perspective that helps keep my writing from becoming too narrow or constrained by its genre. Finally, my editors at Permuted Press—Jacob Kier and D.L. Snell—gave the manuscript its final tweaking and tightened up a lot of loose points.
The city in the story is based very loosely on Grand Rapids, Michigan, where I spent two recent summers studying at Calvin College as part of the Seminars in Christian Scholarship (see <http://www.calvin.edu/scs>). I saw no point in naming it explicitly in the story, as the factor of local color would not carry the kind of weight it does with more famous locales, and I would then be endlessly criticized for any necessary departures from the details of the real-life city. For maps and description of local landmarks like the Van Andel Museum, see <http://www.visitgrandrapids.org>.
People are always curious about one's literary influences. I don't think most of my exemplars are typical of horror writers, but I suspect that what we think of as "typical" is an unfair generalization, and horror writers are as diverse in their tastes as doctors, lawyers, or college professors. However, the influence of—and even direct allusions to—several books that have haunted me for decades are scattered throughout the present work. Alert readers will readily spot images taken from Dante's _Inferno_ , Melville's _Moby-Dick_ , several of Shakespeare's plays, and, of course, the Bible. What work of literature could be truly horrifying without drawing on the accumulated depictions of evil throughout Western literature?
But beyond details or allusions, I would have to say many of the ideas in this book ultimately go back to two men whose works I first encountered about a quarter century ago. First, St. Augustine, for his most profound and influential ruminations on human depravity that still, to this day, fifteen centuries after he lived, do not merely influence discussions of sin and evil, they more or less determine what will be discussed, the terms in which the concepts can be discussed, and, to a large extent, the conclusions that can be drawn.
And second, of course, no zombie book or film today can fail to acknowledge George A. Romero, for completely determining the identity, meaning, and importance of zombies in our culture. Influences from and allusions to his films abound in the present work, as they must in any zombie film or fiction today. I believe the continuing ability of Romero's zombie hordes both to scare and challenge us in our arrogance makes them as potent and relevant as St. Augustine's more overtly theological work. They continue to have "bite," as it were.
Finally, thanks are always due to my family, during and after any writing project. My obsession with monsters is tolerated by my wife, Marlis, and our daughter, Sophia, while it is enthusiastically encouraged by our son, Charles.
Kim Paffenroth
Cornwall on Hudson, NY
June 2010
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\title{A residue formula for the fundamental Hochschild $3$-cocycle for $SU_{q}(2)$}
\author{Ulrich
Kr\"{a}hmer\dag\thanks{email:
\texttt{ulrich.kraehmer@glasgow.ac.uk},
\texttt{adam.rennie@anu.edu.au},
\texttt{roger.senior@anu.edu.au}},\
\ Adam Rennie\ddag,\
\ Roger Senior\ddag \\[6pt]
\dag School of Mathematics \&
Statistics, University of Glasgow\\
15 University Gardens, G12 8QW Glasgow, Scotland\\[6pt]
\ddag Mathematical Sciences Institute,
Australian National University\\
Acton, ACT, 0200, Australia\\[6pt]
}
\begin{document}
\maketitle
\vspace{-8pt}
\begin{abstract}
An analogue of a spectral triple over $SU_{q}(2)$ is constructed for which the usual assumption
of bounded commutators with the Dirac operator fails.
An analytic expression analogous to that for the Hochschild
class of the Chern character for spectral triples
yields a non-trivial twisted Hochschild 3-cocycle.
The problems arising from the unbounded commutators
are overcome by defining a residue functional using projections to cut
down the Hilbert space.
\end{abstract}
\section{Introduction}
\label{sec:intro}
This paper studies the homological dimension of the quantum group
$SU_q(2)$ from the perspective of Connes' spectral triples.
We use an analogue of a spectral triple to construct, by a
residue formula, a nontrivial Hochschild 3-cocycle.
Thus we obtain finer dimension information
than is provided by the nontriviality of a
$K$-homology class, which is sensitive only to dimension modulo 2.
The position of quantum groups
within noncommutative geometry has been studied intensively
over the last 15 years.
In particular, Chakraborty
and Pal \cite{ChP1} introduced a spectral triple for $SU_q(2)$, and this construction was subsequently refined
in \cite{DLSSV} and generalised by
Neshveyev and Tuset in \cite{NT2} to all compact Lie groups $G$.
These spectral triples have analytic dimension $\dim G$
and nontrivial $K$-homology class.
However, when Connes computed the Chern character for Chakraborty and
Pal's spectral triple \cite{C1},
he found that it had cohomological dimension $1$ in the sense that
the degree $\dim SU(2)=3$ term
in the local index formula is a Hochschild coboundary.
Analogous results for the spectral triple from
\cite{DLSSV} were obtained in \cite{DLSSV2}.
Contrasting these `dimension drop' results,
Hadfield and the first author \cite{HK1,HK2}
showed that $SU_q(2)$ is a twisted Calabi-Yau algebra of dimension 3 whose twist is the inverse of the modular automorphism for the Haar state on this compact quantum group, cf. Section \ref{sec:suq2}.
They also computed a cocycle
representing a generator of the nontrivial degree $3$
Hochschild cohomology groups (which we call the fundamental cocycle), and
a dual degree $3$ Hochschild cycle
which we denote $dvol$.
The starting point of the present paper is
the concept of a `modular' spectral triple \cite{CNNR}. These are analogous to
ordinary spectral triples except for the use of twisted traces.
The examples considered in \cite{CNNR} arise from
KMS states of circle actions on $C^*$-algebras, and yield nontrivial
$KK$-classes with $1$-dimensional Chern characters
in twisted cyclic cohomology.
In \cite{KW} it was then shown that they also can be used to obtain
the fundamental cocycle of the standard Podle\'s quantum 2-sphere.
Motivated by this, our
construction here extends the modular spectral triple on
the Podle\'s sphere to all of $SU_q(2)$. This extension is not a
modular spectral triple,
but as our main theorem shows, still
captures the homological dimension 3:
we give a residue formula for a twisted Hochschild $3$-cocycle which
is a nonzero multiple of the fundamental cocycle. We obtain this formula
by analogy with Connes' formula for the Hochschild class of the Chern
character of spectral triples,
\cite[Theorem 8, IV.2.$\gamma$]{C} and \cite{BeF,CPRS1}. A natural next question that
arises is whether our constructions provide a representative of a
nontrivial $K$-homological class.
The organisation of the paper is as follows. In Section \ref{sec:suq2} we recall the definitions of
$SU_q(2)$, the Haar state on $SU_q(2)$ and the associated GNS representation, and finally the modular theory
of the Haar state. In Section \ref{sec:homology} we recall the homological constructions of \cite{HK1,HK2},
and prove some elementary results
we will need when we come to show that our residue cocycle does indeed
recover the class of the fundamental cocycle.
Section \ref{sec:mero} contains all the key analytic results
on meromorphic extensions of certain functions
that allow us to prove novel summability type results for operators whose
eigenvalues have mixed polynomial and exponential growth,
see Lemma \ref{lem:general-pole-3}.
Section \ref{sec:spec} constructs an analogue of a spectral triple $(\A,\H,\D)$
over the algebra $\A$ of polynomials in the
standard generators
of the $C^*$-algebra $SU_q(2)$. The key requirement of
bounded commutators fails, and this `spectral triple' fails to be finitely summable in the usual sense
(however, it is $\theta$-summable). Using an ultraviolet cutoff
we can recover finite summability of the operator $\D$ on a subspace
of $\H$ with respect to a suitable twisted trace.
However, our representation of $\A$ does not restrict to this
subspace, and so we are prevented from obtaining a genuine spectral triple.
In Section \ref{sec:res-hochs} we define a residue functional $\tau$. Heuristically, for an operator $T$,
$$
\tau(T)=\mbox{Res}_{s=3}\mbox{Trace}(\Delta^{-1} QT(1+\D^2)^{-s/2}).
$$
Here $\Delta$ implements the modular
automorphism of the Haar state, $\D$ is our Dirac operator and $Q$ is
a suitable projection that implements the cutoff. The existence, first
of the trace, and then the residue, are both nontrivial matters.
The main properties of $\tau$ are described in
Theorem \ref{thm_res_gives_int}, and in particular we show that
the domain of $\tau$ contains the products of commutators $a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}]$ for
$a_i\in\A$. In addition, $\tau$ is a twisted trace on a suitable subalgebra of the domain containing these products.
The main result, Theorem \ref{thm:big-fish}, proves that the map
$a_{0}, \ldots, a_{3} \mapsto \tau(a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}])$ is a twisted Hochschild
3-cocycle, whose cohomology class is non-trivial and coincides with (a multiple of) the fundamental class.
\medskip
{\bf Acknowledgements} We would like to thank our colleagues Alan
Carey, Victor Gayral, Jens Kaad, Andrzej Sitarz and Joe V\'arilly for stimulating
discussions on these topics.
The second and third authors were supported by the Australian Research Council.
The first author was supported by
the EPSRC fellowship EP/E/043267/1 and partially by the
Polish Government Grant N201 1770 33 and
the Marie Curie PIRSES-GA-2008-230836
network.
\section{Background on $SU_{q}(2)$}
\label{sec:suq2}
The notations and conventions of \cite{KS} will be used throughout for consistency. We
recall that $\A:=\ensuremath{\mathcal{O}}(SU_q(2))$, for $q \in (0, 1)$, is the unital Hopf $*$-algebra with generators
$a, b, c, d$ satisfying the relations
\envaln{
ab = qba, \ \ \ ac = qca, \ \ \ bd &= qdb, \ \ \ cd = qdc, \ \ \ bc = cb \\
ad = 1 + qbc, \ & \ \ da = 1 + q^{-1}bc
}
and carrying the usual Hopf structure,
as in e.g. \cite{KS}. The involution is
given by
\enveqn{
a^{\ast} = d, \ \ \ b^{\ast} = -qc, \ \ \ c^{\ast} = -q^{-1}b, \ \ \ d^{\ast} = a.
}
We choose to view $\A$ as being generated by $a, b, c, d$ explicitly,
rather than just $a, b$, in order to make formulae more readable.
\begin{prop}[{\cite[Proposition 4.4]{KS}}]
The set $\{ a^{n}b^{m}c^{r}, \ b^{m}c^{r}d^{s} \ | \ m, r, s \in \ensuremath{\mathbb{N}}_{0}, \ n \in \ensuremath{\mathbb{N}} \}$
is a vector space basis of $\A$. These monomials will be referred to as the polynomial basis.
\end{prop}
Recall that for each $l \in \frac{1}{2} \ensuremath{\mathbb{N}}_0$, there is a
unique (up to unitary equivalence) irreducible corepresentation $V_l$ of
the coalgebra $\A$ of dimension $2l+1$, and that $\A$ is
cosemisimple. That is, if we fix a vector space basis in
each of the $V_l$ and denote by $t^l_{i,j} \in \A$ the corresponding
matrix coefficients, then we have the following analogue of the
Peter-Weyl theorem.
\begin{thm}[{\cite[Theorem 4.13]{KS}}] \label{thm_rep_basis}
Let
$ I_{l} := \{ -l, -l+1, \ldots, l-1,l \} $.
Then the set $\{ \re{l}{i}{j} \ | \ l \in
\frac{1}{2} \ensuremath{\mathbb{N}}_{0}, \ i, j \in I_{l} \}$ is a vector space basis of $\A$.
\end{thm}
This will be referred to as the
Peter-Weyl basis. With a suitable choice of basis in
$V_\frac{1}{2}$, one has
\envaln{
a &= \re{\frac{1}{2}}{-\frac{1}{2}}{-\frac{1}{2}}, & b &= \re{\frac{1}{2}}{-\frac{1}{2}}{\frac{1}{2}}, & c &= \re{\frac{1}{2}}{\frac{1}{2}}{-\frac{1}{2}},
& d &= \re{\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}.
}
The expressions
for the Peter-Weyl basis elements as
linear combinations of the polynomial
basis elements can be found in
\cite[Section 4.2.4]{KS}.
The quantized universal enveloping algebra $U_{q}(\mathfrak{sl}(2))$ is a
Hopf algebra which is generated by $k, k^{-1}, e, f$ with relations
\enveqn{
kk^{-1} = k^{-1}k = 1, \quad kek^{-1} = qe, \quad kfk^{-1} = q^{-1}f,
\quad [e, f] = \frac{k^{2} - k^{-2}}{q - q^{-1}}.
}
Note that in \cite{KS} this algebra is
denoted by
$\breve{U}_{q}(\mathrm{sl}_{2})$
and $U^\mathrm{ext}_{q}(\mathrm{sl}_{2})$. The algebra
$U_{q}(\mathfrak{sl}(2))$ carries the
following Hopf structure
\envaln{
\Delta(k) = k \otimes k, \quad \Delta(e) = e \otimes k &+ k^{-1} \otimes e, \quad \Delta(f) = f \otimes k + k^{-1} \otimes f \\
S(k) = k^{-1}, \quad S(e) &= -qe, \quad S(f) = -q^{-1}f \\
\varepsilon(k) = 1, \quad & \varepsilon(e) = \varepsilon(f) = 0.
}
Adding the following involution
\enveqn{
k^{\ast} = k, \quad e^{\ast} = f, \quad f^{\ast} = e
}
we obtain a Hopf $\ast$-algebra which
we denote by $U_{q}(\mathfrak{su}(2))$.
\begin{thm}[{\cite[Theorem 4.21]{KS}}]
\label{theorem_dual_pairing}
There exists a unique dual pairing $\left\langle \cdot , \cdot \right\rangle$ of the
Hopf algebras $U_{q}(\mathfrak{sl}(2))$ and $\A$ such that
\envaln{
\left\langle k, a \right\rangle = q^{-\frac{1}{2}}, &\quad \left\langle k, d \right\rangle = q^{\frac{1}{2}}, \quad \left\langle e, c \right\rangle = \left\langle f, b \right\rangle = 1 \\
\left\langle k, b \right\rangle = \left\langle k, c \right\rangle = \left\langle e, a \right\rangle = &\left\langle e, b \right\rangle = \left\langle e, d \right\rangle = \left\langle f, a \right\rangle = \left\langle f, c \right\rangle = \left\langle f, d \right\rangle = 0.
}
This pairing is compatible with the $*$-structures on $U_{q}(\mathfrak{sl}(2))$ and $\A$, \cite[Chapter 1]{KS}.
\end{thm}
The dual pairing between the Hopf algebras
$\left\langle \cdot, \cdot \right\rangle \colon U_{q}(\mathfrak{sl}(2)) \times \A \rightarrow \ensuremath{\mathbb{C}}$
defines left and right actions of $U_{q}(\mathfrak{sl}(2))$ on $\A$.
Using Sweedler notation ($\Delta(x) = \sum x_{(1)} \otimes x_{(2)}$) these actions are given by
\envaln{
g \triangleright x &:= \sum x_{(1)} \left\langle g, x_{(2)} \right\rangle & x \triangleleft g &:= \sum x_{(2)} \left\langle g, x_{(1)} \right\rangle, & &
\text{for all} \ \ x \in \A, \ g \in U_{q}(\mathfrak{sl}(2)).
}
The left and right actions make $\A$ a $U_{q}(\mathfrak{sl}(2))$-bimodule
\cite[Proposition 1.16]{KS}.
Our definition of the $q$-numbers is
\enveqn{
[a]_{q} := \frac{q^{-a} - q^{a}}{q^{-1} - q}=Q(q^{-a} - q^{a}) \qquad \text{for any} \ a \in \ensuremath{\mathbb{C}},
}
where we abbreviated
$Q := (q^{-1} - q)^{-1} \in (0, \infty)$.
The following lemma recalls the explicit formulas for the action of the generators on the
Peter-Weyl basis.
\begin{lemma}
For all $n \in \ensuremath{\mathbb{Z}}$,
\envaln{
k^{n} \triangleright \re{l}{i}{j} &= q^{nj} \re{l}{i}{j} & \re{l}{i}{j} \triangleleft k^{n} &= q^{ni} \re{l}{i}{j} \\
e \triangleright \re{l}{i}{j} &= \sqrt{\left[ l + \tfrac{1}{2} \right]_{q}^{2} - \left[ j + \tfrac{1}{2} \right]_{q}^{2} } \,\,\re{l}{i}{j+1} & f \triangleright \re{l}{i}{j} &= \sqrt{\left[ l + \tfrac{1}{2} \right]_{q}^{2} - \left[ j - \tfrac{1}{2} \right]_{q}^{2} } \,\, \re{l}{i}{j-1}.
}
\end{lemma}
Later we will use the notation
\enveqn{
\ensuremath{\p_{k}} := k \triangleright \cdot\,, \qquad \ensuremath{\p_{e}} := e \triangleright \cdot\,, \qquad \ensuremath{\p_{f}} := f \triangleright \cdot\,,
}
especially when we extend these
operators from $\A$ to
suitable completions.
Also observe that
$\Delta(k^{n}) = k^{n} \otimes k^{n}$ for all $n \in \ensuremath{\mathbb{Z}}$, hence $k^{n} \triangleright \cdot$
and $\cdot \triangleleft k^{n}$ are
algebra automorphisms on
$\A$. They are not $*$-algebra automorphisms since for $\alpha\in\A$ we have
$(k \triangleright \alpha)^*=k^{-1} \triangleright
\alpha^*,\ (\alpha \triangleleft k)^*=\alpha^*
\triangleleft k^{-1}$. Finally, we
introduce
$$
\partial_{H}(\re{l}{i}{j}) = j \re{l}{i}{j},
$$
and we note that formally $\partial_k=q^{\partial_H}$.
\subsection{The GNS representation for the Haar state}
We denote by $A:=C^*(SU_q(2))$ the
universal $C^*$-completion of the
$*$-algebra $\A$ \cite[Section~4.3.4]{KS}.
Let $h$ be the Haar state
of $A$ whose values on basis
elements are
\enveqn{
h(a^{i}b^{j}c^{k}) = h(d^{i}b^{j}c^{k}) = \delta_{i, 0} \delta_{j, k} (-1)^{k} [k+1]_q^{-1}, \quad
h(\re{l}{i}{j}) = \delta_{l0}.
}
Let $\ensuremath{\mathcal{H}}_{h}$ denote the GNS space $L^{2}(A, h)$,
where the inner product $\left\langle x, y \right\rangle = h(x^{\ast} y)$ is conjugate linear in the first variable.
The representation of $A$ on $\ensuremath{\mathcal{H}}_{h}$ is is induced by left
multiplication in $A$.
The set $\{ \re{l}{i}{j} \ | \ l \in
\frac{1}{2} \ensuremath{\mathbb{N}}_{0}, \ i, j \in I_{l} \}$ is
an orthogonal basis for $\H_h$ with
\enveqn{
\left\langle \re{l}{i}{j}, \re{l'}{i'}{j'} \right\rangle = \delta_{l,l'} \delta_{i,i'} \delta_{j,j'} q^{-2i} [2l+1]_{q}^{-1}.
}
\subsection{Modular Theory}
Following Woronowicz, we call the automorphism
$$
\vartheta (\alpha) := k^{-2}
\triangleright \alpha \triangleleft
k^{-2},\quad
\alpha \in \A
$$
the modular automorphism of $\A$.
The action of $\vartheta$ on the
generators of $\A$ and the Peter-Weyl
basis is given by
\enveqn{
\vartheta(a) = q^{2}a, \ \ \vartheta(b) = b, \ \ \vartheta(c) = c, \ \ \vartheta(d) = q^{-2}d, \ \
\vartheta(\re{l}{r}{s}) = q^{-2(r+s)} \re{l}{r}{s}.
}
The modular automorphism is a (non $\ast$-) algebra automorphism; more precisely for any
$\alpha \in \A$
\enveqn{
\vartheta(\alpha)^{\ast} = \vartheta^{-1}(\alpha^{\ast}).
}
The Haar state is related to the modular automorphism by the following proposition.
\begin{prop}[{\cite[Proposition 4.15]{KS}}]
\label{prop_haar_twisted}
For $\alpha,\,\beta \in \A$, we have $h(\alpha\beta) = h(\vartheta(\beta)\alpha)$.
\end{prop}
In fact, $h$ extends to a KMS state on
$A$ for the strongly
continuous one-parameter group
$\vartheta_{t}$, $t \in \mathbb{R}$, of
$*$-automorphisms of
$A$
which is given on the generators by
\envaln{
\vartheta_{t}(a) &:= q^{-2it} a\,, &
\vartheta_{t}(b) &:= b\,, &
\vartheta_{t}(c) &:= c\,, &
\vartheta_{t}(d) &:= q^{2it} d\,.
}
We extend this to an
action $\vartheta_{\cdot}\, \colon \ensuremath{\mathbb{C}} \times \A \rightarrow \A$
by algebra (not $\ast$-) automorphisms that is defined on generators
by
\envaln{
\vartheta_{z}(a) &:= q^{-2iz} a\,, &
\vartheta_{z}(b) &:= b\,, &
\vartheta_{z}(c) &:= c\,, &
\vartheta_{z}(d) &:= q^{2iz} d\,,
}
so that the modular automorphism
$\vartheta$ is $\vartheta_i$.
We can implement $\vartheta_t$ in
the GNS representation on $\ensuremath{\mathcal{H}}_h$.
To do this, we define an unbounded
linear operator $\Delta_{F}$
on $\A\subset \ensuremath{\mathcal{H}}_h$ by
\enveqn{
\Delta_{F}(\re{l}{i}{j}) := q^{2i+2j} \re{l}{i}{j}
}
and call this the full modular
operator. Then we have
\enveqn{
\vartheta_{t}(x) \xi = \Delta_F^{it} x
\Delta_F^{-it} \xi\,, \qquad \text{for all}
\ x \in A \ \
\text{and} \ \xi \in \ensuremath{\mathcal{H}}_h.
}
The subscript $F$ denotes that this operator is
associated to the full modular
automorphism $\vartheta$. In addition,
we define the left and the right
modular operators on $\A\subset \H_h$ by
\envaln{
\Delta_{L}(\re{l}{i}{j}) &:= q^{2j} \re{l}{i}{j}, & \Delta_{R}(\re{l}{i}{j}) &:= q^{2i} \re{l}{i}{j},
}
so $\Delta_{F} =
\Delta_{L} \Delta_{R} = \Delta_{R}
\Delta_{L}$. Just as $\Delta_F$
implements the modular automorphism
group, the left and right modular
operators implement one-parameter groups
of automorphisms of $A$:
\envaln{
& \sigma_{L, t}(\re{l}{r}{s}) = q^{2its} \re{l}{r}{s} = \Delta_{L}^{it} \re{l}{r}{s} \Delta_{L}^{-it}\,, & & \sigma_{R, t}(\re{l}{r}{s}) = q^{2itr} \re{l}{r}{s} = \Delta_{R}^{it} \re{l}{r}{s} \Delta_{R}^{-it}.
}
As with the full action, the left and
right actions are periodic and hence
give rise to actions of $\T$ on $A$.
These
may be extended to a complex action
on the $\ast$-subalgebra $\A$ which we will
denote $\sigma_{L, z}$ and $\sigma_{R,
z}$. In particular, we obtain for $z=i$
the algebra automorphisms
\envaln{
\sigma_{L} &:= k^{-2} \triangleright \cdot & \sigma_{R} &:= \cdot \triangleleft k^{-2} & \vartheta &= \sigma_{L} \sigma_{R} = \sigma_{R} \sigma_{L} \\
& & \sigma_{L}(\re{l}{r}{s}) = q^{-2s} \re{l}{r}{s} \qquad & \qquad \sigma_{R}(\re{l}{r}{s}) = q^{-2r} \re{l}{r}{s}\\
\vartheta(\alpha) \xi &= \Delta_{F}^{-1} \alpha \Delta_{F} \xi &
\sigma_{L}(\alpha) \xi &= \Delta_{L}^{-1} \alpha \Delta_{L} \xi &
\sigma_{R}(\alpha) \xi &= \Delta_{R}^{-1} \alpha \Delta_{R} \xi.
}
The fixed point algebra for the left
action on $\A$ is isomorphic to the
standard Podle\'s quantum 2--sphere
$\ensuremath{\mathcal{O}}(S_{q}^{2})$.
We will denote its $C^{\ast}$-completion
by $B$.
As the left action is periodic, we may define a positive faithful expectation
$\Phi \colon A \rightarrow B$ by
\enveqn{
\Phi(x) = \frac{\ln(q^{-2})}{2\pi } \int_{0}^{2\pi/\ln (q^{-2})}
\sigma_{L, t}(x) dt.
}
More generally, given $n \in \ensuremath{\mathbb{Z}}$ and $x \in A$ we define
\enveqn{
\Phi_{n}(x) = \frac{\ln(q^{-2})}{2\pi } \int_{0}^{2\pi/\ln(q^{-2})} t^{-n} \sigma_{L, t}(x) dt.
}
Since $\sigma_{L, t}$ is a strongly
continuous action on $A$, the $\Phi_{n}$ are
continuous maps on $A$. Observe that $\Phi = \Phi_{0}$ and
\enveqn{
\Phi_{n}(\re{l}{i}{j}) = \delta_{n, 2j} \re{l}{i}{j}
}
Hence the $\Phi_{n}$ can be extended to bounded operators on the GNS space $\ensuremath{\mathcal{H}}_{h}$,
and in fact the $\Phi_{n}$ are projections onto the spectral subspaces of the left circle action.
So we make explicit the decomposition of $A$ into the left spectral subspaces by defining
\envaln{
B_{n} &:= \Phi_{n}(A) = \{ \alpha \in A \ | \ \sigma_{L, t}(\alpha) = q^{2int} \alpha \} \quad\mbox{and}\quad
\ensuremath{\mathcal{H}}_{n} := L^{2}(B_{n}, h)
}
where $h$ is the Haar state (restricted to $B_{n}$). This leads to the following decomposition
for the GNS space
\enveqn{
\ensuremath{\mathcal{H}}_{h} = \bigoplus_{n = -\infty}^{\infty} \ensuremath{\mathcal{H}}_{n}.
}
The commutation relations for the projections $\Phi_{n}$ and the operators
$\ensuremath{\p_{k}}$, $\ensuremath{\p_{e}}$ and $\ensuremath{\p_{f}}$ are found from the definitions on the Peter-Weyl basis to be
\envaln{
\ensuremath{\p_{k}} \Phi_{n} &= \Phi_{n} \ensuremath{\p_{k}} =
q^{\frac{n}{2}} \Phi_{n}
& \partial_{H} \Phi_{n} &= \Phi_{n} \partial_{H} = \frac{n}{2} \Phi_{n}
& \Delta_{L} \Phi_{n} &= \Phi_{n} \Delta_{L} = q^{n} \Phi_{n} \\
\ensuremath{\p_{e}} \Phi_{n} &= \Phi_{n+2} \ensuremath{\p_{e}} & \ensuremath{\p_{f}} \Phi_{n} &= \Phi_{n-2} \ensuremath{\p_{f}}\,.
}
The left actions of $e$ and $f$ are twisted derivations in the sense that for $\alpha,\,\beta \in \A$
\envaln{
\ensuremath{\p_{e}}(\alpha\beta) &= \ensuremath{\p_{e}}(\alpha) \ensuremath{\p_{k}}(\beta) + \ensuremath{\p_{k}^{-1}}(\alpha) \ensuremath{\p_{e}}(\beta) \\
\ensuremath{\p_{f}}(\alpha\beta) &= \ensuremath{\p_{f}}(\alpha) \ensuremath{\p_{k}}(\beta) + \ensuremath{\p_{k}^{-1}}(\alpha) \ensuremath{\p_{f}}(\beta)\,.
}
More generally, given $\alpha \in A$ and $\xi \in \ensuremath{\mathcal{H}}_{h}$
\enval{
\ensuremath{\p_{e}}(\alpha \xi) &= \ensuremath{\p_{e}}(\alpha) \Delta_{L}^{\frac{1}{2}} \xi + \sigma_{L}^{\frac{1}{2}}(\alpha) \ensuremath{\p_{e}}(\xi) &
\ensuremath{\p_{f}}(\alpha \xi) &= \ensuremath{\p_{f}}(\alpha) \Delta_{L}^{\frac{1}{2}} \xi + \sigma_{L}^{\frac{1}{2}}(\alpha) \ensuremath{\p_{f}}(\xi) \label{eqn_twisted_ef} \\
&= \ensuremath{\p_{e}}(\alpha) \Delta_{L}^{\frac{1}{2}} \xi + \Delta_{L}^{-\frac{1}{2}} \alpha \Delta_{L}^{\frac{1}{2}} \ensuremath{\p_{e}}(\xi) &
&= \ensuremath{\p_{f}}(\alpha) \Delta_{L}^{\frac{1}{2}} \xi + \Delta_{L}^{-\frac{1}{2}} \alpha \Delta_{L}^{\frac{1}{2}} \ensuremath{\p_{f}}(\xi). \nonumber
}
See e.g.~\cite{scream} and the references therein
for background on the generalisation of this setting in terms of
Hopf-Galois extensions.
\section{Twisted homology and cohomology}
\label{sec:homology}
We recall that the algebra $\A$ is a $\vartheta^{-1}$-twisted
Calabi-Yau algebra of dimension 3,
see~\cite{HK2} and the references therein
for this result and some background. Since the centre of $\A$
consists only of the scalar multiples of
$1_\A$, this means
in particular that the
cochain complex
$C^\bullet:=\mathrm{Hom}_\mathbb{C}
(\A^{\otimes_\mathbb{C}
\bullet+1},\mathbb{C})$, with differential
$b_{\vartheta^{-1}} : C^n \rightarrow C^{n+1}$
given by
\envaln{
(b_{\vartheta^{-1}}\varphi)(a_{0}, \ldots, a_{n}, a_{n+1})
&= \sum_{i = 0}^{n} (-1)^{n} \varphi(a_{0}, \ldots, a_{i}a_{i+1}, \ldots, a_{n+1}) \\
& \quad + (-1)^{n+1} \varphi(\vartheta^{-1}(a_{n+1})a_{0}, a_{1}, \ldots, a_{n}),
}
is exact in degrees $n>3$ and has third cohomology
$H^3(C,b_{\vartheta^{-1}}) \simeq \mathbb{C}$.
An explicit cocycle whose cohomology class generates
$H^3(C,b_{\vartheta^{-1}})$ can be
constructed using the following
incarnation of the cup product $\smallsmile$ in
Hochschild cohomology:
\begin{lemma}\label{tuna}
Let
$\sigma_0,\ldots,\sigma_3$
be automorphisms of $\A$,
$\int : \A \rightarrow \mathbb{C}$ be a
$\sigma_0 \circ \vartheta^{-1} \circ \sigma_3^{-1}$-twisted trace, that is,
$$
\int \alpha\beta=\int \sigma_0(\vartheta^{-1}(\sigma_3^{-1}(\beta)))\alpha,
$$
and $\partial_i : \A \rightarrow \A$, $i=1,2,3$,
be $\sigma_{i-1}$-$\sigma_i$-twisted
derivations, that is,
$$
\partial_i(\alpha\beta)=\sigma_{i-1}(\alpha)
\partial_i (\beta)+
\partial_i(\alpha) \sigma_i(\beta).
$$
Then the functional defined via the cup product by
$$
\left(\int \smallsmile \partial_1
\smallsmile \partial_2
\smallsmile
\partial_3\right)(a_0,a_1,a_2,a_3):=
\int \sigma_0(a_0) \partial_1(a_1)
\partial_2(a_2) \partial_3(a_3)
$$
is a $\vartheta^{-1}$-twisted cocycle, $b_{\vartheta^{-1}} (\int \smallsmile \partial_1
\smallsmile \partial_2
\smallsmile
\partial_3)=0$.
\end{lemma}
\begin{proof}
This is a straightforward computation:
\begin{align*}
&\quad \left(b_{{\vartheta^{-1}}}\int \smallsmile \partial_1
\smallsmile \partial_2
\smallsmile
\partial_3\right)(a_0,a_1,a_2,a_3,a_4)\\
&=\ \int \sigma_0(a_0a_1) \partial_1(a_2)
\partial_2(a_3) \partial_3(a_4)
-\int \sigma_0(a_0) \partial_1(a_1a_2)
\partial_2(a_3) \partial_3(a_4)\\
&\quad\ +\int \sigma_0(a_0) \partial_1(a_1)
\partial_2(a_2a_3)
\partial_3(a_4)
-\int \sigma_0(a_0) \partial_1(a_1)
\partial_2(a_2) \partial_3(a_3a_4)\\
&\quad\ +\int \sigma_0({{\vartheta^{-1}}}(a_4)a_0) \partial_1(a_1)
\partial_2(a_2) \partial_3(a_3)\\
&=\ -\int \sigma_0(a_0) \partial_1(a_1)
\partial_2(a_2) \partial_3(a_3)
\sigma_3(a_4)
+\int \sigma_0({{\vartheta^{-1}}}(a_4)) \sigma_0(a_0) \partial_1(a_1)
\partial_2(a_2) \partial_3(a_3)\\
&=\ 0.\qedhere
\end{align*}
\end{proof}
Less straightforward is that when
applying the above result with
$$
\sigma_0=\sigma_1=k^{-4}
\triangleright \cdot\,,\quad
\sigma_2=k^{-2}
\triangleright \cdot\,,\quad
\sigma_3=\mathrm{id},
$$
$$
\partial_1=(k^{-4}
\triangleright \cdot) \circ \partial_H
,\quad
\partial_2=(k^{-3} \triangleright
\cdot) \circ \partial_e,\quad
\partial_3=(k^{-1} \triangleright
\cdot) \circ \partial_f
$$
and a suitable twisted trace, one obtains a
cohomologically nontrivial
$\vartheta^{-1}$-twisted cocycle.
\begin{lemma}[{\cite[Corollary~3.8]{HK2}}]\label{haddock}
Define a linear functional
$\int_{[1]} : \A \rightarrow \mathbb{C}$ by
$$
\int_{[1]} a^{n} b^{m} c^{r} :=
\delta_{n,0} \delta_{m,0} \delta_{r,0},\quad
\int_{[1]} b^{m} c^{r} d^{s} :=
\delta_{m,0} \delta_{r,0} \delta_{s,0}.
$$
Then $\int_{[1]}$ is a $\sigma_{L}^{2} \circ \vartheta^{-1}$-twisted trace, and
the cochain $\varphi \in C^3$ given by
\enveqn{
\varphi(a_{0}, \ldots, a_{3}) =
\int_{[1]} \left(k^{-4} \triangleright (a_{0} \,\partial_H( a_{1})) \right) \left(k^{-3} \triangleright \ensuremath{\p_{e}}(a_{2}) \right) \left(k^{-1} \triangleright \ensuremath{\p_{f}}(a_{3}) \right)
}
is a cocycle,
$b_{\vartheta^{-1}} \varphi = 0$, whose
cohomology class is nontrivial,
$b_{\vartheta^{-1}} \psi \neq \varphi$
for all $\psi \in C^2$.
\end{lemma}
Later, we will also have to consider the
cocycles that are obtained by using
the (twisted) derivations
$\partial_H,\partial_e,\partial_f$ in a different
order. Explicitly,
this is handled by the following result.
\begin{lemma}\label{salmon}
In the situation of Lemma~\ref{tuna},
define
$$
\tilde \partial_3 =
\sigma_1 \circ \sigma_2^{-1}
\circ \partial_3,\quad
\tilde \partial_2:=
\partial_2 \circ \sigma_2^{-1}
\circ \sigma_3,\quad
\hat \partial_2:=\sigma_0 \circ
\sigma_1^{-1} \circ \partial_2,\quad
\hat \partial_1:=\partial_1
\circ \sigma_1^{-1} \circ \sigma_2.
$$
Then we have
\begin{align*}
\int \smallsmile \partial_1
\smallsmile \partial_2 \smallsmile
\partial_3 +
\int \smallsmile \partial_1
\smallsmile \tilde \partial_3 \smallsmile
\tilde \partial_2
&= b_{\vartheta^{-1}} \psi_{132},\\
\int \smallsmile \partial_1
\smallsmile \partial_2 \smallsmile
\partial_3 +
\int \smallsmile \hat\partial_2
\smallsmile \hat \partial_1 \smallsmile
\partial_3
&= b_{\vartheta^{-1}} \psi_{213},
\end{align*}
where
\begin{align*}
\psi_{132} (a_0,a_1,a_2)
&:=
\int \sigma_0(a_0)
\partial_1(a_1)
\partial_2
(\sigma_2^{-1}(\partial_3(a_2))),\\
\psi_{213} (a_0,a_1,a_2)
&:=
-\int \sigma_0(a_0)
\partial_1 (\sigma_1^{-1}(\partial_2(a_1)))\partial_3(a_2)
.
\end{align*}
\end{lemma}
\begin{proof}
Straightforward computation.
\end{proof}
Applying Lemma \ref{salmon} repeatedly to the cocycle
$\varphi$ from Lemma~\ref{haddock} gives cohomologous cocycles.
\begin{corl}\label{cor:perms}
The cocycle $\varphi$ from
Lemma~\ref{haddock} is cohomologous to each of
$$
\varphi_{132}
(a_0,a_1,a_2,a_3):=
-q^{-2}
\int_{[1]}
\left(k^{-4} \triangleright (a_{0} \,
\partial_H ( a_{1})) \right)
\left(k^{-3} \triangleright
\ensuremath{\p_{f}}(a_{2}) \right)
\left(k^{-1} \triangleright \ensuremath{\p_{e}}(a_{3}) \right),
$$
$$
\varphi_{213} (a_0,a_1,a_2,a_3):=
-\int_{[1]}
\left(k^{-4} \triangleright
a_{0}\right)
\left(k^{-3} \triangleright
\partial_e(a_1)\right)
\left(k^{-2} \triangleright \partial_H (a_2)\right)
\left(k^{-1} \triangleright
\partial_f(a_3)\right),
$$
$$
\varphi_{312}
(a_0,a_1,a_2,a_3):=
q^{-2}
\int_{[1]}
\left(k^{-4} \triangleright
a_{0}\right)
\left(k^{-3} \triangleright
\partial_f(a_1)\right)
\left(k^{-2} \triangleright
\partial_H ( a_2)\right)
\left(k^{-1} \triangleright
\partial_e(a_3)\right),
$$
$$
\varphi_{231} (a_0,a_1,a_2,a_3):=
\int_{[1]}
\left(k^{-4} \triangleright
a_{0}\right)
\left(k^{-3} \triangleright
\partial_e(a_1)\right)
\left(k^{-1} \triangleright
\partial_f(a_2)\right)
\left(\partial_H (a_3)\right)
$$
and
$$
\varphi_{321}
(a_0,a_1,a_2,a_3):=
-q^{-2}
\int_{[1]}
\left(k^{-4} \triangleright
a_{0}\right)
\left(k^{-3} \triangleright
\partial_f(a_1)\right)
\left(k^{-1} \triangleright
\partial_e(a_2)\right)\left(
\partial_H ( a_3)\right).
$$
\end{corl}
\begin{proof}
To begin, one applies
Lemma~\ref{salmon} to $\varphi$ with
$$
\tilde \partial_3 =
(k^{-3}
\triangleright \cdot) \circ \partial_f,\quad
\tilde \partial_2=
(k^{-3} \triangleright
\cdot) \circ \partial_e \circ (k^{2}
\triangleright \cdot),\quad
\hat \partial_2=(k^{-3} \triangleright
\cdot) \circ \partial_e,\quad
\hat \partial_1:=(k^{-4} \triangleright
\cdot) \circ \partial_H
(\cdot)
\circ (k^{2} \triangleright
\cdot)\,.
$$
The formulae for these derivations can be simplified by commuting $\partial_e$ and
$k\,\triangleright$ to obtain
$$
\tilde \partial_3 =
(k^{-3}
\triangleright \cdot) \circ \partial_f,\quad
\tilde \partial_2=
q^{-2} (k^{-1} \triangleright
\cdot) \circ \partial_e,\quad
\hat \partial_2=(k^{-3} \triangleright
\cdot) \circ \partial_e,\quad
\hat \partial_1:=(k^{-2} \triangleright
\cdot) \circ \partial_H(
\cdot).
$$
This gives $\varphi_{132}$ and $\varphi_{213}$.
Then we can apply Lemma~\ref{salmon}
again to $\varphi_{213}$.
Going from $\varphi_{213}$ to $\varphi_{312}$ is easy, since it only involves
exchanging $e$ and $f$.
Next we obtain $\varphi_{231}$ from $\varphi_{213}$ by applying
Lemma~\ref{salmon}
with
$$
\sigma_0=k^{-4}
\triangleright \cdot,\quad\sigma_1=
\sigma_2=k^{-2}
\triangleright \cdot,\quad
\sigma_3=\mathrm{id},
$$
$$
\partial_1=(k^{-3} \triangleright
\cdot) \circ \partial_e,\quad
\partial_2=(k^{-2}
\triangleright \cdot) \circ \partial_H
,\quad
\partial_3=(k^{-1} \triangleright
\cdot) \circ \partial_f
$$
which gives
$$
\tilde \partial_3=\sigma_1 \circ
\sigma_2^{-1} \circ \partial_3=\partial_3=
(k^{-1} \triangleright
\cdot) \circ \partial_f,
$$
$$
\tilde \partial_2=\partial_2
\circ \sigma_2^{-1} \circ \sigma_3=
(k^{-2}
\triangleright \cdot) \circ \partial_H
\circ (k^{2}
\triangleright \cdot)=\partial_H\,.
$$
The last cocycle is obtained analogously from $\varphi_{312}$.
\end{proof}
A homologically nontrivial 3-cycle $dvol$ in the (pre)dual chain complex
$C_\bullet:=\A^{\otimes_\mathbb{C}\bullet+1}$ (with differential dual to $b_{\vartheta^{-1}}$)
has been computed in \cite{HK1,HK2}:
\enval{
dvol &:= d \otimes a \otimes b \otimes c - d \otimes a \otimes c \otimes b + q \, d \otimes c \otimes a \otimes b \nonumber \\
& \quad - q^{2} \, d \otimes c \otimes b \otimes a + q^{2} \, d \otimes b \otimes c \otimes a - q \, d \otimes b \otimes a \otimes c \nonumber \\
& \quad + c \otimes b \otimes a \otimes d - c \otimes b \otimes d \otimes a + q \, c \otimes d \otimes b \otimes a \nonumber \\
& \quad - c \otimes d \otimes a \otimes b + c \otimes a \otimes d \otimes b - q^{-1} \, c \otimes a \otimes b \otimes d \nonumber \\
& \quad + (q^{-1}-q) \, c \otimes b \otimes c \otimes b \label{eqn_dvol}
}
With this normalisation, we have
$\varphi(dvol)=1$.
\section{Some meromorphic functions}
\label{sec:mero}
In this section
we demonstrate that certain functions have meromorphic
continuations. These functions arise in the residue formula for the Hochschild cocycle in
the next two sections. We require the following notation.
For any $l \in \frac{1}{2} \mathbb{N}_0$
and $-(2l+1) \leq n \leq (2l+1)$ define
\enveq{
\lambda_{l,n} :=
\sqrt{\left(\frac{n}{2}\right)^{2} + q^{n}\left(\left[ l+\frac{1}{2} \right]_{q}^{2} - \left[\frac{n}{2} \right]_{q}^{2} \right)}\,.
}
We also define the finite sets
\enveqn{
\mathcal{J}_{l} := \begin{cases}
\{ 0, 2, \ldots, 2l-1 \} & \quad l \in (\ensuremath{\mathbb{N}}_{0} + \tfrac{1}{2}) \\
\{ 1, 3, \ldots, 2l-1 \} & \quad l \in \ensuremath{\mathbb{N}}
\end{cases}.
}
\begin{lemma}
\label{lemma_holomorphic}
The formulas
\enveqn{
z \mapsto f_1(z):=\sum_{2l = 1}^{\infty} \sum_{i = -l}^{l} \sum_{n \in \mathcal{J}_{l}}
\frac{q^{2l-2i}}{ (1 + \lambda_{l, n}^{2})^{z/2} }
}
\enveqn{
z \mapsto f_2(z):=\sum_{2l = 1}^{\infty} \sum_{i = -l}^{l} \sum_{n \in \mathcal{J}_{l}}
\frac{q^{2l-n}}{ (1 + \lambda_{l, n}^{2})^{z/2} }
}
define holomorphic functions on $\ensuremath{\mathrm{Dom}}_2$, where we abbreviate
$$
\ensuremath{\mathrm{Dom}}_t := \{z \in \mathbb{C} \mid
\mathrm{Re}(z) > t\},\quad t \in \mathbb{R}.
$$
\end{lemma}
\begin{proof}
We will show that the sums converge uniformly on compacta.
To begin with, we take
$z=t \in (2,\infty)$, and compute the summation over the $i$ parameter for $f_{1}$ and $f_{2}$ giving
\enval{
f_{1}(t)
&= \sum_{2l = 1}^{\infty} \sum_{n \in \mathcal{J}_{l}}
\frac{q^{2l} [2l+1]_{q} }{ (1 + \lambda_{l, n}^{2})^{t/2} }, &
f_{2}(t)
&= \sum_{2l = 1}^{\infty} \sum_{n \in \mathcal{J}_{l}} \frac{(2l+1) q^{2l-n}}{ (1 + \lambda_{l, n}^{2})^{t/2} }.
\label{eq:pacific-giant-octopus}
}
For $l \in \frac{1}{2}\mathbb{N}_0$ and $n\in \mathcal{J}_l$ we have the inequality
$$
\left[l+\frac{1}{2}
\right]^2_q-\left[\frac{n}{2}
\right]^2_q \ge
[2l]_q
$$
with equality attained for $n=2l-1$. This inequality implies
\begin{equation}
1 + \lambda_{l, n}^{2} \geq 1 + \left(\frac{n}{2}\right)^{2} + q^{n}[2l]_{q}
\geq 1 + \left(\frac{n}{2}\right)^{2} + q^{n-2l+1}.
\label{eq:mullet}
\end{equation}
Since the summands in Equation \eqref{eq:pacific-giant-octopus} are positive, we
may invoke Tonelli's theorem to rearrange the order of summation
$$
\sum_{2l = 1}^{\infty} \sum_{n \in \mathcal{J}_{l}}
\rightarrow \sum_{n = 0}^{\infty} \sum_{l = (n+1)/2}^{\infty}.
$$
Combining the elementary inequality $q^{2l} [2l+1]_{q} \leq q^{-1}Q$ with
Equation \eqref{eq:mullet} gives the inequalities
\envaln{
f_{1}(t) &\leq q^{-1} Q \sum_{n = 0}^{\infty}
\sum_{l = \frac{n+1}{2}}^{\infty} \frac{1 }{ (1 + \left(\frac{n}{2}\right)^{2} + q^{n-2l+1})^{t/2} }, &
f_{2}(t) &\leq \sum_{n = 0}^{\infty}
\sum_{l = \frac{n+1}{2}}^{\infty} \frac{(2l+1) q^{2l-n}}{ (1 + \left(\frac{n}{2}\right)^{2} + q^{n-2l+1})^{t/2} }.
}
We reparameterise the sums defining $f_1$ and $f_2$ using $y = 2l-1-n$ with
summation range $y=0$ to $y=\infty$. This yields
\enval{
f_{1}(t) &\leq q^{-1} Q \sum_{n = 0}^{\infty} \sum_{y = 0}^{\infty} \frac{1 }{ (1 + \left(\frac{n}{2}\right)^{2} + q^{-y})^{t/2} }, &
f_{2}(t) &\leq \sum_{n = 0}^{\infty} \sum_{y = 0}^{\infty} \frac{(y+n+2) q^{y+1}}{ (1 + \left(\frac{n}{2}\right)^{2} + q^{-y})^{t/2} }.
\label{eq:barramundi}
}
Next we employ the inequality $\alpha^{2} + \beta^{2} \geq \alpha \beta$, valid for
any positive real numbers $\alpha$ and $\beta$, to $f_{1}(t)$. This yields
\enveqn{
f_{1}(t) \leq q^{-1} Q \sum_{n =
0}^{\infty} \sum_{y = 0}^{\infty} \,q^{yt/4}\,
\left(1 + \left(\frac{n}{2}\right)^{2}\right)^{-t/4} < \infty \qquad \text{for all} \ t > 2.
}
For the function $f_{2}(t)$, we evaluate the sums over $y$ on the right hand side to obtain, for some
positive constants $C_{1}$ and $C_{2}$,
\enveqn{
f_{2}(t) \leq \sum_{n = 0}^{\infty}
\sum_{y = 0}^{\infty} \frac{(y+n+2)
q^{y+1}}
{ \left(1 + \left(\frac{n}{2}\right)^{2}\right)^{t/2} }
= \sum_{n = 0}^{\infty} \frac{C_{1} + C_{2}n }{ \left(1 + \left(\frac{n}{2}\right)^{2}\right)^{t/2} }.
}
This last sum is finite for all $t > 2$, and bounded uniformly for $t\geq 2+\epsilon$ for any $\epsilon>0$.
This establishes that
$f_1,f_2$ are finite for all $\mathrm{Re}(z) > 2$, and the sums defining
them converge uniformly on vertical strips, and so on compacta.
Finally, to show that $f_1,f_2$
are holomorphic in the half-plane ${\mathrm Re}(z)>2$, we invoke the
Weierstrass convergence theorem.
\end{proof}
\begin{lemma}
\label{lem:general-pole-3}
For any positive reals $x,\, y,\, r > 0$, $w\in \N$, and $z \in
\ensuremath{\mathrm{Dom}}_3$, define
\enveqn{
h(z) :=
\sum_{n=1}^{\infty}\sum^\infty_{m=w}
\frac{e^{rm} }{ (x^{2} n^{2} + y^{2} e^{rm})^{z/2} }
}
Then we have:
\begin{enumerate}
\item $h$ is a holomorphic function on $\ensuremath{\mathrm{Dom}}_3$;
\item $h$ has a meromorphic continuation to
$\ensuremath{\mathrm{Dom}}_2$ with a simple pole
at $z=3$;
\item This continuation can be written as
\enveqn{
h(z) = \frac{\sqrt{\pi}}{2xy^{z-1}} \frac{\Gamma(\frac{z-1}{2})}{\Gamma(\frac{z}{2})}
\frac{e^{-rw(z-3)/2}}{1 - e^{-r(z-3)/2}} - \frac{1}{2y^{z}} \frac{e^{-rw(z-2)/2}}{1 - e^{-r(z-2)/2}} + err(z)
}
where $err$ is a holomorphic function on $\ensuremath{\mathrm{Dom}}_2$
that satisfies
$$
|err(z)| \leq \frac{1}{2y^{\mathrm{Re}(z)}} \frac{e^{-rw(\mathrm{Re}(z) - 2)/2}}{1 - e^{-r(\mathrm{Re}(z) - 2)/2}}.
$$
\end{enumerate}
\end{lemma}
\begin{proof}
Until further notice, we take $z$ real and positive. Later we will extend our
results to complex $z$ as in Lemma \ref{lemma_holomorphic}.
Inserting the Mellin transform of $f(t)=e^{-(x^2n^2+y^2e^{rm})t}$ gives
\enveqn{
h(z) = \sum_{n=1}^\infty \sum_{m=w}^{\infty}
\frac{e^{rm} }{\Gamma(\frac{z}{2})} \int_{0}^{\infty} t^{\frac{z}{2} - 1}e^{-tx^{2} n^{2}} e^{-t y^{2} e^{rm}} dt.
}
For $z$ real, all terms above are positive. Therefore we can apply Tonelli's theorem to
exchange the order of integration with summation.
Having done this, we consider the sum $\sum_{n = 1}^{\infty} e^{-tx^{2} n^{2}}$. The Poisson summation
formula provides the identity
\enveqn{
\sum_{n = 1}^{\infty} e^{-tx^{2} n^{2}} = \frac{1}{2} \left( \sqrt{\frac{\pi}{tx^{2}}}
\left( 1 + 2 \sum_{n = 1}^{\infty} e^{-\frac{n^{2}\pi^{2}}{tx^{2}}} \right) - 1 \right).
}
Substituting this identity into the expression for $h(z)$ we find
\envaln{
h(z) &= \frac{1}{2} \sum_{m=w}^{\infty} \frac{e^{rm}}{(y^{2} e^{rm})^{\frac{z}{2}}} \left( \frac{\sqrt{\pi}}{x}
\frac{\Gamma(\frac{z-1}{2})}{\Gamma(\frac{z}{2})} (y^{2} e^{rm})^{\frac{1}{2}} - 1 \right) \\
& \quad + \frac{\sqrt{\pi}}{x} \sum_{n=1}^\infty \sum_{m=w}^{\infty} \frac{e^{rm} }{\Gamma(\frac{z}{2})}
\int_{0}^{\infty} t^{\frac{z-1}{2} - 1}e^{-\frac{n^{2}\pi^{2}}{tx^{2}}} e^{-t y^{2} e^{rm}} dt.
}
To explore the convergence of the double sum we denote
\enveqn{
g_{n}(s) := \int_{0}^{\infty} t^{\frac{z-1}{2} - 1}e^{-\frac{n^{2}\pi^{2}}{tx^{2}}} e^{-t s} dt.
}
Later we will set $s = y^{2} e^{rm} > 0$, so we consider only positive,
real $s$, making $g_{n}(s)$ a positive real function.
Using
\cite[Section 26:14]{OS} to evaluate this Laplace transform gives
\enveqn{
g_{n}(s) =
2 \left( \frac{n \pi}{x \sqrt{s}} \right)^{\frac{z-1}{2}} K_{\frac{z-1}{2}} \left( \frac{2 n \pi \sqrt{s}}{x} \right)
}
where $u\mapsto K_{\nu}(u)$ is the modified Bessel function of the second kind. For $u > 0$ and real $\nu > 1/2$,
$u^{\nu}K_{\nu}(u)$ is positive, as both $u^{\nu}$ and $K_{\nu}(u)$ are positive. Also,
the derivative (referring again to \cite{OS}) is given by
\enveqn{
\frac{\partial}{\partial u}\left( u^{\nu}K_{\nu}(u) \right) = -u^{\nu} K_{\nu - 1}(u) \leq 0 \qquad \text{for all} \ u \geq 0.
}
Thus the function $u\mapsto u^{\nu}K_{\nu}(u)$ is positive and monotonically decreasing
for all $u > 0$. Hence for all $\epsilon > 0$ we have the bound
\enveq{
\epsilon \sum_{n = 1}^{\infty} (\epsilon n)^{\nu} K_{\nu}(\epsilon n) \leq \int_{0}^{\infty} u^{\nu} K_{\nu}(u) du.
\label{eq:bound}
}
Evaluating the integral (using \cite[Chapter 51]{OS}) yields
\enveqn{
\sum_{n = 1}^{\infty} (\epsilon n)^{\nu} K_{\nu}(\epsilon n) \leq \frac{1}{\epsilon} 2^{\nu - 1}
\Gamma(\tfrac{1}{2}) \Gamma(\nu + \tfrac{1}{2}).
}
If we now set $s = y^{2} e^{rm}$, we obtain the bound
\enveqn{
\sum_{n=1}^\infty \sum_{m=w}^{\infty}\frac{e^{rm} }{\Gamma(\frac{z}{2})}
\int_{0}^{\infty} t^{\frac{z-1}{2} - 1}e^{-\frac{n^{2}\pi^{2}}{tx^{2}}} e^{-t y^{2} e^{rm}} dt
\leq 2\sum_{n=1}^\infty \sum_{m=w}^{\infty} \frac{e^{rm} }{\Gamma(\frac{z}{2})}
\left( \frac{n \pi}{x ye^{rm/2}} \right)^{\frac{z-1}{2}} K_{\frac{z-1}{2}} \left( \frac{2 n \pi ye^{rm/2}}{x} \right).
}
Now estimating the sum over $n$ on the right using Equation \eqref{eq:bound} gives us
\envaln{
2\sum_{n=1}^{\infty} \left( \frac{n \pi}{x \sqrt{s}} \right)^{\frac{z-1}{2}} K_{\frac{z-1}{2}}
\left( \frac{2 n \pi \sqrt{s}}{x} \right) &= 2 \left( \frac{1}{2s} \right)^{\frac{z-1}{2}}
\sum_{n=1}^{\infty} \left( \frac{2n \pi \sqrt{s}}{x} \right)^{\frac{z-1}{2}} K_{\frac{z-1}{2}}
\left( \frac{2 n \pi \sqrt{s}}{x} \right) \\
& \leq 2 \left( \frac{1}{2s} \right)^{\frac{z-1}{2}} \frac{x}{2 \pi \sqrt{s}} 2^{\frac{z-1}{2} - 1}
\Gamma(\tfrac{z}{2}) \Gamma(\tfrac{1}{2}) \\
&= \frac{x \Gamma(\tfrac{1}{2})\Gamma(\tfrac{z}{2})}{2 \pi } \frac{1}{s^{z/2}}
=\frac{x \Gamma(\tfrac{1}{2})\Gamma(\tfrac{z}{2})}{2 \pi } \frac{1}{y^ze^{zrm/2}}.
}
Hence by summing the remaining geometric series in $m$ we obtain the bound
\envaln{
\sum_{n=1}^\infty \sum_{m=w}^{\infty} \frac{e^{rm} }{\Gamma(\frac{z}{2})}
\int_{0}^{\infty} t^{\frac{z-1}{2} - 1}e^{-\frac{n^{2}\pi^{2}}{tx^{2}}} e^{-t y^{2} e^{rm}} dt
&\leq \frac{\Gamma \left(\frac{z}{2} \right)}{\Gamma(\frac{z}{2})}
\frac{x \Gamma(\tfrac{1}{2})}{y^{z} 2 \pi } \sum_{m=w}^{\infty} \frac{e^{rm}}{e^{rmz/2}} \\
& \leq \frac{x \Gamma(\tfrac{1}{2})}{y^{z} 2 \pi }
\frac{e^{-rw(z - 2)/2}}{1 - e^{-r(z - 2)/2}}.
}
Evaluating the remaining geometric series in $h(z)$ as above, we arrive at
\enveq{
h(z) = \frac{\sqrt{\pi}}{2xy^{z-1}} \frac{\Gamma(\frac{z-1}{2})}{\Gamma(\frac{z}{2})}
\frac{e^{-rw(z-3)/2}}{1 - e^{-r(z-3)/2}} - \frac{1}{2y^{z}} \frac{e^{-rw(z-2)/2}}{1 - e^{-r(z-2)/2}} + err(z)
\label{eq:box-jelly}
}
where
\envgan{
err(z) := \frac{\sqrt{\pi}}{x} \sum_{n=1}^\infty \sum_{m=w}^{\infty} \frac{e^{rm} }{\Gamma(\frac{z}{2})}
\int_{0}^{\infty} t^{\frac{z-1}{2} - 1}e^{-\frac{n^{2}\pi^{2}}{tx^{2}}} e^{-t y^{2} e^{rm}} dt, \\
err(z) \leq \frac{1}{2y^{z}} \frac{e^{-rw(z - 2)/2}}{1 - e^{-r(z - 2)/2}}.
}
Thus the sum defining the function $err$ converges for all $z > 2$, and this
convergence is uniform on compact intervals. Now we observe that for $z\in\C$ we
have $|h(z)|\leq h(|z|)$ and
similarly $|err(z)|\leq err(|z|)$. Hence the sums defining $h$ converge uniformly on closed vertical
strips in the half-plane $\ensuremath{\mathrm{Dom}}_3$, and so on compacta. Similarly the sums and integral defining
$err$ converge uniformly on compact subsets of the half-plane $\ensuremath{\mathrm{Dom}}_2$.
Hence the Weierstrass convergence theorem implies
that $err$ is holomorphic on the half-plane $\ensuremath{\mathrm{Dom}}_2$ and that $h$ is holomorphic
on $\ensuremath{\mathrm{Dom}}_3$.
Moreover the formula for $h$, Equation \eqref{eq:box-jelly}, provides a meromorphic continuation
of $h$ to the half-plane $\ensuremath{\mathrm{Dom}}_2$.
\end{proof}
\begin{lemma}
\label{lemma_tau_residue}
The formula
\enveqn{
f(z):= \sum_{n=0}^{\infty} \sum_{l = \frac{n+1}{2}}^{\infty} \frac{q^{n-2l} }{ (1 + \lambda_{l, n}^{2})^{z/2} }
}
defines a holomorphic function on $\ensuremath{\mathrm{Dom}}_3$. Moreover $f$ has a meromorphic continuation
to $\ensuremath{\mathrm{Dom}}_2$, a simple pole at $z = 3$ with
residue $4qQ^{-2}/\ln(q^{-1})$.
\end{lemma}
\begin{proof}
First we write
\enveqn{
1 + \lambda_{l, n}^{2} = 1 +\tfrac{n^{2}}{4} + q^{n} \left( \left[ l + \tfrac{1}{2} \right]^{2} - \left[ \tfrac{n}{2} \right]^{2} \right) = \tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{n-2l} + C_{n, l}
}
where $C_{n,l}$ is uniformly bounded in $n,l$, and is given by
\enveqn{
C_{n, l} = 1 + Q^{2} q^{n} (q^{2l+1} - 2) - q^{n} \left[ \tfrac{n}{2} \right]^{2},\qquad |C_{n,l}|\leq 1+3Q^2.
}
Now we reparametrise the summation by letting $m = 2l-n$, yielding
\enveqn{
f(z) =
\sum_{n=0}^{\infty} \sum_{m = 1}^{\infty}
\frac{q^{-m} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m} + C_{n, m})^{z/2} }
}
where we understand $C_{n, m} = C_{n, l=(n+m)/2}$. The function
\enveqn{
z \mapsto \sum_{m = 1}^{\infty} \frac{q^{-m} }{ (Q^{2} q^{-1} q^{-m} + C_{0, m})^{z/2} } = \sum_{m = 1}^{\infty} \frac{q^{m(\frac{z}{2} - 1)} }{ (Q^{2} q^{-1} + q^{m} C_{0, m})^{z/2} }
}
has summands with absolute value bounded by $M q^{m(\frac{z}{2} - 1)}$, $M>0$ constant, and so
by the Weierstrass convergence theorem is holomorphic for $\mathrm{Re}(z) > 2$. Hence for
some holomorphic function $holo$ on $\ensuremath{\mathrm{Dom}}_2$ we have
\enval{
f(z) &= \sum_{n, m=1}^{\infty} \frac{q^{-m} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m} + C_{n, m})^{z/2} }
+ holo(z) \nonumber \\
&= \sum_{n, m=1}^{\infty} \frac{q^{-m} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m})^{z/2}} \left( 1 + \frac{C_{n, m}}{\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m}} \right)^{-z/2} + holo(z). \label{eqn_fz_expand}
}
The strategy now is to perform a binomial expansion on
$$
\left( 1 + \frac{C_{n, m}}{\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m}} \right)^{-z/2}
$$
ending up with a new sum of functions $\sum_{n,m,k}D_{n,m,k} \,h(z+2k)$ where $h$ is
as in Lemma \ref{lem:general-pole-3}.
The binomial expansion requires the inequality
\enveqn{
\frac{C_{n, m}}{\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m}} < 1
}
which holds for sufficiently large $m$.
Recall that $|C_{n,m}|\leq 1+3Q^2=:C$ uniformly in
$n,\,m$, and so we may choose $p \in \ensuremath{\mathbb{N}}$ such that
\enveqn{
q^{-p} > q Q^{-2}C \qquad \implies \qquad \frac{|C_{n, m}|}{\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m}} < 1
\quad \forall n \geq 1,\ m \geq p.
}
Now, for any fixed $p$, sums of the form
\enveqn{
\sum_{n=1}^{\infty} \sum_{m=1}^{p-1}
\frac{q^{-m} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m} + C_{n, m})^{z/2}}
}
can immediately be seen to be holomorphic for $\mathrm{Re}(z) > 2$
as the sum can be bounded by a constant multiple of the Riemann zeta function.
Hence for such a choice of $p\in \N$ and for some holomorphic function
$holo$ on $\ensuremath{\mathrm{Dom}}_2$ we have
\envaln{
f(z) &= \sum_{n=1}^{\infty} \sum_{m=p}^{\infty} \frac{q^{-m} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m})^{z/2}} \left( 1 + \frac{C_{n, m}}{\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m}} \right)^{-z/2} + holo(z).
}
Now we perform the binomial expansion, separating the
resulting infinite sum $\sum_{k=0}^\infty$ into the
$k=0$ term and $\sum_{k=1}^\infty$. This gives
\envaln{
f(z)&= \sum_{n=1}^{\infty} \sum_{m=p}^{\infty} \frac{q^{-m} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m})^{z/2}} + \sum_{k = 1}^{\infty} \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right) \sum_{n=1}^{\infty} \sum_{m=p}^{\infty} \frac{q^{-m} (C_{n, m})^{k} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m})^{\frac{z+2k}{2}}} + holo(z)\\
&= h(z) + \sum_{k = 1}^{\infty} \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right) \sum_{n=1}^{\infty}
\sum_{m=p}^{\infty} \frac{q^{-m} (C_{n, m})^{k} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m})^{\frac{z+2k}{2}}}
+ holo(z),
}
where $h$ is as in Lemma \ref{lem:general-pole-3}, with $x=1/2$, $y=q^{-1/2}Q$,
$r=\ln(q^{-1})$ and $w=p$. Our aim now is to show that $f-h$ is a
holomorphic function on $\ensuremath{\mathrm{Dom}}_2$. We need to show that the remaining summation converges
to such a function. This remaining sum is bounded by
\begin{align*}
&\left|\sum_{k = 1}^{\infty} \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right) \sum_{n=1}^{\infty}
\sum_{m=p}^{\infty} \frac{q^{-m} (C_{n, m})^{k} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m})^{\frac{z+2k}{2}}}\right|\\
&\qquad\leq \sum_{k = 1}^{\infty} \left| \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right)\right| \,C^k
\sum_{n=1}^{\infty}
\sum_{m=p}^{\infty} \frac{q^{-m}}{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m})^{\frac{{\mathrm Re}(z)+2k}{2}}}\\
&\qquad\qquad= \sum_{k = 1}^{\infty} \left| \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right)\right| \,C^k h({\mathrm Re}(z)+2k).
\end{align*}
To estimate this sum of functions, we infer from Lemma \ref{lem:general-pole-3} that there exists
a positive function $M$ which is defined for $\mathrm{Re}(z) > 3$ and such that
\enveqn{
|h(z)| \leq M(z) \frac{e^{-\mathrm{Re}(z)rp/2}}{y^{\mathrm{Re}(z)}}
= M(z) (q^{\frac{1}{2}(p+1)}Q^{-1})^{\mathrm{Re}(z)}.
}
Hence
\enveqn{
\left| \sum_{k = 1}^{\infty} \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right) \sum_{n=1}^{\infty}
\sum_{m=p}^{\infty} \frac{q^{-m} (C_{n, m})^{k} }{ (\tfrac{1}{4} n^{2} + Q^{2} q^{-1} q^{-m})^{\frac{z+2k}{2}}} \right| \leq \sum_{k = 1}^{\infty} \left| \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right) \right|
C^{k} M(z+2k) (q^{\frac{1}{2}(p+1)}Q^{-1})^{\mathrm{Re}(z)+2k}.
}
Recall that $p$ was chosen such that $q^{-p} > q Q^{-2}C$. Also
the function $z\mapsto M(z)$ is uniformly bounded for ${\mathrm Re}(z)\geq 4$.
Hence, for all $z$ with ${\mathrm Re}(z)\geq 2$, the function $k\mapsto M(z+2k)$ is uniformly
bounded in $k$, by ${\bf M}$ say.
It thus follows that the sum
\begin{align*}
\sum_{k = 1}^{\infty} \left| \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right) \right| C^k
M(z+2k)(q^{\frac{1}{2}(p+1)}Q^{-1})^{\mathrm{Re}(z)+2k}
&\leq {\bf M} \sum_{k = 1}^{\infty} \left| \left(\begin{array}{c} -\frac{z}{2} \\ k \end{array}\right) \right|(q^{p+1}Q^{-2}C)^{k}
\end{align*}
converges for $\mathrm{Re}(z) > 2$, by comparing with the binomial expansion on the right hand side.
The convergence is again uniform on compacta, so invoking
Weierstrass' convergence theorem
we conclude that $f(z) - h(z)$ is holomorphic for $\mathrm{Re}(z) > 2$. Hence there exists a function
$holo$ which is defined and holomorphic for ${\mathrm Re}(z)>2$ such that
\enveqn{
f(z) = \frac{\sqrt{\pi}}{(q^{-\frac{1}{2}}Q)^{z-1}}
\frac{\Gamma(\frac{z-1}{2})}{\Gamma(\frac{z}{2})} \frac{q^{p(z-3)/2}}{1 - q^{(z-3)/2}} + holo(z)
}
So we see $f(z)$ is holomorphic for $\mathrm{Re}(z) > 3$, meromorphic for $\mathrm{Re}(z)>2$ and has a a simple pole
at $z = 3$ with residue $4qQ^{-2}/\ln(q^{-1})$.
\end{proof}
\section{An analogue of a spectral triple}
\label{sec:spec}
We now introduce an analogue of a
spectral triple over $\A$.
Let $\ensuremath{\mathcal{H}} := \ensuremath{\mathcal{H}}_{h} \oplus
\ensuremath{\mathcal{H}}_{h}$ be the Hilbert space given by two copies of the
GNS space $\H_h=L^{2}(A, h)$. We define
a grading on $\H$ by
$\Gamma = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1
\end{array}\right)$. For any operator $\omega$ on $\H$
we abbreviate
\begin{equation}
\omega^+:=\frac{1+\Gamma}{2}
\omega \frac{1+\Gamma}{2},\quad
\omega^-:=\frac{1-\Gamma}{2}
\omega \frac{1-\Gamma}{2}.
\label{eq:giant-squid}
\end{equation}
The algebra $\A$ is represented on $\ensuremath{\mathcal{H}}$ by
\enveqn{
\alpha \mapsto \left(\begin{array}{cc} \pi_{h}(\alpha) & 0 \\ 0 & \pi_{h}(\alpha) \end{array}\right)
}
for $\alpha \in \A$. Here $\pi_{h}$ denotes the GNS representation by left
multiplication on each copy of the
space. In the sequel we will omit
the symbol $\pi_h$.
We now introduce some unbounded operators and
projections
\envaln{
\hat \Delta_{R} &= \left(\begin{array}{cc} \Delta_{R} & 0 \\ 0 &
\Delta_{R} \end{array}\right) & \hat\Delta_{L} &= \left(\begin{array}{cc}
q^{-1} \Delta_{L} & 0 \\ 0 & q
\Delta_{L} \end{array}\right) &
\Psi_{n} = \left(\begin{array}{cc} \Phi_{n+1} & 0 \\ 0 & \Phi_{n-1} \end{array}\right)
}
on $\A \oplus
\A \subset \ensuremath{\mathcal{H}}$ and use
them to define (on the same domain)
\enveqn{
\ensuremath{\mathcal{D}} = \frac{1}{2} \sum_{n = -\infty}^{\infty} \Psi_{n} \left(\begin{array}{cc} n & 0 \\ 0 & -n \end{array}\right) + \hat\Delta_{L}^{\frac{1}{2}} \left(\begin{array}{cc} 0 & \ensuremath{\p_{e}} \\ \ensuremath{\p_{f}} & 0 \end{array}\right).
}
We will see in the following lemma that
the commutators $[\ensuremath{\mathcal{D}}, \alpha]$ of $\ensuremath{\mathcal{D}}$ with algebra elements
are not necessarily bounded, yet
unbounded in a very controlled manner.
Even though $(\A,\H,\D)$ thus fails to
be a spectral triple, we will still be
able to construct
an analytic expression for a residue Hochschild cocycle from the commutators.
\begin{lemma}
\label{lemma_triple_properties}
The triple $(\A, \ensuremath{\mathcal{H}}, \ensuremath{\mathcal{D}})$ has the following properties:
\begin{enumerate}
\item The unbounded operator $\ensuremath{\mathcal{D}}$ is
essentially self-adjoint.
\item The commutator $[\ensuremath{\mathcal{D}}, \alpha]$ is
given by $\tilde{S}(\alpha) + \tilde{T}(\alpha) \hat\Delta_{L}$,
where the linear maps $\tilde{S}, \tilde{T} \colon
\A \rightarrow B(\ensuremath{\mathcal{H}})$
are given by
\envaln{
\tilde{S}(\alpha) &= \partial_{H}(\alpha) \Gamma &
\tilde{T}(\alpha) &= \left(\begin{array}{cc} 0 & q^{-\frac{1}{2}}
\ensuremath{\p_{e}}(\sigma_{L}^{-\frac{1}{2}}(\alpha))
\\ q^{\frac{1}{2}}
\ensuremath{\p_{f}}(\sigma_{L}^{-\frac{1}{2}}(\alpha)) &
0 \end{array}\right).
}
\end{enumerate}
\end{lemma}
\begin{proof}
First we recall from Section \ref{sec:mero} the numbers
\enveq{
\lambda_{l,n} :=
\sqrt{\left(\frac{n}{2}\right)^{2} + q^{n}\left(\left[ l+\frac{1}{2} \right]_{q}^{2} - \left[\frac{n}{2} \right]_{q}^{2} \right)},
}
where $l \in \frac{1}{2} \mathbb{N}_0$
and $-(2l+1) \leq n \leq (2l+1)$. Also recall $ I_{l} := \{ -l, -l+1, \ldots, l-1,l \}$.
Then the set
\enveqn{
\left\{ \begin{pmatrix} 0 \\ \re{l}{i}{l} \end{pmatrix} ,
\begin{pmatrix} \re{l}{i}{-l} \\ 0 \end{pmatrix}, \begin{pmatrix} \re{l}{i}{j} \\ \stackrel{}{C^{l}_{j, \pm}} \re{l}{i}{j-1}
\end{pmatrix} \colon l \in \frac{1}{2} \ensuremath{\mathbb{N}}_{0}, \ i \in I_{l}, \ j \in I_{l} \backslash \{-l\} \right\},
}
\enveqn{
where \ \ C^{l}_{j, \pm} = \frac{\pm \lambda_{l, 2j-1} - (j-\tfrac{1}{2}) }{ q^{j-\frac{1}{2}} \sqrt{\left[ l+\tfrac{1}{2} \right]_{q}^{2} - \left[j - \tfrac{1}{2} \right]_{q}^{2}} }
}
is an orthogonal basis for $\ensuremath{\mathcal{H}}$ comprised of eigenvectors of $\ensuremath{\mathcal{D}}$.
The corresponding eigenvalues are $-(l+\tfrac{1}{2}), -(l+\tfrac{1}{2})$ and $\pm \lambda_{l, 2j-1}$ respectively.
This spectral representation establishes that $\ensuremath{\mathcal{D}}$ is essentially self-adjoint.
Next, the commutator of $\ensuremath{\mathcal{D}}$ with a homogeneous algebra element $\alpha = \Phi_{p} (\alpha)$,
for some $p \in \ensuremath{\mathbb{Z}}$, is computed directly. It is sufficient to consider just this case, because $\A$ consists of
finite linear combinations of homogeneous elements (the generators are homogeneous). For such an element $\alpha$ we have
\envaln{
[\ensuremath{\mathcal{D}}, \alpha] &= \frac{1}{2} \sum_{n =
-\infty}^{\infty} \Psi_{n} \alpha \left(\begin{array}{cc}
n & 0 \\ 0 & -n \end{array}\right) +
\hat\Delta_{L}^{\frac{1}{2}} \left(\begin{array}{cc} 0 & \ensuremath{\p_{e}} \\
\ensuremath{\p_{f}} & 0 \end{array}\right) \alpha \\
& \quad - \frac{1}{2} \sum_{n = -\infty}^{\infty} \alpha \Psi_{n} \left(\begin{array}{cc} n & 0 \\ 0 & -n \end{array}\right) - \alpha \hat\Delta_{L}^{\frac{1}{2}} \left(\begin{array}{cc} 0 & \ensuremath{\p_{e}} \\ \ensuremath{\p_{f}} & 0 \end{array}\right).
}
It follows from the definition of the projections $\Phi_{n}$, now regarded as a linear operator on $\H_h$, that
$\alpha \Phi_{n} = \Phi_{n+p} \alpha$ for any $n \in \ensuremath{\mathbb{Z}}$. Using this, together
with the definition of the derivations $\partial_e$ and $\partial_f$ in Equation \ref{eqn_twisted_ef}, the commutator simplifies to
\envaln{
[\ensuremath{\mathcal{D}}, \alpha] &= \frac{1}{2} \alpha \sum_{n =
-\infty}^{\infty} \Psi_{n}
\left(\left(\begin{array}{cc} n+p & 0 \\ 0 & -n-p \end{array}\right) - \left(\begin{array}{cc} n & 0 \\ 0 & -n \end{array}\right) \right) \\
& \quad + \hat\Delta_{L}^{\frac{1}{2}} \left(\begin{array}{cc} 0 & (\ensuremath{\p_{e}}(\alpha) \Delta_{L}^{\frac{1}{2}} + \sigma_{L}^{\frac{1}{2}}(\alpha) \ensuremath{\p_{e}}) \\ (\ensuremath{\p_{f}}(\alpha) \Delta_{L}^{\frac{1}{2}} + \sigma_{L}^{\frac{1}{2}}(\alpha) \ensuremath{\p_{f}}) & 0 \end{array}\right) - \alpha \hat\Delta_{L}^{\frac{1}{2}} \left(\begin{array}{cc} 0 & \ensuremath{\p_{e}} \\ \ensuremath{\p_{f}} & 0 \end{array}\right).
}
Since $\sigma_{L}^{\frac{1}{2}}(\alpha) = \hat\Delta_{L}^{-\frac{1}{2}} \alpha \hat\Delta_{L}^{\frac{1}{2}}$ as operators on $\A\oplus\A\subset\H$, the last expression for the commutator simplifies to
\enveqn{
\hat\Delta_{L}^{\frac{1}{2}} \left(\begin{array}{cc} 0 &
\sigma_{L}^{\frac{1}{2}}(\alpha) \ensuremath{\p_{e}} \\
\sigma_{L}^{\frac{1}{2}}(\alpha) \ensuremath{\p_{f}} & 0 \end{array}\right) = \alpha \hat\Delta_{L}^{\frac{1}{2}} \left(\begin{array}{cc} 0 & \ensuremath{\p_{e}} \\ \ensuremath{\p_{f}} & 0 \end{array}\right),
}
and hence
\envaln{
[\ensuremath{\mathcal{D}}, \alpha] &= \frac{p}{2} \alpha \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) + \hat\Delta_{L}^{\frac{1}{2}} \left(\begin{array}{cc} 0 & \ensuremath{\p_{e}}(\alpha) \Delta_{L}^{\frac{1}{2}} \\ \ensuremath{\p_{f}}(\alpha) \Delta_{L}^{\frac{1}{2}} & 0 \end{array}\right) \\
&= \partial_{H}(\alpha) \Gamma + \left(\begin{array}{cc} 0 & q^{-\frac{1}{2}} \ensuremath{\p_{e}}(\sigma_{L}^{-\frac{1}{2}}(\alpha)) \\ q^{\frac{1}{2}} \ensuremath{\p_{f}}(\sigma_{L}^{-\frac{1}{2}}(\alpha)) & 0 \end{array}\right) \hat\Delta_{L}. \qedhere
}
\end{proof}
\section{The residue Hochschild cocycle}
\label{sec:res-hochs}
The main step in
the definition of the residue Hochschild cocycle
is the construction of a functional that plays the role of an integral. In
the situations considered in the literature thus far, \cite{C, BeF, GVF, CNNR, CPRS1, KW},
functionals of the form
$$
T\mapsto \tau(T(1+\D^2)^{-z/2})
$$
were used, where $z\in \C$ and $\tau$ is
a faithful normal semifinite trace, or at worst a weight, on a von Neumann algebra containing the
algebra of interest. Often, the von Neumann algebra is just $\B(\H)$, and
the functional $\tau$ is the operator trace.
In this example, we need to apply our functional to
products of commutators $[\D,\alpha]\sim \hat{\Delta}_L$ with
$\alpha\in\A$, so it has to be defined on an algebra of unbounded
operators. We will deal with this using a cutoff that is defined
by the projections
\enveqn{
L_k:=\tilde{L}_k\oplus\tilde{L}_k,\quad \tilde{L}_{k}(\re{l}{i}{j}) := \begin{cases}
\re{l}{i}{j} & \quad l \leq k \\
0 & \quad \mbox{otherwise}
\end{cases}
}
and
\envaln{
P_{1} &= \sum_{n = 0}^{\infty} \Psi_{n} &
P_{2} \left(\begin{array}{c} \re{l}{i}{j} \\ 0 \end{array}\right) = (1-\delta_{j,-l}) \left(\begin{array}{c} \re{l}{i}{j} \\ 0 \end{array}\right) \quad & \quad
P_{2} \left(\begin{array}{c} 0 \\ \re{l}{i}{j} \end{array}\right) = (1-\delta_{j,l}) \left(\begin{array}{c} 0 \\ \re{l}{i}{j} \end{array}\right).
}
Observe $P_{2}$ is the
projection onto
$\left(\mathrm{ker}\left(\begin{array}{cc} 0 & \ensuremath{\p_{e}} \\ \ensuremath{\p_{f}} & 0
\end{array}\right) \right)^\perp$, and that
the projections $L_{k}$ converge strongly to the identity in $\B(\ensuremath{\mathcal{H}})$.
For $s\in \R^+$ we now define a functional $\Upsilon_{s}$
on positive operators $\omega \in\B(\ensuremath{\mathcal{H}}) $
in the following way:
\enveqn{
\Upsilon_{s} (\omega) :=
\sup_{k \in \N} \mathrm{Tr}
\left( P_{1} P_{2} L_{k} (1 +
\ensuremath{\mathcal{D}}^{2})^{-s/4} \hat\Delta_{F}^{-\frac{1}{2}}
\omega
\hat\Delta_{F}^{-\frac{1}{2}} (1 +
\ensuremath{\mathcal{D}}^{2})^{-s/4} P_{1} P_{2}
L_{k}\right),\quad
\hat \Delta _F=\hat \Delta_R \hat
\Delta _L
}
where $\mathrm{Tr}$ is the operator
trace on $\B(\ensuremath{\mathcal{H}})$.
This expression
continues to make sense for possibly
unbounded positive operators defined on
and preserving the subspace
$\A \oplus \A \subset \H$.
\begin{lemma}
\label{lemma_operator_trace}
For each $s\in \R_+$ the functional
$\Upsilon_s$ is positive and normal on
$\B(\H)_+$.
It is faithful and semifinite on $P_1P_2\B(\ensuremath{\mathcal{H}})_+P_1P_2$.
\end{lemma}
\begin{proof}
We will compute the operator trace using the Peter-Weyl basis
$\left\{ \left(\begin{array}{c} \re{l}{i}{j} \\ 0 \end{array}\right), \left(\begin{array}{c} 0 \\ \re{l}{i}{j} \end{array}\right) \right\}$ for $\ensuremath{\mathcal{H}}$.
The operators $(1 + \ensuremath{\mathcal{D}}^{2})$, $\hat\Delta_{F}$, $P_{1}$, $P_{2}$ and $L_{k}$ are
all positive and diagonal in this basis. By using the definition of the operator trace,
the value of the operators $\hat\Delta_{F}^{-1}$ and $(1 + \ensuremath{\mathcal{D}}^{2})^{-s/4}$ on this basis,
and the symmetry property for self-adjoint operators, we compute $\Upsilon_s(\omega)$ for
$\omega\in \B(\H)_+$ (or even $\omega\geq 0$ and affiliated to $\B(\H)$) by
\envaln{
& \mathrm{Tr} \left( P_{1} P_{2} L_{k}(1 + \ensuremath{\mathcal{D}}^{2})^{-s/4} \hat\Delta_{F}^{-\frac{1}{2}} \omega \hat\Delta_{F}^{-\frac{1}{2}} (1 + \ensuremath{\mathcal{D}}^{2})^{-s/4} P_{1} P_{2} L_{k}\right) = \\
&= \sum_{2l = 0}^{\infty}
\sum_{i=-l}^{l} \sum_{j = -l}^{l}
\frac{q^{-2i-(2j-1)}}{(1 + \lambda_{l,
2j-1}^{2})^{s/2}} \frac{\left\langle P_{1}^{+} P_{2}^{+} L_{k} \re{l}{i}{j}, \omega^+P_{1}^{+} P_{2}^{+} L_{k} \re{l}{i}{j} \right\rangle}{\left\langle \re{l}{i}{j}, \re{l}{i}{j} \right\rangle} \\
& \quad + \sum_{2l = 0}^{\infty}
\sum_{i=-l}^{l} \sum_{j = -l}^{l}
\frac{q^{-2i-(2j+1)}}{(1 + \lambda_{l,
2j+1}^{2})^{s/2}} \frac{\left\langle
P_{1}^{-} P_{2}^{-} L_{k}\re{l}{i}{j}, \omega^- P_{1}^{-} P_{2}^{-} L_{k} \re{l}{i}{j} \right\rangle}{\left\langle \re{l}{i}{j}, \re{l}{i}{j} \right\rangle},
}
where $\omega^+$ and $\omega^-$ are as in Equation \eqref{eq:giant-squid}.
Now,
\envaln{
P_{1}^{+} P_{2}^{+} L_{k} \re{l}{i}{j} &= \begin{cases}
\re{l}{i}{j} & \quad \frac{1}{2} \leq j \leq l, \ \frac{1}{2} \leq l \leq k \\
0 & \quad \mbox{otherwise}
\end{cases} \\
P_{1}^{-} P_{2}^{-} L_{k} \re{l}{i}{j} &= \begin{cases}
\re{l}{i}{j} & \quad -\frac{1}{2} \leq j \leq l-1, \ \frac{1}{2} \leq l \leq k \\
0 & \quad \mbox{otherwise}.
\end{cases}
}
So if we set $n = 2j \pm 1$ and recall the sets
\enveqn{
\mathcal{J}_{l} := \begin{cases}
\{ 0, 2, \ldots, 2l-1 \} & \quad l \in (\ensuremath{\mathbb{N}}_{0} + \tfrac{1}{2}) \\
\{ 1, 3, \ldots, 2l-1 \} & \quad l \in \ensuremath{\mathbb{N}}
\end{cases}
}
we may express the trace as
\enval{
& \mathrm{Tr} \left( P_{1} P_{2} L_{k} (1 + \ensuremath{\mathcal{D}}^{2})^{-s/4} \hat\Delta_{F}^{-\frac{1}{2}} \omega \hat\Delta_{F}^{-\frac{1}{2}} (1 + \ensuremath{\mathcal{D}}^{2})^{-s/4} P_{1} P_{2} L_{k}\right) = \nonumber \\
&= \sum_{2l = 1}^{2k} \sum_{i=-l}^{l} \sum_{n \in \mathcal{J}_{l}} \frac{q^{-2i-n}}{(1 + \lambda_{l, n}^{2})^{s/2}} \left( \frac{\left\langle \re{l}{i}{\frac{n+1}{2}}, \omega^+\re{l}{i}{\frac{n+1}{2}} \right\rangle}{\left\langle \re{l}{i}{\frac{n+1}{2}}, \re{l}{i}{\frac{n+1}{2}} \right\rangle} + \frac{\left\langle \re{l}{i}{\frac{n-1}{2}}, \omega^- \re{l}{i}{\frac{n-1}{2}} \right\rangle}{\left\langle \re{l}{i}{\frac{n-1}{2}}, \re{l}{i}{\frac{n-1}{2}} \right\rangle} \right). \label{eqn_trace_formula}
}
This shows that $\Upsilon_s$ is a supremum of a sum of positive vector states and so
automatically positive and normal. To see that it is faithful on $P_1P_2\B(\ensuremath{\mathcal{H}})_+P_1P_2$
we observe that the operator trace is faithful and that $P_1P_2\hat\Delta^{-1/2}_F(1+\D^2)^{-s/4}$ is injective
on $P_1P_2\ensuremath{\mathcal{H}}$.
The semifiniteness comes from the fact that finite rank operators
are in the domain of $\Upsilon_s$.
\end{proof}
We extend $\Upsilon_s$ to an unbounded positive
normal linear functional on $\B(\ensuremath{\mathcal{H}})$ as
usual. In fact, we extend it also to
unbounded operators $\omega$ defined on and
preserving $\A \oplus \A$ by decomposing
$L_k \omega L_k$ for each $k$ into a linear combination of
positive bounded operators.
If for
an operator $\omega$ (not necessarily bounded) the function $s\mapsto \Upsilon_s(\omega)$
has a meromorphic continuation to $\ensuremath{\mathrm{Dom}}_{3-\delta}$ for some $\delta>0$, then we define
\enveqn{
\tau(\omega) :=
\mathrm{Res}_{z=3} \Upsilon_{z} (\omega). }
\begin{lemma}
\label{lemma_tau_bc_zero}
The functional $\tau$ is defined on the positive operator $c^{\ast}c$, and $\tau(c^*c)=0$.
Indeed, for all $m \geq 1$,
\enveqn{
\tau \left( \left(\begin{array}{cc} (c^{\ast}c)^{m} & 0 \\ 0 & 0 \end{array}\right) \right) = \tau \left( \left(\begin{array}{cc} 0 & 0 \\ 0 & (c^{\ast}c)^{m} \end{array}\right) \right) = 0.
}
\end{lemma}
\begin{proof}
The action of the operator $c = c^{+} + c^{-}$ may be described using
the Clebsch-Gordan coefficients (see for example \cite{DLSSV},
\cite{KS}): we have
\envaln{
c^{+} \re{l}{i}{j} &= c^{l+}_{ij} \re{l+\frac{1}{2}}{i+\frac{1}{2}}{j-\frac{1}{2}} &
c^{-} \re{l}{i}{j} &= c^{l-}_{ij} \re{l-\frac{1}{2}}{i+\frac{1}{2}}{j-\frac{1}{2}},
}
where
\envaln{
c^{l+}_{ij} &= q^{(i+j)/2} \frac{\left( [l+i+1]_q [l-j+1]_q \right)^{1/2}}{[2l+1]_q}, &
c^{l-}_{ij} &= - q^{(i+j)/2} \frac{\left( [l-i]_q[l+j]_q \right)^{1/2}}{[2l+1]_q}.
}
Using this description of $c$ to compute the action of $c^{\ast} c$, we find
\envaln{
(c^{\ast}c) \re{l}{i}{j}
&= q^{i+j-1} \left( \frac{[l+i+1]_q[l-j+1]_q}{[2l+1]_q[2l+2]_q} + \frac{[l-i]_q[l+j]_q}{[2l]_q[2l+1]_q} \right) \re{l}{i}{j} \\
& \quad- q^{i+j-1} \left(
\frac{([l+i+1]_q[l-i+1]_q[l+j+1]_q[l-j+1]_q)^{\frac{1}{2}}}{[2l+1]_q[2l+2]_q}
\re{l+1}{i}{j}\right. \\
&\left.\quad+ \frac{([l+i]_q[l-i]_q[l+j]_q[l-j]_q)^{\frac{1}{2}}}{[2l]_q[2l+1]_q} \re{l-1}{i}{j} \right)
}
Let $\epsilon_{k} = Q(1-q^{2k})$, so that $[k]_{q} = q^{-k} \epsilon_{k}$. Then the
above expression can be written as
\envaln{
(c^{\ast}c) \re{l}{i}{j} &= q^{2l} \left( q^{2j} \frac{\epsilon_{l+i+1}\epsilon_{l-j+1}}{\epsilon_{2l+1} \epsilon_{2l+2}} + q^{2i} \frac{\epsilon_{l-i} \epsilon_{l+j}}{\epsilon_{2l} \epsilon_{2l+1}} \right) \re{l}{i}{j} \\
& - q^{2l+i+j} \left( \frac{(\epsilon_{l+i+1} \epsilon_{l-i+1} \epsilon_{l+j+1} \epsilon_{l-j+1})^{\frac{1}{2}}}{\epsilon_{2l+1} \epsilon_{2l+2}} \re{l+1}{i}{j} + \frac{(\epsilon_{l+i} \epsilon_{l-i} \epsilon_{l+j} \epsilon_{l-j})^{\frac{1}{2}}}{\epsilon_{2l} \epsilon_{2l+1}} \re{l-1}{i}{j} \right).
}
Define the scalars $C_{1}(l, i, j)$ and $C_{2}(l, i, j)$ to be
\envaln{
C_{1}(l, i, j) &:= \frac{\epsilon_{l+i+1}\epsilon_{l-j+1}}{\epsilon_{2l+1} \epsilon_{2l+2}} &
C_{2}(l, i, j) &:= \frac{\epsilon_{l-i} \epsilon_{l+j}}{\epsilon_{2l} \epsilon_{2l+1}}.
}
The definition of $\epsilon_{k}$ implies that $C_{1}$ and $C_{2}$ are uniformly
bounded for all $l,\,i,\,j$ appearing in the formula for $\Upsilon_z(c^*c)$.
As in the proof of Lemma \ref{lemma_operator_trace} we compute for $z\in \mathbb{R}
\envaln{
& \mathrm{Tr} \left( P_{1} P_{2} L_{k}
(1 + \ensuremath{\mathcal{D}}^{2})^{-z/4}
\hat\Delta_{F}^{-\frac{1}{2}}
\left(\begin{array}{cc} c^{\ast}c & 0 \\ 0 & 0 \end{array}\right) \hat\Delta_{F}^{-\frac{1}{2}} (1 + \ensuremath{\mathcal{D}}^{2})^{-z/4} P_{1} P_{2} L_{k} \right) \\
&\qquad\qquad = \sum_{2l = 1}^{2k} \sum_{i=-l}^{l} \sum_{n \in \mathcal{J}_{l}} \frac{q^{-2i-n}}{(1 + \lambda_{l, n}^{2})^{z/2}} \left( q^{2l+n+1}C_{1}(l, i, \tfrac{n+1}{2}) + q^{2l+2i}C_{2}(l, i, \tfrac{n+1}{2}) \right),
}
\envaln{
& \mathrm{Tr} \left( P_{1} P_{2} L_{k}
(1 + \ensuremath{\mathcal{D}}^{2})^{-z/4}
\hat\Delta_{F}^{-\frac{1}{2}}
\left(\begin{array}{cc} 0 & 0 \\ 0 & c^{\ast}c \end{array}\right) \hat\Delta_{F}^{-\frac{1}{2}} (1 + \ensuremath{\mathcal{D}}^{2})^{-z/4} P_{1} P_{2} L_{k} \right) \\
&\qquad\qquad = \sum_{2l = 1}^{2k} \sum_{i=-l}^{l} \sum_{n \in \mathcal{J}_{l}} \frac{q^{-2i-n}}{(1 + \lambda_{l, n}^{2})^{z/2}} \left( q^{2l+n+1}C_{1}(l, i, \tfrac{n-1}{2}) + q^{2l+2i}C_{2}(l, i, \tfrac{n-1}{2}) \right).
}
The uniform boundedness of $C_{1}$ and $C_{2}$, together with Lemma \ref{lemma_holomorphic},
demonstrate that the limits as $k \rightarrow \infty$ of the two sums above exist for $z>2$. Hence
$$
z\mapsto \Upsilon_z\left(\left(\begin{array}{cc} c^{\ast}c & 0 \\ 0 & 0 \end{array}\right)\right),\quad
z\mapsto \Upsilon_z\left(\left(\begin{array}{cc} 0 & 0 \\ 0 & c^{\ast}c \end{array}\right)\right)
$$
are well-defined functions for $z>2$.
Indeed the arguments of Lemma \ref{lemma_holomorphic}, together with the Weierstrass convergence
theorem, show that these functions extend to holomorphic functions
on $\ensuremath{\mathrm{Dom}}_2$.
In particular, these functions are holomorphic at $z = 3$ and hence
\enveqn{
\tau \left( \left(\begin{array}{cc} c^{\ast}c & 0 \\ 0 & 0 \end{array}\right) \right) = \tau \left( \left(\begin{array}{cc} 0 & 0 \\ 0 & c^{\ast}c \end{array}\right) \right) = 0.
}
By linearity it follows that $\tau(c^{\ast}c) = 0$ also. Using the normality of $c$, for any
operator $X$ we have the operator inequality
$$
X^*(c^*c)^mX\leq \Vert c^*c\Vert ^{m-1} X^*c^*cX,
$$
and so for $z>2$ real, we have $\Upsilon_{z}((c^{\ast}c)^{m}) \leq \Vert c \Vert^{2m-2} \Upsilon_{z}(c^{\ast}c)$.
Thus for $z>2$, the sum defining $\Upsilon_{z}((c^{\ast}c)^{m})$ converges. Once more invoking
the Weierstrass convergence theorem shows that $z\mapsto \Upsilon_{z}((c^{\ast}c)^{m})$ extends to
a holomorphic function for ${\mathrm Re}(z)>2$. Similar estimates now show that
\enveqn{
\tau \left( \left(\begin{array}{cc} (c^{\ast}c)^{m} & 0 \\ 0 & 0 \end{array}\right) \right)
= \tau \left( \left(\begin{array}{cc} 0 & 0 \\ 0 & (c^{\ast}c)^{m} \end{array}\right) \right) = 0. \qedhere
}
\end{proof}
\begin{thm}
\label{thm_res_gives_int}
Let $\alpha \in \A$ and $X, Y$ be any closed linear operators
on $\ensuremath{\mathcal{H}}_{h}$ which are defined on
and preserve $\A$. Then we have the following well-defined evaluations of $\tau$:
\begin{enumerate}
\item $\tau \left( \left(\begin{array}{cc} 0 & X \\ 0 & 0 \end{array}\right) \right) = \tau \left( \left(\begin{array}{cc} 0 & 0 \\ Y & 0 \end{array}\right) \right) = 0$
\item $\tau(\alpha \Gamma) = 0$
\item $\tau \left(\hat{\Delta}_{L}^{2}
\left(\begin{array}{cc} \alpha & 0 \\ 0 & 0 \end{array}\right)
\right) =
\tau\left(\hat{\Delta}_{L}^{2} \left(\begin{array}{cc}
0 & 0 \\ 0 & \alpha \end{array}\right) \right) = R \int_{[1]} \alpha$
\end{enumerate}
where $\int_{[1]} \colon \A \rightarrow \ensuremath{\mathbb{C}}$ is the functional defined in Lemma \ref{haddock}
and $R = 4(q^{-1}-q)/\ln(q^{-1})$.
\end{thm}
\begin{proof}
Throughout this proof we assume without loss of
generality that any element of $\A$ is homogeneous with respect to both the left and right actions
(that is $\sigma_{L}(\alpha) = q^{p} \alpha$, $\sigma_{R}(\alpha) = q^{p'} \alpha$ for some $p, p'$). This is because finite linear combinations of homogeneous elements span $\ensuremath{\mathcal{A}}$ (cf. Theorem \ref{thm_rep_basis}).
Indeed, if $\alpha \in \ensuremath{\mathcal{A}}$ is homogeneous of a non-zero degree for either the left or right action,
then $\langle \re{l}{i}{j},\alpha\, \re{l}{i}{j}\rangle=0$ and so for any linear operator
$C$ that is diagonal in the
Peter-Weyl basis, $\Upsilon_{s}(C \alpha) = 0$ for all $s \in \ensuremath{\mathbb{R}}_{+}$. Hence,
we need only consider those elements of $\ensuremath{\mathcal{A}}$ that are homogeneous of
degree zero for the left and right actions.
A convenient spanning set for these algebra elements is $\{ 1_\A, (c^{\ast}c)^{m} \colon m \in \ensuremath{\mathbb{N}} \}$.
{\em 1.} By definition
$\Upsilon_s\left(\left(\begin{array}{cc}
0 & X \\
0 & 0
\end{array}\right) \right)=0$ for all
$s>0$, and similarly for
$\left(\begin{array}{cc} 0 &0 \\Y &0
\end{array}\right) $.
{\em 2.} Lemma \ref{lemma_tau_bc_zero} has established that for all $m \geq 1$,
\enveqn{
\tau \left( \left(\begin{array}{cc} (c^{\ast}c)^{m} & 0 \\ 0 & 0 \end{array}\right) \right) = \tau \left( \left(\begin{array}{cc} 0 & 0 \\ 0 & (c^{\ast}c)^{m} \end{array}\right) \right) = 0.
}
By linearity we can extend this to conclude that $\tau((c^{\ast}c)^{m} \Gamma) = 0$.
Finally, for $z$ large and real we compute $\Upsilon_{z}(\Gamma)$ using the
proof of Lemma \ref{lemma_operator_trace}. Now
\envaln{
& \mathrm{Tr} \left( P_{1} P_{2} L_{k} (1 + \ensuremath{\mathcal{D}}^{2})^{-z/4} \hat\Delta_{F}^{-\frac{1}{2}} \Gamma \hat\Delta_{F}^{-\frac{1}{2}} (1 + \ensuremath{\mathcal{D}}^{2})^{-z/4} P_{1} P_{2} L_{k} \right) \\
&\qquad = \sum_{2l=1}^{2k} \sum_{i=-l}^{l} \sum_{n \in \mathcal{J}_{l}} \frac{q^{-2i-n} }{ (1 + \lambda_{l, n}^{2})^{z/2} } - \sum_{2l=1}^{2k} \sum_{i=-l}^{l} \sum_{n \in \mathcal{J}_{l}} \frac{q^{-2i-n} }{ (1 + \lambda_{l, n}^{2})^{z/2} },
}
and for each $k$ the summands above are finite and hence subtract to give zero.
Hence $\Upsilon_{z}(\Gamma) = 0$ for all $z$ and so $\tau(\Gamma) = 0$.
{\em 3.} For $z$ large and real, the evaluation of $\Upsilon_{z}$ as sums of
positive real numbers (as in the proof of Lemma \ref{lemma_operator_trace})
implies the numerical inequality
\enveqn{
\Upsilon_{z}(\hat{\Delta}_{L}^{2} (c^{\ast}c)^{m}) \leq \Upsilon_{z}((c^{\ast}c)^{m}).
}
This is because the introduction of $\hat{\Delta}_{L}^{2}$ multiplies each
summand by $q^{2n} \leq 1$ (cf. Equation \eqref{eqn_trace_formula}).
Lemma \ref{lemma_tau_bc_zero} demonstrates that $\Upsilon_{z}((c^{\ast}c)^{m})$
extends to a function that is holomorphic in a neighbourhood of
$z = 3$, and together with the Weierstrass
convergence theorem the result follows.
Finally we analyse $\Upsilon_{z}\left(\hat\Delta_{L}^{2} \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right) \right)$
and $\Upsilon_{z}\left(\hat\Delta_{L}^{2} \left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right) \right)$. Again
using the proof of Lemma \ref{lemma_operator_trace} we find
\envaln{
& \mathrm{Tr} \left( P_{1} P_{2} L_{k} (1 + \ensuremath{\mathcal{D}}^{2})^{-z/4} \hat\Delta_{F}^{-\frac{1}{2}} \hat\Delta_{L}^{2} \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right) \hat\Delta_{F}^{-\frac{1}{2}} (1 + \ensuremath{\mathcal{D}}^{2})^{-z/4} P_{1} P_{2} L_{k} \right) \\
& \qquad = \mathrm{Tr} \left( P_{1} P_{2} L_{k} (1 + \ensuremath{\mathcal{D}}^{2})^{-z/4} \hat\Delta_{F}^{-\frac{1}{2}} \hat\Delta_{L}^{2} \left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right) \hat\Delta_{F}^{-\frac{1}{2}} (1 + \ensuremath{\mathcal{D}}^{2})^{-z/4} P_{1} P_{2} L_{k} \right) \\
& \qquad\qquad = \sum_{2l=1}^{2k} \sum_{i=-l}^{l} \sum_{n \in \mathcal{J}_{l}} \frac{q^{-2i+n} }{ (1 + \lambda_{l, n}^{2})^{z/2} } \\
& \qquad\qquad\qquad = Qq^{-1} \sum_{2l=1}^{2k} \sum_{n \in \mathcal{J}_{l}} \frac{q^{n-2l} }{ (1 + \lambda_{l, n}^{2})^{z/2} } - Q q\sum_{2l=1}^{2k} \sum_{n \in \mathcal{J}_{l}} \frac{q^{n+2l} }{ (1 + \lambda_{l, n}^{2})^{z/2} }
}
For $z$ real, the sum $\sum_{2l=1}^{2k} \sum_{n \in \mathcal{J}_{l}} q^{n+2l}/(1 + \lambda_{l, n}^{2})^{z/2}$ is bounded above by $f_{2}(z)$ from Lemma \ref{lemma_holomorphic} for all $k$. By the Weierstrass convergence theorem we conclude that as $k \rightarrow \infty$, this sum converges to a function with a holomorphic extension about $z = 3$. Next, when considering the sum $\sum_{2l=1}^{2k} \sum_{n \in \mathcal{J}_{l}} q^{n-2l}/(1 + \lambda_{l, n}^{2})^{z/2}$, observe by rearranging the order of summation
$$
\sum_{2l = 1}^{2k} \sum_{n \in \mathcal{J}_{l}}
\rightarrow \sum_{n = 0}^{2k} \sum_{l = (n+1)/2}^{k},
$$
that Lemma \ref{lemma_tau_residue} proves that the sum has a limit as
$k \rightarrow \infty$ and the corresponding function of $z$ extends to a meromorphic function
with a simple pole at $z = 3$. The residue at $z = 3$ is $4qQ^{-2}/\ln(q^{-1})$
and from the definition of $\tau$ we conclude that for $R = 4(q^{-1}-q)/\ln(q^{-1})$,
\enveqn{
\tau\left(\hat\Delta_{L}^{2} \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right) \right) = \tau\left(\hat\Delta_{L}^{2} \left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right) \right) = R.
}
Finally, we compare the definition of $R \int_{[1]}$ in Lemma \ref{haddock}
to the evaluation of $\tau$ on $\A$ derived here
and observe that they agree on $\A$.
\end{proof}
\begin{lemma}
\label{lemma_twisted_trace}
Given any matrix $M \in \M_{2}(\ensuremath{\mathcal{A}})$ and any $\alpha \in \ensuremath{\mathcal{A}}$ then $\tau(M \hat{\Delta}_{L}^{2} \alpha) = \tau(\vartheta^{-1}(\alpha) M \hat{\Delta}_{L}^{2})$.
\end{lemma}
\begin{proof}
From Lemma \ref{haddock}, the linear functional $\int_{[1]}$ is a $\sigma_{L}^{2} \circ \vartheta^{-1}$-twisted trace. That is, given any $\alpha, \beta \in \ensuremath{\mathcal{A}}$
\enveqn{
\int_{[1]} \alpha \beta = \int_{[1]} \sigma_{L}^{2}(\vartheta^{-1}(\beta)) \alpha.
}
Now we separate the matrix $M = M_{d} + M_{o}$ into diagonal and off-diagonal matrices respectively.
Then by Theorem \ref{thm_res_gives_int},
$\tau(M_{d} \hat{\Delta}_{L}^{2} \alpha)$ and $\tau(M_{o} \hat{\Delta}_{L}^{2} \alpha)$
are both well-defined, so by linearity
\enveqn{
\tau(M \hat{\Delta}_{L}^{2} \alpha) = \tau(M_{d} \hat{\Delta}_{L}^{2} \alpha) + \tau(M_{o} \hat{\Delta}_{L}^{2} \alpha) = \tau(M_{d} \hat{\Delta}_{L}^{2} \alpha) + 0.
}
Since $M_{d}$ is diagonal, we may write
\enveqn{
M_{d} \hat{\Delta}_{L}^{2} = \hat{\Delta}_{L}^{2} \sigma_{L}^{2}(M_{d})
}
where $\sigma_{L}$ acts componentwise on the matrix. Using the value of
$\tau(\hat{\Delta}_{L}^{2} \sigma_{L}^{2}(M_{d}) \alpha)$ from Theorem \ref{thm_res_gives_int}, we have
\envaln{
\tau(M \hat{\Delta}_{L}^{2} \alpha) = \tau(\hat{\Delta}_{L}^{2} \sigma_{L}^{2}(M_{d}) \alpha) &= R \int_{[1]} \sigma_{L}^{2}(M_{d}^{+}) \alpha + R \int_{[1]} \sigma_{L}^{2}(M_{d}^{-}) \alpha \\
&= R \int_{[1]} \sigma_{L}^{2}(\vartheta^{-1}(\alpha)) \sigma_{L}^{2}(M_{d}^{+}) + R \int_{[1]} \sigma_{L}^{2}(\vartheta^{-1}(\alpha)) \sigma_{L}^{2}(M_{d}^{-}),
}
by the twisted trace property of $\int_{[1]}$. Recombining these two terms yields
\envaln{
\tau(\hat{\Delta}_{L}^{2} \sigma_{L}^{2}(M_{d}) \alpha)
= \tau(\hat{\Delta}_{L}^{2} \sigma_{L}^{2}(\vartheta^{-1}(\alpha)) \sigma_{L}^{2}(M_{d}))
= \tau(\vartheta^{-1}(\alpha) \hat{\Delta}_{L}^{2} \sigma_{L}^{2}(M_{d}))
= \tau(\vartheta^{-1}(\alpha) M_{d} \hat{\Delta}_{L}^{2}).
}
Now, $\tau(\vartheta^{-1}(\alpha) M_{o} \hat{\Delta}_{L}^{2})$ is well defined and has value zero, so we can write
\enveqn{
\tau(M \hat{\Delta}_{L}^{2} \alpha) = \tau(\vartheta^{-1}(\alpha) M_{d} \hat{\Delta}_{L}^{2}) + \tau(\vartheta^{-1}(\alpha) M_{o} \hat{\Delta}_{L}^{2}) = \tau(\vartheta^{-1}(\alpha) M \hat{\Delta}_{L}^{2}). \qedhere
}
\end{proof}
\begin{thm}
\label{thm:big-fish}
Given any $a_{0}, \ldots, a_{3} \in \ensuremath{\mathcal{A}}$, the map
$\phi_{\mathrm{res}}:a_{0}, \ldots, a_{3} \mapsto \tau(a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}])$ is a $\vartheta^{-1}$-twisted Hochschild 3-cocycle, whose cohomology class is
non-trivial. The cocycle $\phi_{\mathrm{res}}$ has non-zero
pairing with the $\vartheta^{-1}$-twisted 3-cycle $dvol$ defined in \eqref{eqn_dvol}, giving
\enveqn{
\left\langle \phi_{\mathrm{res}}, dvol \right\rangle = 3R(q^{-1} + q)=4!\,\frac{q^{-1}+q}{2}\,\frac{q^{-1}-q}{\ln(q^{-1})}.
}
The cocycle may be written as
\enveqn{
\phi_{\mathrm{res}} = q^{2}R(\varphi + \varphi_{213} + \varphi_{231}) + R(\varphi_{132} + \varphi_{312} + \varphi_{321})
}
where $\varphi$ and $\varphi_{ijk}$ are the cocycles described in Lemma \ref{haddock}
and Corollary \ref{cor:perms}.
\end{thm}
\begin{proof}
First consider $\pi_{\ensuremath{\mathcal{D}}}(a_{0}, a_{1}, a_{2}, a_{3}) = a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}]$
as an unbounded operator on $\A\oplus\A\subset\ensuremath{\mathcal{H}}$. Using the equality
$[\ensuremath{\mathcal{D}}, \alpha] = \tilde{S}(\alpha) + \tilde{T}(\alpha)\hat\Delta_{L}$, we see that
$\pi_{\ensuremath{\mathcal{D}}}(a_{0}, a_{1}, a_{2}, a_{3})$ can be expanded into 8 terms.
Recall that by Theorem \ref{thm_res_gives_int} the functional $\tau$ vanishes on
off-diagonal operators. Four of the eight terms in the expansion of
$\pi_{\ensuremath{\mathcal{D}}}(a_{0}, a_{1}, a_{2}, a_{3})$ are off-diagonal since, for all $\alpha\in\A$, $\tilde{S}(\alpha)$ is
diagonal and $\tilde{T}(\alpha)$ is off-diagonal. Thus
\envaln{
& \tau \left( a_{0} \left( \tilde{T}(a_{1})\hat\Delta_{L} \tilde{S}(a_{2}) \tilde{S}(a_{3}) + \tilde{S}(a_{1}) \tilde{T}(a_{2})\hat\Delta_{L} \tilde{S}(a_{3}) \right. \right. \\
& \quad \left. \left. + \tilde{S}(a_{1}) \tilde{S}(a_{2}) \tilde{T}(a_{3})\hat\Delta_{L} + \tilde{T}(a_{1}) \hat\Delta_{L} \tilde{T}(a_{2}) \hat\Delta_{L} \tilde{T}(a_{3}) \hat\Delta_{L} \right) \right) = 0.
}
Therefore, $\phi_{\mathrm{res}}(a_{0},a_{1}, a_{2}, a_{3})$ reduces to
\enval{
\phi_{\mathrm{res}}(a_{0}, a_{1}, a_{2}, a_{3})
&= \tau \left( a_{0} \left( \tilde{S}(a_{1}) \tilde{S}(a_{2}) \tilde{S}(a_{3})
+ \tilde{S}(a_{1}) \tilde{T}(a_{2}) \hat\Delta_{L} \tilde{T}(a_{3}) \hat\Delta_{L} \right. \right. \nonumber \\
& \quad \left. \left. + \tilde{T}(a_{1}) \hat\Delta_{L} \tilde{S}(a_{2}) \tilde{T}(a_{3}) \hat\Delta_{L}
+ \tilde{T}(a_{1}) \hat\Delta_{L} \tilde{T}(a_{2}) \hat\Delta_{L} \tilde{S}(a_{3}) \right) \right).
\label{eq:three-not-four}
}
From Lemma \ref{lemma_triple_properties} it follows that
\enveqn{
a_{0} \tilde{S}(a_{1}) \tilde{S}(a_{2}) \tilde{S}(a_{3})
= a_{0} \partial_{H}(a_{1}) \partial_{H}(a_{2}) \partial_{H}(a_{3}) \Gamma
}
and recall that from Theorem \ref{thm_res_gives_int}, $\tau(\alpha \Gamma) = 0$
for all $\alpha \in \A$. Since
$a_{0} \partial_{H}(a_{1}) \partial_{H}(a_{2}) \partial_{H}(a_{3}) \in \A$ we have
\enveqn{
\tau (a_{0} \tilde{S}(a_{1}) \tilde{S}(a_{2}) \tilde{S}(a_{3})) = 0.
}
We now move all the $\hat{\Delta}_L$'s to the right in the remaining terms in Equation \eqref{eq:three-not-four}.
For $\alpha\in\A$, we use
$\hat\Delta_{L} \tilde{S}(\alpha) = \tilde{S}(\sigma_{L}^{-1}(\alpha)) \hat\Delta_{L}$ , and
\envaln{
\hat\Delta_{L} \tilde{T}(\alpha)
&= \left(\begin{array}{cc} 0 & q^{-1} \Delta_{L} q^{-\frac{1}{2}} \ensuremath{\p_{e}}(\sigma_{L}^{-\frac{1}{2}}(\alpha)) \\
q \Delta_{L} q^{\frac{1}{2}} \ensuremath{\p_{f}}(\sigma_{L}^{-\frac{1}{2}}(\alpha)) & 0 \end{array}\right) \\
&= \left(\begin{array}{cc} 0 & q^{-2} q^{-\frac{1}{2}} \sigma_{L}^{-1}(\ensuremath{\p_{e}}(\sigma_{L}^{-\frac{1}{2}}(\alpha))) \\
q^{2} q^{\frac{1}{2}} \sigma_{L}^{-1}(\ensuremath{\p_{f}}(\sigma_{L}^{-\frac{1}{2}}(\alpha))) & 0 \end{array}\right) \hat{\Delta}_{L} \\
&= \left(\begin{array}{cc} 0 & q^{-\frac{1}{2}} \ensuremath{\p_{e}}(\sigma_{L}^{-\frac{3}{2}}(\alpha)) \\
q^{\frac{1}{2}} \ensuremath{\p_{f}}(\sigma_{L}^{-\frac{3}{2}}(\alpha)) & 0 \end{array}\right) \hat{\Delta}_{L} \\
&= \tilde{T}(\sigma_{L}^{-1}(\alpha)) \hat{\Delta}_{L}.
}
This yields
\envaln{
\phi_{\mathrm{res}}(a_{0}, \,& a_{1}, a_{2}, a_{3})
= \tau \left( a_{0} \left( \tilde{S}(a_{1}) \tilde{T}(a_{2}) \tilde{T}(\sigma_{L}^{-1}(a_{3})) \right. \right. \\
& \left. \left. + \tilde{T}(a_{1}) \tilde{S}(\sigma_{L}^{-1}(a_{2})) \tilde{T}(\sigma_{L}^{-1}(a_{3}))
+ \tilde{T}(a_{1}) \tilde{T}(\sigma_{L}^{-1}(a_{2})) \tilde{S}(\sigma_{L}^{-2}(a_{3})) \right) \hat{\Delta}_{L}^{2} \right).
}
In this form Theorem \ref{thm_res_gives_int} tells us that $\phi_{\mathrm{res}}$
is a well defined, multilinear functional on $\ensuremath{\mathcal{A}}^{\otimes 4}$.
In order to demonstrate that this cochain is indeed a twisted Hochschild cocycle,
it remains only to show that the boundary operator maps the cochain to zero.
This result follows from the Leibniz property of the commutators together with
Lemma \ref{lemma_twisted_trace}. Explicitly,
\envaln{
& (b_{3}^{\vartheta^{-1}} \phi_{\mathrm{res}}) (a_{0}, \ldots, a_{4}) = \tau(a_{0}a_{1} [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}] [\ensuremath{\mathcal{D}}, a_{4}]) - \tau(a_{0} [\ensuremath{\mathcal{D}}, a_{1}a_{2}] [\ensuremath{\mathcal{D}}, a_{3}] [\ensuremath{\mathcal{D}}, a_{4}]) \\
& \qquad + \tau(a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}a_{3}] [\ensuremath{\mathcal{D}}, a_{4}]) - \tau(a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}a_{4}]) + \tau(\vartheta^{-1}(a_{4})a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}]) \\
& \qquad \qquad = - \tau(a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}]a_{4}) + \tau(\vartheta^{-1}(a_{4})a_{0} [\ensuremath{\mathcal{D}}, a_{1}] [\ensuremath{\mathcal{D}}, a_{2}] [\ensuremath{\mathcal{D}}, a_{3}]) = 0,
}
where the last equality follows from Lemma \ref{lemma_twisted_trace}.
In order to identify $\phi_{\mathrm{res}}$, we use Lemma \ref{lemma_triple_properties} to write,
for $a_{0}, \ldots, a_{3} \in \A$,
\envaln{
a_{0} & \left( \tilde{S}(a_{1}) \tilde{T}(a_{2}) \tilde{T}(\sigma_{L}^{-1}(a_{3}))
+ \tilde{T}(a_{1}) \tilde{S}(\sigma_{L}^{-1}(a_{2})) \tilde{T}(\sigma_{L}^{-1}(a_{3})) \right. \\
& \left. + \tilde{T}(a_{1}) \tilde{T}(\sigma_{L}^{-1}(a_{2})) \tilde{S}(\sigma_{L}^{-2}(a_{3})) \right)
= \left(\begin{array}{cc} \pi_{1}(a_{0}, \ldots, a_{3}) & 0 \\ 0 & \pi_{2}(a_{0}, \ldots, a_{3}) \end{array}\right)
}
for some multi-linear maps $\pi_{1}, \pi_{2} \colon \A^{\otimes 4} \rightarrow \A$.
Again using Lemma \ref{lemma_triple_properties}, we have
\enval{
\pi_{1}(a_{0}, \ldots, a_{3})
&= a_{0} \partial_{H}(a_{1}) \ensuremath{\p_{e}}(\sigma_{L}^{-\frac{1}{2}}(a_{2})) \ensuremath{\p_{f}}(\sigma_{L}^{-\frac{3}{2}}(a_{3})) - a_{0} \ensuremath{\p_{e}}(\sigma_{L}^{-\frac{1}{2}}(a_{1})) \partial_{H}(\sigma_{L}^{-1}(a_{2}))
\ensuremath{\p_{f}}(\sigma_{L}^{-\frac{3}{2}}(a_{3})) \nonumber \\
& \quad + a_{0} \ensuremath{\p_{e}}(\sigma_{L}^{-\frac{1}{2}}(a_{1})) \ensuremath{\p_{f}}(\sigma_{L}^{-\frac{3}{2}}(a_{2}))
\partial_{H}(\sigma_{L}^{-2}(a_{3})), \label{eqn_pi_one}
}
\enval{
\pi_{2}(a_{0}, \ldots, a_{3})
&= - a_{0} \partial_{H}(a_{1}) \ensuremath{\p_{f}}(\sigma_{L}^{-\frac{1}{2}}(a_{2})) \ensuremath{\p_{e}}(\sigma_{L}^{-\frac{3}{2}}(a_{3})) + a_{0} \ensuremath{\p_{f}}(\sigma_{L}^{-\frac{1}{2}}(a_{1})) \partial_{H}(\sigma_{L}^{-1}(a_{2}))
\ensuremath{\p_{e}}(\sigma_{L}^{-\frac{3}{2}}(a_{3})) \nonumber \\
& \quad - a_{0} \ensuremath{\p_{f}}(\sigma_{L}^{-\frac{1}{2}}(a_{1})) \ensuremath{\p_{e}}(\sigma_{L}^{-\frac{3}{2}}(a_{2}))
\partial_{H}(\sigma_{L}^{-2}(a_{3})). \label{eqn_pi_two}
}
Then by Theorem \ref{thm_res_gives_int}, and the $\sigma_L$ invariance of $\int_{[1]}$, we have
\enveq{
\phi_{\mathrm{res}}(a_{0} , a_{1}, a_{2}, a_{3})
= R \int_{[1]} \pi_{1}(a_{0}, \ldots, a_{3}) + R \int_{[1]} \pi_{2}(a_{0}, \ldots, a_{3}). \label{eqn_cocycle_pi}
}
Comparing Equations \eqref{eqn_pi_one}, \eqref{eqn_pi_two}, \eqref{eqn_cocycle_pi} with the expressions for the cocycles identified in Lemma \ref{haddock} and Corollary \ref{cor:perms} we find
\enveqn{
\phi_{\mathrm{res}} =
q^{2}R (\varphi + \varphi_{213} + \varphi_{231}) + R (\varphi_{132} + \varphi_{312} + \varphi_{321}).
}
The evaluation of this cocycle on the cycle $dvol$ (see Equation \eqref{eqn_dvol}) is a straightforward computation using the explicit expressions obtained. The result is
\enveqn{
\left\langle \phi_{\mathrm{res}} , dvol \right\rangle = 3R(q^{-1} + q). \qedhere
}
\end{proof}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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\section{Introduction}
\label{sec:intro}
Vocoders were originally used for speech compression in the field of communication. Recently, vocoders have been utilized in various fields such as text-to-speech\cite{shen2018natural, ren2020fastspeech, lancucki2020fastpitch, lim2020jdi} and voice conversion\cite{liu2018wavenet} or speech-to-speech translation\cite{jia2019direct}. Neural vocoders generate human-like voices using neural networks, instead of using traditional methods that contain audible artifacts \cite{griffin1984signal, kawahara1999restructuring, morise2016world}.
Recently, it has been demonstrated that vocoders exhibit superior performances in generation speed and audio fidelity when trained with single speaker utterances. However, some models face difficulty when generating natural sounds in multiple domains such as speakers, language, or expressive utterances. The ability of these models can be evaluated by the sound quality when the model is trained on data of multiple speakers and the sound quality of the unseen domain (from an out-of-domain source). A vocoder that can generate high-fidelity audio in various domains, regardless of whether the input has been encountered during training or has come from an out-of-domain source, is usually called a \textit{universal vocoder}\cite{lorenzo2018towards, hsu2019towards}.
MelGAN\cite{kumar2019melgan} is a vocoder based on generative adversarial networks (GANs)\cite{goodfellow2014generative}. It is a lightweight and robust model for unseen speakers but yields lower fidelity than popularly employed models \cite{oord2016wavenet, kalchbrenner2018efficient, prenger2019waveglow}. MelGAN alleviates the metallic sound that occurs mainly in unvoiced and breathy speech segments through multi-scale discriminators that receive different scale waveforms as inputs. However, it has not been implemented efficiently for learning with multiple speakers for a universal vocoder.
In this study, we propose \textit{Universal MelGAN}. The generated waveform of the original MelGAN with audible artifacts appears as an over-smoothing problem with a non-sharp spectrogram. We added \textit{multi-resolution spectrogram discriminators} to the model to address this problem in the frequency domain. Our multi-scale discriminators enable fine-grained spectrogram prediction by discriminating waveforms and spectrograms. In particular, they alleviate the over-smoothing problem in the high frequency band of the large footprint model, enabling the generation of realistic multi-speaker waveforms.
To evaluate the performance of the proposed model, we compare with full-band MelGAN (FB-MelGAN)\cite{yang2020multi} as a baseline and two other vocoders: WaveGlow\cite{prenger2019waveglow} and WaveRNN\cite{lorenzo2018towards}. We designed experiments in both Korean and English for language independency. For evaluation, we prepared multiple speaker utterances that included unseen domain scenarios, such as new speakers, emotions, and languages.
The evaluation results indicate that the proposed model achieved the best mean opinion score (MOS) in most scenarios and efficiently preserved the fidelity in unseen speakers. In addition, the evaluations show that the model efficiently preserves the original speech, even in challenging domains such as expressive utterances and unseen languages. In multi-speaker text-to-speech scenarios, our model can generate high-fidelity waveforms with high MOS, and the model outperforms compared vocoders. This results without any external domain information suggest the possibility of the proposed model as a universal vocoder.
\section{Related work}
\label{sec:prior}
MelGAN\cite{kumar2019melgan} is a GAN-based lightweight vocoder in which the generator comprises transposed convolution layers for upsampling and a stack of residual blocks for effective conversion. Multiple discriminators are trained with different scale waveforms to operate in different ranges. Recently, a modified architecture called FB-MelGAN\cite{yang2020multi} with improved fidelity has been proposed.
WaveGlow\cite{prenger2019waveglow} can directly maximize the likelihood of data based on a normalizing flow. A chain of flows transforms simple distributions (e.g. isotropic Gaussian) into the desired data distribution. It has been shown\cite{maiti2020speaker} that the speaker generalization of WaveGlow obtained high objective scores than other models.
WaveRNN\cite{kalchbrenner2018efficient} is an autoregressive model that generates waveforms using recurrent neural network (RNN) layers. It has been demonstrated that WaveRNN preserves sound quality in unseen domains\cite{lorenzo2018towards}. The robustness of WaveRNN using speaker representations has been discussed\cite{paul2020speaker}.
\begin{figure*}[htb]
\begin{minipage}[b]{0.4\linewidth}
\centering
\centerline{\includegraphics[height=6.8cm]{figure1a.png}}
\centerline{(a) Generator}\medskip
\end{minipage}
\begin{minipage}[b]{0.55\linewidth}
\centering
\centerline{\includegraphics[height=6.8cm]{figure1b.png}}
\centerline{(b) Multi-scale waveform and spectrogram discriminators}\medskip
\end{minipage}
\caption{An architecture of the Universal MelGAN.}
\label{fig:model}
\end{figure*}
\section{Description of the proposed model}
\label{sec:proposed}
\subsection{FB-MelGAN}
\label{ssec:fb_melgan}
FB-MelGAN\cite{yang2020multi} has been used as the baseline in this study. The advantages of FB-MelGAN include pre-training of the generator, increased receptive field of residual stacks, and application of multi-resolution short-time Fourier transform (STFT) loss\cite{yamamoto2020parallel} as auxiliary loss. These modifications are effective to achieve better fidelity and training stability.
The multi-resolution STFT loss is the sum of multiple spectrogram losses calculated with different STFT parameter sets. It comprises the spectral convergence loss ${\cal L}^{m}_{\mathrm{sc}}(\cdot, \cdot)$ and the log STFT magnitude loss ${\cal L}^{m}_{\mathrm{mag}}(\cdot, \cdot)$. These objectives are defined as follows:
\begin{equation}
\label{eq:sc}
{\cal L}^{m}_{\mathrm{sc}}({\mathbf x}, \hat{{\mathbf x}}) = \frac{\Vert{}\vert{}\mathit{STFT}_{m}({\mathbf x})\vert{}-\vert{}\mathit{STFT}_{m}(\hat{{\mathbf x}})\vert{}\Vert{}_{F}}{\Vert{}\vert{}\mathit{STFT}_{m}({\mathbf x})\vert{}\Vert{}_{F}}
\end{equation}
\begin{equation}
\label{eq:mag}
{\cal L}^{m}_{\mathrm{mag}}({\mathbf x}, \hat{{\mathbf x}}) = \frac{1}{N}\Vert{}\log{}\vert{}\mathit{STFT}_{m}({\mathbf x})\vert{}-\log{}\vert{}\mathit{STFT}_{m}(\hat{{\mathbf x}})\vert{}\Vert{}_{1}
\end{equation}
\begin{equation}
\label{eq:stft}
{\cal L}_{\mathrm{aux}}(G)=\frac{1}{M}\sum_{m=1}^M\mathbb{E}_{{\mathbf x}, \hat{{\mathbf x}}}\Big{[}{\cal L}^{m}_{\mathrm{sc}}({\mathbf x}, \hat{{\mathbf x}})+{\cal L}^{m}_{\mathrm{mag}}({\mathbf x}, \hat{{\mathbf x}})\Big{]}
\end{equation}
where $\Vert\cdot\Vert_{F}$ and $\Vert\cdot\Vert_{1}$ denote the Frobenius and L1 norms, and $M$ and $N$ denote the number of STFT parameter sets and the number of elements in the STFT magnitude, respectively. $\vert\mathit{STFT}_{m}(\cdot)\vert$ denotes the STFT magnitude of the $m$-th STFT parameter set. The loss is used to minimize distances between the real data ${\mathbf x}$ and predicted data $\hat{{\mathbf x}}=G({\mathbf c})$, using the generator $G$. The overall objectives with auxiliary loss are defined as follows:
\begin{equation}
\label{eq:generator}
{\cal L}_{\mathrm{G}}(G, D)={\cal L}_{\mathrm{aux}}(G)+\frac{\lambda}{K}\sum_{k=1}^{K}\mathbb{E}_{\hat{{\mathbf x}}}[(D_{k}(\hat{{\mathbf x}})-1)^2]
\end{equation}
\begin{equation}
\label{eq:discriminator}
{\cal L}_{\mathrm{D}}(G, D)=\frac{1}{K}\sum_{k=1}^{K}(\mathbb{E}_{{\mathbf x}}[(D_{k}({\mathbf x})-1)^2]+\mathbb{E}_{\hat{{\mathbf x}}}[D_{k}(\hat{{\mathbf x}})^2])
\end{equation}
where ${\mathbf c}$ denotes the mel-spectrogram, and $K$ refers to the number of discriminators $D_{k}$. There is a balancing parameter $\lambda$ that optimizes the adversarial and auxiliary loss simultaneously.
\begin{figure*}[htb]
\begin{minipage}[b]{1.0\linewidth}
\centering
\centerline{\includegraphics[width=17.5cm]{figure2.png}}
\medskip
\end{minipage}
\caption{Comparison of spectrograms of the generated waveforms in the 6 to 12 kHz bands. (Each waveform has 24 kHz sampling rate.)}
\label{fig:spectrogram}
\end{figure*}
\subsection{Universal MelGAN: Improvements}
\label{ssec:improvements}
We confirmed that FB-MelGAN trained on the utterances of hundreds of speakers generated unsatisfactory sound quality waveforms, unlike when trained on a single speaker's utterances. In FB-MelGAN, multi-scale discriminators alleviate this degradation problem. However, in our multi-speaker experiment, metallic sounds were discernable, especially in unvoiced and breathy segments.
To trade inference speed and quality, we increased the sizes of the hidden channel in the generator by four times and added gated activation units (GAUs)\cite{van2016conditional} to the last layers of each residual stack. Although the expansion of the nonlinearity improved the average quality of the multi-speaker, the over-smoothing problem appeared in the high frequency band, accompanied by audible artifacts.
We assumed that discriminators in the temporal domain may not be sufficient for the problem in the frequency domain. To solve the problem, we propose \textit{multi-resolution spectrogram discriminators} that expand the spectrogram discriminator\cite{su2020hifi}. The objecties for \textit{Universal MelGAN} can be updated as follows:
\vspace{-0.2cm}
\begin{equation}
\label{eq:ours_generator}
\begin{split}
{\cal L}^{\prime}_{\mathrm{G}}(G, D)={\cal L}_{\mathrm{aux}}(G)+\frac{\lambda}{K\!+\!M}\big{(}\sum_{k=1}^{K}\mathbb{E}_{\hat{{\mathbf x}}}[(D_{k}(\hat{{\mathbf x}})-1)^2]\\[-3pt]+\sum_{m=1}^{M}\mathbb{E}_{\hat{{\mathbf x}}}[(D^{s}_{m}(\vert\mathit{STFT}_{m}(\hat{{\mathbf x}})\vert)-1)^2]\big{)}
\end{split}
\end{equation}
\vspace{-0.2cm}
\begin{equation}
\label{eq:ours_discriminator}
\begin{split}
{\cal L}^{\prime}_{\mathrm{D}}(G, D)=\frac{1}{K\!+\!M}\big{(}\sum_{k=1}^{K}(\mathbb{E}_{{\mathbf x}}[(D_{k}({\mathbf x})-1)^2]+\mathbb{E}_{\hat{{\mathbf x}}}[D_{k}(\hat{{\mathbf x}})^2]) \\[-3pt]\!+\!\sum_{m=1}^{M}\!(\mathbb{E}_{{\mathbf x}}[(D^{s}_{m}(\vert\mathit{STFT}_{m}({\mathbf x})\vert)\!-\!1)^2]\!+\!\mathbb{E}_{\hat{{\mathbf x}}}[D^{s}_{m}(\vert\mathit{STFT}_{m}(\hat{{\mathbf x}})\vert)^2])\!\big{)}
\end{split}
\end{equation}
where $D^{s}_{m}$ denotes a \textit{spectrogram discriminator} attached to the multi-resolution STFT module. Each $m$-th model uses the previously calculated spectrogram to minimize the $m$-th STFT loss.
Fig. 1. shows the architecture of the Universal MelGAN. The waveforms generated by the large footprint generator in Fig. 1(a) are discriminated by multiple scales for both the waveform and spectrogram in Fig. 1(b).
In Fig. 2, we compared the high frequency bands of the generated waveforms. Although the large foorprint model of the baseline produces improved resolution, it still faces an over-smoothing problem, especially in high bands above 9 kHz. The proposed model not only alleviates this problem (red) but also generates harmonic shapes that there are not in the ground-truth in some segments (yellow).
\section{Experiments}
\label{sec:experiments}
We designed experiments in both Korean and English. During training, we used reading style and studio-quality internal datasets with 62 speakers and 265k utterances in Korean. We also used the reading style open datasets, LJSpeech-1.1\cite{ito2017lj} and LibriTTS(train-clean-360)\cite{zen2019libritts}, with 905 speakers and 129k utterances in English. To accurately evaluate the robustness, we prepared test sets that included various seen and unseen domains, as shown in Table 1.
To evaluate the seen domain (i.e. speaker), we considered two scenarios: a speaker with relatively more utterances and a speaker with relatively fewer utterances in the total training set. To evaluate whether the generated waveforms of the speaker trained with fewer utterances also preserved the sound quality well, the training set comprised a relatively large set of single speakers (K1 for Korean, E1 for English) and a multi-speaker small set for each speaker (K2 for Korean, E2 for English). Each speaker's test set was used to evaluate the sound quality.
To evaluate the unseen domains, three scenarios that were not included in the training data were considered: speaker, emotion, and language. We prepared utterances of unseen speakers to evaluate the robustness of the speaker (K3 for Korean, E3 for English). We included the following variations of emotional utterances, such as happiness, sadness, anger, fear, disgust, or sports casting in the evaluation sets (K4 for Korean, E4 for English). Ten unseen languages were used to evaluate the language robustness (M1 for both languages). We evaluated the universality of each model using the utterances of these unseen domains.
\begin{table}[ht]
\caption{Datasets for evaluation.}
\label{tab:evaluation_set}
\centering
\begin{tabular}{l|l|l}
\hline
\multicolumn{3}{c}{Korean datasets} \\
\hline
Index & Name & Remarks \\ \hline
K1 & Internal dataset \#1 & Seen: single speaker \\
K2 & Internal dataset \#2 & Seen: multi-speaker \\
K3 & Internal dataset \#3 & Unseen: speaker \\
\multirow{2}{*}{K4} & Internal dataset \#4 & \multirow{2}{*}{Unseen: emotion} \\
& \,+AICompanion-Emotion\tablefootnote{Available at: \url{ http://aicompanion.or.kr/nanum/tech/data_introduce.php?idx=45}} & \\ \hline
\multicolumn{3}{c}{English datasets} \\
\hline
Index & Name & Remarks \\ \hline
E1 & LJSpeech-1.1\cite{ito2017lj} & Seen: single speaker \\
E2 & LibriTTS(train-clean-360)\cite{zen2019libritts} & Seen: multi-speaker \\
E3 & LibriTTS(test-clean)\cite{zen2019libritts} & Unseen: speaker \\
E4 & BlizzardChallenge2013\cite{king2013blizzard} & Unseen: emotion \\ \hline
\multicolumn{3}{c}{Multiple language dataset} \\
\hline
Index & Name & Remarks \\ \hline
M1 & CSS10\cite{park2019css10} & Unseen: language \\ \hline
\end{tabular}
\end{table}
\subsection{Data configurations}
\label{ssec:data_configurations}
All speech samples were resampled to a rate of 24 kHz. We used a high-pass filter at 50 Hz and normalized the loudness to -23 LUFS. 100-band log-mel-spectrograms with 0 to 12 kHz frequency bands were extracted by using a 1024-point Fourier transform, 256 sample frame shift, and 1024 sample frame length. All spectrograms were normalized utterance-wise to have an average of 0 and variance of 1.
\subsection{Model settings}
\label{ssec:settings}
The settings that are not specified follow the original papers or implementations in the footnotes.
When training WaveGlow\cite{prenger2019waveglow}\footnote{\url{https://github.com/NVIDIA/waveglow}} and WaveRNN\cite{lorenzo2018towards}\footnote{\url{https://github.com/bshall/UniversalVocoding}}, GitHub implementations were used for reproducibility. In WaveGlow, we trained up to 1M steps with all settings same as those in the implementation. For inference, a noise sampling parameter $\sigma=0.6$ was used. In WaveRNN, we set up predicting 10-bit $\mu$-law samples. At every 100k step, we reduced the learning rate to half of the initial value of $4e-4$ and trained up to 500k steps.
In FB-MelGAN\cite{yang2020multi}, each upsampling rate was set to 8, 8, and 4 to match the hopping size of 256. In our dataset, stable training was effectively realized by reducing the initial learning rate of the discriminators to $5e-5$. The batch size was set to 48. We trained up to 700k steps using the Adam optimizer.
In Universal MelGAN, the channel sizes of the input layers and the layers inside the three residual stacks in the generator were increased by four times; therefore, they were changed to 2048, 1024, 512, and 256, respectively. GAU was added inside each residual stack, and the channel size of the previous layer was doubled to maintain the same output size after GAU processing. Each spectrogram discriminator has a structure similar to the waveform discriminator (same as the discriminator used in FB-MelGAN), as shown in Fig. 1(b). 1-d convolutions were replaced with 2-d, and all layers have a channel size of 32, 1 group, and 1 dilation. The last two layers have a kernel size of 3, and all other layers have kernel sizes of 9. During operation, the length of the temporal domain was reduced to a stride of 2 over 3 times. Leaky ReLU with $\alpha=0.2$ was used for each activation. The balancing parameter $\lambda$ for all discriminators was set to 2.5. We trained up to 700k steps with the same learning rate, training strategy, batch size, and optimizer as in FB-MelGAN.
\section{Results}
\label{sec:results}
We implemented MOS assessments in which listeners scored naturalness from 1 (negative) to 5 (positive) for each sample. Ten randomly sampled utterances were prepared for each scenario and model, and 150 scores were collected to calculate the mean and 95\% confidence intervals. Fifteen native listeners participated in the Korean evaluation. For the English evaluation, we used a crowed-sourced evaluation via Amazon Mechanical Turk, with more than 15 workers from the US for each scenario.
\let\thefootnote\relax\footnote{Audio samples are available at the following URL: \\ \url{https://kallavinka8045.github.io/icassp2021/}}
\subsection{Seen speakers}
\label{ssec:seen_speakers}
These scenarios consist of evaluation sets for a single speaker with a relatively large training set for each model and for multiple speakers using a relatively small dataset.
Table 2 shows that the proposed model scores higher than most models, and the difference between the two scenarios is the smallest. This is the first result that represents the robustness of the proposed model that generates high-fidelity speech, regardless of whether the speaker's utterances were used frequently during training.
\begin{table}[t]
\centering
\caption{MOS results of each model for seen speakers.}
\label{tab:seen_speakers}
\begin{tabular}{l|cc}
\hline
\multicolumn{3}{c}{Trained with Korean utterances} \\ \hline
Model & Single speaker(K1) & Multi-speaker(K2) \\ \hline
WaveGlow & 3.42$\pm$0.08 & 3.10$\pm$0.07 \\
WaveRNN & 3.97$\pm$0.08 & 3.42$\pm$0.08 \\
FB-MelGAN & 3.25$\pm$0.09 & 2.72$\pm$0.09 \\
Ours & \textbf{4.19$\pm$0.09} & \textbf{4.05$\pm$0.08} \\ \hline
Recordings & 4.33$\pm$0.08 & 4.23$\pm$0.08 \\ \hline
\multicolumn{3}{c}{Trained with English utterances} \\ \hline
Model & Single speaker(E1) & Multi-speaker(E2) \\ \hline
WaveGlow & 3.65$\pm$0.15 & 3.27$\pm$0.19 \\
WaveRNN & \textbf{3.85$\pm$0.14} & 3.70$\pm$0.15 \\
FB-MelGAN & 3.62$\pm$0.16 & 3.37$\pm$0.17 \\
Ours & 3.81$\pm$0.15 & \textbf{3.71$\pm$0.15} \\ \hline
Recordings & 3.89$\pm$0.16 & 3.79$\pm$0.16 \\ \hline
\end{tabular}
\end{table}
\subsection{Unseen domains: speaker, emotion, language}
\label{ssec:unseen_domains}
These scenarios comprise evaluation sets of utterances from domains that were never used for training.
Table 3 shows that the results of our model are the closest to the recordings in most scenarios. Note that most models are not efficient in maintaining the performance in the emotion and language sets, compared to the score of speaker set. However, our model maintains a relatively small score difference. This result represents that our model can preserve sound quality in various unseen domains.
\begin{table}[t]
\centering
\caption{MOS results of each model for unseen domains.}
\label{tab:ood_scenarios}
\begin{tabular}{l|ccc}
\hline
\multicolumn{4}{c}{Trained with Korean utterances} \\ \hline
Model & Speaker(K3) & Emotion(K4) & Language(M1) \\ \hline
WaveGlow & 3.27$\pm$0.07 & 3.11$\pm$0.06 & 3.27$\pm$0.07 \\
WaveRNN & 3.83$\pm$0.08 & 2.46$\pm$0.09 & 2.85$\pm$0.07 \\
FB-MelGAN & 2.90$\pm$0.08 & 2.60$\pm$0.09 & 2.71$\pm$0.08 \\
Ours & \textbf{4.15$\pm$0.08} & \textbf{3.91$\pm$0.08} & \textbf{3.67$\pm$0.07} \\ \hline
Recordings & 4.32$\pm$0.08 & 4.31$\pm$0.07 & 3.99$\pm$0.07 \\ \hline
\multicolumn{4}{c}{Trained with English utterances} \\ \hline
Model & Speaker(E3) & Emotion(E4) & Language(M1) \\ \hline
WaveGlow & 3.51$\pm$0.16 & 3.13$\pm$0.18 & 3.35$\pm$0.18 \\
WaveRNN & \textbf{3.80$\pm$0.15} & 3.54$\pm$0.16 & 3.41$\pm$0.18 \\
FB-MelGAN & 3.48$\pm$0.17 & 3.11$\pm$0.16 & 3.14$\pm$0.18 \\
Ours & \textbf{3.80$\pm$0.16} & \textbf{3.76$\pm$0.16} & \textbf{3.58$\pm$0.18} \\ \hline
Recordings & 3.95$\pm$0.15 & 4.01$\pm$0.16 & 3.71$\pm$0.17 \\ \hline
\end{tabular}
\end{table}
\subsection{Multi-speaker text-to-speech}
\label{ssec:multi_tts}
To evaluate this scenario, we trained the JDI-T\cite{lim2020jdi} acoustic model with a pitch and energy predictor\cite{ren2020fastspeech, lancucki2020fastpitch} using a dataset with four speakers (containing K1 and subsets of K2). Each trained vocoder was fine-tuned by 100k steps using a pair of the ground-truth waveforms and the predicted mel-spectrograms. Note that we prepared the predicted mel-spectrograms of JDI-T by using the text, reference duration, ground-truth pitch, and energy.
Table 4 shows that Universal MelGAN achieved a real-time synthesis rate of 0.028 real time factor (RTF), with a higher MOS that outperforms other models. We used an NVIDIA V100 GPU to compare the inference speed of each model. All figures were measured without any hardware optimizations (i.e. mixed precision) or methods for accelerating the inference speed with a decrease in sound quality (i.e. batched sampling\cite{kalchbrenner2018efficient} or multi-band generation strategy\cite{yu2019durian}). This result indicates that the proposed model has the ability to synthesize high-fidelity waveforms from text in real-time.
\begin{table}[t]
\centering
\caption{MOS results of multi-speaker text-to-speech and inference speed of each vocoder.}
\label{tab:multi_TTS}
\begin{tabular}{l|c|c}
\hline
\multicolumn{3}{c}{Trained with Korean utterances} \\ \hline
Model & TTS(K1+K2) & Inference speed \\ \hline
WaveGlow & 3.36$\pm$0.06 & 0.058 RTF \\
WaveRNN & 3.06$\pm$0.10 & 10.12 RTF \\
FB-MelGAN & 3.43$\pm$0.09 & 0.003 RTF \\
Ours & \textbf{4.22$\pm$0.06} & 0.028 RTF \\ \hline
\end{tabular}
\end{table}
\section{Conclusion}
\label{sec:conclusion}
In this study, we propose Universal MelGAN, a robust neural vocoder for high-fidelity synthesis in multiple domains. We solved the over-smoothing problem that causes a metallic sound, by attaching multi-resolution spectrogram discriminators to the model. Our model is stable while generating waveforms with fine-grained spectrograms in large footprint models. The evaluation results indicate that the proposed model achieved the highest MOS in most seen and unseen domain scenarios. The result demonstrates the universality of the proposed model. For more general use of the model, we will study a lightweight model in the future and apply the multi-band strategy to reduce the complexity while preserving the sound quality.
\bibliographystyle{IEEEbib}
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Ellen Ripley, discovered floating in space, is awakened after 57 years of hypersleep near the same planet from Alien (1979), now undergoing terraforming in the early stages of colonization. Contact was lost with the colonists; Marines land and are killed off one-by-one. Our heroes barely escape as they destroy the remains of the colony with a thermonuclear device, but the alien has attached itself to their ship. Yes, it's the same plot as Alien. Best watched as the Director's Cut, as the film needs those extra 17 minutes of scenes. Won Oscars for Best Visual Effects and Best Sound Effects Editing; received nominations for 5 others including Best Actress, Best Score, and Best Set Decoration. Won Hugo for Best Dramatic Presentation.
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Q: Run a macro on a sheet, while working on a other I have a wokrbook with muliple sheets.
On my second sheet I have a little macro which works pretty good, but if want to run it, while i'm on a other sheet, excel vba shows me a bug and its not working.
This is my code:
Dim myDate As Variant, rng As Range
Dim lastRow As Integer, penultimateRow As Integer
Set rng = Range("A1:A207")
Set myDate = rng.Find(What:=Int(Date), LookIn:=xlFormulas)
Worksheets("Lernzeit-Tracking").Cells(myDate.Row, myDate.Column + 1).Value = _
Worksheets("Übersicht").Range("L20").Value
With Worksheets("Lernzeit-Tracking")
penultimateRow = .Cells(.Cells(Rows.Count, 2).End(xlUp).Row, 2).End(xlUp).Row
lastRow = .Cells(Rows.Count, 2).End(xlUp).Row
End With
If IsEmpty(Sheets("Lernzeit-Tracking").Cells(lastRow - 1, 2)) = True Then
Sheets("Lernzeit-Tracking").Range(Cells(penultimateRow + 1, 2), Cells(lastRow - 1, 2)).Value = 0
End If
The Error message says:
"Object variable not set"
and the line
Worksheets("Lernzeit-Tracking").Cells(myDate.Row, myDate.Column + 1).Value == Worksheets("Übersicht").Range("L20").Value
is marked
How can i run my macro while wokring on a other sheet?
A: You are missing an object qualifier on the following line
Set rng = Range("A1:A207")
Using Range without an object qualifier will return the range "A1:A207" on the activesheet. Instead try the below line or adjust the sheet name accordingly.
Set rng = Thisworkbook.Worksheets("Lernzeit-Tracking").Range("A1:A207")
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"redpajama_set_name": "RedPajamaStackExchange"
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Герб СР Сербии (Социалистической Республики Сербия) — один из государственных символов Социалистической Республики Сербии, автономной республики в составе СФРЮ. Утверждён был 17 января 1947 года Народной скупщиной Народной Республики Сербии (художником герба был Джордже Андреевич). Упразднён герб был в 2004 году, когда Парламент Сербии утвердил современный герб Сербии. Герб использовался на протяжении 62 лет — дольше, чем любой другой герб Сербии.
Описание герба
Герб представлял собой поле, окружённое с одной стороны пшеничными колосьями, с другой дубовыми листьями. Колосья снизу были перевязаны лентой красного цвета с цифрами 1804 и 1941, между колосьями располагался щит красного цвета с белым оцилом (без креста), над щитом располагалась красная пятиконечная звезда, а под щитом находилось солнце и колесо с шестерёнками на фоне солнца. Колосья символизировали крестьянский труд и благополучие страны, солнце символизировало возрождение Югославии, красная звезда — символ правящей партии, колесо с шестерёнками символизировало развивающуюся промышленность, а цифры 1804 и 1941 означали соответственно даты Первого сербского восстания против турок и начала войны против нацистской Германии. Щит с оцилом являлся малым гербом православной Сербии, однако коммунистическая власть убрала крест с щита в рамках секуляризации, дабы не препятствовать свободе вероисповеданий.
В конституции от 1947 года говорилось следующее:
Статья 4.Государственный герб Народной Республики Сербия представляет собой поле, окружённое с одной стороны колосьями пшеницы, а с другой стороны дубовыми листьями. Колосья пшеницы и дубовые ветви связаны лентой, на которой изображены даты сербских национальных восстаний 1804 и 1941 годов. Вверху между колосьями и листьями изображена пятиконечная звезда. Под звездой располагается колесо на фоне солнца, а чуть выше них — щит с четырьмя оцилами.
В конституции от 1963 года говорилось:
Статья 8.Герб Социалистической Республики Сербия представляет собой поле, окружённое с одной стороны колосьями пшеницы, а с другой стороны дубовыми листьями.Колосья пшеницы и дубовые ветви связаны лентой, на которой изображены даты сербского восстания 1804 года и начала Народно-освободительной войны и социалистической революции 1941 года. Вверху между колосьями и листьями изображена пятиконечная звезда. В поле под звездой располагается солнце, а на его фоне колесо с шестерёнками. Над ними располагается щит с четырьмя оцилами, расположенными правильно по углам.
В конституции от 1974 года то же самое говорилось в статье 5, а само описание герба осталось прежним. Герб оставался официальным гербом Сербии после распада Югославии вплоть до 2004 года. 31 мая 1992 года на референдуме о государственных символах большинство избирателей проголосовали за утверждение флага с красной звездой и традиционный герб Сербии в виде щита с сербским крестом без двуглавого орла. Однако результаты референдума не были признаны, поскольку в нём не приняло участие более половины избирателей (с учётом Косово). Герб СР Сербии оставался гербом Сербии вплоть до 17 августа 2004 года, пока не было принято в Скупщине Постановление о использовании герба, флага и гимна Республики Сербии — использовании герба, утверждённого ещё 16 июня 1882 года. Окончательно флаг, герб и гимн юридически были утверждены 19 мая 2009 года в новом «Законе об утверждении и использовании герба, флага и гимна Республики Сербия».
Примечания
См. также
Гербы бывших югославских республик
Герб Югославии
Герб Сербии
Сербия, Социалистическая Республика
Государственные символы Югославии
Гербы с венком из пшеничных колосьев
Гербы с изображением дубового венка
Гербы с изображением пятиконечных звёзд
Гербы с изображением солнца
Гербы, содержащие зубчатое колесо
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\section{Introduction}
Let $k=\mathbb F_q$ be a finite field of characteristic $p$ and $V\subseteq\mathbb A^{n}_k$ a geometrically irreducible affine variety of dimension $r$. To fix ideas, suppose that its $\ell$-adic cohomology groups $\mathrm H^i_c(V\otimes\bar k,\bar{\mathbb Q}_\ell)$ vanish for $i\neq r,2r$ (e.g. $V$ a smooth hypersurface with smooth section at infinity). If $\dim\mathrm H^r_c(V\otimes\bar k,\bar{\mathbb Q}_\ell)=d$, the famous Deligne-Weil bound \cite[Th\'eor\`eme 8.1]{deligne1974conjecture} gives an estimate
$$
\left|\#V(k)-q^r\right|\leq d\cdot q^\frac{r}{2}
$$
for the number of rational points of $V$. Suppose now that we have a large finite abelian group $G$ acting on $V$. Then $G$ has an induced Frobenius invariant action on $\mathrm H^r_c(V\otimes\bar k,\bar{\mathbb Q}_\ell)$, so this vector space splits as a direct sum $\bigoplus_\psi\mathrm H^r_c(V\otimes\bar k,\bar{\mathbb Q}_\ell)^\psi$, where the sum is indexed by the set of characters of $G$ and $\mathrm H^r_c(V\otimes\bar k,\bar{\mathbb Q}_\ell)^\psi$ is the subspace of $\mathrm H^r_c(V\otimes\bar k,\bar{\mathbb Q}_\ell)$ on which $G$ acts via $\psi$. In particular, we get a decomposition
$$
\#V(k)-q^r=(-1)^r\mathrm {Tr}(\mathrm{Frob}_k|\mathrm H^r_c(V\otimes\bar k,\bar{\mathbb Q}_\ell))=(-1)^r\sum_\psi\mathrm {Tr}(\mathrm{Frob}_k|\mathrm H^r_c(V\otimes\bar k,\bar{\mathbb Q}_\ell)^\psi)
$$
If the situation is generic enough, one would expect that there should be some cancellation among the terms of the sum, thus giving a significative improvement of the Deligne-Weil bound if $\#G$ (and thus the number of terms in the sum) is large.
For instance, if $k'={\mathbb F}_{q'}$ is a subfield of $k$, the Artin-Schreier curve $y^{q'}-y=f(x)$ for $f\in k[x]$ has a natural action of the additive group $k'$ (where $t$ acts by $(x,y)\mapsto (x,y+t)$. In \cite{rlwan2010} this fact was used to give an improvement of the Weil bound for the curve of order of magnitude $\sqrt{q'}$.
In the same vein, suppose that $\mathcal F$ is an $\ell$-adic sheaf on $\mathbb A^1_k$ which is invariant under translation by elements of $k'$. Equivalently, $\mathcal F$ is the pull-back by the \'etale map $x\mapsto x^{q'}-x$ of a sheaf $\mathcal G$ on $\mathbb A^1_k$. Then
$$
\sum_{t\in k}\mathrm {Tr}(\mathrm{Frob}_{k,t}|\mathcal F_{\bar t})=\sum_{t\in k}\mathrm {Tr}(\mathrm{Frob}_{k,t^{q'}-t}|\mathcal G_{\bar t^{q'}-\bar t})=q'\cdot\sum_{\mathrm {Tr}_{k/k'}(u)=0}\mathrm {Tr}(\mathrm{Frob}_{k,u}|\mathcal G_{\bar u})
$$
Similarly, if $\mathcal F$ on $\mathbb G_{m,k}$ is invariant under the group of homotheties with ratios in $k'^\star$ (so $\mathcal F$ is the pull-back of a sheaf $\mathcal G$ on $\mathbb G_{m,k}$ under the $(q'-1)$-th power map), we can write
$$
\sum_{t\in k^\star}\mathrm {Tr}(\mathrm{Frob}_{k,t}|\mathcal F_{\bar t})=(q'-1)\cdot\sum_{\mathrm{N}_{k/k'}(u)=1}\mathrm {Tr}(\mathrm{Frob}_{k,u}|\mathcal G_{\bar u})
$$
If $\mathcal F$ is pure of weight $0$ and has no geometrically constant components, the Weil bound for the sum has order $O(d\sqrt q)$, where $d=\dim\mathrm H^1_c(\mathbb A^1_{\bar k},\mathcal F)$ (resp. $\dim\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)$). On the other hand, for the right hand side we expect an estimate of the form $O(e\sqrt{q'}^{[k:k']+1})$ where $e=\dim\mathrm H^1_c(\mathbb A^1_{\bar k},\mathcal G)$. Now since $\mathcal F$ is the pull-back of $\mathcal G$ under a map of degree $\sim q'$, for general $\mathcal F$ we should have $d\sim q'e$. This gives an estimate $O(d\sqrt{q'}^{[k:k']-1})$ for the second sum, which improves the Weil estimate by a factor of $\sqrt{q'}$.
For instance, if $\psi:k\to\bar{\mathbb Q}_\ell^\star$ (respectively $\chi:k^\star\to\bar{\mathbb Q}_\ell^\star$) is an additive (resp. multiplicative) character of $k$, and $f\in k[x]$ is a polynomial of the form $g(x^{q'}-x)$ or of the form $g(x^{q'-1})$ with $g\in k[x]$ of degree $d$, the classical Weil bound for the exponential sum $\sum_{x\in k}\psi(f(x))$ (resp. $\sum_{x\in k}\chi(f(x))$) is $\cong (dq'-1)\sqrt{q}\cong dq'^{\frac{[k:k']}{2}+1}$. Writing it as a sum over a ``trace set'' or a ``norm set'' we should obtain (for generic $f$) an estimate of the form $C_dq'^{\frac{[k:k']+1}{2}}$, where $C_d$ depends only on $d$ and $[k:k']$. See the examples in sections 6 and 7 for explicit conditions on $f$ that imply this estimate.
So we reduce the Frobenius trace sum of $\mathcal F$ on $\mathbb A^1$ (or on $\mathbb G_m$) to a sum of Frobenius traces of a simpler object $\mathcal G$ but on a more complicated space, defined by non-algebraic equations of the form $\mathrm {Tr}_{k/k'}(u)=\lambda$ or $\mathrm{N}_{k/k'}(u)=\mu$. Sums of this type have been previously studied in the literature (cf. \cite{katz1993estimates}, \cite{katz1995note}, \cite{chai2004character}, \cite{li2006character}) mainly using the method of Weil descent. This method consists of identifying the set of elements of $k$ with a given trace or norm over $k'$ with the set of rational points on a $([k:k']-1)$-dimensional variety over $k'$, and thus reducing the sum to a more classical sum over the rational points of a variety.
One disadvantage of this method is that one may lose some information when identifying those two sets. As a rather crude example of this phenomenon, consider the sum of the constant $1$ over the set of elements of $k$ with norm $1$ over $k'$. This sum is obviously equal to $\frac{q-1}{q'-1}=1+q'+\cdots+q'^{n-1}$, where $n=[k:k']$. When applying Weil descent, the given set is identified with an $(n-1)$-dimensional torus over $k'$, where $n=[k:k']$. So its cohomology has dimension ${{n-1}\choose{j}}$ and weight $2j$ in degree $j+n-1$ for every $j=0,\ldots,n-1$, and we obtain an estimate $\sum_{j=0}^{n-1}{{n-1}\choose{j}}q'^j=(1+q')^{n-1}$ which is worse than the actual value $\sum_{j=0}^{n-1} q'^j $. See remark \ref{compare} for a more elaborated example of this issue.
In this article we introduce another method to systematically study these kinds of sums. For a given $\ell$-adic sheaf (or, more generally, a derived category object) $\mathcal F$ on a geometrically connected commutative algebraic group $G$ over $k$, an integer $m\geq 1$ and a point $t\in G(k_m)$ (where $k_m$ is the extension of $k$ of degree $m$ in a fixed algebraic closure $\bar k$) we define the $r$-th local \emph{norm $L$-function} of $\mathcal F$ at $t$ as
$$
L^{\mathrm{N},r}(\mathcal F,k_m,t;T):=\exp\sum_{s\geq 1}f^{\mathrm{N},r}_\mathcal F(k_{ms},t)\frac{T^s}{s}
$$
where
$$
f^{\mathrm{N},r}_\mathcal F(k_m,t):=\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}\mathrm {Tr}(\mathrm{Frob}_{k_{mr},u}|\mathcal F_{\bar u})
$$
and $\mathrm{N}_{k_{mr}/k_m}:G(k_{mr})\to G(k_m)$ is the norm map.
These functions can be used to estimate sums defined on sets given by trace and norm conditions in the same way that classical $L$-functions are used to obtain information about usual sums over the set of rational points of a variety. The main result of this article is the fact that these functions are rational:
\begin{thm}
For every object $\mathcal F\in{\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$, every $m\geq 1$ and every $t\in G(k_m)$, the $r$-th norm $L$-function $L^{\mathrm{N},r}(\mathcal F,k_m,t;T)$ is rational. If $\mathcal F$ is mixed of integral weights, all its reciprocal roots and poles are pure of integral $q^m$-weight.
\end{thm}
We give some explicit estimates in the cases where $G=\mathbb A^1_k$ or $G=\mathbb G_{m,k}$. In order to obtain good estimates for the sums, we need information on the degree and the weights of the roots and poles of these functions. We will see that, in both cases, there are special objects (extensions of Artin-Schreier sheaves in the additive case, extensions of Kummer sheaves in the multiplicative case) for which the weights reach their maximal value. For these objects there are explicit formulas for the trace and norm $L$-functions, so they can be easily controlled. For the remaining objects, there are good estimates for the weights of the reciprocal roots and poles of the $L$-functions at $t$ for all $t$ in a certain dense open subset of $G$ that depends on $r$, which can be explicitely computed in some cases. In many examples we will also be able to obtain explicit bounds for the total degree of the $L$-functions.
The additive and multiplicative cases can be studied in parallel. However, in the additive case there is a great advantage thanks to the existence of the $\ell$-adic Fourier transform. This allows to reduce the study of the trace $L$-functions to the study of the Fourier transform of the object $\mathcal F$, and more precisely of its geometric monodromy (section 6). In the multiplicative case we lack this shortcut, and instead we rely on recent work by Katz \cite{katz2010mellin} on the tensor category of perverse sheaves on $\mathbb G_{m,k}$ under convolution in order to obtain explicit results for some important examples (section 7).
The author would like to thank the referee for his careful reading of an earlier version of the article and his many useful suggestions for improvement.
\section{$\bar{\mathbb Q}_\ell$-representable functions}
Let $k=\mathbb F_q$ be a finite field of characteristic $p>0$ and $\bar k={\bar{\mathbb F}_q}$ a fixed algebraic closure. For each positive integer $m$, we denote by $k_m={\mathbb F}_{q^m}$ the unique extension of $k$ of degree $m$ inside $\bar k$. Fix a prime $\ell\neq p$ and a field isomorphism $\iota:\bar{\mathbb Q}_\ell\to{\mathbb C}$. We will use this isomorphism to identify $\bar{\mathbb Q}_\ell$ and ${\mathbb C}$ without making any further mention to it. Let $X$ be a separated scheme of finite type over $k$. We define $\mathcal C_X$ to be the set
$$
\mathcal C_X:=\{f:\coprod_{m\geq 1} X(k_m)\to \bar{\mathbb Q}_\ell\}=\prod_{m\geq 1}\{f: X(k_m)\to\bar{\mathbb Q}_\ell\}
$$
of $\bar{\mathbb Q}_\ell$-valued functions defined on the disjoint union $\coprod_{m\geq 1}X(k_m)$. It is a commutative ring with the obvious point-wise operations.
Let ${\mathcal Sh}(X,\bar{\mathbb Q}_\ell)$ be the abelian category of constructible $\bar{\mathbb Q}_\ell$-sheaves on $X$, ${\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$ the corresponding derived category and $K_0(X,\bar{\mathbb Q}_\ell)$ its Grothendieck group. That is, the free abelian group generated by the isomorphism classes of elements of ${\mathcal Sh}(X,\bar{\mathbb Q}_\ell)$ with relations $[{\mathcal F}]=[{\mathcal G}]+[{\mathcal H}]$ for every short exact sequence $0\to{\mathcal G}\to{\mathcal F}\to{\mathcal H}\to 0$. It is also the Grothendieck group of ${\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$, and for every $K\in {\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$ we have $[K]=\sum_i(-1)^i[{\mathcal H}^i(K)]$.
From now on we will only consider sheaves and derived category objects which are mixed of integral $q$-weights (either with respect to the isomorphism $\iota$, or with respect to \emph{every} isomorphism $\bar{\mathbb Q}_\ell\to{\mathbb C}$). For every $\mathcal F\in{\mathcal Sh}(X,\QQ)$ (or, more generally, in ${\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$) we define an element $f_\mathcal F$ of $\mathcal C_X$ in the following way (cf. \cite[1.1]{laumon1987transformation}): for every $m\geq 1$ and every $t\in X(k_m)$, $f_\mathcal F(k_m,t):=\mathrm {Tr}(\mathrm{Frob}_{k_m,t}|\mathcal F_{\bar t})$ is the trace of the action of a geometric Frobenius element at $t$ on the stalk of $\mathcal F$ at a geometric point $\bar t$ over $t$. Given an exact sequence $0\to{\mathcal G}\to{\mathcal F}\to{\mathcal H}\to 0$ in ${\mathcal Sh}(X,\QQ)$ it is clear that $f_\mathcal F=f_\mathcal G+f_{\mathcal H}$ (since taking stalks at a given geometric point is an exact functor), therefore the application $\mathcal F\mapsto f_\mathcal F$ extends to a homomorphism of abelian groups $\Phi:K_0(X,\bar{\mathbb Q}_\ell)\to\mathcal C_X$, which is actually a homomorphism of rings if we endow $K_0(X,\bar{\mathbb Q}_\ell)$ with the multiplication defined by $[\mathcal F]\times[\mathcal G]=[\mathcal F\otimes\mathcal G]$ for sheaves $\mathcal F$, $\mathcal G$ and extended by linearity. The homomorphism $\Phi$ is injective \cite[Th\'eor\`eme 1.1.2]{laumon1987transformation}.
\begin{defn}
A function $f\in\mathcal C_X$ is called \emph{$\bar{\mathbb Q}_\ell$-representable} if it is in the image of $\Phi$, that is, if there exists some (necessarily unique) $F\in K_0(X,\bar{\mathbb Q}_\ell)$ such that $f=f_F:=\Phi(F)$. In that case $f$ is said to be \emph{represented by} $F$. The set of all such functions is denoted by $\mathcal C_{X,rep}$.
\end{defn}
Thus $\mathcal C_{X,rep}$ is a subring of $\mathcal C_X$ isomorphic to $K_0(X,\bar{\mathbb Q}_\ell)$. The following results are easy consequences of the definitions and Grothendieck's trace formula:
\begin{prop}\cite[1.1.1.4]{laumon1987transformation}
For every $k$-morphism $\phi:X\to Y$ of separated schemes of finite type over $k$ and every $f\in\mathcal C_{Y,rep}$ the function $\phi^\star f$ given by $\phi^\star f(k_m,t)=f(k_m,\phi(t))$ is in $\mathcal C_{X,rep}$.
\end{prop}
\begin{prop}\cite[1.1.1.3]{laumon1987transformation}
For every $k$-morphism $\phi:X\to Y$ of separated schemes of finite type over $k$ and every $f\in\mathcal C_{X,rep}$ the function $\phi_! f$ given by $\phi_! f(k_m,t)=\sum_{u\in X(k_m),\phi(u)=t}f(k_m,u)$ is in $\mathcal C_{Y,rep}$.
\end{prop}
\begin{defn}
For every $\ell$-adic unit $\alpha\in\bar{\mathbb Q}_\ell$ of integral $q$-weight, the \emph{constant function} $\kappa_\alpha\in\mathcal C_X$ defined by $\alpha$ is given by $(k_m,t)\mapsto \alpha^m$.
\end{defn}
The constant function $\kappa_\alpha$ is represented by the geometrically constant sheaf $\alpha^{\mathrm{deg}}$, that is, the pull-back of the character $\mathrm{Gal}(\bar k/k)\to\bar{\mathbb Q}_\ell^\star$ mapping the geometric Frobenius element to $\alpha$. The subset of constant functions is a multiplicative subgroup of $\mathcal C_{X,rep}$, but it is not closed under addition.
\begin{defn}
Let $f\in\mathcal C_X$, $m\geq 1$ and $t\in X(k_m)$. The \emph{local $L$-function of $f$ at $t$} is defined as
$$
L(f,k_m,t;T):=\exp\sum_{s\geq 1} f(k_{ms},t)\frac{T^s}{s}\in 1+T\bar{\mathbb Q}_\ell[[T]]
$$
\end{defn}
It is clear that for every $f$ and $g$ in $\mathcal C_X$ we have
$$
L(f+g,k_m,t;T)=L(f,k_m,t;T)\cdot L(g,k_m,t;T),
$$
for every $\ell$-adic unit $\alpha$ of integral $q$-weight
$$
L(\kappa_\alpha f,k_m,t;T)=L(f,k_m,t;\alpha^m T)
$$
and, for every $k$-morphism $\phi:X\to Y$
$$
L(\phi^\star f,k_m,t;T)=L(f,k_m,\phi(t);T).
$$
\begin{prop}
For every $f\in\mathcal C_{X,rep}$, every $m\geq 1$ and every $t\in X(k_m)$ the $L$-function $L(f,k_m,t;T)$ is rational and all its reciprocal roots and poles have integral $q^m$-weight.
\end{prop}
\begin{proof}
By additivity, it suffices to prove if when $f$ is represented by a sheaf $\mathcal F\in{\mathcal Sh}(X,\QQ)$. But in that case it is well known that
$$
L(f,k_m,t;T)=\exp\sum_{s\geq 1}\mathrm {Tr}(\mathrm{Frob}_{k_m,t}^s|\mathcal F_{\bar t})\frac{T^s}{s}=\det(1-T\cdot\mathrm{Frob}_{k_m,t}|\mathcal F_{\bar t})^{-1}.
$$
\end{proof}
\section{The convolution Adams operation}
Let $S=\mathrm{Spec\:} k$ be the spectrum of a field (or, more generally, a base scheme such that the derived category of $\ell$-adic sheaves is well defined on ${\mathcal Sch}/S$, e.g. a regular scheme of dimension $\leq 1$ \cite[1.1.2]{deligne1980conjecture}). Let $X$ be a separated scheme of finite type over $S$ and $H$ a finite group (regarded as acting trivially on $X$), and consider the category $\mathcal Sh(X,\bar{\mathbb Q}_\ell)_H$ of $\bar{\mathbb Q}_\ell$-sheaves on $X$ endowed with an action of $H$ and its derived category ${\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)_H$.
Given a representation $\rho:H\to\mathrm{GL}(V)$ (where $V$ is a finite dimensional vector space over $\bar{\mathbb Q}_\ell$) we get a functor $\mathcal Sh(X,\bar{\mathbb Q}_\ell)_H\to\mathcal Sh(X,\bar{\mathbb Q}_\ell)$ given by $\mathcal F\to\mathcal F^\rho:={\mathcal Hom}_H(V,\mathcal F)$, where $V$ is regarded as a constant sheaf on $X$ with an $H$-action given by $\rho$. If ${\mathbf 1}$ is the trivial representation then $\mathcal F^{\mathbf 1}=\mathcal F^H$ is the $H$-invariant part, and in general $\mathcal F^\rho={\mathcal Hom}(V,\mathcal F)^H$. Since $\bar{\mathbb Q}_\ell$ has characteristic zero, the functor $\mathcal F\to\mathcal F^\rho$ is exact, and it commutes with passage to fibres: for every geometric point $\bar x\in X(\bar k)$, we have $(\mathcal F^\rho)_{\bar x}=\mathrm{Hom}_H(V,\mathcal F_{\bar x})$. In particular, it extends to the derived category and we get a functor ${\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)_H\to{\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$, $K\mapsto K^\rho$ such that ${\mathcal H}^i(K^\rho)={\mathcal H}^i(K)^\rho$ for every $i\in{\mathbb Z}$. If $\rho':H\to\mathrm{GL}(V')$ is another finite dimensional $\bar{\mathbb Q}_\ell$-representation, it is clear that $K^{\rho\oplus\rho'}=K^\rho\oplus K^{\rho'}$.
\begin{lem}\label{directimage}
Let $f:X\to Y$ be an $S$-morphism of separated schemes of finite type over $S$ and $K\in{\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$ an object with an $H$-action. Then for every finite dimensional representation $\rho$ of $H$ $(\mathrm R f_\star K)^\rho=\mathrm R f_\star (K^\rho)$ and $(\mathrm R f_! K)^\rho=\mathrm R f_! (K^\rho)$.
\end{lem}
\begin{proof}
It suffices to prove it for $\mathrm R f_\star$, since clearly $j_! (K^\rho)=(j_! K)^\rho$ for an open immersion $j$. Let $\mathcal F$ be a $\bar{\mathbb Q}_\ell$-sheaf on $X$ with an $H$-action. Since $f_\star$ is left exact, we have $(f_\star\mathcal F)^H=f_\star(\mathcal F^H)$. Therefore for every representation $\rho:H\to\mathrm{GL}(V)$ of $H$
$$
(f_\star\mathcal F)^\rho={\mathcal Hom}_H(V,f_\star\mathcal F)={\mathcal Hom}(V,f_\star\mathcal F)^H=(f_\star{\mathcal Hom}(f^\star V,\mathcal F))^H=
$$
$$
=f_\star({\mathcal Hom}(V,\mathcal F)^H)=f_\star({\mathcal Hom}_H(V,\mathcal F))=f_\star(\mathcal F^\rho).
$$
By the exactness of $(-)^\rho$ we deduce that $(\mathrm R f_\star\mathcal F)^\rho=\mathrm R f_\star(\mathcal F^\rho)$ for every sheaf $\mathcal F$, and then also for every object $K\in{\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$.
\end{proof}
We will be mainly interested in the following situation: $K\in{\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$ is any object, $r\geq 1$ is an integer, and the symmetric group in $r$ letters ${\mathfrak S}_r$ acts on $K^{\otimes r}$ via permutation of the factors. Then for every representation $\rho:{\mathfrak S}_r\to\mathrm{GL}(V)$ we get an object $\mathrm R(\rho)K:=(K^{\otimes r})^\rho\in{\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$. We write $\mathrm{Sym}^r K$ (respectively $\wedge^r K$) for $\mathrm R(\rho)K$ if $\rho$ is the trivial representation (resp. the sign character).
If $G$ is a geometrically connected commutative group scheme of finite type over $S$ we will also use the following variant. Recall that, for any two objects $K,L\in{\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$, their \emph{$!$-convolution} (which we will simply call convolution) is the object
$K\ast L :=\mathrm R\mu_!(K\boxtimes L)\in{\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$, where $\mu:G\times G\to G$ is the multiplication map. It is an associative and commutative triangulated bifunctor \cite[2.5]{katz1996rls}.
For any $r\geq 1$, the multiplication map $G^r\to G$ factors through $\mathrm{Sym}^r G$, so $K^{\ast r}:=K\ast\cdots\ast K$ ($r$ factors) is endowed with a natural action of ${\mathfrak S}_r$, induced by its action on $\pi_!K^{\boxtimes r}$ above $\mathrm{Sym}^rG$ by permutation of the factors (where $\pi:G^r\to\mathrm{Sym}^r G$ is the natural projection). We denote $\mathrm R(\ast\rho)K:=(K^{\ast r})^\rho\in{\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$, and we write $\mathrm{Sym}^{\ast r} K$ (respectively $\wedge^{\ast r} K$) for $\mathrm R(\ast\rho)K$ if $\rho$ is the trivial representation (resp. the sign character).
The following result generalizes \cite[Lemme 1.3]{deligne569fonction} (for $A$ a field of characteristic $0$):
\begin{prop}\label{mainprop}
Let $a:G\to S$ be the structural map. Then for every $K\in{\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$ and every finite dimensional $\bar{\mathbb Q}_\ell$-representation $\rho$ of ${\mathfrak S}_r$ we have a quasi-isomorphism
$$
\mathrm R a_!(\mathrm R(\ast\rho)K)\cong\mathrm R(\rho)(\mathrm R a_!K).
$$
In particular, if $S=\mathrm{Spec\:} k$ is the spectrum of a separably closed field we have $$\mathrm R\Gamma_c(G,\mathrm R(\ast\rho)K)\cong\mathrm R(\rho)(\mathrm R\Gamma_c(G,K)).$$
\end{prop}
\begin{proof}
By lemma \ref{directimage} we have $\mathrm R a_!(\mathrm R(\ast\rho)K)=\mathrm R a_!((K^{\ast r})^\rho)\cong(\mathrm R a_!(K^{\ast r}))^\rho$. If $\mu:G^r\to G$ denotes the multiplication map, $K^{\ast r}=\mathrm R\mu_! K^{\boxtimes r}$, so
$$
(\mathrm R a_!(K^{\ast r}))^\rho\cong(\mathrm R a_!(\mathrm R\mu_!K^{\boxtimes r}))^\rho\cong(\mathrm R(a\mu)_! K^{\boxtimes r})^\rho.
$$
Now by K\"unneth, there is a ${\mathfrak S}_r$-equivariant quasi-isomorphism $\mathrm R(a\mu)_! K^{\boxtimes r}\cong(\mathrm R a_! K)^{\otimes r}$ (where ${\mathfrak S}_r$ acts on the right by permutation of the factors), so
$$
(\mathrm R(a\mu)_! K^{\boxtimes r})^\rho\cong ((\mathrm R a_! K)^{\otimes r})^\rho=\mathrm R(\rho)(\mathrm R a_! K).
$$
\end{proof}
We also have the following shift formulas:
\begin{prop}\label{shift}
Let $\sigma$ be the sign character and $\rho$ any finite dimensional representation of ${\mathfrak S}_r$. For every $K\in{\mathcal D}^b_c(X,\bar{\mathbb Q}_\ell)$ we have
$$
\mathrm R(\rho)(K[1])\cong(\mathrm R(\rho\otimes\sigma)K)[r]
$$
and for every $K\in{\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$
$$
\mathrm R(\ast\rho)(K[1])\cong(\mathrm R(\ast(\rho\otimes\sigma))K)[r].
$$
\end{prop}
\begin{proof}
We will prove the first formula, the second one is similar. With the obvious notations, we have
$$
\mathrm R(\rho)(K[1])\cong{\mathcal Hom}_{{\mathfrak S}_r}(\rho,K[1]^{\otimes r})
$$
and
$$
(\mathrm R(\rho\otimes\sigma)K)[r]\cong{\mathcal Hom}_{{\mathfrak S}_r}(\rho\otimes\sigma,K^{\otimes r}[r])
$$
so it suffices to show that the action of ${\mathfrak S}_r$ on $K^{\otimes r}[r]$ is the same as its action on $K[1]^{\otimes r}(\cong K^{\otimes r}[r])$ twisted by $\sigma$.
Suppose that $r=2$, and let $\tau\in{\mathfrak S}_2$ be the transposition. There is a natural isomorphism $\phi:K[1]\otimes K[1]\cong (K\otimes K)[2]$ given by $a_i\otimes b_j\mapsto (-1)^{i}a_i\otimes b_j$ for $a_i\in K^i=K[1]^{i-1}$ and $b_j\in K^j=K[1]^{j-1}$. On $K[1]\otimes K[1]$, the transposition $\tau$ acts by $a_i\otimes b_j\mapsto (-1)^{(i-1)(j-1)}b_j\otimes a_i$ (since $a_i,b_j$ live on degrees $i-1$ and $j-1$ respectively). On $(K\otimes K)[2]$ it acts by $a_i\otimes b_j\mapsto (-1)^{ij}b_j\otimes a_i$. So for $a_i\otimes b_j\in K[1]^{i-1}\otimes K[1]^{j-1}$ we have $\phi\circ\tau(a_i\otimes b_j)=\phi((-1)^{(i-1)(j-1)}b_j\otimes a_i)=(-1)^{ij-i+1}b_j\otimes a_i$ and $\tau\circ\phi(a_i\otimes b_j)=\tau((-1)^ia_i\otimes b_j)=(-1)^{i+ij}b_j\otimes a_i$, so $\phi\circ\tau(a_i\otimes b_j)=-\tau\circ\phi(a_i\otimes b_j)$. In other words, the action of $\tau$ on $K[1]\otimes K[1]$ is the negative of its action on $(K\otimes K)[2]$ via the isomorphism $\phi$.
Now let $r>2$, and let $\tau\in{\mathfrak S}_r$ be any transposition. Without loss of generality, we may assume that $\tau=(1\; 2)$. Then we have an isomorphism $\phi_r:K[1]^{\otimes r}=K[1]^{\otimes 2}\otimes K[1]^{\otimes(r-2)}\cong K^{\otimes 2}\otimes K^{\otimes (r-2)}[r]=K^{\otimes r}[r]$, where $\phi_r=\phi\otimes\phi_{r-2}$ inductively. Since $\tau$ acts trivially on $K[1]^{\otimes (r-2)}$ and $K^{\otimes (r-2)}$, by the $r=2$ case the actions of $\tau$ on $K[1]^{\otimes r}$ and $K^{\otimes r}[r]$ differ by sign. We conclude that the actions of ${\mathfrak S}_r$ on $K[1]^{\otimes r}$ and $K^{\otimes r}[r]$ are twists of each other by the sign character.
\end{proof}
If $\rho':{\mathfrak S}_r\to V'$ is another finite dimensional $\bar{\mathbb Q}_\ell$-representation of ${\mathfrak S}_r$, it is clear that $\mathrm R(\rho\oplus\rho')K=\mathrm R(\rho)K\oplus\mathrm R(\rho')K$ and $\mathrm R(\ast(\rho\oplus\rho'))K=\mathrm R(\ast\rho)K\oplus\mathrm R(\ast\rho')K$. Therefore it makes sense to define $\mathrm R(\tau)K$ (respectively $\mathrm R(\ast\tau)K$) as an element of the Grothendieck group $K_0(X,\bar{\mathbb Q}_\ell)$ (resp. $K_0(G,\bar{\mathbb Q}_\ell)$) for any virtual $\bar{\mathbb Q}_\ell$-representation $\tau$ of ${\mathfrak S}_r$: if $\tau=\rho-\rho'$, we set $\mathrm R(\tau)K:=[\mathrm R(\rho)K]-[\mathrm R(\rho')K]$ (resp. $\mathrm R(\ast\tau)K:=[\mathrm R(\ast\rho)K]-[\mathrm R(\ast\rho')K]$). Proposition \ref{mainprop} implies that
\begin{equation}\label{adamscommute}
\mathrm R a_!(\mathrm R(\ast\tau)K)=\mathrm R(\tau)(\mathrm R a_!K)
\end{equation}
in $K_0(S,\bar{\mathbb Q}_\ell)$ for $a:G\to S$ the structural map.
Let $\rho:{\mathfrak S}_r\to \mathrm{GL}(L)$ be the standard representation, where $L\subset\bar{\mathbb Q}_\ell^r$ is the hyperplane defined by $\sum x_i=0$, and let $\tau_r$ be the virtual representation $\sum_{i=0}^{r-1} (-1)^i\wedge^i\rho$.
\begin{prop}\label{young}
Let $\mathcal F\in{\mathcal Sh}(X,\QQ)$. Then
$$
\mathrm R(\tau_r)\mathcal F=\mathcal F^{[r]}:=\sum_{i=1}^r(-1)^{i-1} i [\mathrm{Sym}^{r-i}\mathcal F\otimes\wedge^i\mathcal F]\in K_0(X,\bar{\mathbb Q}_\ell)
$$
is the $r$-th Adams power of $\mathcal F$.
\end{prop}
\begin{proof}
For every $i=1,\ldots,r$ we have
$$
\mathrm{Sym}^{r-i}\mathcal F\otimes\wedge^ i\mathcal F ={\mathcal Hom}_{{\mathfrak S}_{r-i}\times{\mathfrak S}_i}({\mathbf 1}\times\sigma,\mathcal F^{\otimes r})
$$
where $\sigma$ is the sign character of ${\mathfrak S}_i$. By Frobenius reciprocity, this is the same as $\mathrm R(\eta)\mathcal F$, where $\eta=\mathrm{Ind}^{{\mathfrak S}_r}_{{\mathfrak S}_{r-i}\times{\mathfrak S}_i}({\mathbf 1}\times\sigma)$.
By Pieri's formula (cf. \cite[4.41, A.7]{fulton1996representation}) for $i<r$ $\eta$ is the sum of two irreducible representations $\eta_i$ and $\eta_{i+1}$ with inverted-L shaped Young diagrams
$$
\yng(5,1,1)
$$
of vertical lengths $i$ and $i+1$ respectively, that is, the $(i-1)$-th and $i$-th exterior powers of the standard representation of ${\mathfrak S}_r$ (cf. \cite[4.6]{fulton1996representation}). Therefore
$$
\sum_{i=1}^r(-1)^{i-1} i[\mathrm{Sym}^{r-i}\mathcal F\otimes\wedge^ i\mathcal F]=\mathrm R\left(\left(\sum_{i=1}^{r-1}(-1)^{i-1} i (\wedge^{i-1}\rho+\wedge^{i}\rho)\right)+(-1)^{r-1} r\sigma\right)\mathcal F=
$$
$$
=\mathrm R\left(\sum_{i=0}^{r-1} (-1)^{i} \wedge^i\rho\right)\mathcal F=\mathrm R(\tau_r)\mathcal F.
$$
\end{proof}
Since the Adams power is additive, we deduce
\begin{cor}
The map $\mathcal Sh(X,\bar{\mathbb Q}_\ell)\to K_0(X,\bar{\mathbb Q}_\ell)$ given by $\mathcal F\mapsto \mathrm R(\tau_r)\mathcal F$ extends to a homomorphism of abelian groups $\mathrm R(\tau_r):K_0(X,\bar{\mathbb Q}_\ell)\to K_0(X,\bar{\mathbb Q}_\ell)$.
\end{cor}
\begin{rem}\label{rank}\emph{
The sum $\sum_{i=0}^{r-1}(-1)^i\mathrm R(\wedge^i\rho)\mathcal F$ gives an ``optimal'' expression for the Adams power, in the sense that there is no further cancellation among different sign terms. If $\mathcal F$ has rank $n$ then for every $i=0,\ldots,r-1$ the rank of $\mathrm R(\wedge^i\rho)\mathcal F$ is the dimension of the Weyl module corresponding to the partition $(r-i,1,\cdots,1)$ (with $i$ $1$'s), that is,
$$
{{n+r-i-1}\choose{r}}{{r-1}\choose{i}}
$$
by \cite[Theorem 6.3, 6.4]{fulton1996representation}.}
\end{rem}
\begin{defn}
Let $K\in {\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$. The $r$-th \emph{convolution Adams power} of $K$ is the object $K^{[\ast r]}:=\mathrm R(\ast\tau_r)K\in K_0(G,\bar{\mathbb Q}_\ell)$.
\end{defn}
\begin{prop}
If $K\to M\to L\to K[1]$ is a distinguished triangle in ${\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$ then $M^{[\ast r]}=K^{[\ast r]}+L^{[\ast r]}$ for every $r\geq 1$. In particular, the $r$-th convolution Adams power extends to a homomorphism of abelian groups $K_0(G,\bar{\mathbb Q}_\ell)\to K_0(G,\bar{\mathbb Q}_\ell)$.
\end{prop}
\begin{proof}
For every $r\geq 1$ $M^{\ast r}=\mathrm R\mu_!(M^{\boxtimes r})$ has a filtration with quotients $\mathrm R\mu_! P_k$ for $k=0,1,\ldots,r$, where
$$
P_k=\bigoplus_{J\subseteq\{1,\ldots,r\},|J|=k}N_{J,1}\boxtimes\cdots\boxtimes N_{J,r}
$$
and $N_{J,j}=K$ (respectively $N_{J,j}=L$) if $j\in J$ (resp. $j\notin J$). The action of ${\mathfrak S}_r$ preserves this filtration, and acts transitively on the set $\{\mathrm R\mu_!(N_{J,1}\boxtimes\cdots\boxtimes N_{J,r})|J\subseteq\{1,\ldots,r\},|J|=k\}$ for each $k$, with stabilizer ${\mathfrak S}_k\times{\mathfrak S}_{r-k}$ for $\mathrm R\mu_!(K^{\boxtimes k}\boxtimes L^{\boxtimes(r-k)})$. In other words, the action of ${\mathfrak S}_r$ on $\mathrm R\mu_!P_k$ is the action induced by that of ${\mathfrak S}_k\times{\mathfrak S}_{r-k}$ on $\mathrm R\mu_!(K^{\boxtimes k}\boxtimes L^{\boxtimes(r-k)})=K^{\ast k}\ast L^{\ast(r-k)}$ by permutation of the first $k$ and the last $r-k$ factors. Therefore
$$
[\mathrm R(\ast\rho)M]=[{\mathcal Hom}_{{\mathfrak S}_r}(\rho,M^{\ast r})]=\sum_{k=0}^r[{\mathcal Hom}_{{\mathfrak S}_r}(\rho,\mathrm{Ind}^{{\mathfrak S}_r}_{{\mathfrak S}_k\times{\mathfrak S}_{r-k}}K^{\ast k}\ast L^{\ast(r-k)})]=
$$
$$
=\sum_{k=0}^r[{\mathcal Hom}_{{\mathfrak S}_k\times{\mathfrak S}_{r-k}}(\rho_{|{\mathfrak S}_k\times{\mathfrak S}_{r-k}},K^{\ast k}\ast L^{\ast(r-k)})]
$$
for any finite dimensional $\bar{\mathbb Q}_\ell$ representation $\rho$ of ${\mathfrak S}_r$. In particular, if $\rho$ is the standard representation and $1\leq k\leq r-1$, the Littlewood-Richardson formula gives (cf. \cite[4.43, A.8]{fulton1996representation}):
$$
(\wedge^i\rho)_{|{\mathfrak S}_k\times{\mathfrak S}_{r-k}}=\bigoplus_{j+l=i}(\wedge^j\rho_1)\times(\wedge^l\rho_2)\oplus\bigoplus_{j+l=i-1}(\wedge^j\rho_1)\times(\wedge^l\rho_2)
$$
where $\rho_1$ and $\rho_2$ are the standard representations of ${\mathfrak S}_k$ and ${\mathfrak S}_{r-k}$ respectively and $j\leq k-1$, $l\leq r-k-1$ in the sums, so we obtain
$$
M^{[\ast r]}=\sum_{i=0}^{r-1}(-1)^i[\mathrm R(\ast\wedge^i\rho)M]=
$$
$$
=\sum_{i=0}^{r-1}(-1)^i\sum_{k=0}^r[{\mathcal Hom}_{{\mathfrak S}_k\times{\mathfrak S}_{r-k}}((\wedge^i\rho)_{|{\mathfrak S}_k\times{\mathfrak S}_{r-k}},K^{\ast k}\ast L^{\ast(r-k)})]=
$$
$$
=\sum_{i=0}^{r-1}(-1)^i[\mathrm R(\ast\wedge^i\rho)K]+\sum_{i=0}^{r-1}(-1)^i[\mathrm R(\ast\wedge^i\rho)L]+
$$
$$
+\sum_{k=1}^{r-1}\sum_{i=0}^{r-1}(-1)^i\left(\sum_{j+l=i}[{\mathcal Hom}_{{\mathfrak S}_k\times{\mathfrak S}_{r-k}}((\wedge^j\rho_1)\times(\wedge^l\rho_2),K^{\ast k}\ast L^{\ast(r-k)})]+\right.
$$
$$
\left.+\sum_{j+l=i-1}[{\mathcal Hom}_{{\mathfrak S}_k\times{\mathfrak S}_{r-k}}((\wedge^j\rho_1)\times(\wedge^l\rho_2),K^{\ast k}\ast L^{\ast(r-k)})]\right).
$$
The last sum clearly vanishes, so we conclude that
$$
M^{[\ast r]}=\sum_{i=0}^{r-1}(-1)^i[\mathrm R(\ast\wedge^i\rho)K]+\sum_{i=0}^{r-1}(-1)^i[\mathrm R(\ast\wedge^i\rho)L]=K^{[\ast r]}+L^{[\ast r]}.
$$
\end{proof}
\section{Norm $L$-functions}
We go back to the case where $k=\mathbb F_q$ is a finite field. Let $G$ be a geometrically connected commutative group scheme of finite type over $k$. For every positive integers $m,r$ there is a norm map $\mathrm{N}_{k_{mr}/k_m}:G(k_{mr})\to G(k_m)$ given by
$$
\mathrm{N}_{k_{mr}/k_m}(u)=\prod_{\sigma\in\mathrm {Gal}(k_{mr}/k_m)}\sigma(u).
$$
\begin{defn}
Let $f\in\mathcal C_G$ and $r\geq 1$ an integer. The \emph{$r$-th norm power} of $f$ is the function $f^{\mathrm{N},r}\in\mathcal C_G$ given by $$f^{\mathrm{N},r}(k_m,t)=\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}f(k_{mr},u)$$
\end{defn}
The following properties are immediate consequences of the definitions:
\begin{prop}\label{easy}
Let $f,g\in\mathcal C_G$, $\alpha\in\bar{\mathbb Q}_\ell$ an $\ell$-adic unit of integral $q$-weight and $r\geq 1$ an integer. Then
\begin{enumerate}
\item $(f+g)^{\mathrm{N},r}=f^{\mathrm{N},r}+g^{\mathrm{N},r}$
\item $(\kappa_\alpha\cdot f)^{\mathrm{N},r}=\kappa_{\alpha^r}\cdot f^{\mathrm{N},r}$
\end{enumerate}
\end{prop}
The goal of this section is to show that $\mathcal C_{G,rep}$ is invariant under these operations. More precisely, the Frobenius trace function of the $r$-th convolution Adams power of $K$ is the $r$-th norm power of the Frobenius trace function of $K$:
\begin{thm}
For every $K\in K_0(G,\bar{\mathbb Q}_\ell)$ and every $r\geq 1$ we have
$$
\Phi(K^{[\ast r]})=\Phi(K)^{\mathrm{N},r}.
$$
\end{thm}
Let $\chi:G(k)\to \bar{\mathbb Q}_\ell^\star$ be a character. By \cite[1.4-1.8]{deligne569application} there is a rank $1$ smooth $\bar{\mathbb Q}_\ell$-sheaf ${\mathcal L}_\chi$ on $G$ such that, for every $m\geq 1$ and every $t\in G(k_m)$,
$$
\mathrm {Tr}(\mathrm{Frob}_{k_m,t}|{\mathcal L}_{\chi,\bar t})=\chi(\mathrm{N}_{k_m/k}(t)).
$$
Tensoring with ${\mathcal L}_\chi$ is an autoequivalence of the triangulated category ${\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$. By \cite[8.1.10]{katz1990esa} there is a quasi-isomorphism $K^{\ast r}\otimes{\mathcal L}_\chi\cong(K\otimes{\mathcal L}_\chi)^{\ast r}$ for every $r\geq 1$, which is compatible with the natural ${\mathfrak S}_r$ actions. In particular, for every (virtual) finite dimensional $\bar{\mathbb Q}_\ell$-representation $\rho$ of ${\mathfrak S}_r$ we have
$$
(\mathrm R(\ast\rho)K)\otimes{\mathcal L}_\chi\cong\mathrm R(\ast\rho)(K\otimes{\mathcal L}_\chi).
$$
Taking $\rho=\tau_r$, we get
\begin{equation}\label{tensorchi}
K^{[\ast r]}\otimes{\mathcal L}_\chi=(K\otimes{\mathcal L}_\chi)^{[\ast r]}
\end{equation}
in $K_0(G,\bar{\mathbb Q}_\ell)$ for every $r\geq 1$.
\begin{proof}[Proof of theorem 4.3]
We have to show that, for every $m\geq 1$ and every $t\in G(k_m)$,
$$
\Phi(K^{[\ast r]})(k_m,t)=\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}\Phi(K)(k_{mr},u).
$$
Since the operation $K\mapsto K^{[\ast r]}$ commutes with extending scalars to a finite extension of $k$, we may assume without loss of generality that $m=1$. The (ordinary finite group) Fourier transform gives a bijection between the set of $\bar{\mathbb Q}_\ell$-valued maps defined on $G(k)$ and the set of $\bar{\mathbb Q}_\ell$-valued maps defined on the set of characters of $G(k)$, so the previous equality for every $t\in G(k)$ is equivalent to the equality
$$
\sum_{t\in G(k)}\chi(t)\Phi(K^{[\ast r]})(k,t)=\sum_{t\in G(k)}\chi(t)\sum_{\mathrm{N}_{k_{r}/k}(u)=t}\Phi(K)(k_r,u).
$$
for every character $\chi:G(k)\to\bar{\mathbb Q}_\ell^\star$. The left hand side is, by the Grothendieck-Lefschetz trace formula,
$$
\sum_{t\in G(k)}\Phi({\mathcal L}_\chi)(k,t)\Phi(K^{[\ast r]})(k,t)=\sum_{t\in G(k)}\Phi(K^{[\ast r]}\otimes{\mathcal L}_\chi)(k,t)=
$$
$$
=\sum_{t\in G(k)}\Phi((K\otimes{\mathcal L}_\chi)^{[\ast r]})(k,t)=\mathrm {Tr}(\mathrm{Frob}_k|\mathrm R\Gamma_c(G\otimes\bar k,(K\otimes{\mathcal L}_\chi)^{[\ast r]})).
$$
The right hand side is, again by the trace formula,
$$
\sum_{u\in G(k_r)}\chi(\mathrm{N}_{k_r/k}(u))\Phi(K)(k_r,u)=\sum_{u\in G(k_r)}\Phi(K\otimes{\mathcal L}_\chi)(k_r,u)=
$$
$$
=\mathrm {Tr}(\mathrm{Frob}_{k_r}|\mathrm R\Gamma_c(G\otimes\bar k,K\otimes{\mathcal L}_\chi))=\mathrm {Tr}(\mathrm{Frob}_k^r|\mathrm R\Gamma_c(G\otimes\bar k,K\otimes{\mathcal L}_\chi))=
$$
$$
=\mathrm {Tr}(\mathrm{Frob}_k|\mathrm R\Gamma_c(G\otimes\bar k,K\otimes{\mathcal L}_\chi)^{[r]})
$$
since $\mathrm {Tr}(\phi|V^{[r]})=\mathrm {Tr}(\phi^r|V)$ for any endomorphism $\phi$ of a vector space $V$ (cf. \cite[Theorem 1.1]{fu2004moment}). The result follows then from equation (\ref{adamscommute}) applied to $\tau=\tau_r$, which tells us that the virtual $\mathrm{Frob}_k$-modules $\mathrm R\Gamma_c(G\otimes\bar k,(K\otimes{\mathcal L}_\chi)^{[\ast r]})$ and $\mathrm R\Gamma_c(G\otimes\bar k,K\otimes{\mathcal L}_\chi)^{[r]}$ are isomorphic.
\end{proof}
\begin{cor}\label{maincor}
For every $K\in K_0(G,\bar{\mathbb Q}_\ell)$, every $r\geq 1$ and every $t\in G(k_r)$ the $r$-th norm $L$-function of $K$ at $t$ $L^{\mathrm{N},r}(K,k_m,t;T)$ is rational and all its reciprocal roots and poles have integral $q^m$-weight.
\end{cor}
Using proposition \ref{easy} and the injectivity of $\Phi$ we deduce
\begin{cor}\label{twist}
For every $\ell$-adic unit $\alpha$ of integral $q$-weight we have
$$
(\alpha^{deg}\otimes K)^{[\ast r]}=\alpha^{r\cdot deg}\otimes K^{[\ast r]}.
$$
\end{cor}
We conclude this section with a useful formula relating the usual Adams powers and the convolution Adams powers in the case where $G$ is the additive group $\mathbb A^n_k$. Fix a non-trivial character $\psi:k\to\bar{\mathbb Q}_\ell^\star$. Recall that the \emph{Fourier transform} with respect to $\psi$ is the functor ${\mathcal D}^b_c(\mathbb A^n_k,\bar{\mathbb Q}_\ell)\to{\mathcal D}^b_c(\mathbb A^n_k,\bar{\mathbb Q}_\ell)$ given by (cf. \cite{katz1985transformation}):
$$
\mathrm{FT}_\psi(K)=\mathrm R\pi_{2!}(\pi_1^\star K\otimes\mu^\star {\mathcal L}_\psi)[n]
$$
where $\pi_1,\pi_2:\mathbb A^n_k\times\mathbb A^n_k\to\mathbb A^n_k$ are the projections, $\mu:\mathbb A^n_k\times\mathbb A^n_k\to\mathbb A^1_k$ is given by $\mu((x_1,\ldots,x_n),(y_1,\ldots,y_n))=x_1y_1+\cdots+x_ny_n$ and ${\mathcal L}_{\psi}$ is the Artin-Schreier smooth sheaf on $\mathbb A^1_k$ associated to $\psi$. It is an autoequivalence of the triangulated category ${\mathcal D}^b_c(\mathbb A^n_k,\bar{\mathbb Q}_\ell)$. In particular, any action of a finite group $H$ on an object $K\in{\mathcal D}^b_c(\mathbb A^n_k,\bar{\mathbb Q}_\ell)$ induces an action on $\mathrm{FT}_\psi(K)$.
\begin{lem}\label{lemmafourier}
Let $K\in{\mathcal D}^b_c(\mathbb A^n_k,\bar{\mathbb Q}_\ell)$ be an object with an action of the finite group $H$ and $\rho:H\to\mathrm{GL}(V)$ a finite dimensional representation of $H$. Then $(\mathrm{FT}_\psi(K))^\rho=\mathrm{FT}_\psi(K^\rho)$.
\end{lem}
\begin{proof}
By lemma \ref{directimage} we have
$$
(\mathrm{FT}_\psi(K))^\rho=\mathrm R\pi_{2!}(\pi_1^\star K\otimes\mu^\star {\mathcal L}_\psi)^\rho[n]=\mathrm R\pi_{2!}((\pi_1^\star K\otimes\mu^\star {\mathcal L}_\psi)^\rho)[n]=
$$
$$
=\mathrm R\pi_{2!}((\pi_1^\star K)^\rho\otimes\mu^\star {\mathcal L}_\psi)[n]
$$
since $\mu^\star{\mathcal L}_\psi$ is smooth of rank $1$ on $\mathbb A^n_k\times\mathbb A^n_k$. On the other hand,
$$
(\pi_1^\star K)^\rho={\mathcal Hom}_\rho(V,\pi_1^\star K)={\mathcal Hom}(\pi_1^\star V,\pi_1^\star K)^H=
$$
$$
=(\pi_1^\star{\mathcal Hom}(V,K))^H=\pi_1^\star({\mathcal Hom}(V,K)^H)=\pi_1^\star(K^\rho)
$$
since $\pi_1^\star$ is exact. We conclude that
$$
(\mathrm{FT}_\psi(K))^\rho=\mathrm R\pi_{2!}(\pi_1^\star (K^\rho)\otimes\mu^\star {\mathcal L}_\psi)[n]=\mathrm{FT}_\psi(K^\rho).
$$
\end{proof}
\begin{prop}\label{fourier}
For every $K\in{\mathcal D}^b_c(\mathbb A^n_k,\bar{\mathbb Q}_\ell)$, every integer $r\geq 1$ and every finite dimensional $\bar{\mathbb Q}_\ell$-representation $\rho$ of ${\mathfrak S}_r$ there is a quasi-isomorphism
$$
\mathrm{FT}_\psi(\mathrm R(\ast\rho)K)\cong\mathrm R(\rho\otimes\sigma^{n})\mathrm{FT}_\psi(K)[-n(r-1)].
$$
where $\sigma$ is the sign character of ${\mathfrak S}_r$. That is, the Fourier transform interchanges the operations $\mathrm R(\ast\rho)$ and $\mathrm R(\rho\otimes\sigma^{n})$ (up to a shift).
\end{prop}
\begin{proof}
By \cite[Corollaire 9.6]{brylinski1986transformations}, for every $K,L\in\mathbb A^n_k$ we have the formula
$$
\mathrm{FT}_\psi(K\ast L)\cong\mathrm{FT}_\psi(K)\otimes\mathrm{FT}_\psi(L)[-n]
$$
and, in particular, there is a natural quasi-isomorphism
$$
\mathrm{FT}_\psi(K^{\ast r})\cong\mathrm{FT}_\psi(K)^{\otimes r}[-(r-1)n]
$$
for every $r\geq 1$. The natural ${\mathfrak S}_r$-actions on $\mathrm{FT}_\psi(K^{\ast r})[(r-1)n]$ and $\mathrm{FT}_\psi(K)^{\otimes r}$ differ by a twist by the $n$-th power of the sign character $\sigma$ (at a geometric point $\bar t$ over $t=(t_1,\ldots,t_n)\in k_m^n$ the stalks are $\mathrm R\Gamma_c(\mathbb A^n_{\bar k},K^{\ast r}\otimes{\mathcal L}_{\psi_t})[rn]\cong\mathrm R\Gamma_c(\mathbb A^n_{\bar k},(K\otimes{\mathcal L}_{\psi_t})^{\ast r})[rn]\cong\mathrm R\Gamma_c(\mathbb A^n_{\bar k},K\otimes{\mathcal L}_{\psi_t})^{\otimes r}[rn]$ and $\mathrm R\Gamma_c(\mathbb A^n_{\bar k},K\otimes{\mathcal L}_{\psi_t})[n]^{\otimes r}$ respectively, where ${\mathcal L}_{\psi_t}$ is the rank $1$ smooth sheaf corresponding to the character of $k_m^n$ $x\mapsto \psi(\mathrm {Tr}_k(t_1x_1+\cdots+t_nx_n))$, and the proof of proposition \ref{shift} applied $n$ times shows that the actions of $\mathfrak S_r$ on these differ by a twist by $\sigma^n$). For any finite dimensional $\bar{\mathbb Q}_\ell$-representation $\rho$ of ${\mathfrak S}_r$ we have then
$$
\mathrm{FT}_\psi(\mathrm R(\ast\rho)K)=\mathrm{FT}_\psi((K^{\ast r})^\rho)=\mathrm{FT}_\psi(K^{\ast r})^\rho\cong(\mathrm{FT}_\psi(K)^{\otimes r}[-(r-1)n])^{\rho\otimes\sigma^n}=
$$
$$
=(\mathrm{FT}_\psi(K)^{\otimes r})^{\rho\otimes\sigma^{n}}[-(r-1)n]=\mathrm R(\rho\otimes\sigma^{n})\mathrm{FT}_\psi(K)[-n(r-1)]
$$
by lemma \ref{lemmafourier}.
\end{proof}
\begin{cor}\label{fourieradams}
For every $K\in{\mathcal D}^b_c(\mathbb A^n_k,\bar{\mathbb Q}_\ell)$ and every integer $r\geq 1$ we have
$$
\mathrm{FT}_\psi(K^{[\ast r]})=(\mathrm{FT}_\psi K)^{[r]}.
$$
\end{cor}
\begin{proof}
Let $\rho$ be the standard representation of ${\mathfrak S}_r$. By the proposition, we have
$$
\mathrm{FT}_\psi(K^{[\ast r]})=\sum_{i=0}^{r-1}(-1)^i[\mathrm{FT}_\psi(\mathrm R(\ast\wedge^i\rho)K)]=
$$
$$
=(-1)^{n(r-1)}\sum_{i=0}^{r-1}(-1)^i[\mathrm R((\wedge^i\rho)\otimes\sigma^{n})\mathrm{FT}_\psi K].
$$
If $n$ is even this proves the statement. If $n$ is odd, then $\sigma^{n}=\sigma$ and $(\wedge^i\rho)\otimes\sigma=\wedge^{r-1-i}\rho$ so
$$
\mathrm{FT}_\psi(K^{[\ast r]})=(-1)^{r-1}\sum_{i=0}^{r-1}(-1)^i[\mathrm R(\wedge^{r-1-i}\rho)\mathrm{FT}_\psi K]=
$$
$$
=\sum_{i=0}^{r-1}(-1)^{r-1-i}[\mathrm R(\wedge^{r-1-i}\rho)\mathrm{FT}_\psi K]=\sum_{i=0}^{r-1}(-1)^{i}[\mathrm R(\wedge^{i}\rho)\mathrm{FT}_\psi K]=(\mathrm{FT}_\psi K)^{[r]}.
$$
\end{proof}
\section{The dimension 1 case}
From now on we will assume that $G$ is affine of dimension $1$ (so $G\otimes\bar k$ is either the affine line $\mathbb A^1_{\bar k}$ or the torus $\mathbb G_{m,\bar k}$). We will describe more precisely the operation $K\mapsto K^{[\ast r]}$ in this situation by splitting $K$ into its perverse cohomology sheaves. Since $G$ is a smooth curve, perverse sheaves have an easy description \cite[5.2.2]{beilinson1982faisceaux}: they are objects $\mathcal P\in{\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$ which have non-zero cohomology only in degrees $0$ and $-1$, ${\mathcal H}^0(\mathcal P)$ is punctual (that is, $j^\star{\mathcal H}^0(\mathcal P)=0$ for some dense open set $j:U\hookrightarrow G$) and ${\mathcal H}^{-1}(\mathcal P)$ has no punctual sections (that is, the adjunction map ${\mathcal H}^{-1}(\mathcal P)\to j_\star j^\star{\mathcal H}^{-1}(\mathcal P)$ is injective for any dense open set $j:U\hookrightarrow G$). The full subcategory ${\mathcal P}erv(G,\bar{\mathbb Q}_\ell)\subset{\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$ of perverse sheaves on $G$ is an abelian category in which exact sequences are just distinguished triangles. Irreducible objects in this category are of two types: punctual objects $i_{x\star}\mathcal F[0]$ where $i_x:\{x\}\hookrightarrow G$ is the inclusion of a closed point, and middle extensions $j_{\star !}(\mathcal F[1])\cong (j_\star\mathcal F)[1]$ where $j:U\hookrightarrow G$ is the inclusion of a dense open subset and $\mathcal F$ is an irreducible smooth $\bar{\mathbb Q}_\ell$-sheaf on $U$. If $G=\mathbb A^1_k$, the Fourier transform functor (with respect to any non-trivial character $\psi:k\to\bar{\mathbb Q}_\ell^\star$) preserves perverse objects and is an autoequivalence of the category of perverse sheaves \cite[Corollaire 2.1.5]{katz1985transformation}.
The derived category of the category of perverse sheaves is again ${\mathcal D}^b_c(G,\bar{\mathbb Q}_\ell)$. In particular, its Grothendieck group is $K_0(G,\bar{\mathbb Q}_\ell)$. Therefore, by additivity it suffices to study $\mathcal P^{[\ast r]}$ for $\mathcal P$ an irreducible perverse sheaf. Since we are assuming that everything is mixed of integral $q$-weights, such a perverse sheaf is pure of some integral $q$-weight \cite[Corollaire 5.3.4]{beilinson1982faisceaux} and, in particular, geometrically semisimple \cite[Th\'eor\`eme 5.3.8]{beilinson1982faisceaux}. By corollary \ref{twist} we can further assume that it is pure of weight $1$.
\begin{lem}\label{negligible}
Let ${\mathcal P}\in{\mathcal P}erv(G,\bar{\mathbb Q}_\ell)$ be irreducible. The following conditions are equivalent:
\begin{enumerate}
\item $\mathcal P\otimes\bar k$ contains a sub-object isomorphic to ${\mathcal L}_\chi[1]$ for some $r\geq 1$ and some character $\chi:G(k_r)\to\bar{\mathbb Q}_\ell^\star$ (cf. \cite[1.4-1.8]{deligne569application}).
\item ${\mathcal P}\otimes\bar k$ is a direct sum of objects of the form ${\mathcal L}_{\chi_i}[1]$ for some $r\geq 1$ and some characters $\chi_i:G(k_r)\to\bar{\mathbb Q}_\ell^\star$.
\end{enumerate}
In that case, if $r$ is the smallest positive integer such that ${\mathcal L}_\chi$ is defined over $k_r$, then $\mathcal P\cong\alpha^{deg}\otimes\pi_{r\star}{\mathcal L}_\chi[1]$ for some $\ell$-adic unit $\alpha$, where $\pi_r:G\otimes k_r\to G$ is the projection.
\end{lem}
\begin{proof}
(2)$\Rightarrow$(1) is trivial. Suppose that (1) holds, and let $r$ be the smallest positive integer such that ${\mathcal L}_\chi$ is defined over $k_r$. Then we have an injective map $\alpha^{deg}\otimes{\mathcal L}_{\chi}[1]\to\pi_r^\star\mathcal P$ for some $\ell$-adic unit $\alpha$ and, by adjunction, a non-zero map $\alpha^{deg}\otimes\pi_{r\star}{\mathcal L}_{\chi}[1]\to\mathcal P$. Since $\mathcal P$ is irreducible, this map is surjective. In particular ${\mathcal H}^0(\mathcal P)=0$ and $\mathcal P$ is of the form $\mathcal F[1]$ for some sheaf $\mathcal F$ without punctual sections.
Let $\sigma\in\mathrm {Gal}(k_r/k)$, then $\alpha^{deg}\otimes{\mathcal L}_{\chi\circ\sigma}[1]=\sigma^\star(\alpha^{deg}\otimes{\mathcal L}_\chi)[1]\hookrightarrow \sigma^\star\pi_r^\star\mathcal P\cong\pi_r^\star\mathcal P$. Moreover $\chi\circ\sigma\neq\chi$ if $\sigma\neq Id$ since otherwise $\chi$ would be the pull-back of a character of $G(k')$, where $k'$ is the subfield of $k_r$ fixed by $\sigma$, and ${\mathcal L}_\chi$ would be defined over $k'$, contrary to the hypothesis. Therefore $\pi^\star_r\mathcal F$ contains at least $r$ non-isomorphic smooth subsheaves of rank $1$, so its rank (which is the rank of $\mathcal F$) must be at least $r$. We conclude that the map $\alpha^{deg}\otimes\pi_{r\star}{\mathcal L}_{\chi}[1]\to\mathcal P$ is an isomorphism. In particular, $\mathcal P\otimes\bar k$ is the direct sum of ${\mathcal L}_{\chi\circ\sigma}[1]$ for all $\sigma\in\mathrm {Gal}(k_r/k)$.
\end{proof}
Following \cite{katz2010mellin} we will say that an irreducible perverse sheaf $\mathcal P$ on $G$ is \emph{negligible} if it satisfies the equivalent conditions in the previous lemma.
\begin{prop}
Let $\mathcal P$ be an irreducible perverse sheaf on $G$ of weight $1$, and suppose that $\mathcal P$ is non-negligible. Then for every $r\geq 1$, $\mathcal P^{[\ast r]}\in K_0(G,\bar{\mathbb Q}_\ell)$ is an integral combination of classes of perverse sheaves of weights $\leq r$.
\end{prop}
\begin{proof}
By \cite[2.6.4, 2.6.8, 2.6.13, 2.6.14]{katz1996rls} the $r$-fold convolution $\mathcal P^{\ast r}$ is a perverse sheaf, of weights $\leq r$ (since $K$ is perverse if and only if $K\otimes \bar k$ is). For every representation $\rho$ of ${\mathfrak S}_r$, $\mathrm R(\ast\rho)\mathcal P=(\mathcal P^{\ast r})^\rho$ is then also perverse of weights $\leq r$ (since ${\mathcal H}^i((\mathcal P^{\ast r})^\rho)={\mathcal H}^i(\mathcal P^{\ast r})^\rho$ is a subsheaf of ${\mathcal H}^i(\mathcal P^{\ast r})$ for every $i$, and is therefore zero for $i\neq 0,-1$, punctual for $i=0$ and without punctual sections for $i=-1$). In particular $\mathcal P^{[\ast r]}=\sum_{i=0}^{r-1}(-1)^i\mathrm R(\ast\wedge^i\rho)\mathcal P$ is an integral combination of classes of such perverse sheaves, where $\rho$ is the standard representation of ${\mathfrak S}_r$.
\end{proof}
\begin{cor}\label{UC}
Let $\mathcal P$ be a non-negligible irreducible perverse sheaf on $G$ of weight $1$. For every integer $r\geq 1$ there exists a dense open set $U_{\mathcal P,r}\subseteq G$ and a constant $C_{\mathcal P,r}$ such that for every integer $m\geq 1$ and every $t\in U_{\mathcal P,r}(k_m)$ the $L$-function $L^{\mathrm{N},r}({\mathcal P},k_m,t;T)$ has total degree bounded by $C_{\mathcal P,r}$ and all its reciprocal roots and poles have $q^m$-weight $\leq r-1$. In particular, for every $m\geq 1$ and every $t\in U_{\mathcal P,r}(k_m)$ we have the estimate
$$
|f^{\mathrm{N},r}_\mathcal P(k_m,t)|\leq C_{\mathcal P,r}q^{\frac{m(r-1)}{2}}
$$
If $t\notin U_{\mathcal P,r}(k_m)$, then all reciprocal roots and poles of $L^{\mathrm{N},r}({\mathcal P},k_m,t;T)$ have $q^m$-weight $\leq r$.
\end{cor}
\begin{proof} For every $i=0,\ldots,r-1$, let ${\mathcal Q}_i$ be the perverse sheaf $\mathrm R(\ast\wedge^i\rho)\mathcal P$, where $\rho$ is the standard representation of ${\mathfrak S}_r$. Let $U_i\subseteq G$ be the largest open set on which ${\mathcal H}^0({\mathcal Q}_i)=0$, and $C_i$ the generic rank of ${\mathcal H}^{-1}({\mathcal Q}_i)$. We define $U_{\mathcal P,r}=U_0\cap\cdots\cap U_{r-1}$ and $C_{\mathcal P,r}=C_0+\cdots+C_{r-1}$.
For every integer $m\geq 1$ and every $t\in U_{\mathcal P,r}(k_m)$, we have
$$
L^{\mathrm{N},r}({\mathcal P},k_m,t;T)=L(\mathcal P^{[\ast r]},k_m,t;T)=\prod_{i=0}^{r-1}L({\mathcal Q}_i,k_m,t;T)^{(-1)^i}
$$
$$
=\prod_{i=0}^{r-1}L({\mathcal H}^{-1}({\mathcal Q}_i),k_m,t;T)^{(-1)^{i+1}}
$$
The result follows from the fact that ${\mathcal H}^{-1}({\mathcal Q}_i)$ is mixed of weights $\leq r-1$ and does not have punctual sections for any $i$, so its rank at any point of $U_{\mathcal P,r}$ is less than or equal to its generic rank. If $t\notin U_{\mathcal P,r}(k_m)$ we would also get factors of the form $L({\mathcal H}^{0}({\mathcal Q}_i),k_m,t;T)$ which are mixed of weight $\leq r$.
\end{proof}
The result extends to any perverse $\mathcal P$ pure of weight $1$ as long as no irreducible component of $\mathcal P$ is negligible. As an easy example, we compute $U_{\mathcal P,r}$ and $C_{\mathcal P,r}$ when $\mathcal P=\delta_a$ is a punctual object supported on $a\in G(k)$ (placed in degree $0$).
\begin{prop}
For every $a\in G(k)$ and $r\geq 1$, $\delta_a^{[\ast r]}=\delta_{a^r}$. In particular, $U_{\mathcal P,r}(\bar k)=G(\bar k)-\{a\}$ and $C_{\mathcal P,r}=0$.
\end{prop}
\begin{proof}
For every $r\geq 1$, $\delta_a^{\ast r}=\delta_{a^r}$, with trivial ${\mathfrak S}_r$ action \cite[2.5.3]{katz1996rls}. Therefore $\mathrm R(\ast\wedge^i\rho)\delta_a=\delta_{a^r}$ for $i=0$ and $0$ for $i>0$. We conclude that $\delta_a^{[\ast r]}=[\delta_{a^r}]$.
\end{proof}
\begin{cor}
Let $\mathcal P$ be a punctual perverse sheaf on $G$ supported on $Z\subseteq G$. Let $S=\{z^r|z\in Z(\bar k)\}$. Then in corollary \ref{UC} one can take $U_{\mathcal P,r}(\bar k)=G(\bar k)-S$ and $C_{\mathcal P,r}=0$.
\end{cor}
\begin{proof}
It is an immediate consequence of the previous proposition and the additivity of the convolution Adams power, since the operation $\mathcal P\mapsto\mathcal P^{[\ast r]}$ commutes with extension of scalars to $\bar k$.
\end{proof}
\begin{rem}{\rm
This shows that, in general, the $U_{\mathcal P,r}$ and $C_{\mathcal P,r}$ defined in the proof of corollary \ref{UC} are not the best possible ones, since there may be some cancellation among the ${\mathcal Q}_i$'s when taking the alternating product. For instance, if $k={\mathbb F}_3$, $G=\mathbb A^1_k$ and $\mathcal P$ is punctual supported on $\mathrm{Spec\:} k[t]/(t^2+1)=\{\pm i\}$, the proof gives $U_{\mathcal P,2}=G-\{0,\pm 2i\}=G-\{0,\pm i\}$ (since $\mathcal P^{\ast 2}=\mathrm{Sym}^{\ast 2}\mathcal P\oplus\wedge^{\ast 2}\mathcal P$, so the union of the supports of $\mathrm{Sym}^{\ast 2}\mathcal P$ and $\wedge^{\ast 2}\mathcal P$ is the support of $\mathcal P^{\ast 2}$), while the previous corollary shows that we could take $U_{\mathcal P,2}=G-\{\pm i\}$.
In the remainder of the article we will assume, unless otherwise stated, that $U_{\mathcal P;r}$ and $C_{\mathcal P,r}$ are the ones defined in the proof of \ref{UC}.}
\end{rem}
\section{Examples on $\mathbb A^1_k$}
In this section we further specialize to the case $G=\mathbb A^1_k$. Here the ``norm'' map $G(k_{mr})=k_{mr}\to G(k_m)=k_m$ is just the trace, so we will write $L^{\mathrm {Tr},r}$ and $f^{\mathrm {Tr},r}$ instead of $L^{\mathrm{N},r}$ and $f^{\mathrm{N},r}$. Fix a non-trivial character $\psi:k\to\bar{\mathbb Q}_\ell^\star$. Since the Fourier transform with respect to $\psi$ preserves perversity and interchanges punctual objects and (shifted) Artin-Schreier sheaves, an irreducible perverse object $\mathcal P$ on $\mathbb A^1_k$ is negligible if and only if its Fourier transform is punctual. For those objects we can explicitely determine $\mathcal P^{[\ast r]}$:
\begin{prop}\label{ASformula}
Let ${\mathcal P}\in{\mathcal P}erv(\mathbb A^1_k,\bar{\mathbb Q}_\ell)$ be irreducible, negligible and pure of weight $1$. Then
$$
\mathcal P^{[\ast r]}=[(\alpha q)^{(r-1) deg}\otimes\mathcal P]
$$
for some $\ell$-adic unit $\alpha\in\bar{\mathbb Q}_\ell$ of weight $0$.
\end{prop}
\begin{proof} After taking Fourier transform with respect to $\psi$ on both sides, the equality is equivalent by corollary \ref{fourieradams} to
$$
{\mathcal Q}^{[r]}=[(\alpha q)^{(r-1) deg}\otimes{\mathcal Q}]
$$
where $\mathcal Q$ is the Fourier transform of $\mathcal P$, which is punctual and pure of weight $2$ by hypothesis.
Since ${\mathcal Q}$ is punctual and irreducible, there exists a closed point $x\in\mathbb A^1_k$ and an irreducible sheaf $\mathcal F$ on $\{x\}=\mathrm{Spec\:} k(x)$ such that ${\mathcal Q}=i_{x\star}\mathcal F[0]$. Since $\mathcal F$ is irreducible, it must be equal to $(\alpha q)^\mathrm{deg}$ for some $\ell$-adic unit $\alpha$ of weight $0$. We will show that ${\mathcal Q}^{[r]}=[(\alpha q)^{(r-1) deg}\otimes{\mathcal Q}]$ by comparing their trace functions and using the injectivity of $\Phi$.
For any $m\geq 1$ and every $t\in k_m$, $\Phi({\mathcal Q}^{[r]})(k_m,t)=\mathrm {Tr}(\mathrm{Frob}_{k_m,t}|i_{x\star}\mathcal F_{\bar t}^{[r]})=\mathrm {Tr}(\mathrm{Frob}_{k_{mr},t}|i_{x\star}\mathcal F_{\bar t})=(\alpha q)^{mr}$ if $t\in\{x\}(k_{mr})$ and $0$ otherwise. But $t\in\{x\}(k_{mr})$ if and only if $t\in\{x\}(k_m)$ (if and only if $t$ is a root of the irreducible polynomial that defines $x$), so in any case $\Phi({\mathcal Q}^{[r]})(k_m,t)=(\alpha q)^{m(r-1)}\Phi({\mathcal Q})(k_m,t)=\Phi((\alpha q)^{(r-1) deg}\otimes{\mathcal Q})(k_m,t)$.
\end{proof}
The $\ell$-adic unit $\alpha$ can be determined from ${\mathcal P}$ in the following way: Let $d\geq 1$ be an integer such that ${\mathcal H}^{-1}({\mathcal P})\otimes k_d$ splits as an extension of Artin-Schreier sheaves. Then $\alpha$ is any $d$-th root of
$$
\frac{\mathrm {Tr}(\mathrm{Frob}_{k_d,0}|({\mathcal H}^{-1}({\mathcal P})\otimes k_d)_0)}{\mathrm{rank}\,{\mathcal H}^{-1}({\mathcal P})}.
$$
\begin{cor}\label{AScase}
If ${\mathcal P}\in{\mathcal P}erv(\mathbb A^1_k,\bar{\mathbb Q}_\ell)$ is irreducible, negligible and pure of weight $1$, then there exists some $\ell$-adic unit $\alpha\in\bar{\mathbb Q}_\ell$ of weight $0$ such that for every integer $m\geq 1$ and every $t\in k_m$
$$
L^{\mathrm {Tr},r}({\mathcal P},k_m,t;T)=L({\mathcal P},k_m,t;(\alpha q)^{m(r-1)}T).
$$
\end{cor}
For non-negligible $\mathcal P$ we can give the following characterization of $U_{\mathcal P,r}$:
\begin{prop}\label{ULr}
Let $\mathcal P\in{\mathcal P}erv(\mathbb A^1_k,\bar{\mathbb Q}_\ell)$ be irreducible and non-negligible. Let $\mathrm{FT}_\psi\mathcal P=\mathcal G[1]$, with $\mathcal G\in\mathcal Sh(\mathbb A^1_k,\bar{\mathbb Q}_\ell)$ an irreducible middle extension sheaf. Let $a\in \bar k$, and ${\mathcal L}_{\psi_{-a}}$ the Artin-Schreier sheaf on $\mathbb A^1_{\bar k}$ associated to the character $t\mapsto\psi(-at)$. Then $a\in U_{\mathcal P,r}(\bar k)$ if and only if the following equivalent conditions hold:
\begin{enumerate}
\item $\mathrm H^2_c(\mathbb A^1_{\bar k},{\mathcal L}_{\psi_{-a}}\otimes\mathrm R(\wedge^i\rho)\mathcal G)=0$ for every $i=0,\ldots,r-1$, where $\rho$ is the standard representation of ${\mathfrak S}_r$.
\item $\mathrm H^2_c(\mathbb A^1_{\bar k},{\mathcal L}_{\psi_{-a}}\otimes\mathrm{Sym}^{r-i}\mathcal G\otimes\wedge^i\mathcal G)=0$ for every $i=0,\ldots,r$.
\end{enumerate}
\end{prop}
\begin{proof}
For (1) it is a consequence of the definition of $U_{\mathcal P,r}$ in proposition \ref{UC}: since $\mathcal G[1]$ is the Fourier transform of $\mathcal P$, by \ref{fourier} and \ref{shift} the Fourier transform of $\mathrm R(\ast\wedge^i\rho)\mathcal P$ is $\mathrm R((\wedge^i\rho)\otimes\sigma)(\mathcal G[1])[1-r]\cong(\mathrm R(\wedge^i\rho)\mathcal G[r])[1-r]=\mathrm R(\wedge^i\rho)\mathcal G[1]$, so $\mathrm H^2_c(\mathbb A^1_{\bar k},{\mathcal L}_{\psi_{-a}}\otimes\mathrm R(\wedge^i\rho)\mathcal G)$ is (a Tate twist of) the stalk at $a$ of ${\mathcal H}^0({\mathcal Q}_i)$.
The equivalence between (1) and (2) can be deduced from the formulas $\mathrm{Sym}^{r-i}\mathcal G\otimes\wedge^i\mathcal G=\mathrm R(\wedge^{i-1}\rho\oplus\wedge^i\rho)\mathcal G=\mathrm R(\wedge^{i-1}\rho)\mathcal G\oplus\mathrm R(\wedge^i\rho)\mathcal G$ for $i=1,\ldots,r-1$, $\mathrm{Sym}^r\mathcal G=\mathrm R({\mathbf 1})\mathcal G$ and $\wedge^r\mathcal G=\mathrm R(\wedge^{r-1}\rho)\mathcal G$.
\end{proof}
We will now apply these results to some particular examples of sheaves.
\begin{prop}\label{trex1}
Suppose that $\mathcal P=\mathcal F[1]$ where $\mathcal F\in{\mathcal Sh}(\mathbb A^1_{\bar k},\bar{\mathbb Q}_\ell)$ is a geometrically semisimple middle extension sheaf of generic rank $d$, Euler characteristic $-e$ and Swan conductor at infinity $c$ such that $\mathrm H^2_c(\mathbb A^1_{\bar k},\mathcal F)=0$ and all its slopes at infinity are $<1$ (e.g. $\mathcal F$ tamely ramified at infinity). Let $S:=\{a_1,\ldots,a_s\}\subset \bar k$ be the set of ramification points of $\mathcal F\otimes\bar k$, and $S_r:=S+\cdots +S$ ($r$ summands). Then for every $r\geq 1$ $U_{\mathcal P,r}(\bar k)$ contains $\bar k-S_r$, and $C_{\mathcal P,r}$ is bounded by
$$(1+c)\sum_{i=0}^{r-1} \left[{{d+e-c+r-i-1}\choose{r}}-{{e+r-i-1}\choose{r}}\right]{r-1\choose i}.
$$
\end{prop}
\begin{proof}
First of all, $\mathcal P$ does not have negligible subquotients: since $\mathrm H^2_c(\mathbb A^1_{\bar k},\mathcal F)=0$ $\mathcal P\otimes\bar k$ can not have constant subsheaves, and it can not have non-trivial Artin-Schreier subsheaves either since $1$ is not a slope at infinity.
Therefore $\mathrm{FT}_\psi\mathcal P=\mathcal G[1]$, where $\mathcal G\in\mathcal Sh(\mathbb A^1_k,\bar{\mathbb Q}_\ell)$ is a middle extension sheaf. Laumon's local Fourier transform theory \cite[2.4]{laumon1987transformation} implies that $\mathcal G$ is smooth on $\mathbb G_{m,\bar k}$, since $1$ is not a slope of $\mathcal F$ at infinity. Its generic rank is $\dim\mathrm H^1_c(\mathbb A^1_{\bar k},\mathcal F\otimes{\mathcal L}_{\psi_a})$ for any $a\neq 0$, that is, $\dim\mathrm H^1_c(\mathbb A^1_{\bar k},\mathcal F)-\mathrm{Swan}_\infty\mathcal F+\mathrm{Swan}_\infty(\mathcal F\otimes{\mathcal L}_{\psi_a})=e-c+d$ (since all slopes at infinity of $\mathcal F\otimes{\mathcal L}_{\psi_a}$ are equal to $1$). Its rank at $0$ is $\dim\mathrm H^1_c(\mathbb A^1_{\bar k},\mathcal F)=e$. At infinity, it is the direct sum, for $s\in S$, of ${\mathcal F}_s\otimes{\mathcal L}_{\psi_s}$, where ${\mathcal F}_s$ is the local Fourier transform operator $\mathcal F^{(0,\infty)}$ \cite[2.4.2.3]{laumon1987transformation} applied to the local monodromy of $\mathcal F$ at $s$. Its Swan conductor at $0$ is $c$, since for generic $a$ the dimension of $\mathrm H^1_c(\mathbb A^1_{\bar k},\mathcal G\otimes{\mathcal L}_{\psi_a})$ is $d$ (by the involutivity of Fourier transform) and, by the Euler-Poincar\'e formula \cite[Expos\'e X, Corollaire 7.12]{grothendieck1977cohomologie}, it is also equal to $\mathrm{Swan}_\infty(\mathcal G\otimes{\mathcal L}_{\psi_a})+\mathrm{Swan}_0(\mathcal G\otimes{\mathcal L}_{\psi_a})-\dim\mathcal G_0=(e-c+d)+\mathrm{Swan}_0\,\mathcal G-e$.
Then for every $i=0,\ldots,r-1$ the sheaf $\mathrm R(\wedge^i\rho)\mathcal G$ is smooth on $\mathbb G_{m,\bar k}$, has generic rank ${{d+e-c+r-i-1}\choose{r}}{r-1\choose i}$ (cf. remark \ref{rank}), its rank at $0$ is ${{e+r-i-1}\choose{r}}{r-1\choose i}$ and its monodromy action at infinity splits as a direct sum ${\mathcal N}_s\otimes{\mathcal L}_{\psi_s}$ for $s\in S_r$, where ${\mathcal N}_s$ is a representation of $I_\infty$ whose slopes are all $<1$. In particular, for every $a\in\bar k-S_r$, the sheaf ${\mathcal L}_{\psi_{-a}}\otimes\mathrm R(\wedge^i\rho)\mathcal G$ is totally wild at infinity, so $a\in U_{{\mathcal L},r}(\bar k)$ by proposition \ref{ULr}.
Furthermore, the dimension of $\mathrm H^1_c(\mathbb A^1_c,{\mathcal L}_{\psi_{-a}}\otimes\mathrm R(\wedge^i\rho)\mathcal G)$ is, by the Euler-Poincar\'e formula, equal to $$\mathrm{Swan}_0({\mathcal L}_{\psi_{-a}}\otimes\mathrm R(\wedge^i\rho)\mathcal G)+$$
$$+\mathrm{Swan}_\infty({\mathcal L}_{\psi_{-a}}\otimes\mathrm R(\wedge^i\rho)\mathcal G)-{{e+r-i-1}\choose{r}}{r-1\choose i}\leq
$$
$$
\leq c\left[{{d+e-c+r-i-1}\choose{r}}{r-1\choose i}-{{e+r-i-1}\choose{r}}{r-1\choose i}\right]+
$$
$$+{{d+e-c+r-i-1}\choose{r}}{r-1\choose i}-{{e+r-i-1}\choose{r}}{r-1\choose i}\leq$$
$$
\leq (1+c)\left[{{d+e-c+r-i-1}\choose{r}}-{{e+r-i-1}\choose{r}}\right]{r-1\choose i}.
$$
since the Swan conductor of $\mathcal G$ at $0$ (and therefore all its slopes) are less than or equal to $c$. We conclude by applying the formula for $C_{\mathcal P,r}$ in the proof of proposition \ref{UC}.
\end{proof}
Our first example improves \cite[Corollary 4]{rlwan2010}:
\begin{exa}
Let $g\in k[x]$ be a polynomial of degree $d$ prime to $p$ and $\mathcal P=\mathcal F[1]$, where $\mathcal F$ is the kernel of the trace map $g_\star\bar{\mathbb Q}_\ell\to\bar{\mathbb Q}_\ell$. Let $S$ be the set of critical values of $g$, and $S_r=S+\cdots+S$ ($r$ summands). Then $\bar k-S_r\subseteq U_{\mathcal P,r}(\bar k)$ for every $r\geq 1$. Therefore, for every $m\geq 1$ and every $t\in k_m$ which is not the sum of $r$ critical values of $g$ (i.e. such that the affine hypersurface $g(x_1)+\cdots+g(x_r)=t$ in $\mathbb A^r_{\bar k}$ is smooth) we have
$$
\left| \#\{x\in k_{mr}|\mathrm {Tr}_{k_{mr}/k_m}(g(x))=t\}-q^{m(r-1)}\right|
\leq\sum_{i=0}^{r-1} {{d+r-i-2}\choose{r}}{{r-1}\choose i} q^{\frac{m(r-1)}{2}}.
$$
In particular, if the affine hypersurface $g(x_1)+\cdots+g(x_r)=0$ in $\mathbb A^r_{\bar k}$ is smooth we have
$$
\left|\#\{(x,y)\in k_{mr}^2|y^{q^m}-y=g(x)\}-q^{mr}\right|
\leq \sum_{i=0}^{r-1} {{d+r-i-2}\choose{r}}{{r-1}\choose i} q^{\frac{m(r+1)}{2}}.
$$
\end{exa}
\begin{proof}
The left hand side is $|f^{\mathrm {Tr},r}_{g_\star\bar{\mathbb Q}_\ell}(k_m,t)-f^{\mathrm {Tr},r}_{\bar{\mathbb Q}_\ell}(k_m,t)|=|f^{\mathrm {Tr},r}_{\mathcal F}(k_m,t)|$. By Proposition \ref{ASformula} and the comment after it, $f^{\mathrm {Tr},r}_{\bar{\mathbb Q}_\ell}=\kappa_{q^{r-1}}\cdot f_{\bar{\mathbb Q}_\ell}$, so $f^{\mathrm {Tr},r}_{\bar{\mathbb Q}_\ell}(k_m,t)=q^{m(r-1)}$.
On the other hand, $\mathcal F$ has rank $d-1$ and satisfies the hypotheses of proposition \ref{trex1} with $e=c=0$, since it is tamely and totally ramified at infinity (the inertia group at infinity acts via the direct sum of all its non-trivial characters with trivial $d$-th power) and there is an exact sequence $0\to\mathcal F\to g_\star\bar{\mathbb Q}_\ell\to\bar{\mathbb Q}_\ell\to0$ with $\dim\mathrm H^1_c(\mathbb A^1_{\bar k},g_\star\bar{\mathbb Q}_\ell)=\dim\mathrm H^1_c(\mathbb A^1_{\bar k},\bar{\mathbb Q}_\ell)=0$. The first inequality follows. The second one is an easy consequence of the identity
$$
\#\{(x,y)\in k_{mr}^2|y^{q^m}-y=g(x)\}=q^m\cdot\#\{x\in k_{mr}|\mathrm {Tr}_{k_{mr}/k_m}(g(x))=0\}.
$$
\end{proof}
\begin{exa}
Let $\chi:k^\star\to\bar{\mathbb Q}_\ell^\star$ be a non-trivial multiplicative character of order $n$, $g\in k[x]$ a non-constant polynomial which has no roots in $\bar k$ with multiplicity divisible by $n$, $\mathcal F:={\mathcal L}_{\chi(g)}=g^\star{\mathcal L}_\chi$ and $\mathcal P=\mathcal F[1]$. Let $S\subseteq \bar k$ be the set of roots of $g$, and $S_r=S+\cdots+S$ ($r$ summands). Then $\bar k-S_r\subseteq U_{\mathcal P,r}(\bar k)$ for every $r\geq 1$. In particular, for every $m\geq 1$ and every $t\in k_m$ which is not the sum of $r$ roots of $g$,
$$
\left| \sum_{\mathrm {Tr}_{k_{mr}/k_m}(x)=t}\chi(\mathrm{N}_{k_{mr}/k}(g(x)))\right|
\leq\sum_{i=0}^{r-1}{{a+r-i-2}\choose{r-1}}{{r-1}\choose i} q^{\frac{m(r-1)}{2}}.
$$
where $a$ is the number of distinct roots of $g$ in $\bar k$.
\end{exa}
\begin{proof}
The sheaf $\mathcal F$ is a middle extension of rank $1$ ramified at the roots of $g$ and therefore it is not isomorphic to an Artin-Schreier sheaf. We can then apply Proposition \ref{trex1} to it, where $d=1$, $e=a-1$ by the Euler-Poincar\'e formula and $c=0$ (since the inertia group at infinity acts on $\mathcal F$ via a power of the tame character $\chi$).
\end{proof}
\begin{prop}\label{trex2}
Under the hypotheses of proposition \ref{trex1}, suppose further that the action of the inertia group at every point of $\mathbb A^1_{\bar k}$ on the generic stalk of $\mathcal F$ (modulo its invariant subspace under this action) is a successive extension of a fixed tame character $\chi$ of order $n$. Then $U_{\mathcal P,r}=\mathbb A^1_k$ for every $r\geq 1$ which is not divisible by $n$.
\end{prop}
\begin{proof}
In this case, since the local Fourier transform of a tame character is its conjugate, the representations ${\mathcal F}_s$ in the proof of proposition \ref{trex1} are successive extensions of the character $\bar\chi$, so the representations ${\mathcal N}_s$ appearing (tensored with Artin-Schreier characters) in the monodromy at infinity of $\mathrm R(\wedge^i\rho)\mathcal G$ are successive extensions of the character $\bar\chi^r$. In particular, the action of the inertia group $I_\infty$ on $\mathrm R(\wedge^i\rho)\mathcal G$ has no invariants if $\chi^r$ is non-trivial (since the tensor product of a non-trivial tame character and a (possibly trivial) Artin-Schreier character can not be trivial).
\end{proof}
\begin{exa}
Let $g\in k[x]$ be a polynomial of degree $d$ prime to $p$, and let $\mathcal F$ be the kernel of the trace map $g_\star\bar{\mathbb Q}_\ell\to\bar{\mathbb Q}_\ell$. Suppose that $p>2$ and the derivative $g'$ has no multiple roots. Then for every odd $r\geq 1$, every $m\geq 1$ and every $t\in k_m$ we have
$$
\left| \#\{x\in k_{mr}|\mathrm {Tr}_{k_{mr}/k_m}(g(x))=t\}-q^{m(r-1)}\right|
\leq\sum_{i=0}^{r-1} {{d+r-i-2}\choose{r}}{{r-1}\choose i} q^{\frac{m(r-1)}{2}}
$$
and, in particular,
$$
\left|\#\{(x,y)\in k_{mr}^2|y^{q^m}-y=g(x)\}-q^{mr}\right|
\leq \sum_{i=0}^{r-1} {{d+r-i-2}\choose{r}}{{r-1}\choose i} q^{\frac{m(r+1)}{2}}.
$$
\end{exa}
\begin{proof}
In this case the monodromy of $\mathcal F$ at each ramified finite point is a successive extension of the quadratic character by the hypothesis on $g'$.
\end{proof}
\begin{exa}
Let $\chi:k^\star\to\bar{\mathbb Q}_\ell^\star$ be a non-trivial multiplicative character of order $n$, $g\in k[x]$ a square-free polynomial of degree $d$ and $\mathcal F:={\mathcal L}_{\chi(g)}=g^\star{\mathcal L}_\chi$. Then for every $r\geq 1$ not divisible by $n$, every $m\geq 1$ and every $t\in k_m$ we have
$$
\left| \sum_{\mathrm {Tr}_{k_{mr}/k_m}(x)=t}\chi(\mathrm{N}_{k_{mr}/k}(g(x)))\right|
\leq\sum_{i=0}^{r-1}{{d+r-i-2}\choose{r-1}}{{r-1}\choose i} q^{\frac{m(r-1)}{2}}.
$$
\end{exa}
\begin{proof}
In this case the inertia groups at all ramified finite points act on $\mathcal F$ via $\chi$. Additionally, since $g$ is square-free, it has $d$ distinct roots on $\bar k$.
\end{proof}
In order to get more precise results, we need to consider the global monodromy. Suppose that $\mathcal P\in{\mathcal P}erv(\mathbb A^1_k,\bar{\mathbb Q}_\ell)$ is pure of weight $1$ and does not have any negligible subquotient. Then $\mathrm{FT}_\psi\mathcal P=\mathcal G[1]$, where $\mathcal G\in\mathcal Sh(\mathbb A^1_k,\bar{\mathbb Q}_\ell)$ is a middle extension sheaf, pure of weight $1$ (as a middle extension). Let $V$ be the generic stalk of $\mathcal G$, $G\subseteq\mathrm{GL}(V)$ its global geometric monodromy group and $G_0\subseteq G$ its unit connected component. Since $\mathcal G$ is pure, its restriction to any open set on which it is smooth is geometrically semisimple \cite[Th\'eor\`eme 3.4.1]{deligne1980conjecture} and therefore $G_0$ is a semisimple algebraic group \cite[Corollaire 1.3.9]{deligne1980conjecture}.
\begin{prop}
Under the previous hypotheses, $0\in U_{\mathcal P,r}$ if and only if for every $i=0,\ldots,r$ the representation $\mathrm{Sym}^{r-i} V\otimes\wedge^i V$ of $G$ has no non-zero invariants.
\end{prop}
\begin{proof}
This is a restatement of proposition \ref{ULr}, since $\mathrm H^2_c(\mathbb A^1_{\bar k},\mathrm{Sym}^{r-i}\mathcal G\otimes\wedge^i\mathcal G)$ is the coinvariant space of $\mathrm{Sym}^{r-i} V\otimes\wedge^i V$ under the action of $\pi_1(U)$ (where $U$ is the largest open subset of $\mathbb A^1_k$ on which $\mathcal G$ is smooth), which has the same dimension as the invariant space.
\end{proof}
\begin{prop}\label{gm}
Under the previous hypotheses, suppose that $G/G_0$ has order prime to $p$. Then $\mathbb G_{m,k}\subseteq U_{\mathcal P,r}$ for every $r\geq 1$.
\end{prop}
\begin{proof}
Otherwise there would exist some $a\in\bar k^\star$ and some $i=0,\ldots,r$ such that the representation ${\mathcal L}_{\psi_{-a}}\otimes\mathrm{Sym}^{r-i} V\otimes\wedge^i V$ of $G$ has non-zero invariants. In other words, the representation $\mathrm{Sym}^{r-i} V\otimes\wedge^i V$ contains a subcharacter ${\mathcal L}_{\psi_a}$ of order $p$. Then its kernel $G'$ would be a closed normal subgroup of $G$ of index $p$, so $G/G_0$ would contain the $p$-group $G/G'$.
\end{proof}
\begin{exa}
Let $g\in k[x]$ be a polynomial of degree $d$ prime to $p$ and $\mathcal P=\mathcal F[1]$ where $\mathcal F$ is the kernel of the trace map $g_\star\bar{\mathbb Q}_\ell\to\bar{\mathbb Q}_\ell$. Suppose that $p>2d-1$ and the $(d-1)(d-2)$ differences between pairs of critical values of $g$ are all distinct. Let $s$ be the sum of the $d-1$ critical values of $g$. Then
\begin{enumerate}
\item $\mathbb A^1_k-\{\frac{rs}{d-1}\}\subseteq U_{\mathcal P,r}$ for any $r$.
\item $\frac{rs}{d-1}\in U_{\mathcal P,r}$ for any $r$ if $d$ is even, and for any $r\neq d-1$ if $d$ is odd.
\end{enumerate}
\end{exa}
\begin{proof}
By \cite[Theorem 7.9.6]{katz1990esa}, in this case $G_0=\mathrm{SL}(V)$, and by \cite[Proposition 4.1]{rlwan2010}, $G/G_0$ has order $1,2,p$ or $2p$ in the cases $s=0, d\text{ odd}; s=0, d\text{ even}; s\neq 0, d\text{ odd and }s\neq 0, d\text{ even}$. For $s=0$, (1) is a consequence of Proposition \ref{gm} and (2) is proven in \cite[Corollary 4.2]{rlwan2010}.
Suppose that $s\neq 0$, and let $h(x)=g(x)-s/(d-1)$ be the translation of $g$ by $-s/(d-1)$. Then the critical values of $h$ add up to $0$. Let $\mathcal F'=\ker\,(\mathrm{tr}:h_\star\bar{\mathbb Q}_\ell\to\bar{\mathbb Q}_\ell)=\tau_{-s/(d-1)}\mathcal F$, then its Fourier transform $\mathcal G'$ is $\mathcal G\otimes{\mathcal L}_{\psi_{-s/(d-1)}}$, so $\mathcal G=\mathcal G'\otimes{\mathcal L}_{\psi_{s/(d-1)}}$. Therefore ${\mathcal L}_{\psi_{-a}}\otimes\mathrm{Sym}^{r-i}\mathcal G\otimes\wedge^i\mathcal G={\mathcal L}_{\psi_{rs/(d-1)-a}}\otimes\mathrm{Sym}^{r-i}\mathcal G'\otimes\wedge^i\mathcal G'$. As seen above, the monodromy group of $\mathcal G'$ is either $\mathrm{SL}(V)$ or $\{\pm 1\}\times\mathrm{SL}(V)$ and does not have characters of order $p$, so ${\mathcal L}_{\psi_{rs/(d-1)-a}}\otimes\mathrm{Sym}^{r-i}\mathcal G'\otimes\wedge^i\mathcal G'$ can only have non-zero invariants for $a=rs/(d-1)$, and in that case only for $r=d-1$ and $d$ odd \cite[Corollary 4.2]{rlwan2010}.
\end{proof}
\begin{exa}
Let $g\in k[x]$ be a polynomial of degree $d\geq 3$ prime to $p$, $\psi:k\to\bar{\mathbb Q}_\ell^\star$ a non-trivial additive character and $\mathcal P={\mathcal L}_{\psi(g)}[1]$. Suppose that $p>2d+1$ and $g(x+a)+b$ is not odd for any $a,b\in\bar k$. Let $a_{d-1}$ be the coefficient of $x^{d-1}$ in $g$. Then
\begin{enumerate}
\item $\mathbb A^1_k-\{\frac{ra_{d-1}}{d}\}\subseteq U_{\mathcal P,r}$ for any $r\geq 1$.
\item $\frac{ra_{d-1}}{d}\in U_{\mathcal P,r}$ for any $r\neq d-1$.
\end{enumerate}
In all such cases, for $t\in U_{\mathcal P,r}(k_m)$,
$$
\left| \sum_{\mathrm {Tr}_{k_{mr}/k_m}(x)=t}\psi(\mathrm {Tr}_{k_{mr}/k}(g(x)))\right|\leq\frac{1}{d-1}\sum_{i=0}^{r-1} {{d+r-i-2}\choose{r}}{{r-1}\choose i} q^{\frac{m(r-1)}{2}}.
$$
\end{exa}
\begin{proof} The Swan conductor of ${\mathcal L}_{\psi(g)}$ at infinity is $d>1$, so $\mathcal P$ is not geometrically an Artin-Schreier object. Let $h(x)=g(x-\frac{a_{d-1}}{d})$. Then the coefficient of $x^{d-1}$ in $h$ is $0$. Let $\mathcal P'={\mathcal L}_{\psi(h)}[1]=\tau_{-a_{d-1}/d}\mathcal P$, $\mathcal G[1]$ and $\mathcal G'[1]$ the Fourier transforms of $\mathcal P$ and $\mathcal P'$ respectively. Then $\mathcal G'=\mathcal G\otimes{\mathcal L}_{\psi_{-a_{d-1}/d}}$ and ${\mathcal L}_{\psi_{-a}}\otimes\mathrm{Sym}^{r-i}\mathcal G\otimes\wedge^i\mathcal G={\mathcal L}_{\psi_{ra_{d-1}/d-a}}\otimes\mathrm{Sym}^{r-i}\mathcal G'\otimes\wedge^i\mathcal G'$. By \cite[Theorem 19]{katz-monodromy}, the monodromy group of $\mathcal G'$ is $G=\mathrm{SL}(V)$. Since $\mathrm{Sym}^{r-i}V\otimes\wedge^i V=\mathrm{Hom}(\wedge^{d-1-i} V,\mathrm{Sym}^{r-i} V)$ only has $\mathrm{SL}(V)$-invariants for $r=d-1$ and $G$ does not have characters of order $p$, we conclude that $\mathrm H^2_c(\mathbb A^1_{\bar k},{\mathcal L}_{\psi_{-a}}\otimes\mathrm{Sym}^{r-i}\mathcal G\otimes\wedge^i\mathcal G)$ vanishes as long as $a\neq \frac{ra_{d-1}}{d}$ or $r\neq d-1$.
In that case, since $\mathrm R(\wedge^i\rho)\mathcal G$ is smooth on $\mathbb A^1_k$ and all its slopes at infinity are $\leq \frac{d}{d-1}$ (because all slopes of $\mathcal G$ at infinity are equal to $\frac{d}{d-1}$ by \cite[Theorem 7.5.4]{katz1990esa}), by the Euler-Poincar\'e formula $\dim\,\mathrm H^1_c(\mathbb A^1_{\bar k},{\mathcal L}_{\psi_a}\otimes\mathrm R(\wedge^i\rho)\mathcal G)=\mathrm{Swan}_\infty({\mathcal L}_{\psi_a}\otimes\mathrm R(\wedge^i\rho)\mathcal G)-\mathrm{rank}({\mathcal L}_{\psi_a}\otimes\mathrm R(\wedge^i\rho)\mathcal G)\leq (\frac{d}{d-1}-1){{d+r-i-2}\choose{r}}{{r-1}\choose i}$ (cf. remark \ref{rank}), so
$$
C_{\mathcal P,r}\leq\frac{1}{d-1}\sum_{i=0}^{r-1} {{d+r-i-2}\choose{r}}{{r-1}\choose i}.
$$
\end{proof}
\begin{exa}
Let $g\in k[x]$ be a polynomial of degree $d\geq 3$ prime to $p$, $\psi:k\to\bar{\mathbb Q}_\ell^\star$ a non-trivial additive character and $\mathcal P={\mathcal L}_{\psi(g)}[1]$. Suppose that $p>2d+1$ and $g(x+a)+b$ is odd for some $a,b\in\bar k$ (so $d$ is necessarily odd). Let $a_{d-1}$ be the coefficient of $x^{d-1}$ in $g$. Then
\begin{enumerate}
\item $\mathbb A^1_k-\{\frac{ra_{d-1}}{d}\}\subseteq U_{\mathcal P,r}$ for any $r\geq 1$.
\item $\frac{ra_{d-1}}{d}\in U_{\mathcal P,r}$ for any $r\neq 2t$ for $t=1,\ldots,\frac{d-1}{2}$.
\end{enumerate}
In all such cases, for $t\in U_{\mathcal P,r}(k_m)$,
$$
\left| \sum_{\mathrm {Tr}_{k_{mr}/k_m}(x)=t}\psi(\mathrm {Tr}_{k_{mr}/k}(g(x)))\right|\leq\frac{1}{d-1}\sum_{i=0}^{r-1} {{d+r-i-2}\choose{r}}{{r-1}\choose i} q^{\frac{m(r-1)}{2}}.
$$
\end{exa}
\begin{proof}
The proof is similar to the previous one. In this case, $G=\mathrm{Sp}(V)$ by \cite[Theorem 19]{katz-monodromy} so, by \cite[Lemma on p.18]{katz2001frobenius}, $\mathrm{Sym}^{r-i}V\otimes\wedge^i V=\mathrm{Hom}(\wedge^{i} V,\mathrm{Sym}^{r-i} V)$ only has $\mathrm{Sp}(V)$-invariants for even $r\leq d-1$.
\end{proof}
\section{Examples on $\mathbb G_{m,k}$}
In this section we will assume $G=\mathbb G_{m,k}$. As in the $\mathbb A^1_k$ case, we will first determine the convolution Adams powers of negligible objects. By lemma \ref{negligible}, such an object is a twist of an object of the form $\pi_{d\star}{\mathcal L}_\chi[1]$ where $\pi_d:\mathbb G_{m,k_d}\to\mathbb G_{m,k}$ is the projection and $\chi:k_d^\star\to\bar{\mathbb Q}_\ell^\star$ is a character which is not the pullback by the norm map of a multiplicative character of a proper subfield. In other words, $d$ is the smallest positive integer such that $q^d-1$ is a multiple of the order of $\chi$.
\begin{prop}\label{negligiblemult}
Let $\mathcal P=\pi_{d\star}{\mathcal L}_\chi[1]$, where $\chi:k_d^\star\to\bar{\mathbb Q}_\ell^\star$ is a character of order $n\geq 1$, $d$ is the smallest integer such that $n|q^d-1$ and $\pi_d:\mathbb G_{m,k_d}\to \mathbb G_{m,k}$ is the projection. Then
$$
\mathcal P^{[\ast r]}=\sum_{i=0}^{r-1}[q^{i\cdot deg}\otimes\mathcal P]
$$
for every $r\geq 1$.
\end{prop}
\begin{proof}
We will show that both sides have the same trace function, that is, that for every $m\geq 1$ and every $t\in k_m^\star$ we have
$$
f^{\mathrm{N},r}_\mathcal P(k_m,t)=\sum_{i=0}^{r-1}q^{mi} f_\mathcal P(k_m,t).
$$
By definition of $\mathcal P$, we have
$$
f_\mathcal P(k_m,t)=\left\{\begin{array}{ll}
0 & \mbox{if }n\not |q^m-1 \\
-\sum_{i=0}^{d-1}\chi^{q^i}(\mathrm{N}_{k_m/k_d}(t)) & \mbox{if }n|q^m-1
\end{array}
\right.
$$
and
$$
f^{\mathrm{N},r}_\mathcal P(k_m,t)=\left\{\begin{array}{ll}
0 & \mbox{if }n\not |q^{mr}-1 \\
-\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}\sum_{i=0}^{d-1}\chi^{q^i}(\mathrm{N}_{k_{mr}/k_d}(u)) & \mbox{if }n|q^{mr}-1
\end{array}
\right.
$$
If $n\not |q^{mr}-1$, the equality is obvious. If $n|q^m-1$, the left hand side is
$$
\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}-\sum_{i=0}^{d-1}\chi^{q^i}(\mathrm{N}_{k_{mr}/k_d}(u))=\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}-\sum_{i=0}^{d-1}\chi^{q^i}(\mathrm{N}_{k_m/k_d}(t))=
$$
$$
=\frac{q^{mr}-1}{q^m-1}\cdot f_\mathcal P(k_m,t)=\left(\sum_{i=0}^{r-1}\kappa_{q^i}(k_m,t)\right)\cdot f_\mathcal P(k_m,t)
$$
so the equality holds. It remains to prove that $f^{r,\times}_\mathcal P(k_m,t)=0$ in the case where $n|q^{mr}-1$ but $n\not|q^m-1$.
In that case, we claim that there is an element $u_0\in k_{mr}$ such that $\mathrm{N}_{k_{mr}/k_m}(u_0)=1$ but $u_0^{\frac{q^{mr}-1}{n}}\neq 1$. Otherwise, the polynomial $x^{\frac{q^{mr}-1}{q^m-1}}-1$ would divide $x^{\frac{q^{mr}-1}{n}}-1$, so $\frac{q^{mr}-1}{q^m-1}$ would divide $\frac{q^{mr}-1}{n}$, which is impossible since $n$ does not divide $q^m-1$. Then $\mathrm{N}_{k_{mr}/k_m}(u)=\mathrm{N}_{k_{mr}/k_m}(uu_0)$, so for every $i=0,\ldots,d-1$
$$
\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}\chi^{q^i}(\mathrm{N}_{k_{mr}/k_d}(u)) =\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}\chi^{q^i}(\mathrm{N}_{k_{mr}/k_d}(uu_0))=
$$
$$
=\chi^{q^i}(\mathrm{N}_{k_{mr}/k_d}(u_0))\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}\chi^{q^i}(\mathrm{N}_{k_{mr}/k_d}(u)).
$$
Now since the character $\chi^{q^i}\circ\mathrm{N}_{k_{mr}/k_d}$ of $k_{mr}^\star$ has order $n$ and $u_0^{\frac{q^{mr}-1}{n}}\neq 1$ it follows that $\chi^{q^i}(\mathrm{N}_{k_{mr}/k_d}(u_0))\neq 1$, and $\sum_{\mathrm{N}_{k_{mr}/k_m}(u)=t}\chi^{q^i}(\mathrm{N}_{k_{mr}/k_d}(u))$ must then be zero. So $f^{\mathrm{N},r}_\mathcal P(k_m,t)=0$.
\end{proof}
In order to compute explicitely the $r$-th norm $L$-function of a given perverse sheaf, we first split the negligible components from the non-negligible ones. For the negligible components proposition \ref{negligible} gives us an exact formula, so let us focus on the non-negligible objects.
\begin{prop}\label{Sr}
Let $\mathcal F\in{\mathcal Sh}(\GG_{m,k},\QQ)$ be a geometrically semisimple middle extension sheaf without negligible components which is tamely ramified at both $0$ and $\infty$, let $S\subseteq\bar k^\star$ be the set of finite ramification points of $\mathcal F$ and $S^r:=S\cdots S$ ($r$ factors). Then for every $r\geq 1$, $\bar k-S^r\subseteq U_{\mathcal F[1],r}(\bar k)$.
\end{prop}
\begin{proof}
By \cite[Lemma 19.5]{katz2010mellin}, ${\mathcal H}^0(\mathcal F[1]^{\ast r})$ vanishes on $\mathbb G_{m,k}-S^r$, so the same is true for ${\mathcal H}^0(\mathrm R(\ast\wedge^i\rho)\mathcal F[1])={\mathcal Hom}_{{\mathfrak S}_r}(\wedge^i\rho,{\mathcal H}^0(\mathcal F[1]^{\ast r}))$ for every $i=0,\ldots,r-1$.
\end{proof}
\begin{prop}\label{tame}
Let $\mathcal F,\mathcal G\in{\mathcal Sh}(\GG_{m,k},\QQ)$ be geometrically semisimple middle extension sheaves without negligible components which are everywhere tamely ramified. Then ${\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])$ is everywhere tamely ramified.
\end{prop}
\begin{proof}
Let $S\subseteq\bar k^\star$ (respectively $T\subseteq \bar k^\star$) be the set of ramification points of $\mathcal F$ (resp. $\mathcal G$). Let $m$ (resp. $n$) be the generic rank of $\mathcal F$ (resp. $\mathcal G$), and for every $s\in S$ (resp. $t\in T$) let $m_s$ (resp. $n_t$) be the rank of $\mathcal F$ at $s$ (resp. the rank of $\mathcal G$ at $t$). By the Euler-Poincar\'e formula,
$$
\chi(\mathcal F[1])=\sum_{s\in S}(m-m_s),
$$
$$
\chi(\mathcal G[1])=\sum_{t\in T}(n-n_t)
$$
and
$$
\chi(\mathcal F[1]\ast\mathcal G[1])=\chi(\mathcal F[1])\cdot\chi(\mathcal G[1])=\left(\sum_{s\in S}(m-m_s)\right)\left(\sum_{t\in T}(n-n_t)\right).
$$
By \cite[Lemma 19.5]{katz2010mellin}, $\mathcal F[1]\ast\mathcal G[1]$ is smooth on $\bar k^\star-ST$. If $u\in \bar k^\star-ST$, the rank of ${\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])$ at $u$ is $-\chi(\mathbb G_{m,\bar k},\mathcal F\otimes\phi_u^\star\mathcal G)$, where $\phi_u:\mathbb G_{m,\bar k}\to\mathbb G_{m,\bar k}$ is the automorphism defined by $t\mapsto u/t$. Since, at every point of $\mathbb G_{m,\bar k}$, at least one of $\mathcal F$, $\phi_u^\star\mathcal G$ is smooth, and each local term in the Euler-Poincar\'e formula \cite[Expos\'e X, Corollaire 7.12]{grothendieck1977cohomologie} gets multiplied by $d$ upon tensoring with a smooth sheaf of rank $d$, we conclude that
$$
-\chi(\mathbb G_{m,\bar k},\mathcal F\otimes\phi_u^\star\mathcal G)=-m\cdot\chi(\mathbb G_{m,\bar k},\mathcal G)-n\cdot\chi(\mathbb G_{m,\bar k},\mathcal F)=
$$
\begin{equation}\label{Euler}
=m\sum_{t\in T}(n-n_t)+n\sum_{s\in S}(m-m_s).
\end{equation}
This is the generic rank of ${\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])$. Now let $u\in ST$, and let $R_u$ be the set of pairs $(s,t)\in S\times T$ such that $u=st$. Then $\dim{\mathcal H}^0(\mathcal F[1]\ast\mathcal G[1])_u-\dim{\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])_u=\chi(\mathbb G_{m,\bar k},\mathcal F\otimes\phi_u^\star\mathcal G)$. Again by the Euler-Poincar\'e formula, we have
$$
\chi(\mathbb G_{m,\bar k},\mathcal F\otimes\phi_u^\star\mathcal G)=-\sum_{(s,t)\in R_u}(mn-m_sn_t)-\sum_{s\in S'_u}(mn-m_sn)-\sum_{t\in T'_u}(mn-mn_t)
$$
where $S'_u$ (resp $T'_u$) is the set of $s\in S$ such that $u/s\notin T$ (resp. the set of $t\in T$ such that $u/t\notin S$).
The Euler characteristic of $\mathcal F[1]\ast\mathcal G[1]$ is then
$$
\chi(\mathbb G_{m,\bar k},{\mathcal H}^0(\mathcal F[1]\ast\mathcal G[1]))-\chi(\mathbb G_{m,\bar k},{\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1]))=
$$
$$
=\sum_{u\in ST}(\dim{\mathcal H}^0(\mathcal F[1]\ast\mathcal G[1])_u-\dim{\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])_u)+
$$
$$
+\#ST\cdot\mathrm{gen.rank}({\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1]))+\sum_{u\in ST\cup\{0,\infty\}}\mathrm{Swan}_u{\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])=
$$
$$
=\sum_{u\in ST}\left(m\sum_{t\in T}(n-n_t)+n\sum_{s\in S}(m-m_s)-\sum_{(s,t)\in R_u}(mn-m_sn_t)\right.
$$
$$
-\left.\sum_{s\in S'_u}(mn-m_sn)-\sum_{t\in T'_u}(mn-mn_t)\right)+\sum_{u\in ST\cup\{0,\infty\}}\mathrm{Swan}_u{\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])=
$$
$$
=\sum_{u\in ST}\sum_{(s,t)\in R_u}(m-m_s)(n-n_t)+\sum_{u\in ST\cup\{0,\infty\}}\mathrm{Swan}_u{\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])=
$$
$$
=\left(\sum_{s\in S}(m-m_s)\right)\left(\sum_{t\in T}(n-n_t)\right)+\sum_{u\in ST\cup\{0,\infty\}}\mathrm{Swan}_u{\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1]).
$$
Comparing with (\ref{Euler}), we conclude that $\sum_{u\in ST\cup\{0,\infty\}}\mathrm{Swan}_u{\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])=0$, that is, ${\mathcal H}^{-1}(\mathcal F[1]\ast\mathcal G[1])$ is everywhere tamely ramified.
\end{proof}
\begin{cor}
Let $\mathcal F\in{\mathcal Sh}(\GG_{m,k},\QQ)$ be an everywhere tamely ramified geometrically semisimple middle extension sheaf without negligible components. Then ${\mathcal H}^{-1}(\mathrm R(\ast\rho)\mathcal F[1])$ is everywhere tamely ramified for every $r\geq 1$ and every representation $\rho$ of ${\mathfrak S}_r$.
\end{cor}
\begin{rem}{\rm
Lemma \cite[Lemma 19.5]{katz2010mellin} is proved for the ``middle convolution'' (i.e. the image of the ``forget supports'' map $\mathcal F[1]\ast_!\mathcal G[1]\to \mathcal F[1]\ast_\star \mathcal G[1]$), while we are using ``regular'' $!$-convolution here. However, by \cite[Proposition 3.6.4]{gabber1996faisceaux}, the mapping cone of the forget supports map (and therefore the kernel of the surjective map $\mathcal F[1]\ast_!\mathcal G[1]\to\mathcal F[1]\ast_{mid}\mathcal G[1]$) is a succesive extension of Kummer objects ${\mathcal L}_\chi[1]$ and, in particular, is smooth on $\mathbb G_{m,\bar k}$, so the ramification points (and the non-trivial part of the inertia action at those points) of $\mathcal F[1]\ast_!\mathcal G[1]$ and $\mathcal F[1]\ast_{mid}\mathcal G[1]$ are the same.
}
\end{rem}
We will now try to find a good estimate for the constant $C_{\mathcal F[1],r}$ for an everywhere tamely ramified middle extension sheaf $\mathcal F$.
\begin{lem}
Let $\mathcal F\in{\mathcal Sh}(\GG_{m,k},\QQ)$ be an everywhere tamely ramified middle extension sheaf without negligible components, pure of weight $0$. For every (possibly trivial) character $\chi$ of $\bar k^\star$ and every $j\geq 1$, let $n_{\chi,j}$ be the number of Jordan blocks of size $j$ with eigenvalue $\chi$ in the local monodromies of $\mathcal F$ at $0$ and $\infty$, $n:=\dim\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)$ and $n_{\chi,0}:=n-\sum_{j\geq 1}n_{\chi,j}$. Let $J_\chi=\{j\geq 0|n_{\chi,j}>0\}$. Then the generic rank of ${\mathcal H}^{-1}(\mathrm{Sym}^{\ast r}\mathcal F[1])$ is bounded by
\begin{equation}\label{A}
A_{\mathcal F,r}:=\sum_\chi\sum_{(i_j)\in{\mathbb Z}_{\geq 0}^{J_\chi},\sum i_j=r}\prod_{j\in J_\chi}{{n_{\chi,j}+i_j-1}\choose{i_j}}\sum_{j\in J_\chi}ji_j.
\end{equation}
\end{lem}
\begin{proof}
Notice that the sum is actually finite, since there are only finitely many characters $\chi$ for which $J_\chi\neq\{0\}$. We will show that for any $\chi$ the sum of the sizes of the Jordan blocks in the monodromies of ${\mathcal H}^{-1}(\mathrm{Sym}^{\ast r}\mathcal F[1])$ at $0$ and $\infty$ associated to the character $\chi$ is bounded above by
$$
2\cdot\sum_{(i_j)\in{\mathbb Z}_{\geq 0}^{J_\chi},\sum i_j=r}\prod_{j\in J_\chi}{{n_{\chi,j}+i_j-1}\choose{i_j}}\sum_{j\in J_\chi}ji_j.
$$
Since, by proposition \ref{tame}, ${\mathcal H}^{-1}(\mathrm{Sym}^{\ast r}\mathcal F[1])$ is everywhere tamely ramified, its monodromy at $0$ and $\infty$ is a direct sum of such Jordan blocks. Therefore the sum of these quantities for all characters $\chi$ is twice the rank of ${\mathcal H}^{-1}(\mathrm{Sym}^{\ast r}\mathcal F[1])$, which proves the lemma.
Fix one such $\chi$. By tensoring $\mathcal F$ with the Kummer sheaf ${\mathcal L}_{\bar\chi}$ (which does not change the hypotheses of the lemma), we can assume without loss of generality that $\chi={\mathbf 1}$ is the trivial character. Let $n_j=n_{{\mathbf 1},j}$ for $j\geq 0$. From the exact sequence of sheaves
$$
0\to j_!\mathcal F\to j_\star\mathcal F\to i_{0\star}\mathcal F^{I_0}\oplus i_{\infty\star}\mathcal F^{I_\infty}\to 0
$$
where $j:\mathbb G_{m,k}\to\mathbb P^1_k$, $i_0:\{0\}\to\mathbb P^1_k$ and $i_\infty:\{\infty\}\to\mathbb P^1_k$ are the inclusions, we get an exact sequence
$$
\mathrm H^0(\mathbb P^1_{\bar k},j_\star\mathcal F)\to\mathcal F^{I_0}\oplus\mathcal F^{I_\infty}\to\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)\to\mathrm H^1(\mathbb P^1_{\bar k},j_\star\mathcal F)\to 0.
$$
Since $\mathcal F$ is a middle extension pure of weight $0$, by \cite[Th\'eor\`eme 3.2.3]{deligne1980conjecture} the latter group is pure of weight $1$. On the other hand $\mathcal F^{I_0}\oplus\mathcal F^{I_\infty}$ is mixed of weights $\leq 0$, and in fact (cf. \cite[1.8]{deligne1980conjecture}, \cite[Theorem 7.0.7]{katz1988gauss}) the Frobenius action has $n_{j}$ eigenvalues of weight $1-j$ for every $j\geq 1$. Finally since $\mathcal F$ has no negligible (and in particular constant) components, the first group vanishes. We conclude that $\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)$ has $n_{j}$ Frobenius eigenvalues of weight $1-j$ for every $j\geq 0$.
Write (the semisimplification of) $\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)$ as $\bigoplus_{j\in J_{\mathbf 1}} W_j$, where $W_j$ is pure of weight $1-j$ and $\dim W_j=n_{j}$. Then by proposition \ref{mainprop}
$$
\mathrm H^0_c(\mathbb G_{m,\bar k},\mathrm{Sym}^{\ast r}\mathcal F[1])=\mathrm{Sym}^r\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)=
$$
$$
=\bigoplus_{(i_j)\in{\mathbb Z}_{\geq 0}^{J_{\mathbf 1}},\sum i_j=r}(\bigotimes_{j\in J_{\mathbf 1}}\mathrm{Sym}^{i_j}W_j)
$$
where $\bigotimes_{j\in J_{\mathbf 1}}\mathrm{Sym}^{i_j}W_j$ is pure of weight $\sum_{j\in J_{\mathbf 1}}i_j(1-j)$ and dimension $\prod_{j\in J_{\mathbf 1}}{{n_j+i_j-1}\choose{i_j}}$.
Let $\mathcal G={\mathcal H}^{-1}(\mathrm{Sym}^{\ast r}\mathcal F[1])$ and consider the exact sequence
$$
0\to\mathrm H^0(\mathbb P^1_{\bar k},j_\star\mathcal G)\to\mathcal G^{I_0}\oplus\mathcal G^{I_\infty}\to\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal G)\to\mathrm H^1(\mathbb P^1_{\bar k},j_\star\mathcal G)\to 0.
$$
Since $\mathcal G$ is mixed of weights $\leq r-1$, every unipotent Jordan block of size $e$ in the monodromy of $\mathcal G$ at $0$ or $\infty$ contributes an eigenvalue of weight $\leq r-e$ to $\mathcal G^{I_0}\oplus\mathcal G^{I_\infty}$. This gives a corresponding eigenvalue of $\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal G)$ \emph{except} when it arises from something in $\mathrm H^0(\mathbb P^1_{\bar k},j_\star\mathcal G)$. But in that case, the eigenvalue appears in both monodromies at $0$ and $\infty$. So in the worst case, every two unipotent Jordan blocks of size $e$ give an eigenvalue of weight $\leq r-e$ in $\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal G)$. We conclude that the sum of the sizes of the unipotent Jordan blocks in the monodromies of $\mathcal G$ at $0$ and $\infty$ is bounded by $2\cdot\sum_\lambda(r-w(\lambda))$, where the sum is taken over all Frobenius eigenvalues of $\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal G)$ and $w(\lambda)$ is the weight of the eigenvalue $\lambda$.
If the trivial character does not appear in the local monodromies of $\mathcal F$ at $0$ and $\infty$ then $\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)$ is pure of weight $1$, and therefore $\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal G)\hookrightarrow\mathrm H^0_c(\mathbb G_{m,\bar k},\mathrm{Sym}^{\ast r}\mathcal F[1])=\mathrm{Sym}^r\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)$ is pure of weight $r-1$. So the trivial character does not appear in the local monodromies of $\mathcal G$ either. Otherwise, for every $(i_j)\in{\mathbb Z}_{\geq 0}^{J_{\mathbf 1}}$ such that $\sum_j i_j=r$ we get at most $\prod_{j\in J_{\mathbf 1}}{{n_j+i_j-1}\choose{i_j}}$ eigenvalues of weight $\sum_{j\in J_{\mathbf 1}}i_j(1-j)$ (since $\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal G)\hookrightarrow\mathrm H^0_c(\mathbb G_{m,\bar k},\mathrm{Sym}^{\ast r}\mathcal F[1])=\mathrm{Sym}^r\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)$). So
$$
2\cdot\sum_\lambda(r-w(\lambda))\leq 2\cdot\sum_{(i_j)\in{\mathbb Z}_{\geq 0}^{J_{\mathbf 1}},\sum_j i_j=r}\prod_{j\in J_{\mathbf 1}}{{n_j+i_j-1}\choose{i_j}}(r-\sum_{j\in J_{\mathbf 1}}i_j(1-j))=
$$
$$
=2\cdot\sum_{(i_j)\in{\mathbb Z}_{\geq 0}^{J_{\mathbf 1}},\sum_j i_j=r}\prod_{j\in J_{\mathbf 1}}{{n_j+i_j-1}\choose{i_j}}\sum_{j\in J_{\mathbf 1}}ji_j.
$$
\end{proof}
In the same way one can prove
\begin{lem}
With the notation and hypotheses of the previous lemma, the generic rank of ${\mathcal H}^{-1}(\wedge^{\ast r}\mathcal F[1])$ is bounded by
\begin{equation}\label{B}
B_{\mathcal F,r}:=\sum_\chi\sum_{(i_j)\in{\mathbb Z}_{\geq 0}^{J_\chi},\sum i_j=r}\prod_{j\in J_\chi}{{n_j}\choose{i_j}}\sum_{j\in J_\chi}ji_j.
\end{equation}
\end{lem}
\begin{prop}
With the notation and hypotheses of the previous lemmas, the generic rank of ${\mathcal H}^{-1}(\mathrm{Sym}^{\ast (r-i)}\mathcal F[1]\ast\wedge^{\ast i}\mathcal F[1])$ is bounded by
\begin{equation}\label{C}
M_{\mathcal F,r,i}:= A_{\mathcal F,r-i}{n\choose i}+B_{\mathcal F,i}{{n-1+r-i}\choose {r-i}}.
\end{equation}
\end{prop}
\begin{proof}
Let ${\mathcal A}_j={\mathcal H}^{-j}(\mathrm{Sym}^{\ast(r-i)}\mathcal F[1])[j]$ and ${\mathcal B}_j={\mathcal H}^{-j}(\wedge^{\ast i}\mathcal F[1])[j]$ for $j=0,1$. Then the generic rank of ${\mathcal H}^{-1}(\mathrm{Sym}^{\ast (r-i)}\mathcal F[1]\ast\wedge^{\ast i}\mathcal F[1])$ is less than or equal to the sum of the generic ranks of ${\mathcal H}^{-1}({\mathcal A}_j\ast{\mathcal B}_k)$ for $j,k\in\{0,1\}$. For $i=j=0$ it is a punctual object, so its ${\mathcal H}^{-1}$ vanishes. For $j=1$, $k=0$, the generic rank of ${\mathcal A}_1$ gets multiplied by the dimension of the punctual object ${\mathcal B}_0$, and similarly for $j=0$, $k=1$.
For $j=k=1$, by (the proof of) \cite[Theorem 26.1]{katz2010mellin}, there is an inequality
$$
\mathrm{gen.rank}\,{\mathcal H}^{-1}({\mathcal A}_{1}\ast{\mathcal B}_{1})\leq
$$
$$
\leq (\mathrm{gen.rank}\,{\mathcal A}_{1})\dim\mathrm H^0_c(\mathbb G_{m,\bar k},{\mathcal B}_{1})
+ (\mathrm{gen.rank}\,{\mathcal B}_{1})\dim\mathrm H^0_c(\mathbb G_{m,\bar k},{\mathcal A}_{1})
$$
so
$$
\mathrm{gen.rank}\,{\mathcal H}^{-1}(\mathrm{Sym}^{\ast (r-i)}\mathcal F[1]\ast\wedge^{\ast i}\mathcal F[1])\leq
$$
$$
\leq (\mathrm{gen.rank}\,\mathrm{Sym}^{\ast(r-i)}\mathcal F[1])(\dim\mathrm H^0_c(\mathbb G_{m,\bar k},{\mathcal B}_{0})+\dim\mathrm H^0_c(\mathbb G_{m,\bar k},{\mathcal B}_{1}))+
$$
$$
+ (\mathrm{gen.rank}\,\wedge^{\ast i}\mathcal F[1])(\dim\mathrm H^0_c(\mathbb G_{m,\bar k},{\mathcal A}_{0})+\dim\mathrm H^0_c(\mathbb G_{m,\bar k},{\mathcal A}_{1}))=
$$
$$
= (\mathrm{gen.rank}\,\mathrm{Sym}^{\ast(r-i)}\mathcal F[1])\dim\mathrm H^0_c(\mathbb G_{m,\bar k},\wedge^{\ast i}\mathcal F[1])
$$
$$
+ (\mathrm{gen.rank}\,\wedge^{\ast i}\mathcal F[1])\dim\mathrm H^0_c(\mathbb G_{m,\bar k},\mathrm{Sym}^{\ast(r-i)}\mathcal F[1])=
$$
$$
=(\mathrm{gen.rank}\,\mathrm{Sym}^{\ast(r-i)}\mathcal F[1])\dim\wedge^i\mathrm H^0_c(\mathbb G_{m,\bar k},\mathcal F[1])
$$
$$
+ (\mathrm{gen.rank}\,\wedge^{\ast i}\mathcal F[1])\dim\mathrm{Sym}^{r-i}\mathrm H^0_c(\mathbb G_{m,\bar k},\mathcal F[1]).
$$
so the result follows from the previous two lemmas.
\end{proof}
\begin{cor}\label{Cmult}
Let $\mathcal F\in{\mathcal Sh}(\GG_{m,k},\QQ)$ be an everywhere tamely ramified middle extension sheaf without negligible components, pure of weight $0$. Then
$$
C_{\mathcal F[1],r}\leq\frac{1}{2}\sum_{i=0}^r M_{\mathcal F,r,i}=\frac{1}{2}\sum_{i=0}^r\left(A_{\mathcal F,r-i}{n\choose i}+B_{\mathcal F,i}{{n-1+r-i}\choose {r-i}}\right).
$$
\end{cor}
\begin{proof}
By definition
$$
C_{\mathcal F[1],r}=\sum_{i=0}^{r-1}\mathrm{gen.rank}\;{\mathcal H}^{-1}(\mathrm R(\ast\wedge^i\rho)\mathcal F[1]).
$$
Using that $\mathrm{Sym}^{\ast (r-i)}\mathcal F[1]\ast\wedge^{\ast i}\mathcal F[1]=\mathrm R(\ast(\wedge^{i-1}\rho\oplus\wedge^i\rho))\mathcal F[1]
=\mathrm R(\ast\wedge^{i-1}\rho)\mathcal F[1]\oplus\mathrm R(\ast\wedge^i\rho)\mathcal F[1]$ we get $$\mathrm{gen.rank}\;{\mathcal H}^{-1}(\mathrm{Sym}^{\ast (r-i)}\mathcal F[1]\ast\wedge^{\ast i}\mathcal F[1])=
$$
$$=\mathrm{gen.rank}\;{\mathcal H}^{-1}(\mathrm R(\ast\wedge^{i-1}\rho)\mathcal F[1])+\mathrm{gen.rank}\;{\mathcal H}^{-1}(\mathrm R(\ast\wedge^i\rho)\mathcal F[1])$$
for $i=0,\ldots,r$. Taking the sum over all $i=0,\ldots,r$ we deduce:
$$
2\cdot C_{\mathcal F[1],r}=\sum_{i=0}^r\mathrm{gen.rank}\;{\mathcal H}^{-1}(\mathrm{Sym}^{\ast (r-i)}\mathcal F[1]\ast\wedge^{\ast i}\mathcal F[1]).
$$
We conclude by the previous proposition.
\end{proof}
\begin{exa} \label{normexa1} Let $g\in k[x]$ be a square-free polynomial of degree $d$ prime to $p$ such that $g'$ has no factors with multiplicity $\geq p$, and let $\mathcal F\in{\mathcal Sh}(\GG_{m,k},\QQ)$ be the kernel of the trace map $g_\star\bar{\mathbb Q}_\ell\to\bar{\mathbb Q}_\ell$. Then $t\in U_{\mathcal F[1],r}(\bar k)$ for every $t$ which is not a product of $r$ critical values of $g$. In particular, for every such $t\in k_m^\star$, we have an estimate
$$
\left|\#\{x\in k_{mr}|\mathrm{N}_{k_{mr}/k_m}g(x)=t\}-\frac{q^{mr}-1}{q^m-1}\right|\leq C_{\mathcal F[1],r} q^{\frac{m(r-1)}{2}}.
$$
Morover, we have a bound
$$C_{\mathcal F[1],r}\leq\frac{1}{2}\sum_{i=0}^r{{d-1}\choose i}\left[(r+i){{d-2+r-i}\choose{r-i}}+(d-1)\sum_{j=0}^{r-i-1}{{d-3+j}\choose{j}}(r-i-j)\right].$$
\end{exa}
\begin{rem}\label{compare}\emph{The given bound for $C_{\mathcal F[1],r}$ is polynomial in $r$ if $d$ is fixed. Compare with \cite[Theorem 3.2]{rl2010number}, where the (exponential in $r$) bound $r(d-1)^r$ was obtained using Weil descent.}
\end{rem}
\begin{proof}
The first part is a consequence of propositions \ref{negligible} and \ref{Sr}, since $\mathcal F$ is tamely ramified everywhere and has no negligible components (because $\mathcal F$ is a middle extension sheaf whose monodromy is trivial at $0$ and splits as the direct sum of all non-trivial characters with trivial $d$-th power at infinity). The estimate follows from the equality $f_{g_\star\bar{\mathbb Q}_\ell}^{\mathrm{N},r}=f_{\bar{\mathbb Q}_\ell}^{\mathrm{N},r}+f_{\mathcal F}^{\mathrm{N},r}$.
From the exact sequence
$$
0\to\mathcal F\to g_\star\bar{\mathbb Q}_\ell\to\bar{\mathbb Q}_\ell\to 0
$$
we deduce $\dim\mathrm H^1_c(\mathbb G_{m,\bar k},\mathcal F)=\dim\mathrm H^1_c(\mathbb A^1_{\bar k},\mathcal F)+\dim(\mathcal F_0)=\dim\mathrm H^1_c(\mathbb A^1_{\bar k},g_\star\bar{\mathbb Q}_\ell)-\dim\mathrm H^1_c(\mathbb A^1_{\bar k})+\dim(\mathcal F_0)=\dim(\mathcal F_0)=d-1$.
If $\chi={\mathbf 1}$ is the trivial character, $n_{\chi,1}=d-1$ and $n_{\chi,j}=0$ for every $j\neq 1$. If $\chi\neq{\mathbf 1}$ but $\chi^d={\mathbf 1}$, $n_{\chi,1}=1$, $n_{\chi,0}=d-2$ and $n_{\chi,j}=0$ for every $j>1$. By equations (\ref{A}) and (\ref{B}) we get, for every $i=0,\ldots,r$:
$$
A_{\mathcal F,r-i}={{d-2+r-i}\choose{r-i}}(r-i)+(d-1)\sum_{j=0}^{r-i-1}{{d-3+j}\choose{j}}(r-i-j)
$$
and
$$
B_{\mathcal F,i}= {{d-1}\choose i}\cdot i+(d-1){{d-2}\choose{i-1}}=2i{{d-1}\choose i}
$$
so
$$
M_{\mathcal F,r,i}={{d-1}\choose i}\left[(r+i){{d-2+r-i}\choose{r-i}}+(d-1)\sum_{j=0}^{r-i-1}{{d-3+j}\choose{j}}(r-i-j)\right].
$$
We conclude by corollary \ref{Cmult}.
\end{proof}
\begin{rem}\label{remark}\emph{If $g$ is square-free, it follows from \cite[Lemma 3.1]{rl2010number} that ${\mathcal H}^0(\mathcal F[1]^{\ast r})=0$. In particular, $U_{\mathcal F[1],r}=\mathbb G_{m,\bar k}$ for every $r\geq 1$, so the estimate holds for every $m\geq 1$ and every $t\in k_m^\star$.}
\end{rem}
\begin{exa} \label{normexa2} Let $g\in k[x]$ be a polynomial of degree $d$ prime to $p$ and $\chi:k^\star\to\bar{\mathbb Q}_\ell^\star$ a non-trivial multiplicative character of order $n$. Suppose that $g$ is not a power of $x$, and no root of $g$ has multiplicity divisible by $n$. Let $\mathcal F={\mathcal L}_{\chi(g)}$. Then $t\in U_{\mathcal F[1],r}(\bar k)$ for every $t$ which is not a product of $r$ roots of $g$. In particular, for every such $t\in k_m^\star$, we have an estimate
$$
\left|\sum_{\mathrm{N}_{k_{mr}/k_m}(x)=t}\chi(\mathrm{N}_{k_{mr}/k}(g(x)))\right|\leq C_{\mathcal F[1],r} q^{\frac{m(r-1)}{2}}.
$$
Let $e$ be the largest power of $x$ that divides $g(x)$. Then if $\chi^e\neq\chi^d$, we have a bound
$$
C_{\mathcal F[1],r}\leq \frac{1}{2}\sum_{i=0}^r2\left({{a-1+r-i}\choose{r-i}}{{a-1}\choose{i-1}}+{a\choose i}\sum_{j=0}^{r-i-1}{{a+j-2}\choose j}(r-i-j)\right)
$$
and, if $\chi^e=\chi^d$,
$$
C_{\mathcal F[1],r}\leq \frac{1}{2}\sum_{i=0}^r\left(2{{a-1+r-i}\choose{r-i}}{{a-1}\choose{i-1}}+{a\choose i}\sum_{j=0}^{r-i-1}{{a+j-3}\choose j}(r+1-i-j)(r-i-j)\right)
$$
where $a$ is the number of distinct roots of $g$ in $\bar k^\star$.
\end{exa}
\begin{proof}
Again this is just applying proposition \ref{Sr} and formulas (\ref{A}), (\ref{B}) and (\ref{C}) for the rank. The hypotheses on $g$ imply that ${\mathcal L}_{\chi(g)}$ is a middle extension and ramified at least at one point of $\mathbb G_{m,\bar k}$, and in particular is not negligible. The dimension of $\mathrm H^1_c(\mathbb G_{m,\bar k},{\mathcal L}_{\chi(g)})$ is $a$ by the Euler-Poincar\'e formula, since ${\mathcal L}_{\chi(g)}$ is everywhere tamely ramified. The monodromy at $0$ is the character $\chi^e$, and the monodromy at infinity is $\chi^d$, hence the different bounds for $C_{\mathcal F[1],r}$ depending on them being equal or not.
\end{proof}
In order to obtain sharper results we will make use of a certain algebraic group, the equivalent to the monodromy group of the Fourier transform in the trace case. This group is defined and studied in \cite{katz2010mellin}. Given a geometrically semisimple (e.g. pure of some weight $w$) object $\mathcal P\in{\mathcal P}erv$ without negligible components, let $\langle\mathcal P\rangle$ be the full subcategory of the Tannakian category of perverse sheaves on $\mathbb G_{m,\bar k}$ modulo negligible sheaves with the convolution operator \cite[Th\'eor\`eme 3.7.5]{gabber1996faisceaux} tensor-generated by $\mathcal P\otimes\bar k$. By the fundamental theorem of Tannakian categories \cite[Theorem 2.11]{deligne1981tannakian}, $\langle\mathcal P\rangle$ is tensor-equivalent to the category of representations of a reductive algebraic group $G\subseteq\mathrm{GL}(V)$, where $V=\mathrm H^0(\mathbb A^1_{\bar k},j_{0!}\mathcal P)$ is the fibre functor evaluated at $\mathcal P$ \cite[Theorem 3.1]{katz2010mellin}. Under this equivalence, the class of $\mathcal P$ corresponds to the ``standard'' representation $G\hookrightarrow\mathrm{GL}(V)$, and the class of $\mathcal P^{\ast r}$ (respectively $\mathrm{Sym}^{\ast r}\mathcal P$, $\wedge^{\ast r}\mathcal P$) corresponds to its $r$-th tensor power (resp. its $r$-th symmetric power, its $r$-th alternating power). More generally, for every finite dimensional representation $\rho:{\mathfrak S}_r\to\mathrm{GL}(W)$ of ${\mathfrak S}_r$, the class of $\mathrm R(\ast\rho)\mathcal P$ corresponds to $\mathrm{Hom}_{{\mathfrak S}_r}(W,V^{\otimes r})$.
\begin{prop}\label{subcharacters}
Let $\mathcal P\in{\mathcal P}erv$ be a perverse sheaf pure of weight $w\in {\mathbb Z}$ without negligible components, $G\subseteq\mathrm{GL}(V)$ the corresponding reductive algebraic group and $r\geq 1$ an integer. Then
\begin{enumerate}
\item $1\in U_{\mathcal P,r}(\bar k)$ if and only the fixed subspace of the representation $\mathrm{Sym}^{r-i}V\otimes\wedge^iV$ of $G$ is zero for every $i=0,\ldots,r$.
\item Given a prime to $p$ integer $n$, the $n$-th roots of unity are in $U_{\mathcal P,r}(\bar k)$ if and only if the representation $\mathrm{Sym}^{r-i}V\otimes\wedge^iV$ of $G$ does not contain a subcharacter with trivial $n$-th power for any $i=0,\ldots,r$.
\item $U_{\mathcal P,r}=\mathbb G_{m,k}$ if and only if the representation $\mathrm{Sym}^{r-i}V\otimes\wedge^iV$ of $G$ does not contain subcharacters of finite prime to $p$ order for any $i=0,\ldots,r$.
\end{enumerate}
\end{prop}
\begin{proof}
(1) and (3) are direct consequences of (2). Let $n$ be a prime to $p$ positive integer, and let $\zeta\in \bar k^\star -U_{\mathcal P,r}(\bar k)$ such that $\zeta^n=1$. Then by definition of $U_{\mathcal P,r}$ there is some $i=0,\ldots,r-1$ such that $\zeta$ is in the support of ${\mathcal H}^0(\mathrm R(\ast\wedge^{i}\rho)\mathcal P)$. Then $\zeta$ is in the support of ${\mathcal H}^0(\mathrm{Sym}^{\ast(r-i)}\mathcal P\ast\wedge^{\ast i}\mathcal P)={\mathcal H}^0(\mathrm R(\ast\wedge^{i-1}\rho)\mathcal P\oplus\mathrm R(\ast\wedge^i\rho)\mathcal P)$ (cf. the proof of proposition \ref{young}). In other words, $\mathrm{Sym}^{\ast(r-i)}\mathcal P\ast\wedge^{\ast i}\mathcal P$ contains the punctual object $\delta_\zeta$ as an irreducible component. Regarding their classes in the Tannakian category ${\mathcal P}erv/{\mathcal N}eg$ as representations of $G$, this means that $\mathrm{Sym}^{r-i}V\otimes\wedge^{ i}V$ contains the irreducible subrepresentation associated to the object $\delta_\zeta$, which is a character with trivial $n$-th power (since $\delta_\zeta^{\ast n}=\delta_{\zeta^n}=\delta_1$ is the identity object).
Conversely, every subrepresentation of $\mathrm{Sym}^{r-i}V\otimes\wedge^{ i}V$ which is a character of order divisible by $n$ gives an irreducible component of $\mathrm{Sym}^{\ast(r-i)}\mathcal P\ast\wedge^{\ast i}\mathcal P$ of the form $\delta_\zeta$ for some $\zeta\in\bar k^\star$ \cite[Theorem 6.4]{katz2010mellin}, and $\zeta^n=1$. So, in that case, $\zeta$ is in the support of ${\mathcal H}^0(\mathrm{Sym}^{\ast(r-i)}\mathcal P\ast\wedge^{\ast i}\mathcal P)={\mathcal H}^0(\mathrm R(\ast\wedge^{i-1}\rho)\mathcal P\oplus\mathrm R(\ast\wedge^i\rho)\mathcal P)$ for some $i$, so it must be in the support of ${\mathcal H}^0(\mathrm R(\ast\wedge^i\rho)\mathcal P)$ for some $i$ and therefore is not in $U_{\mathcal P,r}(\bar k)$.
\end{proof}
\begin{cor}
Let $n$ be the order of $G/G_0$, where $G_0$ is the identity connected component of $G$. Then $\bar k^\star-\mu_n(\bar k)\subseteq U_{\mathcal P,r}(\bar k)$ for every $r\geq 1$, where $\mu_n(\bar k):=\{x\in\bar k|x^n=1\}$. In particular, if $G$ is connected, $\bar k^\star-\{1\}\subseteq U_{\mathcal P,r}(\bar k)$ for every $r\geq 1$.
\end{cor}
\begin{proof}
Let $z\in\bar k^\star$, and suppose that $z\notin U_{\mathcal P,r}(\bar k)$. Then by the proof of the previous proposition, $\mathrm{Sym}^{r-i}V\otimes\wedge^{ i}V$ contains a subcharacter of order $a$ for some $i=0,\ldots,r$, where $a$ is the multiplicative order of $z$. Let $G'$ be the kernel of that subcharacter, then $G'$ is a closed normal subgroup of $G$, and $G/G'$ is a quotient of $G/G_0$ of order $a$. Therefore $a=|G/G'|$ divides $n=|G/G_0|$, so $z$ is an $n$-th root of unity.
\end{proof}
\begin{rem}
\emph{By \cite[Theorem 6.5]{katz2010mellin}, under these hypotheses $G/G_0$ is actually cyclic of prime-to-$p$ order $n$, and this group is isomorphic to the group of $\zeta\in\bar k^\star$ such that $\delta_z$ is in the Tannakian subcategory of ${\mathcal P}erv/{\mathcal N}eg$ tensor-generated by $\mathcal P\otimes\bar k$.}
\end{rem}
\begin{cor}\label{constants}
If $G$ contains the scalars $\bar{\mathbb Q}_\ell^\star$, then $U_{\mathcal P,r}=\mathbb G_{m,k}$ for every $r\geq 1$.
\end{cor}
\begin{proof}
By proposition \ref{subcharacters}(3), it suffices to show that the representation $V^{\otimes r}$ of $G$ does not have subcharacters of finite order. A scalar $\lambda\in\bar{\mathbb Q}_\ell^\star\subseteq G$ acts on $V^{\otimes r}$ by multiplication by $\lambda^r$. In particular, on any $G$-invariant subspace of $V^{\otimes r}$ the action of the quotient $\bar{\mathbb Q}_\ell^\star/{\mathbb \mu}_r$ is faithful, so it can never factor through a finite quotient.
\end{proof}
\begin{exa}
Let $g\in k[x]$ be a Morse polynomial of degree $d$ prime to $p$ such that its set of critical values is not isomorphic to a multiplicative translate of itself. Let $\mathcal F$ be the kernel of the trace map $g_\star\bar{\mathbb Q}_\ell\to\bar{\mathbb Q}_\ell$. Then $U_{\mathcal F[1],r}=\mathbb G_{m,k}$ for every $r\geq 1$ and for every $r\geq 1$ and every $t\in k_m^\star$ we have
$$
\left|\#\{x\in k_{mr}|\mathrm{N}_{k_{mr}/k_m}g(x)=t\}-\frac{q^{mr}-1}{q^m-1}\right|\leq C_{\mathcal F[1],r} q^{\frac{m(r-1)}{2}}
$$
where $C_{\mathcal F[1],r}$ is bounded as in \ref{normexa1} if $g$ is square-free. In particular, for every $e$ dividing $q^m-1$ we have the estimate
$$
\left|\#\{(x,y)\in k_{mr}^2|y^{(q^m-1)/e}=g(x)\}-q^{mr}-(\delta-1)\right|
\leq C_{\mathcal F[1],r} (q^m-1)q^{\frac{m(r-1)}{2}}
$$
where $\delta$ is the number of roots of $g$ in $k_m$.
\end{exa}
\begin{proof}
By \cite[Theorem 17.6]{katz2010mellin}, under these hypotheses $G$ is the entire $\mathrm{GL}(V)$, which contains the scalars, so the result follows from corollary \ref{constants}.
The second estimate is an easy consequence of the identity
$$
\#\{(x,y)\in k_{mr}^2|y^{(q^m-1)/e}=g(x)\}=\delta+\frac{q^m-1}{e}\sum_{\lambda^e=1}\#\{x\in k_{mr}|\mathrm{N}_{k_{mr}/k_m}(g(x))=\lambda\}.
$$
\end{proof}
\begin{rem}
\emph{As noted above (remark \ref{remark}), if $g$ is square-free of degree prime to $p$ the result holds without any further hypotheses on $g$.}
\end{rem}
\begin{exa}
Let $g\in k[x]$ be a square-free polynomial of degree $d$ with $g(0)\neq 0$ which is not of the form $h(x^n)$ for any $n\geq 2$, let $\chi:k^\star\to\bar{\mathbb Q}_\ell^\star$ be a multiplicative character such that $\chi^d$ is non-trivial and $\mathcal F={\mathcal L}_{\chi(g)}$. Then $U_{\mathcal F[1],r}=\mathbb G_{m,k}$ for every $r\geq 1$. In particular, for every $r\geq 1$ and every $t\in k_m^\star$ one has
$$
\left|\sum_{\mathrm{N}_{k_{mr}/k_m}(x)=t}\chi(\mathrm{N}_{k_{mr}/k}(g(x)))\right|\leq C_{\mathcal F[1],r} q^{\frac{m(r-1)}{2}}
$$
where $C_{\mathcal F[1],r}$ is bounded as in \ref{normexa2}.
\end{exa}
\begin{proof}
By \cite[Theorem 17.5]{katz2010mellin}, under these hypotheses $G=\mathrm{GL}(V)$ contains the scalars, so the result follows from corollary \ref{constants}.
\end{proof}
\begin{exa}
Let $g\in k[x]=\sum a_ix^i$ be a square-free polynomial of degree $d$ with $a_0\neq 0$ which is not of the form $h(x^n)$ for any $n\geq 2$ and such that $x^dg(1/x)$ is not a scalar multiple of a multiplicative translate of $g$, let $\chi:k^\star\to\bar{\mathbb Q}_\ell^\star$ be a multiplicative character such that $\chi^d$ is trivial and $\mathcal P={\mathcal L}_{\chi(g)}[1]$. Then $U_{\mathcal P,r}=\mathbb G_{m,k}$ for every $r\neq d$ and $U_{\mathcal P,d}=\mathbb G_{m,k}-\{(-1)^da_0\}$. In particular, for every $r\geq 1$ and every $t\in k_m^\star$ (except $t=(-1)^da_0$ when $r=d$) one has
$$
\left|\sum_{\mathrm{N}_{k_{mr}/k_m}(x)=t}\chi(\mathrm{N}_{k_{mr}/k}(g(x)))\right|\leq C_{\mathcal P,r} q^{\frac{m(r-1)}{2}}
$$
where $C_{\mathcal P,r}$ is bounded as in \ref{normexa2}.
\end{exa}
\begin{proof}
By \cite[Theorem 23.1]{katz2010mellin}, under these hypotheses $G$ contains $\mathrm{SL}(V)$. Since $\mathrm{SL}(V)$ is connected, $\mathrm{Sym}^{r-i}V\otimes\wedge^i V=\mathrm{Hom}(\wedge^{d-i}V,\mathrm{Sym}^{r-i}V)$ has a subcharacter of finite order as a $\mathrm{SL}(V)$-representation if and only if it has non-trivial $\mathrm{SL}(V)$-fixed subspace, which happens only for $d-i=r-i=0$ or $1$. So for $r\neq d$ it has no subcharacters of finite order and therefore $U_{\mathcal P,r}=\mathbb G_{m,k}$ by proposition \ref{subcharacters}, (3).
For $r=d$ and $i=r,r-1$, $\mathrm{Sym}^{r-i}V\otimes\wedge^i V$ has $\mathrm{SL}(V)$-fixed subspace of dimension $1$, on which $G$ acts via the determinant \cite[Corollary 4.2]{rlwan2010}. So $\mathrm{Sym}^{r-i}V\otimes\wedge^i V$ contains the determinant character as its only subcharacter of finite order, which by \cite[Theorem 23.1]{katz2010mellin} is the punctual sheaf $\delta_{(-1)^da_0}[0]$ as an element of ${\mathcal P}erv/{\mathcal N}eg$. In other words, the only possible punctual geometric irreducible component of $\mathrm{Sym}^{\ast(d-i)}\mathcal P\ast\wedge^{\ast i}\mathcal P$ is $\delta_{(-1)^da_0}[0]$. Therefore $U_{\mathcal P,d}=\mathbb G_{m,k}-\{(-1)^da_0\}$.
\end{proof}
\begin{exa}
Let $g\in k[x]$ be an Artin-Schreier-reduced polynomial (i.e. it has no monomials with divisible by $p$ exponent) of degree $d$ prime to $p$ which is not of the form $h(x^n)$ for any $n\geq 2$, let $\psi:k\to\bar{\mathbb Q}_\ell^\star$ be a non-trivial additive character and $\mathcal P$ the shifted Artin-Schreier sheaf ${\mathcal L}_{\psi(g)}[1]$. Then $U_{\mathcal P,r}=\mathbb G_{m,k}$ for every $r\geq 1$. In particular, for every $r\geq 1$ and every $t\in k_m^\star$ one has
$$
\left|\sum_{\mathrm{N}_{k_{mr}/k_m}(x)=t}\psi(\mathrm{Trace}_{k_{mr}/k}(g(x)))\right|\leq C_{\mathcal F,r} q^{\frac{m(r-1)}{2}}.
$$
\end{exa}
\begin{proof}
By \cite[Theorem 17.4]{katz2010mellin}, under these hypotheses $G=\mathrm{GL}(V)$ contains the scalars, so the result follows from corollary \ref{constants}.
\end{proof}
\begin{rem}
\emph{By \cite[Theorem 5.1]{katz1988gauss}, in the previous example $\mathcal P^{\ast r}$, and a fortiori its subobjects $\mathrm R(\ast\wedge^i\rho)\mathcal P$, are smooth on $\mathbb G_{m,k}$, tamely ramified at $0$ and totally wild at $\infty$. If conjecture \cite[7.6]{katz1988gauss} is true, then $\mathcal P^{\ast r}$ has a single slope $d/r$ at infinity. In that case,
$$
\mathrm{gen.rank}(\mathrm R(\ast\wedge^i\rho)\mathcal P)=\frac{r}{d}\mathrm{Swan}_\infty(\mathrm R(\ast\wedge^i\rho)\mathcal P)=
$$
$$
=\frac{r}{d}\dim\mathrm H^0_c(\mathbb G_{m,\bar k},\mathrm R(\ast\wedge^i\rho)\mathcal P)=
$$
$$
=\frac{r}{d}\dim\mathrm R(\wedge^i\rho)\mathrm H^0_c(\mathbb G_{m,\bar k},\mathcal P)=\frac{r}{d}{{d+r-i-1}\choose {r}}{r-1\choose i}
$$
by remark \ref{rank}, so we would obtain a bound
$$
C_{\mathcal P,r}\leq\frac{r}{d}\sum_{i=0}^{r-1}{{d+r-i-1}\choose {r}}{r-1\choose i}.
$$
Without the conjecture, we can only assure that
$$
C_{\mathcal P,r}\leq s^{-1}\sum_{i=0}^{r-1}{{d+r-i-1}\choose {r}}{r-1\choose i}.
$$
where $s$ is the smallest slope of $\mathcal P^{\ast r}$ at infinity.
}
\end{rem}
\section{The situation over $k_r$}
In this last section we will briefly discuss what happens when the sheaf or derived category object $\mathcal F$ is defined over the larger field $k_r$, but not over $k$.
Let $\mathcal F\in K_0(G\otimes{k_r},\bar{\mathbb Q}_\ell)$ be an object defined on the geometrically connected commutative group scheme $G\otimes{k_r}$ over $k_r$. In principle it makes sense to consider its $r$-th local norm $L$-function $L^{\mathrm{N},r}(\mathcal F,k,t;T)$ at any $t\in k$, since $\mathrm {Tr}(\mathrm{Frob}_{k_{rs},u}|\mathcal F_{\bar u})$ is defined for every $s\geq 1$ and every $u\in G(k_{rs})$. So we can ask, is this $L$-function rational? The answer is negative in general, as we can see in the following example for $G=\mathbb A^1$:
Let $q$ be odd, $a\in k_2-k$ such that $\mathrm {Tr}_{k_2/k}(a)=0$ and $\mathcal F=i_{a\star}\bar{\mathbb Q}_\ell$ the skyscraper sheaf supported on $a$. Then
$$\sum_{\mathrm {Tr}_{k_{2s}/k_s}(u)=0} \mathrm {Tr}(\mathrm{Frob}_{k_{2s},u}|\mathcal F_{\bar u})=\left\{\begin{array}{ll}
1 & \mbox{if $\mathrm {Tr}_{k_{2s}/k_s}(a)=0$} \\
0 & \mbox{otherwise}
\end{array}\right.
$$
Since $a^q+a=0$, by induction for every $s$ we have $a^{q^s}=(-1)^sa$, so $\mathrm {Tr}_{k_{2s}/k_s}(a)=0$ if $s$ is odd and $2a\neq 0$ if $s$ is even. Therefore
$$
L^{\mathrm {Tr},2}(\mathcal F,k,0;T)=\exp\sum_{s=1}^\infty \frac{T^{2s-1}}{2s-1}=\left(\frac{1+T}{1-T}\right)^{1/2}
$$
is not rational.
However, we have the following slightly weaker but equally useful result:
\begin{prop}
Let $\mathcal F\in K_0(G\otimes k_r,\bar{\mathbb Q}_\ell)$. Then $L^{\mathrm{N},r}(\mathcal F,k,t;T)^r$ is rational for every $t\in G(k)$, and all its reciprocal roots and poles have integral $q$-weight.
\end{prop}
\begin{proof}
By additivity, we may assume that $\mathcal F$ is a single sheaf. Let $\mathcal G$ be the direct sum of all $\sigma^\star\mathcal F$ for every $\sigma\in\mathrm {Gal}(k_r/k)$. Then $\mathcal G$ is invariant under $\mathrm {Gal}(k_r/k)$, so by descent there is a sheaf $\mathcal G_0\in K_0(G,\bar{\mathbb Q}_\ell)$ such that $\mathcal G=\pi_r^\star \mathcal G_0$, where $\pi_r:G\otimes k_r\to G$ is the projection. Therefore $L^{\mathrm{N},r}(\mathcal G,k,t;T)$ is rational for every $t\in k$ by corollary \ref{maincor}.
Again by additivity, we have
$$
L^{\mathrm{N},r}(\mathcal G,k,t;T)=\prod_\sigma L^{\mathrm{N},r}(\sigma^\star \mathcal F,k,t;T)
$$
where the product is taken over all $\sigma\in\mathrm {Gal}(k_r/k)$. We will conclude by showing that for every $\sigma\in\mathrm {Gal}(k_r/k)$ and every $t\in G(k)$ there is an equality $L^{\mathrm{N},r}(\sigma^\star \mathcal F,k,t;T)=L^{\mathrm{N},r}(\mathcal F,k,t;T)$.
Indeed we have, by definition,
$$
L^{\mathrm{N},r}(\sigma^\star \mathcal F,k,t;T)=\exp\sum_{s\geq 1}f_{\sigma^\star \mathcal F}^{\mathrm{N},r}(k_s,t)\frac{T^s}{s}
$$
and, for every $s\geq 1$,
$$
f_{\sigma^\star \mathcal F}^{\mathrm{N},r}(k_s,t)=\sum_{\mathrm{N}_{k_{sr}/k_s}(u)=t}f_{\sigma^\star \mathcal F}(k_{sr},u)=\sum_{\mathrm{N}_{k_{sr}/k_s}(u)=t}f_{\mathcal F}(k_{sr},\sigma(u))=
$$
$$
=\sum_{\mathrm{N}_{k_{sr}/k_s}(u)=t}f_{\mathcal F}(k_{sr},u)=f_\mathcal F^{\mathrm{N},r}(k_s,t)
$$
since $u\mapsto \sigma(u)$ is a permutation of the set of $u\in G(k_{sr})$ such that $\mathrm{N}_{k_{sr}/k_s}(u)=t$.
\end{proof}
In particular, many results proved in the previous sections can be applied in this case for every $t\in G(k)$ via the equality $f_\mathcal G^{\mathrm{N},r}(k,t)=r\cdot f_\mathcal F^{\mathrm{N},r}(k,t)$. Most notably we have:
\begin{cor}
Let $G$ be a geometrically connected affine commutative algebraic group over $k$ of dimension $1$. Let $\mathcal P\in{\mathcal P}erv(G\otimes k_r,\bar{\mathbb Q}_\ell)$ be pure of weight $1$ without negligible components, and let ${\mathcal Q}$ be the direct sum of $\sigma^\star\mathcal P$ for $\sigma\in\mathrm {Gal}(k_r/k)$ (and also its descent to $G_k$). Then for every $t\in U_{{\mathcal Q},r}(k)$ we have the estimate
$$
|f^{\mathrm{N},r}_\mathcal P(k,t)|\leq \frac{C_{{\mathcal Q},r}}{r}q^{\frac{r-1}{2}}.
$$
\end{cor}
which is a direct consequence of corollary \ref{UC} and the previous proposition.
\bibliographystyle{amsplain}
|
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{"url":"https:\/\/www.lmfdb.org\/ModularForm\/GL2\/Q\/holomorphic\/4032\/2\/c\/p\/","text":"# Properties\n\n Label 4032.2.c.p Level 4032 Weight 2 Character orbit 4032.c Analytic conductor 32.196 Analytic rank 0 Dimension 4 CM no Inner twists 2\n\n# Related objects\n\n## Newspace parameters\n\n Level: $$N$$ $$=$$ $$4032 = 2^{6} \\cdot 3^{2} \\cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\\chi]$$ $$=$$ 4032.c (of order $$2$$, degree $$1$$, minimal)\n\n## Newform invariants\n\n Self dual: no Analytic conductor: $$32.1956820950$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\\Q(\\zeta_{12})$$ Coefficient ring: $$\\Z[a_1, \\ldots, a_{17}]$$ Coefficient ring index: $$2^{3}$$ Twist minimal: no (minimal twist has level 1344) Sato-Tate group: $\\mathrm{SU}(2)[C_{2}]$\n\n## $q$-expansion\n\nCoefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\\zeta_{12}$$. We also show the integral $$q$$-expansion of the trace form.\n\n $$f(q)$$ $$=$$ $$q + ( 1 - 2 \\zeta_{12}^{2} + \\zeta_{12}^{3} ) q^{5} + q^{7} +O(q^{10})$$ $$q + ( 1 - 2 \\zeta_{12}^{2} + \\zeta_{12}^{3} ) q^{5} + q^{7} + ( 3 - 6 \\zeta_{12}^{2} + \\zeta_{12}^{3} ) q^{11} + ( 3 - 2 \\zeta_{12} + \\zeta_{12}^{3} ) q^{17} + ( -2 + 4 \\zeta_{12}^{2} + 4 \\zeta_{12}^{3} ) q^{19} + ( 7 + 2 \\zeta_{12} - \\zeta_{12}^{3} ) q^{23} + ( 1 + 4 \\zeta_{12} - 2 \\zeta_{12}^{3} ) q^{25} + ( -2 + 4 \\zeta_{12}^{2} ) q^{29} + ( 2 - 8 \\zeta_{12} + 4 \\zeta_{12}^{3} ) q^{31} + ( 1 - 2 \\zeta_{12}^{2} + \\zeta_{12}^{3} ) q^{35} + 10 \\zeta_{12}^{3} q^{37} + ( -3 + 2 \\zeta_{12} - \\zeta_{12}^{3} ) q^{41} + ( -4 + 8 \\zeta_{12}^{2} - 6 \\zeta_{12}^{3} ) q^{43} + ( -8 \\zeta_{12} + 4 \\zeta_{12}^{3} ) q^{47} + q^{49} + ( 2 - 4 \\zeta_{12}^{2} ) q^{53} + ( -10 + 8 \\zeta_{12} - 4 \\zeta_{12}^{3} ) q^{55} + ( 4 - 8 \\zeta_{12}^{2} - 4 \\zeta_{12}^{3} ) q^{59} + ( 2 - 4 \\zeta_{12}^{2} + 2 \\zeta_{12}^{3} ) q^{61} + ( 2 - 4 \\zeta_{12}^{2} - 4 \\zeta_{12}^{3} ) q^{67} + ( 9 + 6 \\zeta_{12} - 3 \\zeta_{12}^{3} ) q^{71} + ( 12 \\zeta_{12} - 6 \\zeta_{12}^{3} ) q^{73} + ( 3 - 6 \\zeta_{12}^{2} + \\zeta_{12}^{3} ) q^{77} + ( 6 + 4 \\zeta_{12} - 2 \\zeta_{12}^{3} ) q^{79} + ( 6 - 12 \\zeta_{12}^{2} + 2 \\zeta_{12}^{3} ) q^{83} + ( 4 - 8 \\zeta_{12}^{2} + 6 \\zeta_{12}^{3} ) q^{85} + ( 3 - 2 \\zeta_{12} + \\zeta_{12}^{3} ) q^{89} + ( 2 + 4 \\zeta_{12} - 2 \\zeta_{12}^{3} ) q^{95} + ( -4 + 12 \\zeta_{12} - 6 \\zeta_{12}^{3} ) q^{97} +O(q^{100})$$ $$\\operatorname{Tr}(f)(q)$$ $$=$$ $$4q + 4q^{7} + O(q^{10})$$ $$4q + 4q^{7} + 12q^{17} + 28q^{23} + 4q^{25} + 8q^{31} - 12q^{41} + 4q^{49} - 40q^{55} + 36q^{71} + 24q^{79} + 12q^{89} + 8q^{95} - 16q^{97} + O(q^{100})$$\n\n## Character values\n\nWe give the values of $$\\chi$$ on generators for $$\\left(\\mathbb{Z}\/4032\\mathbb{Z}\\right)^\\times$$.\n\n $$n$$ $$127$$ $$577$$ $$1793$$ $$3781$$ $$\\chi(n)$$ $$1$$ $$1$$ $$1$$ $$-1$$\n\n## Embeddings\n\nFor each embedding $$\\iota_m$$ of the coefficient field, the values $$\\iota_m(a_n)$$ are shown below.\n\nFor more information on an embedded modular form you can click on its label.\n\nLabel $$\\iota_m(\\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$\n2017.1\n \u22120.866025 \u2212 0.500000i 0.866025 + 0.500000i 0.866025 \u2212 0.500000i \u22120.866025 + 0.500000i\n0 0 0 2.73205i 0 1.00000 0 0 0\n2017.2 0 0 0 0.732051i 0 1.00000 0 0 0\n2017.3 0 0 0 0.732051i 0 1.00000 0 0 0\n2017.4 0 0 0 2.73205i 0 1.00000 0 0 0\n $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles\n\n## Inner twists\n\nChar Parity Ord Mult Type\n1.a even 1 1 trivial\n8.b even 2 1 inner\n\n## Twists\n\nBy twisting character orbit\nChar Parity Ord Mult Type Twist Min Dim\n1.a even 1 1 trivial 4032.2.c.p 4\n3.b odd 2 1 1344.2.c.g yes 4\n4.b odd 2 1 4032.2.c.m 4\n8.b even 2 1 inner 4032.2.c.p 4\n8.d odd 2 1 4032.2.c.m 4\n12.b even 2 1 1344.2.c.f 4\n24.f even 2 1 1344.2.c.f 4\n24.h odd 2 1 1344.2.c.g yes 4\n48.i odd 4 1 5376.2.a.s 2\n48.i odd 4 1 5376.2.a.x 2\n48.k even 4 1 5376.2.a.p 2\n48.k even 4 1 5376.2.a.be 2\n\nBy twisted newform orbit\nTwist Min Dim Char Parity Ord Mult Type\n1344.2.c.f 4 12.b even 2 1\n1344.2.c.f 4 24.f even 2 1\n1344.2.c.g yes 4 3.b odd 2 1\n1344.2.c.g yes 4 24.h odd 2 1\n4032.2.c.m 4 4.b odd 2 1\n4032.2.c.m 4 8.d odd 2 1\n4032.2.c.p 4 1.a even 1 1 trivial\n4032.2.c.p 4 8.b even 2 1 inner\n5376.2.a.p 2 48.k even 4 1\n5376.2.a.s 2 48.i odd 4 1\n5376.2.a.x 2 48.i odd 4 1\n5376.2.a.be 2 48.k even 4 1\n\n## Hecke kernels\n\nThis newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\\mathrm{new}}(4032, [\\chi])$$:\n\n $$T_{5}^{4} + 8 T_{5}^{2} + 4$$ $$T_{11}^{4} + 56 T_{11}^{2} + 676$$ $$T_{13}$$ $$T_{17}^{2} - 6 T_{17} + 6$$ $$T_{23}^{2} - 14 T_{23} + 46$$ $$T_{31}^{2} - 4 T_{31} - 44$$\n\n## Hecke characteristic polynomials\n\n$p$ $F_p(T)$\n$2$ 1\n$3$ 1\n$5$ $$1 - 12 T^{2} + 74 T^{4} - 300 T^{6} + 625 T^{8}$$\n$7$ $$( 1 - T )^{4}$$\n$11$ $$1 + 12 T^{2} + 170 T^{4} + 1452 T^{6} + 14641 T^{8}$$\n$13$ $$( 1 - 13 T^{2} )^{4}$$\n$17$ $$( 1 - 6 T + 40 T^{2} - 102 T^{3} + 289 T^{4} )^{2}$$\n$19$ $$1 - 20 T^{2} + 54 T^{4} - 7220 T^{6} + 130321 T^{8}$$\n$23$ $$( 1 - 14 T + 92 T^{2} - 322 T^{3} + 529 T^{4} )^{2}$$\n$29$ $$( 1 - 46 T^{2} + 841 T^{4} )^{2}$$\n$31$ $$( 1 - 4 T + 18 T^{2} - 124 T^{3} + 961 T^{4} )^{2}$$\n$37$ $$( 1 + 26 T^{2} + 1369 T^{4} )^{2}$$\n$41$ $$( 1 + 6 T + 88 T^{2} + 246 T^{3} + 1681 T^{4} )^{2}$$\n$43$ $$1 - 4 T^{2} - 3210 T^{4} - 7396 T^{6} + 3418801 T^{8}$$\n$47$ $$( 1 + 46 T^{2} + 2209 T^{4} )^{2}$$\n$53$ $$( 1 - 94 T^{2} + 2809 T^{4} )^{2}$$\n$59$ $$1 - 108 T^{2} + 6806 T^{4} - 375948 T^{6} + 12117361 T^{8}$$\n$61$ $$1 - 212 T^{2} + 18486 T^{4} - 788852 T^{6} + 13845841 T^{8}$$\n$67$ $$1 - 212 T^{2} + 19446 T^{4} - 951668 T^{6} + 20151121 T^{8}$$\n$71$ $$( 1 - 18 T + 196 T^{2} - 1278 T^{3} + 5041 T^{4} )^{2}$$\n$73$ $$( 1 + 38 T^{2} + 5329 T^{4} )^{2}$$\n$79$ $$( 1 - 12 T + 182 T^{2} - 948 T^{3} + 6241 T^{4} )^{2}$$\n$83$ $$1 - 108 T^{2} + 14966 T^{4} - 744012 T^{6} + 47458321 T^{8}$$\n$89$ $$( 1 - 6 T + 184 T^{2} - 534 T^{3} + 7921 T^{4} )^{2}$$\n$97$ $$( 1 + 8 T + 102 T^{2} + 776 T^{3} + 9409 T^{4} )^{2}$$","date":"2020-08-05 08:34:49","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9733135104179382, \"perplexity\": 6315.696083654267}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-34\/segments\/1596439735916.91\/warc\/CC-MAIN-20200805065524-20200805095524-00535.warc.gz\"}"}
| null | null |
A first-of-its-kind housing project is up and running in the mountains of Colorado. In this episode of Grid Talk, Marty Rosenberg talks with Bryan Hannegan, President and CEO of Holy Cross Energy. Mr. Hannegan explains how a net-zero, all electric community is delivering affordable housing in a Colorado tourism town. He'll rundown the features that are helping homeowners keep energy costs down.
Mr. Hannegan also talks about how this project is an opportunity to take new technologies from the research stage and put them into use in the real world and how unique partnerships are paving the way.
Bryan Hannegan is joined Holy Cross Energy in July of 2017. Before that he was Associate Laboratory Director at the National Renewable Energy Laboratory (NREL), where he co-founded the US Department of Energy's Grid Modernization Initiative and started up the Energy Systems Integration Facility (ESIF), a unique "distribution grid in a box" enabling utilities, entrepreneurs and consumers to work together on cleaner, more affordable and more reliable energy systems.
Mr. Hannegan holds a Ph.D. in Earth Systems Science and a M.S. in Mechanical and Aerospace Engineering, both from the University of California at Irvine, and a B.S. in Meteorology from the University of Oklahoma.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
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| 3,948
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{"url":"https:\/\/jira.lsstcorp.org\/browse\/DM-14259","text":"# Try and document running ap_pipe on the Verification Cluster with SLURM\n\nXMLWordPrintable\n\n#### Details\n\n\u2022 Type: Story\n\u2022 Status: Done\n\u2022 Resolution: Done\n\u2022 Fix Version\/s: None\n\u2022 Component\/s: None\n\u2022 Labels:\nNone\n\u2022 Story Points:\n6\n\u2022 Sprint:\nAP F18-1, AP F18-2\n\u2022 Team:\n\n#### Description\n\nSuggestion is to hand-construct SLURM commands rather than write a new pipe_driver since SuperTask will obviate the latter workflow shortly.\n\n#### Activity\n\nHide\nKrzysztof Findeisen added a comment -\n\nIt might be useful to look at what Eric did on DM-12960.\n\nShow\nKrzysztof Findeisen added a comment - It might be useful to look at what Eric did on DM-12960 .\nHide\nMeredith Rawls added a comment -\n\nI followed Eric's example as Krzysztof suggested. This is my workflow on lsst-dev:\n\n\u2022 Edit the files make_run_ap_pipe_conf.py, run_ap_pipe.sh, and run_ap_pipe.sl as desired\n\u2022 $python make_run_ap_pipe_conf.py \u2022$ sbatch run_ap_pipe.sl\n\nProcessing the entire hits2015 dataset through ap_pipe took just a few hours. The results may be viewed in \/project\/mrawls\/hits2015\/rerun\/slurm1.\n\nOne problem was encountered in which 15 srun tasks exited with code 1. This is due to a problem with ap_association IO (see DM-15114). However, an association.db was created with size 154 MB which suggests most of the DIAObjects were stored correctly. Follow-up ticket DM-15081 will take a closer look at what was stored in the database.\n\nShow\nMeredith Rawls added a comment - I followed Eric's example as Krzysztof suggested. This is my workflow on lsst-dev: Edit the files make_run_ap_pipe_conf.py , run_ap_pipe.sh , and run_ap_pipe.sl as desired $python make_run_ap_pipe_conf.py$ sbatch run_ap_pipe.sl Processing the entire hits2015 dataset through ap_pipe took just a few hours. The results may be viewed in \/project\/mrawls\/hits2015\/rerun\/slurm1 . One problem was encountered in which 15 srun tasks exited with code 1. This is due to a problem with ap_association IO (see DM-15114 ). However, an association.db was created with size 154 MB which suggests most of the DIAObjects were stored correctly. Follow-up ticket DM-15081 will take a closer look at what was stored in the database.\nHide\nMeredith Rawls added a comment -\n\nWould you please take a look at this, Eric Bellm?\n\nShow\nMeredith Rawls added a comment - Would you please take a look at this, Eric Bellm ?\nHide\nEric Bellm added a comment -\n\nI have some suggestions on Github for making this modestly more general to ease use on other datasets.\n\nShow\nEric Bellm added a comment - I have some suggestions on Github for making this modestly more general to ease use on other datasets.\nHide\nMeredith Rawls added a comment -\n\nThanks for your comments and offline discussions, Eric Bellm. I pushed some updates and the scripts seem to be running fine on lsst-dev now. Can you please re-review? The scripts still assume there are ccdnums 1 through 62, but I don't think there is any other hardcoded DECam stuff. A user can pass all the main ap_pipe command-line arguments at runtime via the new prep_ap_pipe.sh\u00a0script.\n\nShow\nMeredith Rawls added a comment - Thanks for your comments and offline discussions, Eric Bellm . I pushed some updates and the scripts seem to be running fine on lsst-dev now. Can you please re-review? The scripts still assume there are ccdnums 1 through 62, but I don't think there is any other hardcoded DECam stuff. A user can pass all the main ap_pipe command-line arguments at runtime via the new prep_ap_pipe.sh \u00a0script.\nHide\nMeredith Rawls added a comment -\n\nThe final form of the prep_ap_pipe.sh\u00a0script generates\u00a0three files for an ap_pipe\u00a0slurm job based on user inputs (rerun, repo, calib and template locations, filter, camera, and desired PPDB location). The three files it creates are run_ap_pipe.conf, run_ap_pipe.sh, and run_ap_pipe.sl.\n\nAfter the user runs prep_ap_pipe.sh, the user should enter\u00a0sbatch run_ap_pipe.sl. The slurm job will then process all CCDs (DECam or HSC) in parallel and\u00a0process each visit in the input repository sequentially.\n\nShow\nMeredith Rawls added a comment - The final form of the prep_ap_pipe.sh \u00a0script generates\u00a0three files for an ap_pipe \u00a0slurm job based on user inputs (rerun, repo, calib and template locations, filter, camera, and desired PPDB location). The three files it creates are run_ap_pipe.conf , run_ap_pipe.sh , and run_ap_pipe.sl . After the user runs prep_ap_pipe.sh , the user should enter\u00a0 sbatch run_ap_pipe.sl . The slurm job will then process all CCDs (DECam or HSC) in parallel and\u00a0process each visit in the input repository sequentially.\n\n#### People\n\nAssignee:\nMeredith Rawls\nReporter:\nJohn Swinbank\nReviewers:\nEric Bellm\nWatchers:\nEric Bellm, John Swinbank, Krzysztof Findeisen, Meredith Rawls","date":"2021-03-06 10:53:42","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.47340428829193115, \"perplexity\": 8593.552656454343}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-10\/segments\/1614178374686.69\/warc\/CC-MAIN-20210306100836-20210306130836-00157.warc.gz\"}"}
| null | null |
iStyles Zune 30GB Skin design of Fractal art, Light, Pattern, Purple, Graphic design, Design, Colorfulness, Electric blue, Art, Neon with black, gray, blue, purple colors. Model ZUNE-STATIC.
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|
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"redpajama_set_name": "RedPajamaC4"
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| 9,489
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\section{Introduction}
The biometric authentication problem has been studied extensively in recent years. In a biometric authentication system, a biometric feature, e.g.,~a finger print, a DNA sequence, etc., of a legitimate user is measured and the measurement, called enrollment, is stored in a database. Later this biometric feature of the same user is measured and compared with the enrollment for authentication. Due to the randomness in the process, different measurements of the same biometric features of the same person can not be exactly the same. Thus, the authentication system needs to tolerate a certain level of distortion between the measurements taken in the enrollment stage and the authentication stage.
In a biometric authentication system, successful deception happens when an adversary impersonates a legitimate user by faking a biometric feature close enough to the enrollment and then deceives the authentication system.
The first study on the deception probability in the biometric authentication system is \cite{Ahlswede:1997}, where the authors studied the deception in an authentication system where the enrollment is compressed. The authors obtained the optimal trade-off between the compression rate and the exponent of the probability of successful deception when the adversary has no side information. In the case with correlated side information at the adversary, achievability and converse results on the optimal trade-off were proposed in the paper, however, they do not meet.
A similar result was obtained in a recent paper \cite{Willems:2012}, where the optimal exponent of the deception probability has been given in both cases of uncompressed and compressed enrollment with no side information at the adversary.
A different direction in studying the performance of a biometric authentication system is to study the maximum number of legitimate users, called capacity, allowed in a biometric authentication system under a given tolerated distortion level. In \cite{Willems:2003}, the capacity is obtained if the enrollment is not compressed. Later, the capacity result is generalized to the case where the enrollment is compressed \cite{Tuncel:2009}, and the trade-off between the compression rate and the capacity of the authentication system was studied. The threat of a deception from an adversary was not considered in this line of work.
In this paper, we study the optimal exponent of the probability of successful deception of a biometric authentication system with uncompressed enrollment and side information at the adversary. It represents the scenario where the adversary may have correlated side information, e.g.,~a partial finger print of the legitimate user or a DNA sequence of a relative of the legitimate user. We provide the optimal exponent of the deception probability by providing the proofs of both the achievability and the converse. Our proofs are based on a connection between the problem of deception with side information and the rate distortion problem with side information at both the encoder and decoder.
The reminder of the paper is as follows. In the next section, we state the problem formulation and the main result. The proofs of the achievability and the converse are given in Section III and IV, respectively. Finally, a conclusion is given in Section V.
\section{Problem Formulation and Main Result}
\subsection{Problem Formulation}
Consider a pair of independent and identically distributed (i.i.d.) random sequences $X^n$ and $Y^n$, generated according to a joint distribution $P$, which is defined on a finite space $\mathcal{X}\times\mathcal{Y}$. Th random sequence $X^n$ represents the biometric enrollment in the system and $Y^n$ represents the side information at the adversary. We define a reconstruction space $\hat{\mathcal{X}}$ and a distortion function $d:\mathcal{X}\times\hat{\mathcal{X}}\mapsto \mathbb{R}^+\cup\{0\}$. The distortion between the sequences $x^n\in\mathcal{X}^n$ and $\hat{x}^n\in\hat{\mathcal{X}}^n$ is defined as
\begin{align}
d(x^n,\hat{x}^n)\triangleq\frac{1}{n}\sum_{i=1}^nd(x_i,\hat{x}_i). \label{seqdis}
\end{align}
The legitimate user is successfully identified if the distortion between the measurements in the enrollment stage and the authentication stage does not exceed a certain level, say $\Delta$. We note that the probability of false rejection in our model is the same as the model without side information at the adversary. Therefore, we refer the readers to \cite{Ahlswede:1997,Willems:2012} for the derivation of the maximal probability of false rejection.
The adversary observes the side information $Y^n$ and tries to impersonate the legitimate user using a deception function
$
f:\mathcal{Y}^n\mapsto\hat{\mathcal{X}}^n
$.
We define the achievable deception exponent as follows.
\begin{Def}
A deception exponent $E$ is achievable under the distortion constraint $\Delta$ if there exists a deception function $f$ such that
\begin{align}
-\frac{1}{n}\log\mathsf{Pr}(d(X^n,f(Y^n))\le \Delta+\delta)\le E+\delta.\label{defi}
\end{align}
\end{Def}
In this paper, we are interested in the minimal achievable deception exponent, which is the best the adversary can do. Based on the minimal achievable deception exponent, the designer of the biometric authentication system can choose an appropriate $\Delta$ value that on one hand, limits the probability of successful deception below the tolerance level, and on the other hand, does not cause too large a probability of false rejection when the legitimate user is authenticated \cite{Ahlswede:1997, Willems:2012}.
\subsection{Rate-distortion with Side Information at Both the Encoder and Decoder}\label{WZ}
It turns out that finding the minimal achievable deception exponent with side information at the adversary is intimately related to the rate distortion problem with side information at both the encoder and decoder \cite{Wyner:1976}. Thus, in this subsection, we review the result for the rate distortion problem.
Assume a pair of i.i.d sequences $X^n$ and $Y^n$ generated according to a joint distribution $Q$ defined on $\mathcal{X}\times\mathcal{Y}$, where $Q$ is not necessarily equal to $P$, which is defined in the previous subsection. The random sequence $X^n$ is the source sequence to be reconstructed at the decoder under a certain distortion constraint and $Y^n$ represents the side information available at both the encoder and decoder. The encoding function at the encoder is defined as
$
g:\mathcal{X}^n\times\mathcal{Y}^n\mapsto\{1,2,\dots,M\},
$
and the decoding function at the decoder is defined as
$
\varphi:\{1,2,\dots,M\}\times\mathcal{Y}^n\mapsto \hat{\mathcal{X}}^n
$.
We denote the minimal achievable rate under distortion constraint $\Delta$ in the rate distortion problem with side information at both the encoder and decoder as $R_{SI}(Q,\Delta)$. From \cite{Wyner:1976}, we have
\begin{align}
R_{SI}(Q,\Delta) = \min_{V(\hat{x}|x,y): \mathsf{E}d(X,\hat{X})\le \Delta} I(X;\hat{X}|Y). \label{WZrate}
\end{align}
\emph{Remark}: The above rate distortion problem with side information at both the encoder and decoder can also be viewed as a special case of the Wyner-Ziv problem, i.e.,~the rate distortion problem with side information only at the decoder, as follows: in the Wyner-Ziv problem, view $(X^n,Y^n)$ jointly as the source sequence available at the encoder, view $Y^n$ as the side information at the decoder, and take the the distortion function in the Wyner-ziv problem as
$d(x,\hat{x})$, which is defined in the previous subsection, i.e., the distortion of $Y^n$ does not matter.
By viewing the rate distortion problem with side information at both the encoder and decoder as a special case of the Wyner-Ziv problem, we can invoke the results of the Wyner-Ziv problem, e.g.,~\cite[Theorem 16.5]{Csiszar:2011}, in later development.
\subsection{Main Result}
The main result of this paper is the following theorem.
\begin{Theo} \label{main}
The deception exponent $E$ is achievable under the distortion constraint $\Delta$ if and only if
\begin{align}
E\ge \min_{Q}\{D(Q||P)+R_{SI}(Q,\Delta)\}, \label{main_eq}
\end{align}
where $R_{SI}(Q,\Delta)$ is given in (\ref{WZrate}), the distribution $Q$ is defined on $\mathcal{X}\times\mathcal{Y}$ and $D(Q||P)$ represents the Kullback-Leibler divergence between the distributions $Q$ and $P$ \cite{Csiszar:2011}.
\end{Theo}
In the next two sections, we will show the proofs of the achievability and the converse of Theorem \ref{main} via the connection between the problem of deception with side information and the rate distortion problem with side information at both the encoder and decoder.
\section{The Achievability}
In this section, we will show that there exists a deception function $f$ that can achieve the the deception exponent $D(Q||P)+R_{SI}(Q,\Delta)$ for the distortion constraint $\Delta$ and the distribution $Q$ defined on $\mathcal{X}\times\mathcal{Y}$. We will construct the deception function $f$ from the rate distortion code with side information at both the encoder and decoder.
First, consider the rate distortion problem with side information at both the encoder and decoder as defined in subsection \ref{WZ}, where $(X^n, Y^n)$ are generated i.i.d. according to the distribution $Q$. Theorem 16.5 in \cite{Csiszar:2011} shows that for any sufficiently large $n$, and $0<\tau<1$, there exists a length $n$ code that achieves the rate $R_{SI}(Q,\Delta)$ and satisfies the distortion constraint with probability larger than $1-\tau$. More specifically, there exists a function pair $(g,\varphi)$ such that
\begin{align}
\frac{1}{n}\log||g||&\le R_{SI}(Q,\Delta)+\delta\label{WZr}\\
\mathsf{Pr}\left(d\left(X^n, \varphi(g(X^n,Y^n),Y^n)\right)\le\Delta+\delta\right)&\ge1-\tau\label{tau}
\end{align}
Define $A\subset\mathcal{X}^n\times\mathcal{Y}^n$ as the set of sequences $(x^n,y^n)$ that satisfies the distortion constraint $\Delta+\delta$ under $(g,\varphi)$, i.e.,
\begin{align}
A\triangleq
\{(x^n,y^n)\in &\mathcal{X}^n\times\mathcal{Y}^n: d(x^n, \varphi(g(x^n,y^n),y^n))\le\Delta+\delta \}.
\end{align}
Thus, the inequality in (\ref{tau}) is equivalent to
\begin{align}
Q^n(A)>1-\tau, \label{tau1}
\end{align}
where $Q^n(A)$ is the probability that an i.i.d. randomly generated $(X^n,Y^n)$ according to the distribution $Q$ falls in the set $A$, i.e.,
\begin{align}
Q^n(A)\triangleq \sum_{(x^n,y^n)\in A}\prod_{i=1}^n Q(x_i,y_i).
\end{align}
We further define $A(Q)$ as the intersection of the set $A$ with the typical set $\mathcal{T}_{[Q]_\delta}^n$, where the definition of a typical set can be found in \cite[Definition 2.8]{Csiszar:2011}, i.e.,
\begin{align}
A(Q)\triangleq A\cap\mathcal{T}_{[Q]_\delta}^n.
\end{align}
From \cite[Lemma 2.12]{Csiszar:2011}, we have
\begin{align}
Q^n\left(\mathcal{T}_{[Q]_\delta}^n\right)\ge1-\epsilon_n, \label{epsilon}
\end{align}
and as a result, from (\ref{tau1}) and (\ref{epsilon}), we have
\begin{align}
Q^n(A(Q))\ge Q^n(A)+Q^n\left(\mathcal{T}_{[Q]_\delta}^n\right)-1\ge1-\epsilon_n-\tau.
\end{align}
Thus, from \cite[Lemma 2.14]{Csiszar:2011}, we have
\begin{align}
\frac{1}{n}\log|A(Q)|\ge H(Q)-\epsilon_n.\label{nonsub}
\end{align}
We further define $A_i(Q)\subset A(Q)$ for $i=1,2,\dots,||g||$ as the set of $(x^n,y^n)$ sequences that satisfy the distortion constraint when mapped to index $i$ at the encoder, i.e.,
\begin{align}
A_i(Q)\triangleq
\{(x^n,y^n)\in\mathcal{T}_{[Q]_\delta}^n:d(x^n, \varphi(i,y^n))\le\Delta+\delta\}. \label{Nan10}
\end{align}
Since all $(x^n, y^n)$ sequence pairs that satisfy the distortion constraint $\Delta+\delta$ under $(g,\varphi)$ has to be mapped to an index $g(x^n,y^n)$, we have
\begin{align}
A(Q)&=\bigcup_{i=1}^{||g||}A_{i}(Q),\label{specii}\\
A_{i}(Q)\cap A_{j}(Q)&=\emptyset\qquad \text{for }i\ne j.
\end{align}
We define $i_o$ as the index of the set $A_{i}(Q)$ with the largest cardinality, i.e.,
\begin{align}
i^o\triangleq\arg\max_{i\in\{1,2,\dots,||g||\}}|A_{i}(Q)|.
\end{align}
Then, we have
\begin{align}
|A_{i^o}(Q)|\ge\frac{|A(Q)|}{||g||}.\label{cov}
\end{align}
Therefore, we have
\begin{align}
-\frac{1}{n}\log P^n(A_{i^o}(Q))
\le& -\frac{1}{n}\log\left(|A_{i^o}(Q)|\min_{(x^n,y^n)\in A_{i^o}(Q)}P^n((x^n,y^n))\right)\nonumber\\
\le&-\frac{1}{n}\log\left(\frac{|A(Q)|\min_{(x^n,y^n)\in A_{i^o}(Q)}P^n((x^n,y^n))}{||g||}\right)\label{Nan02}\\
\le& -H(Q)+\epsilon_n+R_{SI}(Q,\Delta)+\delta -\frac{1}{n}\log\left(\min_{(x^n,y^n)\in A_{i^o}(Q)}P^n((x^n,y^n))\right)\label{Nan03}\\
\le& -H(Q)+\epsilon_n+R_{SI}(Q,\Delta)+\delta+D(Q||P)+H(Q)+\epsilon_n \label{Nan04}\\
=& D(Q||P)+R_{SI}(Q,\Delta)+\delta+2\epsilon_n, \label{Nan07}
\end{align}
where (\ref{Nan02}) follows from (\ref{cov}), (\ref{Nan03}) follows from (\ref{WZr}) and (\ref{nonsub}), and
(\ref{Nan04}) follows from \cite[Lemma 2.6]{Csiszar:2011}, i.e.,~for any $(x^n,y^n) \in \mathcal{T}_{[Q]_\delta}^n$,
\begin{align}
-\frac{1}{n}\log P^n\left((x^n,y^n)\right)\le D(Q||P)-&H(Q)+\epsilon_n
\label{misseq}
\end{align}
Now, we construct the deception function $f$ according to $(g,\varphi)$ described above, i.e.,
\begin{align}
f(y^n)=\varphi(i^o,y^n). \label{Nan06}
\end{align}
Then we have
\begin{align}
-\frac{1}{n}\log\mathsf{Pr}(d(X^n,f(Y^n))\le \Delta+\delta)
=&-\frac{1}{n}\log P^n \left(\{(x^n,y^n): d(x^n,\varphi(i^o,y^n))\le \Delta+\delta\}\right) \label{Nan05}\\
=& -\frac{1}{n}\log P^n(A_{i^o}(Q)) \label{Nan08}\\
\leq& D(Q||P)+R_{SI}(Q,\Delta)+\delta+2\epsilon_n, \label{Nan09}
\end{align}
where (\ref{Nan05}) follows from the construction of $f$ in (\ref{Nan06}), (\ref{Nan08}) follows from the definition of $A_i(Q)$ in (\ref{Nan10}), and (\ref{Nan09}) follows from (\ref{Nan07}).
Thus, we have shown that there exists a deception function $f$, constructed according to the encoding and decoding function of the corresponding rate distortion problem with side information, that can achieve the the deception exponent $D(Q||P)+R_{SI}(Q,\Delta)$ for any distribution $Q$ defined on $\mathcal{X}\times\mathcal{Y}$ and distortion constraint $\Delta$.
This concludes the proof of the achievability.
\section{The Converse}
In this section, we will prove that for any deception function $f$, the deception exponent can not be smaller than $\min_Q\{D(Q||P)+R_{SI}(Q,\Delta)\}$. This will be proven by contradiction, i.e., we will show that if there is a deception function with the deception exponent equal to $\min_Q\{D(Q||P)+R_{SI}(Q,\Delta)\}-\alpha$ for some $\alpha>0$, then we can construct a coding scheme in the rate distortion with side information problem with achievable rate smaller than $R_{SI}(Q,\Delta)$, which is obviously false. In contrast to the achievability, which selects one deception function $f(y^n)$ from many rate distortion decoding functions $\phi(i, y^n)$, $i=1,2,\cdots, \|g\|$, i.e., (\ref{Nan06}), the converse requires us to construct many rate distortion decoding functions $\phi(i, y^n)$, $i=1,2,\cdot, \|g\|$ from one deception function $f(y^n)$, which is more difficult.
We first assume a deception function $f$, which achieves the deception exponent $E$ under distortion constraint $\Delta$, as defined in (\ref{defi}). The proof of the converse includes the following three steps.
\noindent\textbf{Step 1 Type selection:} In this step, we will select one type among all the types, which contributes most to the deception probability for the function $f$.
We define a set $A\subset\mathcal{X}^n\times\mathcal{Y}^n$ as follows
\begin{align}
A\triangleq\left\{(x^n,y^n)\in\mathcal{X}^n\times\mathcal{Y}^n:d(x^n, f(y^n))\le\Delta+\delta\right\}.\label{defA}
\end{align}
Based on the definition of set $A$ and the definition of deception exponent in (\ref{defi}), we have that deception exponent $E$ is achievable is equivalent to
\begin{align}
-\frac{1}{n}\log P^n(A)\le E+\delta. \label{Nan13}
\end{align}
We further define $\mathcal{P}$ as the set of all possible empirical distribution on $\mathcal{X}^n\times\mathcal{Y}^n$, i.e.,
\begin{align}
\mathcal{P}\triangleq\left\{Q:Q(x,y)=\frac{i}{n}, \text{ for all } (x,y)\in\mathcal{X}\times\mathcal{Y}, i\in\{0,1,2,\dots,n\}\right\}.
\end{align}
For any distribution $Q\in\mathcal{P}$, we define the set $A(Q)$ as the intersection between the set $A$ and $\mathcal{T}_{Q}^n$, where $\mathcal{T}_{Q}^n$ is the type of $Q$ as defined in \cite[Definition 2.1]{Csiszar:2011}, i.e.,
\begin{align}
A(Q)=A\cap\mathcal{T}_{Q}^n.\label{defAQ}
\end{align}
Thus, we have
\begin{align}
A = \bigcup_{Q\in\mathcal{P}} A(Q).
\end{align}
Since the type $\mathcal{T}_Q^n$ for different empirical distributions $Q\in\mathcal{P}$ are disjoint, we have
\begin{align}
P^n(A)=\sum_{Q\in\mathcal{P}} P^n(A(Q)). \label{Nan11}
\end{align}
We define the type $Q^o$ as the type which contributes most to the deception probability, i.e.,
\begin{align}
Q^o\triangleq \arg\max_{Q\in\mathcal{P}} P^n(A(Q)). \label{Nan11.3}
\end{align}
Let $\beta$ by the number of different types in $\mathcal{X}^n\times\mathcal{Y}^n$. From (\ref{Nan11}) and (\ref{Nan11.3}), we have
\begin{align}
P^n(A(Q^o)) \geq \frac{P^n(A)}{\beta}.\label{Nan11.2}
\end{align}
Thus, we have
\begin{align}
-\frac{1}{n}\log P^n(A(Q^o)) &\leq -\frac{1}{n}\log \frac{P^n(A)}{\beta} \label{Nan11.1}\\
& \leq E+\delta+\frac{1}{n} \log \beta \label{Nan12}\\
&<E+\epsilon_n+\delta, \label{Nan14}
\end{align}
where (\ref{Nan11.1}) follows from (\ref{Nan11.2}), (\ref{Nan12}) follows from (\ref{Nan13}), and (\ref{Nan14}) follows from the fact that $\beta \leq (n+1)^{|\mathcal{X}||\mathcal{Y}|}$ \cite[Lemma 2.2]{Csiszar:2011}.
Since every sequence in the same type is equally probable, we have
\begin{align}
P^n(A(Q^o))=|A(Q^o)| P^n(x^n,y^n), \qquad \forall (x^n,y^n) \in \mathcal{T}_{Q^o}^n.
\end{align}
Thus, for any sequence pair $(x^n,y^n)\in\mathcal{T}_{Q^o}^n$, we have
\begin{align}
\frac{1}{n}\log|A(Q^o)|&=\frac{1}{n} \log \frac{P^n(A(Q^o)}{P^n(x^n,y^n)} \nonumber\\
& > -E-\epsilon_n-\delta-\frac{1}{n}\log P^n((x^n,y^n)) \label{Nan15}\\
&=D(Q^o||P)+H(Q^o)-E-\epsilon_n-\delta,\label{docsize}
\end{align}
where (\ref{Nan15}) follows from (\ref{Nan14}), and (\ref{docsize}) follows from \cite[Lemma 2.6]{Csiszar:2011}, i.e., for distribution $Q^o$,
\begin{align}
-\frac{1}{n}\log P^n((x^n,y^n))=D(Q^o||P)+H(Q^o), \quad \forall (x^n,y^n)\in\mathcal{T}_{Q^o}^n. \nonumber
\end{align}
\textbf{Step 2 Permutation}: In this step, we will construct a rate distortion code, restricted to the type $Q^o$, with side information at both the encoder and decoder from the deception function $f$ via permutations.
We consider the symmetric group $\mathcal{S}_n$, which consists of all the permutations on $\{1,2,\dots,n\}$. For a set $S\subset\mathcal{X}^n$, and a permutation $\pi\in\mathcal{S}_n$, define the set $\pi(S)$ as the set of sequences that is permuted from the sequences in set $S$ by the permutation $\pi$, i.e.,
\begin{align}
\pi(S)\triangleq\{\bar{x}^n\in\mathcal{X}^n:\exists x^n\in S, \pi(x^n)=\bar{x}^n\}.\label{permS}
\end{align}
Thus, $\pi(S)$ is a permuted version of the set $S$.
We will use the following lemma, proved by Ahlswede in 1980, to obtain a covering of $\mathcal{T}_{Q^o}^n$ by the permuted versions of $A(Q^o)$.
\begin{Lem}[Covering Lemma] \cite[Section 6.1]{Ahlswede:1980}\label{covlemma}
For any set $S\in\mathcal{T}_{Q}^n$, there exist permutations $\pi_1,\pi_2\dots,\pi_k\in\mathcal{S}_n$ with
\begin{align}
\bigcup_{i=1}^k\pi_i(S)=\mathcal{T}_{Q}^n,
\end{align}
if
\begin{align}
k>\frac{|\mathcal{T}_{Q}^n|}{|S|}\log|\mathcal{T}_{Q}^n|.
\end{align}
\end{Lem}
From the above lemma, by letting $Q$ be $Q^o$ and $S$ be $A(Q^o)$, we have that there exist permutations $\pi_1,\pi_2\dots,\pi_k\in\mathcal{S}_n$ such that
\begin{align}
\bigcup_{i=1}^k\pi_i(A(Q^o))=\mathcal{T}_{Q^o}^n,\label{typecov}
\end{align}
where $k$ satisfies
\begin{align}
\frac{1}{n}\log k
&=\frac{1}{n} \log |\mathcal{T}_{Q^o}^n|-\frac{1}{n} \log |A(Q^o)|+\log \left(\log |\mathcal{T}_{Q^o}^n| \right)+\epsilon_n \nonumber\\
& \leq E-D(Q^o||P)+2\epsilon_n+\delta+\frac{\log (n H(Q^o))}{n}, \label{Nan16}
\end{align}
where (\ref{Nan16}) follows from (\ref{docsize}) and the fact that
the size of the type $|\mathcal{T}_{Q^o}^n|$ satisfies \cite[Lemma 2.3]{Csiszar:2011}
\begin{align}
|\mathcal{T}_{Q^o}^n|\le\exp(nH(Q^o)).\nonumber
\end{align}
Based on (\ref{defA}) and (\ref{defAQ}), we have that
\begin{align}
A(Q^o)=\left\{(x^n,y^n)\in\mathcal{T}^n_{Q^o}:d(x^n, f(y^n))\le\Delta+\delta\right\}
\end{align}
Based on the definition in (\ref{seqdis}), we see that the same permutation of the two sequences does not change the distortion between the two sequences. Therefore we have for $i=1,2,\dots,k$
\begin{align}
A(Q^o)=\left\{(x^n,y^n)\in\mathcal{T}^n_{Q^o}:d(\pi_i(x^n), \pi_i(f(y^n)))\le\Delta+\delta\right\}.
\end{align}
From the definition of the permutation of a set in (\ref{permS}), we have
\begin{align}
\pi_i(A(Q^o))=\left\{(\bar{x}^n,\bar{y}^n)\in\mathcal{T}^n_{Q^o}:\exists (x^n,y^n)\in A(Q^o),(\pi_i(x^n),\pi_i(y^n))=(\bar{x}^n,\bar{y}^n)\right\}.
\end{align}
Thus, by combining the above two equations, we can view the set $\pi_i(A(Q^o))$ as
\begin{align}
\pi_i(A(Q^o))
&=\left\{(\bar{x}^n,\bar{y}^n)\in\mathcal{T}_{Q^o}^n:d(\bar{x}^n, \pi_i(f(\pi_i^{-1}(\bar{y}^n))))\le\Delta+\delta\right\}.
\end{align}
In other words, the permuted set $\pi_i(A(Q^o))$ can be characterized by a composite function $\pi_i(f(\pi_i^{-1}(\cdot)))$.
With the sets $\pi_i(A(Q^o))$ and the functions $\pi_i(f(\pi_i^{-1}(\cdot)))$ for $i=1,2,\dots,k$, we are ready to construct a rate distortion code with side information.
We assume that $(X^n, Y^n)$ are generated i.i.d. according to distribution $Q^o$. We will construct an encoding-decoding function pair $(g',\varphi')$ for all $(x^n, y^n) \in \mathcal{T}_{Q^o}^n$ as follows.
We define $g'(x^n,y^n)= i$ if $(x^n,y^n)\in\pi_i(A(Q^o))$. If there exist multiple sets $\pi_i(A(Q^o))$ to which $(x^n,y^n)$ belongs, we can arbitrarily pick one set and assign the index of the set to the output of the function $g'$. Define $Q^o_Y$ as the marginal distribution of $Q^o$ in $\mathcal{Y}$. Then, for $i=1,2,\dots,k$ and $y^n\in\mathcal{T}_{Q^o_Y}^n$, we define the decoding function $\varphi'$ as follows
\begin{align}
\varphi'(i,y^n)=\pi_i(f(\pi_i^{-1}(y^n))),\qquad i=1,2,\dots,k.
\end{align}
Due to the covering in (\ref{typecov}), we obtain an encoding-decoding function pair $(g',\varphi')$ for the rate-distortion problem with side information available at both the encoder and decoder for every $(x^n,y^n)\in\mathcal{T}_{Q^o}^n$, which satisfies
\begin{align}
||g'||&=k,\\
d(x^n,\varphi'(g'(x^n,y^n),y^n))&\le \Delta+\delta.
\end{align}
\textbf{Step 3 Blowing-up}: In the previous step, we have construct a code $(g',\varphi')$ for every sequence pair $(x^n,y^n)$ in the type $\mathcal{T}_{Q^o}^n$. In this step, we will expand the code, first to the neighborhood of the type $\mathcal{T}_{Q^o}^n$, and then to the whole space $\mathcal{X}^n\times\mathcal{Y}^n$. The expansion uses the Blowing-up lemma \cite[Chapter 5]{Csiszar:2011}.
First, we expand the code we described in the previous step to the neighborhood of the type $\mathcal{T}_{Q^o}^n$. To do so, let us introduce the definition of the neighborhood of a set as follows.
\begin{Def}\cite[Chapter 5]{Csiszar:2011}
Given a set $S\subset\mathcal{X}^n$, we define the Hamming $l$ neighborhood of $S$ as
\begin{align}
\mathbf{\Gamma}^{l}(S)\triangleq&\left\{x^n\in\mathcal{X}^n:\exists\bar{x}^n\in S, \text{ s.t. }
d_H(x^n,\bar{x}^n)\le l\right\},
\end{align}
where $d_H$ represents the Hamming distance between two sequence pairs, i.e., the number of positions in which the two sequence pairs differ.
\end{Def}
Assume a sequence of positive integer $l_n$ with
$
\frac{l_n}{n}$ converging to $0$.
We consider the $l_n$ neighborhood of the type $\mathcal{T}_{Q^o}^n$, i.e., $\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o}^n)$. Based on Definition 2 and (\ref{typecov}), we have
\begin{align}
\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o}^n)=\bigcup_{i=1}^k\mathbf{\Gamma}^{l_n}(\pi_i(A(Q^o))). \label{NanLiu01}
\end{align}
For any $(x^n,y^n) \in \mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o}^n)$, we will construct a rate distortion code with side information at both the encoder and decoder $(g,\varphi)$ restricted to the $l_n$ neighborhood of the type $\mathcal{T}_{Q^o}^n$, i.e.,~$\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o}^n)$, from the rate distortion code $(g',\varphi')$ defined on the type $\mathcal{T}_{Q^o}^n$ as described in the previous step. The basic idea is that for $(x^n,y^n)$ in the neighborhood of $(\bar{x}^n,\bar{y}^n)\in\mathcal{T}_{Q^o}^n$, we adopt the encoder-decoder pair $(g',\varphi')$ for $(\bar{x}^n,\bar{y}^n)$ as the encoder-decoder pair for $(x^n,y^n)$. The details are as follows.
The encoding function $g$ on the set $\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o}^n)$ includes two parts, i.e., $g=(g_1,g_2)$. The function $g_1$ can be constructed
in a similar way as we constructed the function $g$ on the type $\mathcal{T}_{Q^o}^n$ in the previous step. More specifically, for any $(x^n,y^n) \in \mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o}^n)$,
we define $g_1(x^n,y^n)= i$ if $(x^n,y^n)\in\mathbf{\Gamma}^{l_n}(\pi_i(A(Q^o)))$. Based on (\ref{NanLiu01}), we know such $i$ always exists. If there exist multiple sets $\mathbf{\Gamma}^{l_n}(\pi_i(A(Q^o)))$ to which $(x^n,y^n)$ belongs, we can arbitrarily pick one set and assign the index of the set to the output of the function $g_1$. Then we have
\begin{align}
||g_1||=k.
\end{align}
Once we determine that $g_1(x^n,y^n)= i$, we can assert that there exists a sequence pair $(\bar{x}^n,\bar{y}^n)\in\pi_i(A(Q^o))$ such that
\begin{align}
d_H((x^n, y^n),(\bar{x}^n,\bar{y}^n))\le l_n.
\end{align}
We would like to adopt $\varphi'(i,\bar{y}^n)$ as the decoding function $\varphi(i,y^n)$. However, $\bar{y}^n$ is determined based on sequence pair $(x^n,y^n)$. Therefore, the decoder, with only the knowledge of $y^n$, can not determine $\bar{y}^n$ by itself. To overcome this problem, we need to design $g_2$, i.e., the second part of encoding function, to inform the decoder of $\bar{y}^n$ as follows.
Let us consider $y^n$, the second sequence in the pair $(x^n,y^n)$. We construct a $l_n$ Hamming neighborhood $\mathbf{\Gamma}^{l_n}(y^n)$ around the sequence $y^n$, which is called a Hamming ball, and give every sequence in $\mathbf{\Gamma}^{l_n}(y^n)$ an index.
Obviously, the sequence $\bar{y}^n$ can be uniquely determined by the sequence $y^n$ together with the index of $\bar{y}^n$ with respect to $y^n$, say $j\in\{1,2,\dots,\left|\mathbf{\Gamma}^{l_n}(y^n)\right|\}$.
Then $g_2$, the second part of the encoding function $g$, can be defined as the above index, i.e.,
\begin{align}
g_2(x^n,y^n)=j,
\end{align}
and we have
\begin{align}
||g_2||=\left|\mathbf{\Gamma}^{l_n}(y^n)\right|,
\end{align}
where the size of the Hamming ball satisfies \cite[Lemma 5.1]{Csiszar:2011}
\begin{align}
\frac{1}{n}\log\left|\mathbf{\Gamma}^{l_n}(y^n)\right|\le h\left(\frac{l_n}{n}\right)+\frac{l_n}{n}\log|\mathcal{Y}|,\label{hball}
\end{align}
and $h(\cdot)$ represents the binary entropy function.
Therefore, the size of the function $g$ is
\begin{align}
||g||=||g_1||\cdot||g_2||=k\left|\mathbf{\Gamma}^{l_n}(y^n)\right|. \label{NanLiu02}
\end{align}
Note that for $(x^n,y^n)\in\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o}^n)$, we have $y^n\in\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o_Y}^n)$.
Thus, we expand the domain of the second argument of the function $\varphi$ from the type $\mathcal{T}_{Q^o_Y}^n$ to its $l_n$ neighborhood $\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o_Y}^n)$.
Assume the decoder observes $y^n \in\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o_Y}^n)$, and obtain $g(x^n,y^n)$ from the encoder as follows
\begin{align}
g(x^n,y^n)=(i,j).
\end{align}
From $y^n$ and $j$, we can determine the sequence $\bar{y}^n$ in the type $\mathcal{T}_{Q^o_Y}^n$. We then define $\varphi((i,j),y^n)=\varphi'(i,\bar{y}^n)$ for $i=1,2,\dots,k$,
where the decoding function $\varphi'(i,\bar{y}^n)$ was defined in the previous step.
Based on the definition of $\mathbf{\Gamma}^{l_n}(\pi_i(A(Q^o)))$ , for every $(x^n, y^n)\in\mathbf{\Gamma}^{l_n}(\pi_i(A(Q^o)))$, there exists a $(\bar{x}^n,\bar{y}^n)\in\pi_i(A(Q^o))$ such that
\begin{align}
d_H((x^n, y^n),(\bar{x}^n,\bar{y}^n))\le l_n,
\end{align}
and from the definition of $\pi_i(A(Q^o))$, we know that
\begin{align}
d(\bar{x}^n, \varphi'(i,\bar{y}^n))
\le\Delta+\delta. \label{Nan50}
\end{align}
Therefore, we have
\begin{align}
d(x^n,\varphi(g(x^n,y^n),y^n))&=d(x^n,\varphi'(g'(\bar{x}^n,\bar{y}^n),\bar{y}^n))\nonumber\\
&\le d(\bar{x}^n,\varphi'(g'(\bar{x}^n,\bar{y}^n),\bar{y}^n))+d_M\frac{l_n}{n},\label{Nan21}
\end{align}
where (\ref{Nan21}) is because $x^n$ and $\bar{x}^n$ at most differ in $l_n$ positions and
\begin{align}
d_M\triangleq \max_{(x,\hat{x})\in\mathcal{X}\times\hat{\mathcal{X}}} d(x,\hat{x}).
\end{align}
Hence, with the above definition of $(g,\varphi)$, we have that
\begin{align}
||g||&=k\left|\mathbf{\Gamma}^{l_n}(y^n)\right|,\\
d(x^n,\varphi(g(x^n,y^n),y^n))&\le \Delta+\delta+d_M\frac{l_n}{n}. \label{Nan51}
\end{align}
where (\ref{Nan51}) follows from (\ref{Nan50}) and (\ref{Nan21}).
Finally, we expand the rate-distortion code $(g, \varphi)$ to all $(x^n,y^n) \in \mathcal{X}^n \times \mathcal{Y}^n$. For $(x^n,y^n)\notin\mathbf{\Gamma}^{l_n}(\mathcal{T}_{Q^o}^n)$, we define $g(x^n,y^n)=0$.
And we define $\varphi(0,y^n)$ to be an arbitrary sequence in $\hat{\mathcal{X}}^n$.
Therefore, for $(g,\varphi)$ defined on $\mathcal{X}^n\times\mathcal{Y}^n$, we have
\begin{align}
\frac{1}{n}\log||g|| &\leq
E-D(Q^o||P)+3\epsilon_n+\delta+\frac{\log (nH(Q^o))}{n} +h\left(\frac{l_n}{n}\right)+\frac{l_n}{n}\log|\mathcal{Y}|,\label{Nan20}
\end{align}
where (\ref{Nan20}) follows from (\ref{Nan16}), (\ref{NanLiu02}) and (\ref{hball}).
We use blowing up lemma to calculate the average distortion for the rate distortion code we constructed above. We restate the blowing up lemma as follows.
\begin{Lem}[Blowing up]\cite[Lemma 5.4]{Csiszar:2011}
To any finite set $\mathcal{X}$ and sequence $\epsilon_n\rightarrow 0$, there exist a sequence of positive integers $l_n$ with $\frac{l_n}{n}\rightarrow 0$ and a sequence $\tau_n\rightarrow 1$ such that for any distribution $Q$ defined on $\mathcal{X}$ and every $n,A\subset\mathcal{X}^n$
\begin{align}
Q^n(A)\ge\exp(-n\epsilon_n)
\end{align}
implies
\begin{align}
Q^n(\mathbf{\Gamma}^{l_n}(A))\ge\tau_n.
\end{align}
\end{Lem}
In our case, from \cite[Lemma 2.3]{Csiszar:2011}, we have that for any sequence $\epsilon_n\rightarrow0$ and sufficiently large $n$
\begin{align}
(Q^{o})^{n}\left(\mathcal{T}_{Q^o}^n\right)\ge (n+1)^{-|\mathcal{X}||\mathcal{Y}|}\ge \exp(-n\epsilon_n).
\end{align}
Thus, from Blowing up lemma, we have that there exists a sequence $\eta_n\rightarrow1$ such that
\begin{align}
(Q^{o})^{n}\left(\mathbf{\Gamma}^{l_n}\left(\mathcal{T}_{Q^o}^n\right)\right)\ge\eta_n.\label{blup}
\end{align}
Therefore, for $(g,\varphi)$ defined on $\mathcal{X}^n\times\mathcal{Y}^n$,
\begin{align}
\mathsf{Pr}\left(d(x^n,\varphi(g(x^n,y^n),y^n))\le \Delta+\delta+d_M\frac{l_n}{n}\right)\ge(Q^{o})^{n}\left(\mathbf{\Gamma}^{l_n}\left(\mathcal{T}_{Q^o}^n\right)\right)
\ge\eta_n,\label{probdis}
\end{align}
where (\ref{probdis}) follows from (\ref{Nan21}) and (\ref{blup}).
The inequality in (\ref{probdis}) leads to
\begin{align}
\mathsf{E}(d(X^n,\varphi(g(X^n,Y^n),Y^n))&\le
\Delta+\delta+d_M\frac{l_n}{n}+d_M(1-\eta_n).
\end{align}
Thus, if we have a deception function, which under the distortion constraint $\Delta$ can achieve the deception exponent
\begin{align}
E=\min_Q D(Q||P)+R_{SI}(Q,\Delta)-\alpha,
\end{align}
for some $\alpha>0$,
then we can construct a rate distortion code with side information at both the encoder and decoder, where $(X^n, Y^n)$ is generated i.i.d. according to the distribution $Q^o$, that satisfies
\begin{align}
&\frac{1}{n}\log||g||\nonumber\\
&\leq
\min_Q \{D(Q||P)+R_{SI}(Q,\Delta)\}-\alpha-D(Q^o||P)+3\epsilon_n+\delta+\frac{\log (nH(Q^o))}{n} +h\left(\frac{l_n}{n}\right)+\frac{l_n}{n}\log|\mathcal{Y}|\nonumber\\
&\leq R_{SI}(Q^o,\Delta)-\alpha+3\epsilon_n+\delta+\frac{\log (nH(Q^o))}{n} +h\left(\frac{l_n}{n}\right)+\frac{l_n}{n}\log|\mathcal{Y}|,
\end{align}
and
\begin{align}
\mathsf{E}(d(X^n,\varphi(g(X^n,Y^n),Y^n))&\le
\Delta+\delta+d_M\frac{l_n}{n}+d_M(1-\eta_n).
\end{align}
Since the rate distortion function $R_{SI}(Q^o,\Delta)$ is a continuous function of $\Delta$, the above result contradicts with the result in the rate distortion problem with side information at both the encoder and decoder. This concludes the proof of the converse.
\section{Conclusion}
In this paper, we studied the probability of successful deception of an uncompressed biometric authentication system with side information at the adversary. We found the optimal exponent of the deception probability by providing the proofs of both the achievability and the converse. The results are proved by exploiting a connection between the problem of deception with side information and the rate distortion problem with side information at both the encoder and decoder.
\bibliographystyle{unsrt}
|
{
"redpajama_set_name": "RedPajamaArXiv"
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{"url":"http:\/\/openstudy.com\/updates\/55cc31bce4b0c5fe9805af47","text":"## Albert0898 one year ago 111. The fastest jet airplane can fly at a top speed of about 3,200 feet per second. Approximately how many miles can the jet travel in 15 hours, if it is traveling at top speed? (Note: There are 5,280 feet in one mile.) 112. If k is a positive integer, then for every value of k, the sum of the k smallest distinct odd positive integers is equal to which of the following?\n\nFirst, convert feet to miles, then seconds to hours. $$\\large\\sf \\frac{3200 ~ft}{second} \\times (\\frac{3600~seconds}{hour}) \\times \\frac{1~mile}{5280 ~ft}$$ Your answer should be in the thousands. Just an FYI.","date":"2017-01-20 06:32:39","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6976771354675293, \"perplexity\": 507.6236720057488}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-04\/segments\/1484560280791.35\/warc\/CC-MAIN-20170116095120-00192-ip-10-171-10-70.ec2.internal.warc.gz\"}"}
| null | null |
{"url":"https:\/\/alpopkes.com\/posts\/machine_learning\/kl_divergence\/","text":"One of the points on my long \u2018stuff-you-have-to-look-at\u2019 list is the Kullback-Leibler divergence. I finally took the time to take a detailed look at this topic.\n\n## Definition\n\nThe KL-divergence is a measure of how similar (or different) two probablity distributions are. When having a discrete probability distribution $P$ and another probability distribution $Q$ the KL-divergence for a set of points $X$ is defined as:\n\n$$D_{KL}(P ,|| ,Q) = \\sum_{x \\in X} P(x) \\log \\big( \\frac{P(x)}{Q(x)} \\big)$$\n\nFor probability distributions over continuous variables the sum turns into an integral:\n\n$$D_{KL}(P ,|| ,Q) = \\int_{-\\infty}^{\\infty} p(x) \\log \\big( \\frac{p(x)}{q(x)} \\big) dx$$\n\nwhere $p$ and $q$ denote the probability density functions of $P$ and $Q$.\n\nSource\n\n## Visual example\n\nThe Wikipedia entry on the KL divergence contains a nice illustration:\n\nOn the left hand side we can see two Gaussian probability density functions $p(x)$ and $q(x)$. The right hand side show the area that is integrated when computing the KL divergence from $p$ to $q$. We know that:\n\n\\begin{align} D_{KL}(P ,|| ,Q) &= \\int_{-\\infty}^{\\infty} p(x) \\log \\big( \\frac{p(x)}{q(x)} \\big) dx \\ &= \\int_{-\\infty}^{\\infty} p(x) \\big( \\log p(x) - \\log q(x) \\big) dx \\end{align}\n\nSo for each point $x_i$ on the x-axis, we compute $\\log p(x_i) - \\log q(x_i)$ and multiply the result by $p(x_i)$. We then plot the resulting y-value in the right hand plot. This is how we get to the curve given in the right hand plot. The KL divergence is now defined as the area under the graph, which is shaded.\n\n## KL divergence in machine learning\n\nIn most cases in machine learning we are given a dataset $X$ which was generated by some unknown probability distribution $P$. In this approach $P$ is considered to be the target distribution (that is, the \u2018true\u2019 distribution) which we are trying to approximate using a distribution $Q$. We can evaluate candidate distributions $Q$ using the KL-divergence from $P$ to $Q$. In many cases, for example in variational inference, the KL divergence is used as an optimization criterion which is minimized in order to find the best candidate\/approximation $Q$.\n\n## Interpreting the Kullback Leibler divergence\n\nNote: Several different intepretations of the KL divergence exist. This interpretation describes a probabilistic perspective which is often useful for machine learning.\n\n### Expected value\n\nIn order to understand how the KL divergence works, remember the formula for the expected value of a function. Given a function $f$ with $x$ being a discrete variable, the expected value of $f(x)$ is defined as\n\n$$\\mathbb{E}\\big[f(x)\\big] = \\sum_x f(x) p(x)$$\n\nwhere $p(x)$ is the probability density function of the variable $x$. For the continuous case we have\n\n$$\\mathbb{E}\\big[f(x)\\big] = \\int_{-\\infty}^{\\infty}f(x) p(x) dx$$.\n\nExample: Suppose you made a very good deal and bought three pairs of the best headphones for a reduced price of 200 USD each. You want to sell them for 350 USD each. We define the probability of selling $X=0, X=1, X=2, X=3$ headphones as follows:\n\n$$p(X=0) = 0.1$$ $$p(X=1) = 0.2$$ $$p(X=3) = 0.3$$ $$p(X=4) = 0.4$$\n\nWe further define a function that measures profit: $f(x) = \\text{revenue} - \\text{cost} = 350*X - 200*X$. For example, when selling two pairs of headphones you will make: $f(X=2) = 700 - 400 = 300$. So what\u2019s our expected profit? We can compute it using the formula for the expected value:\n\n\\begin{align} \\mathbb{E}\\big[f(x)\\big] &= p(X=0)*f(X=0) + p(X=1)*f(X=1) + p(X=2)*f(X=2) + p(X=3)*f(X=3) \\\\ &= 0.1 * 0 + 0.2 * 150 + 0.3 * 300 + 0.4 * 450 \\\\ &= 300 \\end{align}\n\n### Ratio of p(x) \/ q(x)\n\nLooking back at the definition of the KL divergence we can see that it\u2019s quite similar to the definition of the expected value. When setting $f(x) = \\log \\big(\\frac{p(x)}{q(x)}\\big)$ we can see that:\n\n\\begin{align} \\mathbb{E}\\big[f(x)\\big] &= \\mathbb{E}_{x \\sim p(x)}\\big[\\log \\big(\\frac{p(x)}{q(x)}\\big) \\big] \\\\ &= \\int_{-\\infty}^{\\infty} p(x) \\log \\big( \\frac{p(x)}{q(x)} \\big) dx \\\\ &= D_{KL}(P || Q) \\end{align}\n\nBut what does that mean? Let\u2019s start by looking at the quantity $\\frac{p(x)}{q(x)}$. When having some probability density function $p$ and another probability density function $q$ we can compare the two by looking at the ratio of the two densities:\n\n$$ratio = \\frac{p(x)}{q(x)}$$\n\nInsight: We can compare two probability density functions by means of the ratio.\n\nBecause both $p(x)$ and $q(x)$ are probability densities they output values between $0$ and $1$. When $q$ is similar to $p$, $q(x)$ should output values close to $p(x)$ for any input $x$.\n\nExample: For some input $x_i$, $p(x_i)$ might be 0.78, i.e. $p(x_i) = 0.78$. Let\u2019s look at different densities $q$:\n\na) If $q$ and $p$ are identical, $q(x)$ would output the same value and the resulting ratio would be one:\n$ratio = \\frac{p(x)}{q(x)} = \\frac{0.78}{0.78} = 1$\n\nb) When $q(x)$ assigns a lower probability to the input $x$ than $p(x)$, the resulting ratio will be larger than one:\n$ratio = \\frac{p(x)}{q(x)} = \\frac{0.78}{0.2} = 3.9$\n\nc) When $q(x)$ assigns a higher probability to the input $x$ than $p(x)$, the resulting ratio will be smaller than one: $$ratio = \\frac{p(x)}{q(x)} = \\frac{0.78}{0.9} \\approx 0.86$$\n\nThe example provides an important insight:\n\nInsight: For any input $x$ the value of the ratio tells us how much more likely $x$ is to occur under $p(x)$ compared to $q(x)$. A value of the ratio larger than 1 indicates that $p(x)$ is the more likely model. A value smaller than 1 indicates that $q$ is the more likely model.\n\n### Ratio for entire dataset\n\nIf we have a whole dataset $X = x_1, \u2026, x_n$ we can compute the ratio of the entire set by taking the product over the individual ratios. Note: this only holds if the examples $x_i$ are independent of each other.\n\n$$ratio = \\prod_{i=1}^n \\frac{p(x_i)}{q(x_i)}$$\n\nTo make the computation easier we can take the logarithm:\n\n$$\\text{log_ratio} = \\sum_{i=1}^n \\log \\big( \\frac{p(x_i)}{q(x_i)} \\big)$$\n\nWhen taking the logarithm, a log ratio value of 0 indicates that both models fit the data equally well. Values larger than 0 indicate that $p$ is the better model, that is, it fits the data better. Values smaller than 0 indicate that $q$ is the better model:\n\n### Example calculation\n\nLet\u2019s calculate the KL divergence for our headphone example. We already have specified a distribution $P$ over the possible outcomes. Let\u2019s define another distribution $Q$ which expresses our belief that it\u2019s very likely that we sell all three pairs of headphones and less likely that we don\u2019t sell all of them (or none).\n\n## Which type of logarithm to use\n\nIt\u2019s interesting to note that we can use different bases for the logarithm in the definition of the KL divergence, depending on the interpretation. For example, when using the natural logarithm the result of the KL divergence is measured in so called \u2018nats\u2019. When using the logarithm to base 2 the result is measured in bits.","date":"2021-02-27 02:54:41","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9955860376358032, \"perplexity\": 251.60463338330698}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-10\/segments\/1614178358064.34\/warc\/CC-MAIN-20210227024823-20210227054823-00161.warc.gz\"}"}
| null | null |
When implementing a Lean plan, don't leave out safety.
Seiri (sort) focuses on sorting and cleaning up by defining which tools and materials are needed at the job site and throwing away anything that is not needed. This eliminates clutter and the inefficiency that occurs when that clutter makes needed items difficult to locate.
Seiton (set in order) focuses on straightening and arranging necessary information, tools and materials in the correct order for their designated areas. Visual aids that can support this effort include a visual scoreboard, Jidoka lights, work areas designated by taped or painted lines, colorcoded ESD workspaces and Kanban squares.
Seiketsu (standardize) focuses on establishing a discipline of cleaning tools, equipment and the job immediately after use. Routine cleaning becomes a way of life.
Shitsuke (sustain) is the most important discipline because employees must continue to maintain the 5S discipline continuously. This includes following all procedures and work instructions.
Some companies follow a 5S Plus process. The sixth S is normally Safety and Environmental Health. While technically much of this is incorporated in Shitsuke, which mandates following all company rules and regulations, it helps put greater emphasis on worker safety and environmental responsibility.
Minimization of defect opportunities through carefully defined material and work-in-process handling and transport procedures.
Elimination of potential fire hazards.
Minimization of contamination from improperly stored hazardous materials.
Minimization of cross-contamination in production areas running RoHS and leaded production lines.
Maintenance of orderly evacuation routes.
Minimization of workplace injuries and accidents through elimination of the hazards created by clutter or unsafe practices.
Improvement in overall efficiency as employees can more easily locate key information and tools, change over lines faster, and identify and eliminate production constraints more quickly.
A positive impression on visitors touring a clean, orderly factory.
What is the best way to establish a program that will sustain itself over time? There are five key elements: Training, intra-work area audits, internal audits, keeping it relevant, and conveying ownership.
A good training program helps link the disciplines holistically to the overall activities of the work area so that employees constantly evaluate the outcome of their activities and the benefits that come with an orderly, clean workplace. For example, loading printed circuit board assemblies in a carrier correctly not only follows the "set in order" discipline, it also minimizes defect opportunities and eliminates wasted motion. The better employees understand the benefits of maintaining the disciplines, the more likely they are to sustain the program. A good training program also shows clear examples of practices to be avoided and should include photos of the most common violations. This helps ensure that even newer, less experienced employees have a complete understanding of practices to avoid, and also reinforces best practices with older employees who may be resistant to changing the way they've done things in the past.
daily or weekly audit responsibility. This ensures regular reinforcement of desired behaviors.
Compliance with 5S or 5S Plus discipline should also be part of the less frequent internal quality audit program. This helps ensure all work areas are consistently maintained, and can help identify areas where a team leader or supervisor may be too lax in their audit practices.
The final way to motivate sustainability is by linking it to other activities such as Kaizen events or training related to continuous improvement activities. Linking the 5S disciplines to these activities helps keep the program fresh and relevant in employees' minds.
When approached holistically as a foundation discipline for an efficient, Lean manufacturing process, 5S is a powerful tool. The key to unlocking its benefits is engaging employees in ways that motivate them to continue to broaden the way they apply the disciplines in their daily work activities. Conveying "ownership" of sustainability to team leaders and supervisors helps ensure continuous reinforcement of best practices.
James Fowler is director of ISO compliance at SigmaTron International (sigmatronintl.com); james.fowler@sigmatronintl.com.
|
{
"redpajama_set_name": "RedPajamaC4"
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| 3,889
|
\section{Introduction}
Human beings can process vast amounts of visual information in parallel through visual system~\cite{koch2006much} because the attention mechanism can selectively attend to the most informative parts of visual stimuli rather than the whole scene~\cite{eriksen1972temporal}.
A recent trend is to incorporate attention mechanisms into deep neural networks (DNNs), to focus on task-relevant parts of the input automatically.
Attention mechanism has benefited sequence modeling tasks as well as a wide range of computer vision tasks to boost the performance and improve the interpretability.
The attention modules in CNNs can be broadly categorized into \textit{channel-wise attention} and \textit{spatial attention}, which learns channel-wise~\cite{hu2018squeeze} and spatially-aware attention scores~\cite{simonyan2013deep} for modulating the feature maps, respectively.
As channel-wise attention would inevitably lose spatial information essential for localizing the important parts, in this paper, we focus on the spatial attention mechanism.
There are \textit{query-based} and \textit{module-based} spatial attention learning methods.
Query-based attention, or `self-attention', generates the attention scores based on the similarity/compatibility between the query and the key content. Though having facilitated various computer vision tasks, such dense relation measurements would lead to heavy computational overheads~\cite{han2020survey}, which significantly hinders its application scenario.
Module-based attention directly outputs an attention map using a learnable network module that takes as input an image/feature. Such an end-to-end inference structure is more efficient than query-based attention, and has been shown to be effective for various computer vision tasks.
Existing spatial attention mechanisms trained for certain tasks typically generate the attention maps by considering the contextual relations of the inputs. Although the attention maps are beneficial to the tasks, the attention learning process fails to consider the inherent information constraints of the attention mechanism in HVS,
\textit{i}.\textit{e}., to ensure that the information bypassed by the attention maps is of minimal redundancy, and meanwhile being sufficient for the task.
To explicitly incorporate the information constraints of the attention mechanism in HVS into attention learning in DNNs, in this paper, we propose an end-to-end trainable spatial attention mechanism inspired by the `Information Bottleneck (IB)' theory.
The whole framework is derived from an information-theoretic argument based on the IB principle. The resulted variational attention maps can effectively filter out task-irrelevant information, which reduces the overload of information processing while maintaining the performance.
To further restrict the information bypassed by the attention map, an adaptive quantization module is incorporated to round the attention scores to the nearest anchor value.
In this way, previous continuous attention values are replaced by a finite number of anchor values, which further compress the information filtered by the attention maps.
To quantitatively compare the interpretability of the proposed attention mechanism with others,
we start from the definition of interpretability on model decision making, and measure the interpretability in a general statistical sense based on the attention consistency between the original and the modified samples that do not alter model decisions.
In summary, our contributions are three-fold:
\begin{itemize}
\item We propose an IB-inspired spatial attention mechanism for visual recognition, which yields variational attention maps that minimize the MI between the attention-modulated representation and the input while maximizing the MI between the attention-modulated representation and the task label.
\item To further filter out irrelevant information, we design a quantization module to round the continuous attention scores to several learnable anchor values during training.
\item The proposed attention mechanism is shown to be more interpretable for the decision making of the DNNs compared with other spatial attention models.
\end{itemize}
Extensive experiments validate the theoretical intuitions behind the proposed IB-inspired spatial attention mechanism, and show improved performances and interpretability for visual recognition tasks.
\section{Related Work}
\label{section:rw}
\subsection{Spatial Attention Mechanism in DNNs}
Attention mechanisms enjoy great success in sequence modeling tasks such as machine translation~\cite{bahdanau2015neural}, speech recognition~\cite{chorowski2015attention} and image captioning~\cite{xu2015show}. Recently, they are also shown to be beneficial to a wide range of computer vision tasks to boost the performance and improve the interpretability of general CNNs.
The attention modules in CNNs fall into two broad categories, namely channel-wise attention and spatial attention. The former learns channel-wise attention scores to modulate the feature maps by reweighing the channels~\cite{hu2018squeeze}. The latter learns a spatial probabilistic map over the input to enhance/suppress each 2D location according to its relative importance \textit{w}.\textit{r}.\textit{t}. the target task~\cite{simonyan2013deep}.
In this section, we focus on spatial attention modules.
\noindent\textbf{Query-based/ Self-attention.}
Originated from query-based tasks~\cite{bahdanau2015neural,xu2015show}, this kind of attention is generated by measuring the similarity/compatibility between the query and the key content.
Seo~\textit{et al}.~\shortcite{seo2018progressive} use a one-hot encoding of the label to query the image and generate progressive attention for attribute prediction.
Jetley~\textit{et al}.~\shortcite{jetley2018learn} utilize the learned global representation of the input image as a query and calculate the compatibility with local representation from each 2D spatial location.
Hu~\textit{et al}.~\shortcite{hu2019local} adaptively determines the aggregation weights by considering the compositional relationship of visual elements in local areas.
Query-based attention considers dense relations in the space, which can improve the discriminative ability of CNNs. However, the non-negligible computational overhead limits its usage to low-dimensional inputs, and typically requires to downsample the original images significantly.
\noindent\textbf{Module-based attention.}
The spatial attention map can also be directly learned using a softmax-/sigmoid-based network module that takes an image/feature as input and outputs an attention map. Being effective and efficient, this kind of attention module has been widely used in computer vision tasks such as action recognition~\cite{sharma2015action} and image classification~\cite{woo2018cbam}.
Our proposed spatial attention module belongs to this line of research, and we focus on improving the performance of visual recognition over baselines without attention mechanisms.
Previous spatial attention learning works focus on the relations among non-local or local contexts to measure the relative importance of each location, and do not consider the information in the feature filtered by the attention maps. We instead take inspiration from the IB theory~\cite{tishby1999information} which maintains a good trade-off between information compression and prediction accuracy, and propose to learn spatial attention that minimizes the MI between the masked feature and the input, while maximizing the MI between the masked feature and the task label.
Such information constraints could also help remove the redundant information from the input features compared with conventional relative importance learning mechanisms.
\begin{figure*}
\begin{minipage}{\textwidth}
\begin{minipage}[t]{0.245\textwidth}
\centering
\includegraphics[width=0.4\textwidth]{graph_uni}
\captionsetup{font=small}
\makeatletter\def\@captype{figure}\makeatother\caption{\textbf{Graphical model} of the probabilistic neural network with IB-inspired spatial attention mechanism (\S\ref{sec:theory}).
\label{fig:graph}}
\end{minipage}
\hfill
\begin{minipage}[t]{0.74\textwidth}
\centering
\includegraphics[width=1\textwidth]{framework}
\captionsetup{font=small}
\makeatletter\def\@captype{figure}\makeatother\caption{\textbf{Framework of the IB-inspired spatial attention mechanism for visual recognition.}
The input $x$ is passed through an attention module to produce a continuous variational attention map $a$, which is quantized to a discrete attention map $a_q$ using a set of learnable anchor values $v_i$.
Then, $a_q$ and $x$ are encoded to a latent vector $z$, and decoded to a prediction $y$.
Loss function in Eq.~(\ref{eq:loss_func}).
See \S\ref{sec:theory} and \S\ref{sec:quantization}.
\label{fig:framework}}
\end{minipage}
\end{minipage}
\vspace{-16pt}
\end{figure*}
\subsection{IB-Inspired Mask Generation}
IB theory has been explored in tasks such as representation learning~\cite{alemi2017deep,achille2018information} and domain adaptation~\cite{du2020learning}.
Here, we focus on IB-inspired mask generation methods,
which yield additive~\cite{schulz2019restricting} or multiplicative masks~\cite{achille2018information,taghanaki2019infomask,zhmoginov2019information} to restrict the information that flows to the successive layers.
Information Dropout~\cite{achille2018information} multiplies the layer feature with a learnable noise mask to control the flow of information. This is equivalent to optimizing a modified cost function that approximates the IB Lagrangian of~\cite{tishby1999information}, allowing the learning of a sufficient, minimal and invariant representation for classification tasks.
InfoMask~\cite{taghanaki2019infomask} filters out irrelevant background signals using the masks optimized on IB theory, which improves the accuracy of localizing chest disease.
Zhmoginov~\textit{et al}.~\shortcite{zhmoginov2019information} generates IB-inspired Boolean attention masks for image classification models to completely prevent the masked-out pixels from propagating any information to the model output.
Schulz~\textit{et al}.~\shortcite{schulz2019restricting} adopts the IB concept for interpreting the decision-making of a pre-trained neural network, by adding noise to intermediate activation maps to restrict and quantify the flow of information. The intensity of the noise is optimized to minimize the information flow while maximizing the classification score.
In this paper, we propose an IB-inspired spatial attention mechanism based on a new variational approximation of the IB principle derived from a neural network incorporated with attention module ($\mathcal{L}_{\text{AttVIB}}$ in Eq.~(\ref{eq:IB_two})).
Compared with the above works, our optimization objective is not derived from the original IB Lagrangian of~\cite{tishby1999information}, thus the resulted attention mechanism is parameterized with different network architectures.
Besides, we focus on learning continuous spatial attention instead of noise masks~\cite{achille2018information,schulz2019restricting}, which is also different from the Boolean attention in~\cite{zhmoginov2019information}.
\section{Methodology}
\label{sec:method}
\subsection{Overview}
\label{sec:overview}
We derived the framework of the proposed spatial attention mechanism theoretically from the IB principle on a neural network incorporated with an attention module.
Assuming that random variables $X, Y, A$ and $Z$ follow the joint conditional distribution of the model shown in Fig.~\ref{fig:graph}. Here, $X, Y, A$ and $Z$ denote the input, output, spatial attention map, and the latent representation obtained from $X$ and $A$, respectively.
We drive the proposed spatial attention framework based on the joint distribution and the IB principle.
The resulted framework is shown in Fig.~\ref{fig:framework}, which consists of an attention module $p(a|x)$, a set of anchor values $\{v_i\}$ for quantization, an encoder $p(z|x,a)$, and a decoder $q(y|z)$.
The input $x$ is first passed through the attention module $p(a|x)$ to produce a continuous variational attention map $a$, which is quantized to a discrete attention map $a_q$ using a set of learnable anchor values $v_i$. Then, $a_q$ and $x$ are encoded to a latent vector $z$, and decoded to a prediction $y$.
The whole framework is trained end-to-end using the loss function defined in Eq.~(\ref{eq:loss_func}).
The derivation of the framework is shown in~\S\ref{sec:theory}, and more details are provided in the supplementary.
The attention score quantization process is shown in~\S\ref{sec:quantization}.
\subsection{IB-inspired Spatial Attention Mechanism}
\label{sec:theory}
We introduce the IB principle~\cite{tishby1999information} to learn the spatial attention that minimizes the MI between the masked representation and the input, while maximizing the MI between the masked representation and the class label.
Different from~\cite{alemi2017deep} for a standard learning framework, we derive new variational bounds of MI within an \textit{attentive framework} for visual recognition, which lead to an IB-inspired spatial attention mechanism.
Let the random variables $X, Y, A$ and $Z$ denote the input, output, spatial attention map, and the latent representation obtained from $X$ and $A$, respectively. The MI between the latent feature $Z$ and its output $Y$ is defined as:
\begin{equation}\label{eq:ZY_lb}
I(Z; Y)= \int p(y, z) \log \frac{p(y|z)}{p(y)} dydz.
\end{equation}
We introduce $q(y|z)$ to be a variational approximation of the intractable $p(y|z)$. Since the Kullback Leibler (KL) divergence is always non-negative, we have:
\begin{equation}\label{eq:KL_nonneg}
\begin{split}
&D_{\text{KL}}[p(y|z)||q(y|z)]=\int p(y|z)\log \frac{p(y|z)}{q(y|z)} dy \ge 0 \\
&\Rightarrow \int p(y|z)\log p(y|z) dy \ge \int p(y|z)\log q(y|z) dy,
\end{split}
\end{equation}
which leads to
\begin{equation}\label{eq:ZY_lb}
I(Z; Y)\ge \int p(y, z) \log q(y|z)dydz + H(Y),
\end{equation}
where $H(Y)\!=\!-\int p(y) \log p(y) dy$ is the entropy of $Y$, which is independent of the optimization procedure thus can be ignored.
By leveraging the fact that $p(y, z)\!=\!\int p(x, y, z, a)dxda\!=\!\int p(x)p(a|x)p(y|x, a)p(z|x, a)dxda$ (see Fig.~\ref{fig:graph}), and ignoring $H(Y)$, the new variational lower bound is as follows:
\begin{equation}\label{eq:ZY_lb_two}
\resizebox{.91\linewidth}{!}{$
I(Z; Y)\!\ge\!\int p(x, y) p(a|x) p(z|x, a)\!\log q(y|z) dxdadydz,
$}
\end{equation}
where $p(a|x)$ is the attention module, $p(z|x, a)$ is the encoder, and $q(y|z)$ is the decoder.
$I(Z; Y)$ can thus be maximized by maximizing its variational lower bound.
Next, to minimize the MI between the attention-modulated representation and the input, we consider $I(Z; X, A)$, and obtain the following upper bound:
\begin{equation}\label{eq:ZXA_ub}
\resizebox{.91\linewidth}{!}{$
I(Z; X, A)\!\le\!\int p(x)p(a|x)p(z|x, a)\!\log \frac{p(z|x, a)}{r(z)} dxdadz,
$}
\end{equation}
where $r(z)$ is a prior distribution of $z$.
In our experiments, we use
a spherical Gaussian $\mathcal{N}(z|0,I)$ as the prior $r(z)$.
By combining Eq.~(\ref{eq:ZY_lb_two}) and~(\ref{eq:ZXA_ub}), we obtain the lower bound of the attentive variational information bottleneck:
\begin{equation}\label{eq:IB_two}
\begin{split}
\mathcal{L}_{\text{AttVIB}} &\equiv I(Z; Y) - \beta I(Z;X, A) \\
= \int &p(x, y) p(a|x) p(z|x, a) \log q(y|z) dxdadydz \\
- \beta &\int p(x)p(a|x)p(z|x, a) \log \frac{p(z|x, a)}{r(z)} dxdadz,
\end{split}
\end{equation}
which offers an IB-inspired spatial attention mechanism for visual recognition. Here, $\beta\!>\!0$ controls the tradeoff between information compression and prediction accuracy.
We approximate $p(x)$, $p(x, y)$ with empirical distribution $p(x)\!=\!\frac{1}{N}\sum_{n=1}^{N}\delta_{x_n}(x)$, $p(x, y)\!=\!\frac{1}{N}\sum_{n=1}^{N}\delta_{x_n}(x)\delta_{y_n}(y)$ following~\cite{alemi2017deep}, where $N$ is the number of training samples, $x_n$ and $(x_n, y_n)$ are samples drawn from data distribution $p(x)$ and $p(x, y)$, respectively.
The approximated lower bound $\mathcal{L}_{\text{AttVIB}}$ can thus be written as:
\begin{equation}\label{eq:L_IB}
\resizebox{.91\linewidth}{!}{$
\begin{split}
\widetilde{\mathcal{L}}_{\text{AttVIB}} = &\frac{1}{N}\sum_{n=1}^{N} \{\int p(a|x_n) p(z|x_n, a) \log q(y_n|z) dadz \\
-\beta &\int p(a|x_n)p(z|x_n, a) \log \frac{p(z|x_n, a)}{r(z)} dadz\}.
\end{split}
$}
\end{equation}
Similar to~\cite{alemi2017deep}, we suppose to use the attention module of the form $p(a|x)\!=\!\mathcal{N}(a| g_e^{\mu}(x), g_e^{\Sigma}(x))$, and the encoder of the form $p(z| x, a)\!=\!\mathcal{N}(z| f_e^{\mu}(x, a), f_e^{\Sigma}(x, a))$, where $g_e$ and $f_e$ are network modules.
To enable back-propagation, we use the reparametrization trick~\cite{kingma2014auto}, and write $p(a|x)da\!=\!p(\varepsilon)d\varepsilon$ and $p(z|x, a)dz\!=\!p(\epsilon)d\epsilon$, where $a\!=\!g(x, \varepsilon)$ and $z\!=\!f(x, a, \epsilon)$ are deterministic functions of $x$ and the Gaussian random variables $\varepsilon$ and $\epsilon$.
The loss function is:
\begin{equation}\label{eq:L_IB_RP}
\resizebox{.91\linewidth}{!}{$
\begin{split}
\mathcal{L}\!=\!-&\frac{1}{N}\sum_{n=1}^{N}\!\{ \mathbb{E}_{\epsilon\sim p(\epsilon)}[\mathbb{E}_{\varepsilon\sim p(\varepsilon)}[\log q(y_n|f(x_n, g(x_n, \varepsilon), \epsilon))]] \\
+ &\beta \mathbb{E}_{\varepsilon\sim p(\varepsilon)}[D_{\text{KL}}[p(z|x_n, g(x_n, \varepsilon)) \parallel r(z)]]\}.
\end{split}
$}
\end{equation}
The first term of the loss function is the negative log-likelihood of the prediction, where the label $y_n$ of $x_n$ is predicted from the latent encoding $z$, and $z$ is generated from $x_n$ and its attention map $a\!=\!g(x_n, \varepsilon)$. Minimizing this term leads to maximal prediction accuracy.
The second term is the KL divergence between distributions of latent encoding $z$ and the prior $r(z)$, the minimization of which enables the model to learn an IB-inspired spatial attention mechanism for visual recognition.
This is different from the regular IB principle~\cite{tishby1999information,alemi2017deep} which is only for the representation learning without attention module.
\begin{table*}[t]
\begin{center}
\begin{threeparttable}
\resizebox{0.9\textwidth}{!}{
\renewcommand\arraystretch{0.95}
\begin{tabular}{l|| c c| c c| c c}
\hline\thickhline
&\multicolumn{2}{c|}{Image Class.}
&\multicolumn{2}{c|}{Fine-grained Recog.}
&\multicolumn{2}{c}{Cross-domain Class.} \\
\cline{2-7}
\multirow{-2}{*}{Model} &CIFAR-10 &CIFAR-100
&CUB-200-2011 &SVHN
&STL10-train &STL10-test \\
\hline
\multicolumn{3}{l}{-- Existing architectures --} \\
VGG~\cite{simonyan2014very} &7.77 &30.62
&34.64 &4.27
&54.66 &55.09\\
VGG-GAP~\cite{zhou2016learning} &9.87 &31.77
&29.50 &5.84
&56.76 &57.24\\
VGG-PAN~\cite{seo2018progressive} &6.29 &24.35
&31.46 &8.02
&52.50 &52.79\\
VGG-DVIB~\cite{alemi2017deep} &~~4.64$^*$ &~~22.88$^*$
&~~\textbf{23.94}$^*$ &~~3.28$^*$
&~~\textbf{51.40}$^*$ &~~\textbf{51.60}$^*$ \\
WRN~\cite{zagoruyko2016wide} &\textbf{4.00} &\textbf{19.25}
&26.50 &–
&– &– \\
\hline
\multicolumn{3}{l}{-- Architectures with attention --} \\
VGG-att2~\cite{jetley2018learn} &5.23 &23.19
&26.80 &3.74
&51.24 &51.71\\
VGG-att3~\cite{jetley2018learn} &6.34 &22.97
&26.95 &3.52
&51.58 &51.68\\
WRN-ABN~\cite{fukui2019attention} &~~3.92$^*$ &18.12
&– &~~2.88$^*$
&~~50.90$^*$ &~~51.24$^*$ \\
VGG-aib (ours) &4.28 &21.56
&23.73 &3.24
&50.64 &51.24\\
VGG-aib-qt (ours) &\textbf{4.10} &\textbf{20.87}
&\textbf{21.83} &\textbf{3.07}
&\textbf{50.44} &\textbf{51.16}\\
WRN-aib (ours) &3.60 &17.82
&17.26 &2.76
&\textbf{50.08} &50.84\\
WRN-aib-qt (ours) &\textbf{3.43} &\textbf{17.64}
&\textbf{15.50} &\textbf{2.69}
&50.34 &\textbf{50.49}\\
\hline
\end{tabular}
}
\end{threeparttable}
\end{center}
\caption{Top-1 error for image classification~(\S\ref{sec:image_cls}), fine-grained recognition~(\S\ref{sec:fg_cls}), and cross-dataset classification~(\S\ref{sec:cd_cls}). $^*$ denotes re-implementation or re-training. Other values are from the original paper. Best values of different backbones are in \textbf{bold}.
\label{table:visual_recognition}}
\end{table*}
\subsection{Attention Score Quantization}
\label{sec:quantization}
We define the continuous attention space as $A\!\in\!\mathbb{R}^{W_a\!\times\!H_a}$, and the quantized attention space as $A_q\!\in\!\mathbb{R}^{W_a\!\times\!H_a}$, where $W_a, H_a$ are the width and height of the attention map, respectively.
As shown in Fig.~\ref{fig:framework}, the input $x$ is passed through an attention module to produce a continuous variational attention map $a$, which is mapped to a discrete attention map $a_q$ through a nearest neighbour look-up among a set of learnable anchor values $\{v_i\!\in\!\mathbb{R}\}_{i=1}^Q$, which is given by:
\begin{equation}\label{eq:qt_att}
a_q^{(w,h)}=v_k,\;k=\arg\min_j \Vert a^{(w,h)}-v_j \Vert_2,
\end{equation}
where $w\!=\!1\dots W_a, h\!=\!1\dots H_a$ are spatial indices. In this way, each score $a^{(w,h)}$ in the continuous attention map is mapped to the 1-of-$Q$ anchor value.
The quantized attention map $a_q$ and the input $x$ are then encoded into a latent representation $z\!\in\!\mathbb{R}^K$, where $K$ is the dimension of the latent space. Finally, $z$ is mapped to the prediction probabilities $y\!\in\!\mathbb{R}^C$, and $C$ is the number of classes.
The complete model parameters include the parameters of the attention module, encoder, decoder, and the anchor values $\{v_i\!\in\!\mathbb{R}\}_{i=1}^Q$.
As the $\arg\min$ operation in Eq.~(\ref{eq:qt_att}) is not differentiable, we resort to the straight-through estimator~\cite{bengio2013estimating} and approximate the gradient of $a_q$ using the gradients of $a$. Though simple, this estimator worked well for the experiments in this paper.
To be concrete, in the forward process the quantized attention map $a_q$ is passed to the encoder, and during the backwards computation, the gradient of $a$ is passed to the attention module unaltered. Such a gradient approximation makes sense because $a_q$ and $a$ share the same $W_a\!\times\!H_a$ dimensional space, and the gradient of $a$ can provide useful information on how the attention module could change its output to minimize the loss function defined in Eq.~(\ref{eq:L_IB_RP}).
The overall loss function is thus defined as in Eq.~(\ref{eq:loss_func}), which extends Eq.~(\ref{eq:L_IB_RP}) with two terms, namely a quantization objective $\Vert \text{sg}[g(x_n, \varepsilon)]\!-\!a_q^\varepsilon\Vert_2^2$ weighted by $\lambda_q$, and a commitment term $\Vert g(x_n, \varepsilon)\!-\!\text{sg}[a_q^\varepsilon] \Vert_2^2$ weighted by $\lambda_c$, where $\text{sg}[\cdot]$ is the stopgradient operator~\cite{van2017neural}, and $a_g^\varepsilon$ is the quantized version of $a\!=\!g(x_n, \varepsilon)$.
The former updates the anchor values to move towards the attention map $a$, and the latter forces the attention module to commit to the anchor values.
We set $\beta\!=0.01$, $\lambda_g\!=\!0.4$, and $\lambda_c\!=\!0.1$ empirically.
\begin{equation}\label{eq:loss_func}
\begin{split}
\mathcal{L}\!=\!-&\frac{1}{N}\sum_{n=1}^{N}\!\{ \mathbb{E}_{\epsilon\sim p(\epsilon)}[\mathbb{E}_{\varepsilon\sim p(\varepsilon)}[\log q(y_n|f(x_n, a_q^\varepsilon, \epsilon))]] \\
+ &\mathbb{E}_{\varepsilon\sim p(\varepsilon)}[\beta D_{\text{KL}}[p(z|x_n, a_q^\varepsilon) \parallel r(z)] \\
+ \lambda_q &\Vert \text{sg}[g(x_n, \varepsilon)]\!-\!a_q^\varepsilon\Vert_2^2\!+\!\lambda_c \Vert g(x_n, \varepsilon)\!-\!\text{sg}[a_q^\varepsilon] \Vert_2^2]\}.
\end{split}
\end{equation}
An illustration of the whole framework is shown in Fig.~\ref{fig:framework}.
In practice, we first use an feature extractor, \textit{e}.\textit{g}. VGGNet~\cite{simonyan2014very}, to extract an intermediate feature $f$ from the input $x$, then learn the attention $a, a_q$ and the variational encoding $z$ from $f$ instead of $x$.
\begin{table*}[t]
\begin{center}
\begin{threeparttable}
\resizebox{0.95\textwidth}{!}{
\renewcommand\arraystretch{0.95}
\begin{tabular}{l|| c| c c c c| c c c c|| c| c | c }
\hline\thickhline
\multirow{2}{*}{Model} &\multirow{2}{*}{Spatial}
&\multicolumn{4}{c|}{CIFAR-10} &\multicolumn{4}{c||}{CIFAR-100}
&\multirow{2}{*}{Frequency}
&\multirow{2}{*}{CIFAR-10} &\multirow{2}{*}{CIFAR-100}\\
\cline{3-10}
& &$p\!=\!4$ &$p\!=\!8$ &$p\!=\!12$ &$p\!=\!16$
&$p\!=\!4$ &$p\!=\!8$ &$p\!=\!12$ &$p\!=\!16$
& & &\\
\hline
\hline
\multirow{2}{*}{VGG-att3$^*$}
&color &91.46 &79.61 &37.71 &~~7.12
&52.92 &39.93 &25.40 &14.56
&$r\!>\!4$ &83.06 &~~3.69 \\
&svhn &90.97 &75.70 &36.74 &~~6.51
&82.08 &63.16 &39.07 &21.03
&$r\!<\!12$ &49.61 &50.90 \\
\hline
\multirow{2}{*}{VGG-aib}
&color &99.22 &99.69 &\textbf{93.78} &\textbf{20.59}
&98.88 &98.56 &\textbf{95.92} &\textbf{22.70}
&$r\!>\!4$ &99.91 &\textbf{99.78} \\
&svhn &98.59 &97.52 &97.35 &72.86
&98.73 &96.70 &96.69 &93.02
&$r\!<\!12$ &53.26 &79.13 \\
\hline
\multirow{2}{*}{VGG-aib-qt}
&color &\textbf{99.26} &\textbf{99.79} &91.10 &18.36
&\textbf{99.18} &\textbf{99.30} &94.59 &20.54
&$r\!>\!4$ &\textbf{99.96} &99.60 \\
&svhn &\textbf{98.65} &\textbf{98.04} &\textbf{97.85} &\textbf{78.82}
&\textbf{99.12} &\textbf{97.64} &\textbf{97.21} &\textbf{94.79}
&$r\!<\!12$ &\textbf{73.52} &\textbf{79.70} \\
\hline
\hline
\multirow{2}{*}{WRN-ABN}
&color &90.76 &65.01 &33.89 &13.24
&89.74 &61.38 &30.56 &~~9.67
&$r\!>\!4$ &38.14 &14.04 \\
&svhn &90.40 &68.41 &38.46 &16.55
&92.77 &63.67 &30.57 &~~9.47
&$r\!<\!12$ &36.33 &25.43 \\
\hline
\multirow{2}{*}{WRN-aib}
&color &\textbf{99.95} &\textbf{93.94} &\textbf{45.17} &~~6.69
&99.86 &95.35 &64.27 &17.15
&$r\!>\!4$ &78.96 &90.44 \\
&svhn &\textbf{99.94} &\textbf{97.34} &\textbf{81.98} &\textbf{47.65}
&99.90 &97.77 &\textbf{89.37} &62.64
&$r\!<\!12$ &84.58 &\textbf{94.18} \\
\hline
\multirow{2}{*}{WRN-aib-qt}
&color &99.84 &84.18 &28.94 &~~4.64
&\textbf{99.97} &\textbf{97.07} &\textbf{69.51} &\textbf{25.80}
&$r\!>\!4$ &72.05 &\textbf{94.13} \\
&svhn &99.91 &96.05 &70.75 &28.35
&\textbf{99.95} &\textbf{98.52} &89.00 &\textbf{63.10}
&$r\!<\!12$ &76.53 &93.17 \\
\hline
\end{tabular}
}
\end{threeparttable}
\end{center}
\caption{Interpretability scores under spatial and frequency domain modification on CIFAR-10 and CIFAR-100. $p$ is the window size of the modified region. $r$ is the radius in frequency domain. See~\S\ref{sec:interp_analysis} for more details. $^*$ denotes re-implementation. Best values in \textbf{bold}.
\label{table:interp}}
\end{table*}
\section{Experiments}
\label{sec:experiments}
To demonstrate the effectiveness of the proposed IB-inspired spatial attention, we conduct extensive experiments on various visual recognition tasks, including image classification~(\S\ref{sec:image_cls}), fine-grained recognition~(\S\ref{sec:fg_cls}), and cross-dataset classification~(\S\ref{sec:cd_cls}), and achieve improved performance over baseline models.
In~\S\ref{sec:interp_analysis}, we compare the interpretability of our attention maps with those of other attention models both qualitatively and quantitatively.
We conduct an ablation study in~\S\ref{sec:ab_study}.
More details are shown in the supplementary.
\subsection{Image Classification}
\label{sec:image_cls}
\noindent\textbf{Datasets and models.} \textit{CIFAR-10}~\cite{krizhevsky2009learning} contains $60,000$ $32\!\times\!32$ natural images of $10$ classes, which are splited into $50,000$ training and $10,000$ test images. \textit{CIFAR-100}~\cite{krizhevsky2009learning} is similar to CIFAR-10, except that it has $100$ classes.
We extend standard architectures, VGG and wide residual network (WRN), with the proposed IB-inspired spatial attention, and train the whole framework from scratch on CIFAR-10, and CIFAR-100, respectively.
We use original input images after data augmentation (random flipping and cropping with a padding of 4 pixels).
\noindent\textbf{Results on CIFAR.}
As shown in Table~\ref{table:visual_recognition}, the proposed attention mechanism achieves noticeable performance improvement over standard architectures and existing attention mechanisms such as GAP, PAN, \textit{VGG-att}\footnote{\tiny\url{https://github.com/SaoYan/LearnToPayAttention}} and ABN\footnote{\tiny\url{https://github.com/machine-perception-robotics-group/attention_branch_network}}.
To be specific, \textit{VGG-aib} achieves $3.49\%$ and $9.06\%$ decrease of errors over the baseline VGG model on CIFAR-10 and CIFAR-100, respectively.
The quantized attention model \textit{VGG-aib-qt} further decreases the errors over \textit{VGG-aib} by $0.18\%$ and $0.69\%$ on the two datasets.
Compared with other VGG-backboned attention mechanisms, ours also achieve superior classification performances.
Similarly, \textit{WRN-aib} and \textit{WRN-aib-qt} also decrease the top-1 errors on CIFAR-10 and CIFAR-100.
\subsection{Fine-grained Recognition}
\label{sec:fg_cls}
\noindent\textbf{Datasets and models.} \textit{CUB-200-2011} (CUB)~\cite{WahCUB_200_2011} contains $5,994$ training and $5,794$ testing bird images from $200$ classes. \textit{SVHN} collects $73,257$ training, $26,032$ testing, and $531,131$ extra digit images from house numbers in street view images.
For CUB, we perform the same preprocessing as~\cite{jetley2018learn}.
For SVHN, we apply the same data augmentation as CIFAR.
\noindent\textbf{Results on CUB.}
The proposed \textit{VGG-aib} and \textit{VGG-aib-qt} achieves smaller classification error compared with baselines built on VGG, including GAP, PAN, and \textit{VGG-att}.
Specially, \textit{VGG-aib} and \textit{VGG-aib-qt} outperform \textit{VGG-att} for a $3.07\%$ and $4.97\%$, respectively.
\noindent\textbf{Results on SVHN.}
Our proposed method achieves lower errors with VGG-backboned models, and comparative errors for WRN backbone methods.
\subsection{Cross-domain Classification}
\label{sec:cd_cls}
\noindent\textbf{Datasets and models.} \textit{STL-10} contains $5,000$ training and $8,000$ test images of resolution $96\!\times\!96$ organized into $10$ classes. Following~\cite{jetley2018learn}, the models trained on CIFAR-10 are directly tested on STL-10 training and test sets without finetuning to verify the generalization abilities.
\noindent\textbf{Results on STL-10.}
As shown in Table~\ref{table:visual_recognition}, the proposed attention model outperforms other competitors on VGG backbone, and achieves comparative performance for WRN backbone.
\subsection{Interpretability Analysis}
\label{sec:interp_analysis}
An \textit{interpretable} attention map is expected to faithfully highlight the regions responsible for the model decision.
Thus, an \textit{interpretable} attention mechanism is expected to yield consistent attention maps for an original input and a modified input if the modification does not alter the model decision.
Here, we quantify the interpretability of an attention mechanism by calculating the `interpretability score', \textit{i}.\textit{e}.~the percentage of attention-consistent samples in prediction-consistent samples under modifications in the spatial or frequency domain.
Here, the attention consistency is measured by the cosine similarity between two flattened and normalized attention maps.
\begin{figure}[t]
\centering
\includegraphics[width=0.45\textwidth]{interp}
\captionsetup{font=small}
\caption{\small{Visualization of attention maps for interpretability (\S\ref{sec:interp_analysis}).}
}
\label{fig:vis_interp}
\end{figure}
The \textit{spatial domain modification} includes randomly occluding the original images in CIFARs with color blocks or images drawn from distinct datasets. The size of the modified region $p$ ranges from $4$ to $16$.
The $2$nd and $3$rd rows in Fig.~\ref{fig:vis_interp} show exemplar images occluded by a random color block and an image randomly drawn from SVHN, respectively, where $p\!=\!8$. As can be observed, our spatial attention model \textit{VGG-aib(-qt)}, and \textit{WRN-aib(-qt)} yield consistent attention maps with the original images (first row).
The interpretability scores are listed in Table~\ref{table:interp}. Our method consistently outperforms other spatial attention mechanisms with the same backbone by a large margin on two datasets for various $p$.
This is because the IB-inspired attention mechanism minimizes the MI between the attention-modulated features and the inputs, thus mitigating the influence of ambiguous information exerted on the inputs to some extent.
We also conduct \textit{frequency domain modification}, which is done by performing Fourier transform on the original samples, and feeding only high-/low-frequency (HF/LF) components into the model.
To preserve enough information and maintain the classification performance, we focus on HF components of $r\!>\!4$, and LF components of $r\!<\!12$, where $r$ is the radius in frequency domain defined as in~\cite{wang2020high}.
The $4$th and $5$th rows in Fig.~\ref{fig:vis_interp} show exemplar images constructed from HF and LF components, respectively. Our attention maps are more royal to those of the original images compared with other competitors.
Out method also yields much better quantitative results than other attention models, as illustrated in Table~\ref{table:interp}.
This is because our IB-inspired attention mechanism can well capture the task-specific information in the input by maximizing the MI between the attention-modulated feature and the task output, even when part of the input information is missing.
\subsection{Ablation Study}
\label{sec:ab_study}
We conduct ablation studies on \textit{CIFAR-10} with the VGG backbone to assess the effectiveness of each component.
\noindent\textbf{Effectiveness of attention module.} Table~\ref{table:visual_recognition} shows that \textit{VGG-aib} and \textit{VGG-aib-qt} outperform the IB-based representation learning model \textit{VGG-DVIB}.
\noindent\textbf{Information tradeoff.} $\beta$ controls the amount of information flow that bypasses the bottleneck of the network. To measure the influence of $\beta$ on the performance, we plot the accuracy values with varying $\beta$ in Fig.~\ref{fig:ablation_study} (a). As can be observed, the accuracy is largest when $\beta\!=\!0.01$.
\noindent\textbf{Latent vector dimension.} We experiment on $K\!=\!64,128$, $256,512$, $1024$.
As shown in Fig.~\ref{fig:ablation_study} (b), $K\!=\!256$ achieves the best performance.
\noindent\textbf{Attention score quantization.} Fig.~\ref{fig:ablation_study} (c) shows the classification accuracy when varying number of anchor values $Q$, where $Q$ between $20$ and $50$ gives better performance.
Exemplary attention maps of cases that are wrongly classified by \textit{VGG-aib} but are correctly predicted by \textit{VGG-aib-qt} are listed in Fig.~\ref{fig:att_maps}. As can be observed, attention quantization can further help focus on more concentrated regions with important information, thus improve the prediction accuracy.
\begin{figure}[t]
\centering
\includegraphics[width=\columnwidth]{ablation}
\put(-192,-6){\fontsize{7pt}{7pt}\selectfont{$\beta$}}
\put(-116,-6){\fontsize{7pt}{7pt}\selectfont{$K$}}
\put(-40,-6){\fontsize{7pt}{7pt}\selectfont{$Q$}}
\put(-170,20){\fontsize{8pt}{8pt}\selectfont{(a)}}
\put(-94,20){\fontsize{8pt}{8pt}\selectfont{(b)}}
\put(-18,20){\fontsize{8pt}{8pt}\selectfont{(c)}}
\captionsetup{font=small}
\caption{\small{\textbf{Ablation study} on CIFAR-10 with VGG backbone (\S\ref{sec:ab_study}).}
}
\label{fig:ablation_study}
\end{figure}
\begin{figure}[t]
\centering
\includegraphics[width=0.4\textwidth]{cmp_qt_htm}
\captionsetup{font=small}
\caption{\small{Effect of attention score quantization. See~\S\ref{sec:ab_study} for details.}
}
\label{fig:att_maps}
\end{figure}
\section{Conclusion}
\label{sec:conclusion}
We propose a spatial attention mechanism based on IB theory for generating probabilistic maps that minimizes the MI between the masked representation and the input, while maximizing the MI between the masked representation and the task label.
To further restrict the information bypassed by the attention map, we incorporate an adaptive quantization mechanism to regularize the attention scores by rounding the continuous scores to the nearest anchor value during training.
Extensive experiments show that the proposed IB-inspired spatial attention mechanism significantly improves the performance for visual recognition tasks by focusing on the most informative parts of the input.
The generated attention maps are interpretable for the decision making of the DNNs, as they consistently highlight the informative regions for the original and modified inputs with the same semantics.
\section*{Acknowledgments}
\label{sec:acknowledgments}
This work is supported in part by General Research Fund (GRF) of Hong Kong Research Grants Council (RGC) under Grant No. 14205018 and No. 14205420.
\newpage
\bibliographystyle{named}
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| null | null |
\section{INTRODUCTION}
Hadronic final states of $\rm e^+\rm e^-$ annihilations have proven to provide
precise tests of the strong interaction
and of its underlying theory, Quantum Chromodynamics (QCD).
The success and increasing significance of QCD tests which was achieved
in the past few years, especially with the precise data from LEP,
was based on improvements and a deeper understanding of the
theoretical predictions as well as of the experimental techniques,
which both substantially profited from the experience of earlier
studies with data from lower energy colliders.
In fact, in many respects it seems desirable to return to the low
energy $\rm e^+\rm e^-$ data and reanalyse them with the knowledge of today.
One reason behind this demand is that the size as well as the
energy dependence of
the strong coupling parameter $\alpha_{\rm s}$ is largest at lower energies, and
thus the characteristic feature of QCD, asymptotic freedom and the
running of $\alpha_{\rm s}$, can be most significantly tested if reliable data
at `low' energies are available; see \cite{qcd96} for a recent review.
In the light of these remarks, a short historical review of QCD tests
in $\rm e^+\rm e^-$ annihilation will be given in this report.
In Section~2, selected highlights of early QCD studies, at energies
below the $\rm Z^0$ pole, will be reviewed,
and some of the original illustrations will be reproduced.
Section~3 contains a brief overview of QCD tests achieved at LEP and
at the SLC.
In section~4, the latest measurements of $\alpha_{\rm s}$ from LEP at energies
above the $\rm Z^0$ pole will be summarised.
A comparison with measurements of $\alpha_{\rm s}$ from other
processes will be given in Section~5, including an update of the
world summary of $\alpha_{\rm s}$ determinations \cite{qcd96}.
\section{QCD IN $\rm e^+\rm e^-$ ANNIHILATION -- FROM SPEAR TO SLC AND LEP}
In 1975, the first \char'134 Evidence for Jet Structure in Hadron
Production by $\rm e^+\rm e^-$ Annihilation" was reported \cite{firstjets}.
The data, taken with the SLAC-LBL magnetic detector at the SPEAR
storage ring, showed increasing evidence for the production of
two-jet like events when the center of mass energy, $E_{cm}$, was
raised from 3 to 7.4 GeV (see Fig.~\ref{2-jet}).
The jet structure manifested itself as a decrease of the mean
sphericity, which is a measure of the global shape of hadronic
events.
The angular distribution of the jet axes from the same measurement
provided evidence that the underlying partons must have
spin~$\frac{1}{2}$.
These observations, which were further corroborated
by similar measurements
at $E_{cm}$ = 14 to 34 GeV at the $\rm e^+\rm e^-$ storage ring PETRA
\cite{tasso-s}, see Fig.~\ref{tasso-s},
confirmed the basic ideas of the quark-parton model,
which relates the constituents
of the static quark model with partons produced in particle
collisions at high energies.
\begin{figure}
\vskip-20mm
\begin{center}
\epsfxsize6.0cm
\epsffile{2-jet.eps}
\end{center}
\vskip-45mm
\caption{\label{2-jet}
Sphericity distributions observed at $E_{cm} = 3.0$~GeV~(a), 6.2~GeV~(b)
and at 7.4~GeV~(c), compared with a jet (solid curves)
and a phase space model (dashed curves).
Data are from the SLAC-LBL Magnetic Detector at SPEAR
\protect\cite{firstjets}.}
\end{figure}
\begin{figure}
\vskip-8mm
\epsfxsize7.5cm
\epsffile{tasso-s.eps}
\vskip-12mm
\caption{\label{tasso-s}
Sphericity distributions at $E_{cm} = 14$~GeV, 22~GeV
and 34~GeV (from TASSO \protect\cite{tasso-s}).}
\end{figure}
\begin{figure}[!htb]
\epsfxsize6.5cm
\epsffile{tasso-3-jet.eps}
\vskip-10mm
\caption{\label{tasso-3-jet}
The first 3-jet event observed by TASSO at PETRA.
Plotted are the momentum vectors of the charged particles, projected into the
principal event plane; the dotted lines indicate the reconstructed jet
axes (from \protect\cite{slwurep}).}
\end{figure}
In 1979, a small fraction of
planar 3-jet events was observed by the PETRA
experiments around $E_{cm} \approx 30$ GeV \cite{gluon}, which
was attributed to the emission of a third parton
with zero electric charge and spin 1 \cite{gspin}, as
expected for gluon bremsstrahlung predicted by QCD~\cite{egr}.
First evidence for 3-jet like
events came from visual scans of the energy flow within hadronic
events (Fig.~\ref{tasso-3-jet}; see also Fig.~\ref{Jade-3jet}),
followed by statistical analyses of global
event shapes like oblateness (Fig.~\ref{markj-o}) and of
angular correlations which are sensitive to the gluon spin
\cite{gspin}.
At the end of 1979, the MARK-J collaboration reported a first
measurement of $\alpha_{\rm s}$, in first order perturbative QCD, from the
shape of the oblateness distribution \cite{as-mkj}.
The first analysis based on the definition and reconstruction of
resolvable jets was published in 1980 by the PLUTO collaboration
\cite{pluto-jets}, see Fig.~\ref{pluto-njet}.
\begin{figure}[!htb]
\epsfxsize6.5cm
\epsffile{markj-o.eps}
\vskip-10mm
\caption{\label{markj-o}
Oblateness distribution measured at PETRA, compared with the
predictions based on a \protect $q \overline{q} g$ model (full line) and a
$q\overline{q}$ model (from \protect \cite{slwurep}).}
\end{figure}
\begin{figure}[!htb]
\vskip-5mm
\epsfxsize7.5cm
\epsffile{Jade-3jet.eps}
\vskip-10mm
\caption{\label{Jade-3jet}
Detector view of a `typical' 3-jet event recorded with the JADE
detector at PETRA.}
\end{figure}
\begin{figure}[!htb]
\vskip-5mm
\epsfxsize7.5cm
\epsffile{tasso-gspin.eps}
\vskip-12mm
\caption{\label{tasso-gspin}
Distribution of the Ellis-Karliner angle \protect $\tilde{\Theta}$ measured by
TASSO \protect\cite{gspin}, compared with model predictions for a vector
and a scalar gluon.}
\end{figure}
\begin{figure}[!htb]
\epsfxsize6.1cm
\epsffile{pluto-njet.eps}
\vskip-12mm
\caption{\label{pluto-njet}
First jet multiplicity distribution measured at PETRA \protect
\cite{pluto-jets}.}
\end{figure}
In 1981, the JADE collaboration found first evidence
\cite{jade-string} that the string hadroni\-zation model
\cite{stringfrag} provides a better description of the
observed hadron flow in 3-jet events than does an independent jet
hadroni\-zation model.
\begin{figure}[!htb]
\epsfxsize6.5cm
\epsffile{jade-as2.eps}
\vskip-10mm
\caption{\label{jade-as2}
Corrected distribution of the scaled energy of the most energetic jet
of reconstructed 3-jet events, compared with analytic QCD predictions
in leading (dashed) and in next-to-leading order (full line)
\protect \cite{jade-as2}.}
\end{figure}
\begin{figure}[!htb]
\vskip-4mm
\epsfxsize7.5cm
\epsffile{jade-r3.eps}
\vskip-10mm
\caption{\label{jade-r3}
The ratio of 3-jet productions rates at 44 and at 34.6 GeV c.m. energy,
analysed with the JADE jet finder
\protect \cite{jadejet2}.
The data are compared with QCD analytic predictions.
For the hypothesis of an $\alpha_{\rm s}$ which does $not$ run with energy, a
constant ratio of 1 would be expected.}
\end{figure}
The year 1982 brought first evidence for 4-jet like events, observed
by JADE at $E_{cm}$ = 33 GeV \cite{jade-4-jet}, and the first
determination of $\alpha_{\rm s}$ in second order perturbation theory (${\cal O}(\as^2)$)~\cite{ert},
again by JADE \cite{jade-as2}.
The latter analysis was already
based on principles and ideas which nowadays are standard for most
of the $\alpha_{\rm s}$ analyses at SLC and LEP:
An analytic QCD calculation in ${\cal O}(\as^2)$ perturbation theory was
fitted to a measured event shape distribution which had been
corrected for detector resolution and hadronization effects
(Fig.~\ref{jade-as2}).
\begin{figure}[!htb]
\epsfxsize7.5cm
\epsffile{amy-nr.eps}
\vskip-10mm
\caption{\label{amy-nr}
Distribution of the cosine of the modified Nachtmann-Reiter angle
for 4-jet events measured by AMY at TRISTRAN
\protect \cite{amyjet}, compared with the predictions of a QCD (solid line) and
an Abelian (dashed line) model.}
\end{figure}
In 1986 JADE published the first detailed analysis of n-jet event
production rates \cite{jadejet1}, introducing a jet finding
mechanism which has since then been used in many other studies.
The ratio of 4-jet over 3-jet event
production rates was found to be significantly larger than predicted
by ${\cal O}(\as^2)$ QCD, an observation that motivated studies of the
influence of the choice of renormalization scales in finite order
perturbative QCD
\cite{scales}.
In 1988 it was
demonstrated by JADE that the energy dependence of 3-jet event
production rates gives evidence for the running of $\alpha_{\rm s}$
\cite{jadejet2},
see Fig.~\ref{jade-r3}.
First signs of the presence of the gluon self coupling
were observed in a study of 4-jet events by AMY \cite{amyjet}
around $E_{cm}$ = 56 GeV (Fig.~\ref{amy-nr}).
\begin{figure}[htb]
\epsfxsize7.5cm
\epsffile{markii-3jet.eps}
\vskip-0.7cm
\caption{\label{markii-3jet}
One of
the first hadronic $\rm Z^0$-decays, observed with the Mark-II detector
at the SLAC Linear Collider.}
\end{figure}
\baselineskip=12.0pt
More detailed reviews about jet
physics in $\rm e^+\rm e^-$ annihilations below the $Z^0$ resonance
can be found e.g. in
\cite{slwurep,naroska,budapest} .
\section{QCD AT AND ABOVE THE $\rm Z^0$ POLE}
In 1989, both the SLC and LEP started to deliver
$\rm e^+\rm e^-$ collisions at and around the $\rm Z^0$ resonance, $E_{cm} \sim 91$~GeV.
An example of an event of the type $\rm e^+\rm e^- \rightarrow Z^0
\rightarrow 3\ jets$, observed with the Mark-II detector at the SLC, is
shown in Fig.~\ref{markii-3jet}.
There exist several detailed reviews of QCD studies at LEP and SLC,
see e.g. \cite{annrev,hebbeker,scotland,burrows},
and many of the most recent results were presented
at this conference.
Here, only an overview of some of the major
achievements will be given:
\par \noindent
New QCD calculations...
\begin{itemize}
\item
${\cal O}(\as^3)$ QCD predictions for the hadronic branching fraction of the
$\rm Z^0$ boson and the $\tau$ lepton \cite{gorishny,braaten},
\item
resummation of leading and next-to-leading logarithms and their
matching with caluclations in complete
${\cal O}(\as^2)$, for several hadronic event shapes
observables \cite{catani},
\item
${\cal O}(\as^2)$ predictions for massive quarks \cite{wbquarks};
\end{itemize}
\noindent
new observables...
\begin{itemize}
\item
jet finding algorithms \cite{bkss}
like the Durham scheme \cite{durhamjets} or the cone algorithm \cite{conejets},
\item
event shape variables like total and wide jet broadening, $B_T$ and $B_w$
\cite{jetbroad},
\item
observables to study 4-jet kinematics \cite{d-a34};
\end{itemize}
\noindent
and new experimental techniques...
\begin{itemize}
\item
simultaneous use of many observables to study and reduce systematic
and theoretical uncertainties in measurements of $\alpha_{\rm s}$ \cite{lepas2},
\item
variations of renormalisation scale, parton virtuality, quark masses
etc. to estimate theoretical uncertainties \cite{lepas2},
\item
tagging of quark - and gluon-jets \cite{qgtag},
\item
detailed studies of $\tau$-lepton decays (\cite{settles});
\end{itemize}
\noindent
... resulted in:
\begin{itemize}
\item
precise determinations of $\alpha_{\rm s}$ (see Sections~4 and~5),
\item
tests of the QCD group structure (see \cite{dissertori}),
\item
detailed studies of differences between quark- and gluon-jets
(see \cite{qgdiff}),
\item
many details of the hadronisation process (see e.g. \cite{lafferty}),
\item
measurements of the heavy quark masses, $m_c(M_\tau)$ and $m_b(M_z)$
\cite{qmasses}.
\end{itemize}
\section{$\alpha_{\rm s}$ AT LEP ABOVE THE $\rm Z^0$-POLE}
\begin{table*}[htb]
Table 1.\ \
Summary of measurements of $\alpha_{\rm s}$ at LEP-1.5 and at LEP-2.\\
\begin{tabular}{|l|l|l|l|c|}
\hline
Exp. & $\alpha_{\rm s} (\sqrt{s} = 133$~GeV) & $\alpha_{\rm s} (\sqrt{s} = 161$~GeV) &
$\alpha_{\rm s} (\sqrt{s} = 172$~GeV) & refs.\\
\hline \hline
ALEPH & $0.115 \pm 0.008 \pm 0.005$ & $0.111 \pm 0.009 \pm 0.005$ &
$0.105 \pm 0.010 \pm 0.004$ & \cite{a-133} \\
DELPHI & $0.116 \pm 0.007^{\ +0.005}_{\ -0.004}$ & $0.107 \pm
0.008^{\ +0.005}_{\ -0.004}$ & $0.104 \pm 0.013^{\ +0.005}_{\ -0.004}$ &
\cite{d-133} \\
L3 & $0.107 \pm 0.005 \pm 0.006$ & $0.103 \pm 0.005 \pm 0.005$ &
$0.104 \pm 0.006 \pm 0.005$ & \cite{l-as} \\
OPAL & $0.110 \pm 0.009 \pm 0.009$ & $0.101 \pm 0.009 \pm 0.009$ &
$0.093 \pm 0.009 \pm 0.009$ & \cite{o-133} \\
\hline
Average & $0.111 \pm 0.003 \pm 0.007$ & $0.105 \pm 0.004 \pm 0.006$ &
$0.102 \pm 0.004 \pm 0.006$ & \\
\hline
\end{tabular}
\end{table*}
The most recent data from LEP
around $E_{cm}$~=~133~GeV, 161~GeV and 172~GeV - although with rather
limited statistics of only a few hundred hadronic events for each
experiment and at each energy point - were analysed in terms of
hadronic event shape distributions, jet production rates, charged
particle multiplicities, momentum distributions of charged particles
and other observables.
These analyses revealed that the new data are well described by
standard QCD and hadronisation models, which were tuned to the
high statistics data sets at the $\rm Z^0$ pole.
They also provided first measurements of $\alpha_{\rm s}$ at these new energies which
are summarised in Table~1.
It should be noted that several of these measurements are not yet officially
published; those results as well as the combined values of $\alpha_{\rm s}$ shown
in Table~1 are therefore still preliminary.
\section{UPDATE OF THE WORLD SUMMARY OF $\alpha_{\rm s}$}
Significant determinations of $\alpha_{\rm s}$, based on perturbative QCD
predictions which were, at least, complete to next-to-leading
order (NLO), date back to 1979/1980 (from structure functions in deep
inelastic scattering; see e.g. \cite{yndurain}) and to 1982 (from
$\rm e^+\rm e^-$ annihilation \cite{jade-as2}).
Many of the early data and determinations of $\alpha_{\rm s}$, see
e.g. \cite{yndurain-brighton}, have been superseded and/or replaced by
more actual measurements and analyses, c.f.
\cite{altarelli-ac,virchaux,dallas,fernandez}.
\begin{table*}[htb]
Table 2. \ \
World summary of measurements of $\alpha_{\rm s}$.
Entries preceded by $\bullet$
are new or updated since summer 1996~\cite{qcd96}.
Abbreviations:
DIS = deep inelastic scattering; GLS-SR = Gross-Llewellyn-Smith sum rules;
Bj-SR = Bjorken sum rules;
(N)NLO = (next-)next-to-leading order perturbation theory;
LGT = lattice gauge theory;
resum. = resummed next-to-leading order.
\begin{center}
\begin{tabular}{|r l|c|l|l|c c|c|}
\hline
& & Q & & & \multicolumn{2}{c|}
{$\Delta \as(M_{\rm Z^0}) $} & \\
& Process & [GeV] & $\alpha_s(Q)$ &
$ \as(M_{\rm Z^0})$ & exp. & theor. & Theory \\
\hline \hline \normalsize
$\bullet$ & DIS [pol. strct. fctn.] & 0.7 - 8 & & $0.120\ ^{+\ 0.010}
_{-\ 0.008}$ & $^{+0.004}_{-0.005}$ & $^{+0.009}_{-0.006}$ & NLO \\
& DIS [Bj-SR] & 1.58
& $0.375\ ^{+\ 0.062}_{-\ 0.081}$ & $0.122\ ^{+\ 0.005}_{-\ 0.009}$ &
-- & -- & NNLO \\
& DIS [GLS-SR] & 1.73
& $0.32\pm 0.05$ & $0.115\pm 0.006$ & $ 0.005 $ & $ 0.003$ & NNLO \\
& $\tau$-decays
& 1.78 & $0.330 \pm 0.030$ & $0.119 \pm 0.004$
& 0.001 & 0.004 & NNLO \\
$\bullet$ & DIS [$\nu$; ${\rm F_2\ and\ F_3}$] & 5.0
& $0.215 \pm 0.016$
& $0.119\pm 0.005$ &
$ 0.002 $ & $ 0.004$ & NLO \\
& DIS [$\mu$; ${\rm F_2}$]
& 7.1 & $0.180 \pm 0.014$ & $0.113 \pm 0.005$ & $ 0.003$ &
$ 0.004$ & NLO \\
& DIS [HERA; ${\rm F_2}$]
& 2 - 10 & & $0.120 \pm 0.010$ & $ 0.005$ &
$ 0.009$ & NLO \\
& DIS [HERA; jets]
& 10 - 60 & & $0.120 \pm 0.009$ & $ 0.005$ &
$ 0.007$ & NLO \\
$\bullet$ & DIS [HERA; ev.shps.]
& 7 - 100 & & $0.118\ ^{+\ 0.007}_{-\ 0.006}$ & $ 0.001$ &
$^{+0.007}_{-0.006}$ & NLO \\
& ${\rm Q\overline{Q}}$ states
& 4.1 & $0.223 \pm 0.009$ & $0.117 \pm 0.003 $ & 0.000 & 0.003
& LGT \\
& $J/\Psi + \Upsilon$ decays
& 10.0 & $0.167\ ^{+\ 0.015\ }_{-\ 0.011\ }$ & $0.113\ ^{+\ 0.007\ }
_{-\ 0.005\ }$ & 0.001 & $^{+\ 0.007}_{-\ 0.005}$ & NLO \\
$\bullet$ & $\rm e^+\rm e^-$ [ev. shapes] & 22.0 & $0.161\ ^{+\ 0.016}_{-\ 0.011}$ &
$0.124\ ^{+\ 0.009}_{-\ 0.006}$ & $\pm 0.005$ & $^{+0.008}_{-0.003}$
& resum \\
& $\rm e^+\rm e^-$ [$\sigma_{\rm had}$] & 34.0 &
$0.146\ ^{+\ 0.031}_{-\ 0.026}$ &
$0.124\ ^{+\ 0.021}_{-\ 0.019}$ & $^{+\ 0.021}_{-\ 0.019}
$ & -- & NLO \\
$\bullet$ & $\rm e^+\rm e^-$ [ev. shapes] & 35.0 & $ 0.143\ ^{+\ 0.011}_{-\ 0.007}$ &
$0.122\ ^{+\ 0.008}_{-\ 0.006}$ & $\pm 0.002$ & $^{+0.008}_{-0.005}$
& resum \\
$\bullet$ & $\rm e^+\rm e^-$ [ev. shapes] & 44.0 & $ 0.137\ ^{+\ 0.010}_{-\ 0.007}$ &
$0.122\ ^{+\ 0.008}_{-\ 0.006}$ & $\pm 0.003$ & $^{+0.007}_{-0.005}$
& resum \\
& $\rm e^+\rm e^-$ [ev. shapes] & 58.0 & $0.132\pm 0.008$ &
$0.123 \pm 0.007$ & 0.003 & 0.007 & resum \\
& $\rm p\bar{\rm p} \rightarrow {\rm b\bar{b}X}$
& 20.0 & $0.145\ ^{+\ 0.018\ }_{-\ 0.019\ }$ & $0.113 \pm 0.011$
& $^{+\ 0.007}_{-\ 0.006}$ & $^{+\ 0.008}_{-\ 0.009}$ & NLO \\
& ${\rm p\bar{p},\ pp \rightarrow \gamma X}$ & 24.2 & $0.137
\ ^{+\ 0.017}_{-\ 0.014}$ &
$0.112\ ^{+\ 0.012\ }_{-\ 0.008\ }$ & 0.006 &
$^{+\ 0.010}_{-\ 0.005}$ & NLO \\
& ${\sigma (\rm p\bar{p} \rightarrow\ jets)}$ & 30 - 500 & &
$0.121\pm 0.009$ & 0.001 & 0.009 & NLO \\
$\bullet$ & $\rm e^+\rm e^-$ [$\Gamma (\rm Z^0 \rightarrow {\rm had.})$]
& 91.2 & $0.124\pm 0.005$ &
$0.124\pm 0.005$ &
$ 0.004$ & $0.003$ & NNLO \\
& $\rm e^+\rm e^-$ [ev. shapes] &
91.2 & $0.122 \pm 0.006$ & $0.122 \pm 0.006$ & $ 0.001$ & $
0.006$ & resum. \\
$\bullet$ & $\rm e^+\rm e^-$ [ev. shapes] & 133.0 & $0.111\pm 0.008$ &
$0.117 \pm 0.008$ & 0.004 & 0.007 & resum \\
$\bullet$ & $\rm e^+\rm e^-$ [ev. shapes] & 161.0 & $0.105\pm 0.007$ &
$0.114 \pm 0.008$ & 0.004 & 0.007 & resum \\
$\bullet$ & $\rm e^+\rm e^-$ [ev. shapes] & 172.0 & $0.102\pm 0.007$ &
$0.111 \pm 0.008$ & 0.004 & 0.007 & resum \\
\hline
\end{tabular}
\end{center}
\end{table*}
An update of a 1996 review of $\alpha_{\rm s}$ \cite{qcd96} is given in Table~2.
Since last year, new results from polarized structure
functions~\cite{altarelli-ball},
a correction of the results from $\nu$-nucleon deep
inelastic scattering \cite{ccfr}, new results from lattice QCD
\cite{davies}
and from a study of hadronic
event shapes at HERA \cite{h1-shapes}, a reanalysis of event shapes
from PETRA data \cite{fernandez}, an update of $\alpha_{\rm s}$ from the
hadronic width of the $\rm Z^0$ \cite{lep-width} and new results from LEP
data above the $\rm Z^0$ pole (see Section~4) became available.
The data are displayed in Figures~\ref{asq97} and~\ref{asmz97}.
Perhaps the most relevant change of previously reported results is the
one from $\nu$-nucleon scattering \cite{ccfr}, which increased - due
to a new energy calibration of the detector - from $\as(M_{\rm Z^0}) = 0.111 \pm
0.006$ to $0.119 \pm 0.005$.
This partly resolves the outstanding problem that
deep inelastic scattering results preferred lower values of $\as(M_{\rm Z^0})$ then did
measurements in $\rm e^+\rm e^-$ annihilation.
\begin{figure}[htb]
\epsfxsize7.5cm
\epsffile{asq97.eps}
\vskip-1.0cm
\caption{\label{asq97}
World summary of \protect $\alpha_{\rm s} (Q)$ (status of July 1997).}
\end{figure}
\begin{figure}[htb]
\vskip-18.0mm
\epsfxsize8.5cm
\epsffile{asfreq.eps}
\vskip-15.0mm
\caption{\label{asfreq}
Frequency distribution of central values of $\as(M_{\rm Z^0})$
(histogram; data from Table~2) and a gaussian with the same mean and width
(curve).}
\end{figure}
\begin{figure}[htb]
\vskip-18mm
\epsfxsize8.5cm
\epsffile{asint.eps}
\vskip-15.0mm
\caption{\label{asint}
Sum of weights of $\as(M_{\rm Z^0})$
(histogram; calculated from data of Table~2) and
gaussian fit (curve).}
\end{figure}
The distribution of the central values of $\as(M_{\rm Z^0})$ is
displayed in Figure~\ref{asfreq}, whereby the result from lattice QCD
is not included because the relevance of its given
uncertainty is not entirely clear.
The unweighted mean of this distribution is 0.1186, the root mean
squared (r.m.s.) is 0.0042.
This distribution, however, does not have a gaussian shape, and
so the r.m.s. is not regarded as a good measure of the error of the
average value of $\as(M_{\rm Z^0})$.
Owing to the fact that the dominating uncertainties of most $\alpha_{\rm s}$
measurements are theoretical rather than statistical or experimental,
which may then be correlated to an unknown degree,
two alternatives to estimate the uncertainty of the
average $\as(M_{\rm Z^0})$ are applied:
First, it is demanded that at least 90\% of the central values of
all measurements must be inside the error interval.
This gives
$\Delta \as(M_{\rm Z^0}) = \pm 0.006$, which can be regarded as a `90\%
confidence interval' (68\% confidence would correspond to $\pm 0.005$).
Second,
assuming that each result of $\as(M_{\rm Z^0})$ has a boxed-shaped rather than a
gaussian probability
distribution over the interval of its quoted error, and summing
all resulting weights\footnote{The weight of a measurement is defined
to be the inverse of the square of its total
error.} for each numerical value of $\as(M_{\rm Z^0})$, leads to the
distribution shown in Figure~\ref{asint}.
The (weighted) mean of this distribution is almost identical to the
unweighted average (see Fig.~\ref{asfreq}), $\as(M_{\rm Z^0}) = 0.1188$.
This distribution now has a gaussian shape; the
r.m.s. is 0.0061 and the width of a gaussian fit to that distribution
is 0.0057.
Therefore the world average result is quoted to be
$$ \as(M_{\rm Z^0}) = 0.119 \pm 0.006,$$
where the error of 0.006 is perhaps a conservative but well defined
estimate of the overall uncertainties.
This average corresponds to the full line in Figure~\ref{asq97} and to
the dashed line plus the shaded band in Figure~\ref{asmz97}.
The agreement between the different measurements is excellent.
Due
to the many new and recent measurements, the running of $\alpha_{\rm s}$ as
predicted by QCD is significantly proven by the data.
\begin{figure}[htb]
\epsfxsize7.5cm
\epsffile{asmz97.eps}
\vskip-0.9cm
\caption{\label{asmz97}
World summary of $\as(M_{\rm Z^0})$.}
\end{figure}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 3,055
|
var chai = require('chai');
var chaiHttp = require('chai-http');
var express = require('express');
var kraken = require('kraken-js');
chai.use(chaiHttp);
var should = chai.should();
describe('Routes', function(){
var app;
before(function (done) {
app = express();
app.on('start', done);
app.use(kraken({
basedir: process.cwd()
}));
});
describe('/paypal/currencyConversion', function(){
it('should return json with conversion from one currency to another.',
function(itDone){
chai.request(app).get('/paypal/currencyConversion?amount=42&from=CNY&to=EUR')
.end(function(error, response){
should.not.exist(error);
response.should.be.an("object");
response.body.should.be.an("object");
response.should.have.status(200);
response.body.should.not.have.property('error');
response.body.should.have.property('from');
response.body.should.have.property('to');
response.body.should.have.property('amount');
response.body.should.have.property('converted');
response.body.should.have.property('symbols');
itDone();
});
}
);
it('should return json with an error for missing amount',
function(itDone){
chai.request(app).get('/paypal/currencyConversion?from=USD&to=CAD')
.end(function(error, response){
response.should.have.status(400);
response.body.should.have.property('error');
itDone();
});
}
);
it('should return json with an error for missing from',
function(itDone){
chai.request(app).get('/paypal/currencyConversion?amount=42&to=CAD')
.end(function(error, response){
response.should.have.status(400);
response.body.should.have.property('error');
itDone();
});
}
);
it('should return json with an error for missing to',
function(itDone){
chai.request(app).get('/paypal/currencyConversion?from=USD&amount=42')
.end(function(error, response){
response.should.have.status(400);
response.body.should.have.property('error');
itDone();
});
}
);
});
describe('/paypal/currencyRate', function(){
it('should return json with the rate to convert from one currency to another',
function(itDone){
chai.request(app).get('/paypal/currencyRate?from=INR&to=USD')
.end(function(error, response){
should.not.exist(error);
response.should.be.an("object");
response.body.should.be.an("object");
response.should.have.status(200);
response.body.should.not.have.property('error');
response.body.should.have.property('from');
response.body.should.have.property('to');
response.body.should.have.property('symbols');
response.body.should.have.property('rate');
itDone();
});
}
);
it('should return json with an error for missing from',
function(itDone){
chai.request(app).get('/paypal/currencyRate?to=USD')
.end(function(error, response){
should.not.exist(error);
response.body.should.have.property('error');
itDone();
});
}
);
it('should return json with an error for missing to',
function(itDone){
chai.request(app).get('/paypal/currencyRate?from=INR')
.end(function(error, response){
should.not.exist(error);
response.body.should.have.property('error');
itDone();
});
}
);
});
});
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,360
|
This PC shows Henry street where the waxworks were located at No. 30. Curiously, we see at No. 37 the 'Original Pennybazaar' of E. Marks. It may be a branch opened later as Thom's 1904 lists Marks Pennybazaar only at 42 George's street.
Merchant's quay is directly across the Four Courts and King's Inn quay. The tram on this SV advertises Denny's Limerick Bacon & Hams.
"Which example did he adduce to induce Stephen to deduce that originality, though producing its own reward, does not invariably conduce to success?
A mythical Queen's Hotel, not the one in Ennis. Michael Watson, a visitor to the site, identified it on Victoria Street, Windermere, Cumbria, England. Thanks Michael!
"Which domestic problem as much as, if not more than, any other frequently engaged his mind?
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,319
|
This according to Pascale Armand, currently starring in Lauren Gunderson's new one-person play, Natural Shocks, opening November 8th at Off-Broadway's WP Theatre. The challenges of performing in this solo piece, with issues ranging from sexual identity to domestic abuse, calls for a broad range of emotional experience an actor needs to delve into, in order to get to the truth of one woman's story.
As she prepares to open in the play, I engaged with Pascale Armand to ask about her process and to expand on what working on Natural Shocks has been like for her.
Ron Fassler: I'm very interested in what your impressions were when you sat down and read the script for the first time. What were the pros and cons in terms of choosing to take on this role?
Pascale Armand: After getting through it the first time, my eyebrows were raised and I just sat in front of my computer motionless for a few minutes. It's a direct reflection of society today and how so many different aspects of the ills we live with may seem separate, but come together to have massive consequences. The message is so relevant in today's world, so the pro for doing the show was a no-brainer. The big con for me was the short amount of rehearsal time. I want to do the story justice and embrace the challenge, but not at the expense of telling the story blandly, without making impact. On its face, that would be hard to do, but I wanted to stick my foot in this. It's too important.
RF: In what ways do you personally identify with this character? In what ways do you not?
PA: I feel that I personally identify with Angela's sense of humor the most. The way she uses humor to camouflage her pain is something I find myself doing when I'm sad or when loved ones are hurting around me. I try to smile or make them smile. That's that universal language everyone understands. A big difference between me and Angela is how far I will take that masking. You'll know when I'm upset about something and it's hard for me to hold on to making everything seem ok for too long when it's not.
RF: How are current world events (particularly here in the U.S. over the past week) affecting your performance and interpretation of the part?
PA: I would have to live under a rock to not see how art is imitating life, especially this week. I actually feel that it's too close, but these are the times that we live in today and we have to deal with that. My footing is becoming more grounded in the role and that's got me delving deeper. I hope for more of the audience's discomfort towards the end of every show. Maybe that will cause people to be more proactive after exiting the theater's doors.
RF: In what ways do tornados serve as a metaphor in your own life? How do you handle anxiety and fear in a world that feels like its veering off course these days?
PA: Right, they're too small and too quick. These days we're watching it -the news- all happen before our very eyes – through the feeds on our phones, the crawlers at the bottoms of our screens. No matter where you are, or how small the snippet of information is, we're still absorbing it and it's hard to pull yourself out of it, it's so…everywhere. It's hard to handle it all. I've heard people say they tune it out or take breaks from watching the news or their social media and I get that, but because things are moving so fast, I feel the worst I can do for myself is to NOT know. Being uninformed can have dire consequences as well. So I take in the info, breathe, then reach out to those I know or think might may be affected by the event. And sometimes that may be by just seeing that someone I was worried about checked themselves safe on social media, but I just want to know how my loved ones are. We have no excuse not to these days and if anything, I would rather use technology to connect than to isolate myself into the mire of the news.
RF: I was very taken with the story about the little boy in front of the TV's at Best Buy. What does that story personally mean to you, either as a child or as an adult?
PA: Kids are very intuitive and are always watching. Their development, even from the minutia that they observe around them, is a learning process of how to be in the world. We need to be more careful around them. With them is where it all begins.
RF: Besides memorization, what other challenges does a one-person show bring to the forefront of your work as an actress?
PA: Being the only focal point for the entirety of the show, keeping the ball in the air for a sustained period of time, always being front-footed throughout the whole performance. One must be crystal clear about EVERYTHING. There is no slacking off! And…trying to stay hydrated but making sure your bladder is COMPLETELY empty before the show starts. That's a juggling act right there!
RF: Can you discuss your relationship with your director, May Adrales, and how the process of rehearsing and early performances of the play unfolded?
PA: May has been really great from the beginning. She (and WP Theater) set up an environment where I felt completely supported and comfortable enough to tell her exactly what I needed. She has answered my hard-hitting, probing questions with such honesty, articulation and grace. She didn't have to do that, so I really appreciate it. We laughed a lot, too, so striking that balance will always get the best work out of me. Her patience with me and just the sheer repetition of getting all these words in my brain was…SAINTLY. She also kept the healthy snacks, brain food and the chocolate coming in rehearsal, so I like her. A LOT.
RF: Who do you think needs to see this play? Do you think it speaks more profoundly to women or to men? Ultimately, what can you say about the power of the play and its message in today's society?
PA: The power of this play is that it is not a cautionary tale. It is a direct replica, a mirror, of what we learn so quickly about these days in a media frenzy that we all say can never happen again … until it does. Over and over. We have got to do better because we're all in this together. In the end, we can point fingers all day all we want, but if we don't DO something (uh… VOTE!), things will stay the same. Watching this play will let you know that that's not a great place to be. Mindsets have to change all around. So yeah, men should see this play and know that they have to treat women with the Golden Rule. Women should see this play, know we deserve more and flood the space we are making for our voices to be heard. Lawmakers should see this play to remember who they work for. Staunch believers in the 2nd Amendment should see this play. Natural Shocks is not just entertainment.
Natural Shocks is at the WP Theatre, 2162 Broadway, New York, 10024, opening November 8th.
|
{
"redpajama_set_name": "RedPajamaC4"
}
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Q: Is it possible to change which work item is associated with a check-in? In MS Team Foundation Server 2010 is it possible to update the work items that a check-in is associated with? I was working late yesterday and checked in against the wrong items and would like to re-associate with the right work items.
A: Work items are associated by linking them to the changesets. All you have to do is remove the link to the wrong work items and add the links to the new ones.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,677
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Q: There is a code error here, how to solve it。 There is a code error here, how to solve it。
const api = { a: 1, b: { c: 2, d: 3, e: { f: 4, g: 5 } } };
interface apiProps {
[name: string]: number | apiProps;
}
function format<T extends apiProps, S extends keyof T, P extends string>(obj: T, str: P): Record<`${S}${P}`, number> {
const result = {} as Record<`${S}${P}`, number>;
for (const key in obj) {
const item = obj[key];
if (typeof item === "number") {
result[`${str}${key}`] = item;
} else {
const children = format(item as apiProps, (str + key) as P);
Object.assign(result, children);
}
}
return result;
}
const newApi = format(api, "");
console.log(newApi);
I want to be able to use it with hints
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
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Minaminihon Broadcasting Co., Ltd. (), also known as MBC, is a Japanese radio and television station. It founded in 1953 and headquartered in Kagoshima, Japan. Minaminihon Broadcasting commences radio broadcasting in 1953. In 1959, Minaminihon Broadcasting started television broadcasting.
Minaminihon Broadcasting is affiliated with the JNN (TV), JRN, NRN (RADIO). It is the only commercial broadcasting that provides both TV and radio services in Kagoshima prefecture. In 2006, MBC started digital terrestrial television broadcasting.
References
Japan News Network
Television stations in Japan
Radio in Japan
Radio stations established in 1953
Television channels and stations established in 1959
1953 establishments in Japan
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
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{"url":"https:\/\/wavelab.at\/sources\/PLUSVein-Contactless\/","text":"home \u00a0 | \u00a0research \u00a0 | \u00a0members \u00a0 | \u00a0projects \u00a0 | \u00a0publications \u00a0 | \u00a0conferences of interest \u00a0 | \u00a0downloads \u00a0 | \u00a0contact & impressum \u00a0 | \u00a0privacy information\n\nThis is \"The Multimedia Signal Processing and Security Lab\", short WaveLab, website. We are a research group at the Artificial Intelligence and Human Interfaces (AIHI) Department of the University of Salzburg led by Andreas Uhl. Our research is focused on Visual Data Processing and associated security questions. Most of our work is currently concentrated on Biometrics, Media Forensics and Media Security, Medical Image and Video Analysis, and application oriented fundamental research in digital humanities, individualised aquaculture and sustainable wood industry.\nPLUSVein-Contactless Finger and Hand Vein Data Set\n\n## Combined Fully Contact-Less Finger and Hand Vein Scanner\n\nThe PLUSVein-Contactless Finger and Hand Vein Database was acquired with our two custom designed combined fully contact-less finger and hand vein acquisition device. This scanner is designed to capture finger as well as hand vein images from the palmar side and can capture light transmission as well as reflected light images. It is based on an NIR enhanced industrial camera equipped with a 9 mm lens in combination with an NIR pass-through filter. Its light transmission light source consists of 1 stripe of 5 NIR laser modules, situated in the top part of the device. Each laser module is brightness controlled individually and automatically based on a preset brightness value to achieve an optimal image contrast. The reflect light illuminator consists of two rows of 8 LEDs each, one row of 8 850 nm LEDs, and one row of 8 950 nm LEDs which is situated on the bottom of the device next to the camera. This illuminator is automatically brightness controlled as well. To assist in positioning of the finger\/hand, the top front part contains an LCD touchscreen display showing the user a live image of the finger\/hand. Further information on the scanners and its design, including all technical details are available on request and can be found at: Combined Fully Contact-Less Finger and Hand Vein Acquisition Device\n\n## PLUSVein-Contactless Finger and Hand Vein Database\n\nThe data set itself consists of 3 subsets: a palmar finger vein one, acquired using the light transmission illuminator and two hand vein ones, one acquired with the 850 nm reflected light illuminator and the other one with the 950 nm reflected light illuminator. Currently it contains$42$ subjects. $6$ fingers (left and right index, middle and ring finger) and two hands (left and right hand) per subject and $5$ images per finger\/hand in $1$ session were captured for each subsets. Hence, the finger vein subset consists of $252$ individual fingers. Each finger is captured 5 times, hence there are effectively $1260$ images in the finger vein subset. The hand vein subset consists of $84$ individual hands. Each hand is captured 5 times, hence there are effectively $420$ images in each of the two hand vein subsets. The dataset consists of a total of $2100$ images. The raw images have a resolution of $1280\u00d71024$ pixels and are stored in 8 bit greyscale png format. The visible area of the finger or hand inside the images is about $600\u00d7180$ and $850\u00d7850$ pixels per finger\/hand, respectively. The ROI images are extracted with the help of edge detection mechanisms and by masking out the background in the images (setting the pixels to black). The finger and hand vein ROI images have a size of $TODO\u00d7TODO$ pixels and $384\u00d7384$ pixels, respectively.\n\nThe database is publicly available for research purposes and the raw finger vein images as well as the ROI images can be obtained here.\n\nThe database is planned to be extended set by a second session as well as by adding further subjects in the future.\n\n## Filename and Directory Structure\n\nFor the raw finger and hand vein images as well as for the extracted ROI images the same directory\u00a0structure applies:\n\u2022 FingerVein_TI: contains the finger vein images acquired using the laser module based light transmission illumination\n\u2022 01: denotes session 1\n\u2022 001 ... 042: a subdirectory for each user contained in the data set\n\u2022 HandVein_850: contains the hand vein images acquired using the LED based reflected light illumination with 850 nm peak wavelength\n\u2022 01: denotes session 1\n\u2022 001 ... 042: a subdirectory for each user contained in the data set\n\u2022 HandVein_950: contains the hand vein images acquired using the LED based reflected light illumination with 950 nm peak wavelength\n\u2022 01: denotes session 1\n\u2022 001 ... 042: a subdirectory for each user contained in the data set\nThe filename are encoded using the following structure:\n[session ID]_[user ID]_[finger ID\/hand ID]-[DORSAL\/PALMAR]_[image ID]_[illumination type].png\n\u2022 session ID: the session ID, two digits, where 01 denotes the first session\n\u2022 user ID: the user ID, three digits, where 001 denotes the first user and 042 denotes the 42nd user\n\u2022 finger ID: the finger ID, two digits, starting from the left thumb (01) till the right pinky finger (10):\n\u2022 02: left index finger\n\u2022 03: left middle finger\n\u2022 04: left ring finger\n\u2022 07: right index finger\n\u2022 08: right middle finger\n\u2022 09: right ring finger\n\u2022 The left thumb (01), left pinky finger (05), right thumb (06) and right pinky finger (10) have not been captured.\n\u2022 hand ID: the hand ID, one character, where L denotes the left hand and R denotes the right hand.\n\u2022 DORSAL\/PALMAR: denotes if the image is captured from the palmar or the dorsal side\n\u2022 image ID: the image ID, two digits, starting from 01.\n\u2022 illumination ID: the illumination type, where TI indicates the laser module based light transmission for finger vein images, RL850 denotes 850 nm reflected light for hand vein images and RL950 denotes 950 nm reflected light for hand vein images as well.\nAn example finger vein image filename is: 01_002_02-PALMAR_03_TI.png\nAn example hand vein image filename is: 01_001_R-PALMAR_01_RL950.png\n\n## Obtaining the Database\n\nTo obtain the PLUSVein-Contactless Finger and Hand Vein Database you have to agree to our license agreement:\nPLUSVein-Contactless Consent Form\n\nPlease download, fill in and sign the license agreement and send it to A. Uhl or via mail to our department. 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| null | null |
\section*{Introduction}
\label{sec:intro}
A \emph{prehomogeneous vector space} is a rational representation
$\rho:G\to{\mathrm{GL}}(V)$ of a connected complex linear algebraic group $G$
that has a (unique) Zariski open orbit $\Omega$ in $V$.
These representations have been much studied from the viewpoint of
number theory and harmonic analysis (e.g., \cite{kimura}).
A particularly useful tool is to consider
the rational
\emph{characters} $\chi:H\to {\mathbb{G}_\mathrm{m}}:={\mathrm{GL}}({\mathbb C})$
of the abelian quotient
$H:=G/[G,G]\cdot G_{v_0}$, where $G_{v_0}$ is the isotropy subgroup at
any $v_0\in\Omega$.
There is a correspondence between these characters and
`relative invariants';
a \emph{relative invariant} $h$ is a rational function on $V$ such
that on $\Omega$, $h$ is holomorphic and $h\circ\rho$ is merely $h$
multiplied by a character of $G$ (see \eqref{eqn:relinv}).
The relative invariants are constructed from the
irreducible polynomials defining the
hypersurface components of the algebraic set $V\setminus \Omega$.
Thus, for example,
the number of hypersurface components of $V\setminus
\Omega$ equals the rank of the free abelian group of characters of $H$.
In this paper we study the \emph{additive functions} of $H$,
the rational homomorphisms $\Phi:H\to {\mathbb{G}_\mathrm{a}}:=({\mathbb C},+)$.
This complements the classical study of the characters, as
up to isomorphism
${\mathbb{G}_\mathrm{m}}$ and ${\mathbb{G}_\mathrm{a}}$ are the only
$1$-dimensional
connected complex linear algebraic groups.
Since
$H$ is connected and abelian,
$H\cong{\mathbb{G}_\mathrm{m}}^k\times {\mathbb{G}_\mathrm{a}}^\ell$ with
$k$ the rank of the character group of $H$ and
$\ell$ the dimension of the vector space of
additive functions of $H$;
the characters and additive functions thus describe $H$ completely.
Of particular interest for us,
the number of irreducible hypersurface components of
$V\setminus \Omega$ equals
\begin{equation}
\label{eqn:kdimhell}
k=\dim(H)-\ell
\end{equation}
(see Corollary \ref{cor:numcomponents}),
where $\dim(H)$ is easily computable.
The content of the paper is as follows.
After reviewing the basic properties of prehomogeneous vector spaces
in \S\ref{sec:intropvs}
and abelian linear algebraic groups in
\S\ref{sec:abelian}, we study the additive functions of $H$ in
\S\ref{sec:addrelinv}.
Just as the
characters of $H$ are related to relative invariants,
we show that
the additive functions of $H$ are related to `additive relative
invariants'
(see Definition \ref{defn:addrelinv}).
By Proposition \ref{prop:whichrationalfunctions},
a rational function $k$ on $V$
is an \emph{additive relative invariant} if and only if
$k(\rho(g)(v))-k(v)$ is independent of $v\in \Omega$ for all $g\in G$,
and
then this difference gives the additive function $G\to{\mathbb{G}_\mathrm{a}}$ associated to
$k$.
These additive relative invariants are homogeneous of degree
$0$ and of the form $\frac{h}{f}$ for a polynomial relative invariant $f$
(Proposition \ref{prop:homogeneous}).
Geometrically, an additive relative invariant $\frac{h}{f}$ describes a
$G$--invariant subset $f=h=0$ of $f=0$, with $G$ permuting
the sets
$$V_{\epsilon}=\{x: f(x)=h(x)-\epsilon=0\},\qquad \epsilon\in{\mathbb C},$$
by $\rho(g)(V_{\epsilon})=V_{\chi(g)\cdot \epsilon}$ for
$\chi$ the character corresponding to the relative invariant
$f$
(see Proposition \ref{prop:funsagreeon});
the converse also holds if $\deg(\frac{h}{f})=0$ and $f$ is reduced.
In \S\ref{subsec:vanishing} we investigate which
characters and additive functions
vanish on which isotropy subgroups,
a key ingredient for \S\ref{sec:lfds}.
In \S\ref{subsec:algind} we establish the algebraic independence of a
generating set of relative invariants and the numerators of a basis of
the additive relative invariants.
Finally, in $\S\ref{subsec:decomp}$ we conjecture
that
every additive relative invariant may be written as a sum of
additive relative invariants of
the form $\frac{h_i}{f_i}$, where $f_i$ is a power of an irreducible
polynomial relative invariant.
We prove this splitting behavior
when the additive relative invariant may be written as a sum of
appropriate fractions.
In \S\ref{sec:exs} we describe some examples of prehomogeneous
vector spaces and their additive relative invariants.
In \S\ref{sec:lfds}, we study prehomogeneous vector spaces
with the property that
$D:=V\setminus \Omega$ is a type of hypersurface called a
\emph{linear free divisor}.
Such objects have been of much interest recently
(e.g., \cite{mondbuchweitz,gmns,freedivisorsinpvs,kpi1,DP-matrixsingI}),
and were our original motivation.
Using a criterion due to Brion and
the results of \S\ref{subsec:vanishing},
we show in Theorem \ref{thm:homomorphism}
that such prehomogeneous vector spaces have
no nontrivial additive relative invariants or nontrivial additive
functions.
Then
by \eqref{eqn:kdimhell}
the number of irreducible components of $D$
equals $\dim(H)$, and this may be easily computed
using only the Lie algebra ${\mathfrak{g}}$ of $G$
(Theorem \ref{thm:ofhomomorphism}; see also Remark
\ref{rem:worksmoregenerally}).
We also gain insight into the structure of $G$ and
the behavior of the $G$--action
(\S\S\ref{subsec:mainthm}, \ref{subsec:structureg},
\ref{subsec:liealgebra}).
In \S\ref{subsec:specialcases}, we study the special cases
where $G$ is abelian, reductive, or solvable, or where $D$ is
irreducible,
simplifying
proofs of several known results.
In \S\ref{subsec:whatelse}
we observe that, although the number of components of a linear free
divisor is computable from the abstract Lie algebra structure
${\mathfrak{g}}$ of $G$, the degrees of the polynomials defining these components are
not,
and seem to
require the representation ${\mathfrak{g}}\to{\mathfrak{gl}}(V)$.
Lemma \ref{lemma:degrees} can compute these
degrees, but requires data that are themselves difficult to
compute.
Finally, in \S\ref{subsec:homotopy} we use our earlier results to
study the homotopy groups of
the complement of a linear free divisor.
\textbf{Acknowledgements:}
We would like to thank Mathias Schulze for very helpful
discussions that led to this work.
He deserves much of the credit for
Corollary \ref{cor:solvable}.
We also thank Christian Barz for finding an error in an example.
\section{Prehomogeneous vector spaces}
\label{sec:intropvs}
We begin by briefly reviewing
prehomogeneous vector spaces.
Our reference is
\cite{kimura}, although the results we describe
were developed by
M.~Sato in the 1960's (\cite{sato}).
In the whole article, we shall study only complex linear algebraic
groups.
For such a $K$,
let $K^0$ denote the connected component of $K$
containing the identity,
$L(K)$ the Lie algebra of $K$,
and $K_v$ the isotropy subgroup at $v$ of a particular $K$--action.
Let $V$ be a finite-dimensional complex vector space, and $G$ a
connected complex linear algebraic group. Let $\rho:G\to{\mathrm{GL}}(V)$ be a
rational representation of $G$, i.e., a homomorphism of linear algebraic
groups.
When $G$ has an open orbit $\Omega$ in $V$, then we call $(G,\rho,V)$
a \emph{prehomogeneous vector space}. Then $\Omega$ is
unique and Zariski open, so that the complement $\Omega^c=V\setminus \Omega$ is an
algebraic set in $V$. We call $\Omega^c$ the \emph{exceptional orbit
variety} as it is the union of the non-open orbits of $G$;
others use \emph{discriminant} or \emph{singular
set}.
One of the basic theorems of prehomogeneous vector spaces is that the
hypersurface components of $\Omega^c$ may be detected from certain
multiplicative characters of $G$.
More precisely,
for any complex linear
algebraic group $K$ let $X(K)$ denote the set of rational
(multiplicative) \emph{characters}, that is, the
homomorphisms $K\to {\mathbb{G}_\mathrm{m}}:={\mathrm{GL}}({\mathbb C})$ of linear algebraic groups.
Then a rational function $f$ on $V$ that is not identically $0$ is a
\emph{relative invariant}
if there exists a $\chi\in X(G)$ such that
\begin{equation}
\label{eqn:relinv}
f(\rho(g)(v))=\chi(g)\cdot f(v)
\end{equation}
for all $v\in \Omega$ and $g\in G$, in which case
we\footnote{Usually this is written
$f\longleftrightarrow \chi$, but we shall use similar notation for
the relationship between
additive functions and additive relative invariants.}
write
$f\stackrel{\mathrm{m}}{\longleftrightarrow} \chi$.
By \eqref{eqn:relinv}, the zeros and poles of
$f$ may occur only on $\Omega^c$.
Actually, $f$ and $\chi$ provide almost the same information
when $f\stackrel{\mathrm{m}}{\longleftrightarrow} \chi$.
Using $f$, we may
choose any $v_0\in \Omega$ and then recover $\chi$ by defining
$$\chi(g)=\frac{f(\rho(g)(v_0))}{f(v_0)}\in({\mathbb C}^*,\cdot)\cong{\mathbb{G}_\mathrm{m}},\qquad g\in G.$$
Conversely, using $\chi$ we may recover a nonzero constant multiple
$h$ of $f$ by choosing any $v_0\in\Omega$ and any nonzero value
$h(v_0)\in{\mathbb C}$,
defining $h$ on $\Omega$ by $h(\rho(g)(v_0))=\chi(g) h(v_0)$,
and then using the density of $\Omega$ to extend $h$ uniquely to
a rational function on $V$.
Let
$X_1(G)$ be the set of $\chi\in X(G)$ for which there exists an $f$
with $f\stackrel{\mathrm{m}}{\longleftrightarrow}\chi$.
Then $X_1(G)$ is an abelian group: if
$f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i$ and $n_i\in {\mathbb Z}$ for $i=1,2$, then
$(f_1)^{n_1}(f_2)^{n_2}\stackrel{\mathrm{m}}{\longleftrightarrow} (\chi_1)^{n_1}(\chi_2)^{n_2}\in
X_1(G)$.
The group $X_1(G)$ contains information
about the hypersurface
components of $\Omega^c$.
\begin{theorem}[e.g., {\cite[Theorem 2.9]{kimura}}]
\label{thm:pvs1}
Let $(G,\rho,V)$ be a prehomogeneous vector space with exceptional
orbit variety $\Omega^c$, and $S_i=\{x\in V: f_i(x)=0\}$,
$i=1,\ldots,r$ the distinct irreducible hypersurface components of
$\Omega^c$. Then the irreducible polynomials $f_1,\ldots,f_r$ are
relative invariants which are algebraically independent, and any
relative invariant is of the form $c\cdot (f_1)^{m_1}\cdots (f_r)^{m_r}$ for
nonzero
$c\in{\mathbb C}$, and $m_i\in{\mathbb Z}$. Moreover, if each $f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i$,
then $X_1(G)$ is a free abelian group of rank $r$ generated by
$\chi_1,\ldots,\chi_r$.
\end{theorem}
The (homogeneous) irreducible polynomials $f_1,\ldots,f_r$ of the Theorem are called the \emph{basic
relative invariants} of $(G,\rho,V)$.
\begin{remark}
\label{rem:omegac}
Since $v\in\Omega$ if and only if the corresponding orbit map $g\mapsto
\rho(g)(v)$ is a submersion at $e\in G$,
and this derivative depends linearly on $v\in V$,
the set $\Omega^c$ may be defined by an ideal generated by
homogeneous polynomials of degree $\dim(V)$
(see \S\ref{subsec:lfdsintro}).
\end{remark}
There exists another description of $X_1(G)$. Fix an element $v_0\in
\Omega$ with isotropy subgroup $G_{v_0}$.
Since the orbit of
$v_0$ is open, we have $\dim(G_{v_0})=\dim(G)-\dim(V)$.
Define the algebraic groups
$$
G_1=[G,G]\cdot G_{v_0}\qquad\textrm{and}\qquad
H=G/G_1.$$
The group $G_1$ does not depend on the choice of $v_0\in\Omega$ as all
such
isotropy subgroups are conjugate,
and $[G,G]\subseteq G_1$.
By \eqref{eqn:relinv}, every $\chi\in X_1(G)$ has
$G_1\subseteq \ker(\chi)$, and thus factors through the quotient to
give a corresponding
$\widetilde{\chi}\in
X(H)$.
In fact, the map $\chi\mapsto \widetilde{\chi}$ is an isomorphism:
\begin{proposition}[e.g., {\cite[Proposition 2.12]{kimura}}]
\label{prop:x1is}
Let $(G,\rho,V)$ be a prehomogeneous vector space with open orbit
$\Omega$. Then for any $v_0\in \Omega$,
$$X_1(G)\cong X(G/[G,G]\cdot G_{v_0}).$$
\end{proposition}
Consequently, the rank of the character group $X(H)$
equals
the number of irreducible hypersurface components of
$V\setminus \Omega$.
\section{Abelian complex linear algebraic groups}
\label{sec:abelian}
Let $(G,\rho,V)$ be a prehomogeneous vector space, and use the
notation of the previous section
with $H=G/[G,G]\cdot G_{v_0}$ for some $v_0\in\Omega$.
To understand the
rank of $X(H)$ and thus the
number of
irreducible hypersurface components of the exceptional orbit variety
$\Omega^c$, we must understand $H$. Since $H$ is
abelian and connected, its structure is very simple.
Recall that over ${\mathbb C}$ there are exactly two distinct
$1$-dimensional connected linear algebraic
groups, ${\mathbb{G}_\mathrm{m}}={\mathrm{GL}}({\mathbb C}^1)\cong({\mathbb C}^*,\cdot)$ and
${\mathbb{G}_\mathrm{a}}=({\mathbb C},+)$.
\begin{proposition}[e.g., {\cite[\S3.2.5]{ov}}]
\label{prop:ckcl}
An abelian connected complex linear algebraic group $K$ is isomorphic to
$({\mathbb{G}_\mathrm{m}})^k\times ({\mathbb{G}_\mathrm{a}})^\ell$ for nonnegative integers $k$ and $\ell$,
where the exponents denote a repeated direct product.
\end{proposition}
\begin{comment}
\begin{proof}
Since $K$ is an algebraic group,
by the Jordan decomposition there exist (abstract) subgroups $K_s$ and $K_u$
of $K$ so that any element of $K$ may be written as $su$ for $s\in
K_s$, $u\in K_u$ (15.4 of \cite{humphreys}
).
For an abelian group, $K_s$ and $K_u$ are closed, and the
product map $\phi: K_s\times K_u\to K$ is an isomorphism of algebraic
groups (15.5 of \cite{humphreys}
).
Since $K$ is connected, so are $K_s$ and $K_u$ (Proof of 15.5 of
\cite{humphreys}).
Since $K_s$ is commutative and consists entirely of semisimple
elements, it is diagonalizable
(see 16.1 of \cite{humphreys}
). Then $K_s$ is also a d-group
(16.1(a) of \cite{humphreys}), and
thus isomorphic to the direct product (of algebraic groups) of a torus
and a finite group
(Theorem 16.2 of \cite{humphreys}
);
since $K_s$ is connected, it must be a torus, isomorphic to
$({\mathbb{G}_\mathrm{m}})^k$
with $k=\dim(K_s)$ (ibid NODO).
Since $K_u$ is commutative and consists entirely of unipotent
elements, it is an elementary unipotent algebraic group.
In characteristic 0, such groups are vector groups, isomorphic to
$({\mathbb{G}_\mathrm{a}})^\ell$ for $\ell=\dim(K_u)$
(Theorem 3.4.7 of \cite{springer}).
\end{proof}
\end{comment}
Note that this Proposition is false for some other fields.
For any linear algebraic group $K$, let ${\mathscr A}(K)$ be the
complex vector space of rational
homomorphisms $\Phi:K\to {\mathbb{G}_\mathrm{a}}$, sometimes called \emph{additive
functions of $K$} (\cite[\S3.3]{springer}).
When $K$ is connected, $\Phi\in{\mathscr A}(K)$ is determined by $d\Phi_{(e)}$,
and hence
${\mathscr A}(K)$ is
finite-dimensional.
When $K$ has the decomposition of Proposition \ref{prop:ckcl},
the rank of $X(K)$ and the dimension of ${\mathscr A}(K)$ are related
to $k$ and $\ell$.
\begin{lemma}
\label{lemma:rankdim}
If $K=({\mathbb{G}_\mathrm{m}})^k\times({\mathbb{G}_\mathrm{a}})^\ell$, then $X(K)$ is free of rank $k$ and
$\dim_{\mathbb C}({\mathscr A}(K))=\ell$.
\end{lemma}
\begin{proof}
By the Jordan decomposition, any $\chi\in X(K)$ will have
$\{1\}^k\times ({\mathbb{G}_\mathrm{a}})^\ell\subseteq \ker(\chi)$, so $\chi$ factors through to an
element of $X(({\mathbb{G}_\mathrm{m}})^k)$.
Thus $X(K)\cong X(({\mathbb{G}_\mathrm{m}})^k)$, and similarly ${\mathscr A}(K)\cong
{\mathscr A}(({\mathbb{G}_\mathrm{a}})^\ell)$.
Finally, an easy exercise shows that ${\mathrm{rank}}(X(({\mathbb{G}_\mathrm{m}})^k))=k$, and
$\dim({\mathscr A}(({\mathbb{G}_\mathrm{a}})^\ell))=\ell$.
\begin{comment}
(see, e.g., Example 3.2.2 of \cite{springer}
).
Coordinate projections generate ${\mathbb C}[({\mathbb{G}_\mathrm{a}})^\ell]$, so any rational
function will be written in terms of these.
Appropriate projection maps will produce a set of $k$ multiplicatively
independent characters $\chi_1,\ldots,\chi_k\in X(K)$. Thus
${\mathrm{rank}}(X(K))\geq k$.
Recall that the existence of the Jordan decomposition and its
preservation by $\chi$ (see, e.g., NODO) implies that
any $\chi\in X(K)$ has $\{1\}^k\times ({\mathbb{G}_\mathrm{a}})^\ell\subseteq \ker(\chi)$
since ${\mathbb{G}_\mathrm{m}}\setminus \{1\}$ consists entirely of semisimple elements,
while ${\mathbb{G}_\mathrm{a}}\setminus \{0\}$
consists entirely of unipotent ones
(the identity element in each group is both).
Since $\chi\in X(K)$ has $\{1\}^k\times ({\mathbb{G}_\mathrm{a}})^\ell$ in
$\ker(\chi)$, it
NODO
\end{comment}
\end{proof}
Thus, for a prehomogeneous vector space we have:
\begin{corollary}
\label{cor:numcomponents}
Let $(G,\rho,V)$ be a prehomogeneous vector space with open orbit
$\Omega$. Let $v_0\in\Omega$ and $H=G/[G,G]\cdot G_{v_0}$.
Then the number of
irreducible hypersurface components of $\Omega^c$ equals
$\dim(H)-\dim({\mathscr A}(H))$.
\end{corollary}
\begin{proof}
By Theorem \ref{thm:pvs1}, Proposition \ref{prop:x1is},
Proposition \ref{prop:ckcl}, and Lemma \ref{lemma:rankdim},
the number of irreducible components equals
${\mathrm{rank}}(X_1(G))={\mathrm{rank}}(X(H))=\dim(H)-\dim({\mathscr A}(H))$.
\end{proof}
It is natural, then, to study
these additive functions and their geometric meaning.
\section{Additive relative invariants}
\label{sec:addrelinv}
In this section, we develop for additive functions the analogue of
the multiplicative theory from
\S \ref{sec:intropvs}.
We encourage the reader to consult the examples of \S\ref{sec:exs} as
needed.
Let $(G,\rho,V)$ be a
prehomogeneous vector space with open orbit $\Omega$.
Let $v_0\in\Omega$
and $H=G/[G,G]\cdot G_{v_0}$.
\subsection{Definition}
We define additive relative invariants
similarly to (multiplicative) relative invariants.
\begin{definition}
\label{defn:addrelinv}
A rational function $h$ on $V$ is an
\emph{additive relative invariant} of $(G,\rho,V)$ if there exists a $\Phi\in {\mathscr A}(G)$
so that
\begin{equation}
\label{eqn:star}
h(\rho(g)(v))-h(v)=\Phi(g)
\end{equation}
for all $v\in\Omega$ and $g\in G$. In this situation, we write
$h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$.
\end{definition}
By \eqref{eqn:star}, the poles of such an $h$ may occur
only on $\Omega^c$, and $h$ is constant on the orbits of $\ker(\Phi)$.
\subsection{Basic properties}
We now establish some basic facts about additive relative invariants.
First we investigate the uniqueness of the relationship $h\stackrel{\mathrm{a}}{\longleftrightarrow}
\Phi$.
\begin{proposition}
\label{prop:basicproperties}
Let $(G,\rho,V)$ be a prehomogeneous vector space.
\begin{enumerate}
\item
\label{enit:hdifferbyalpha}
If $h_1\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$ and $h_2\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$, then
there exists an $\alpha\in {\mathbb C}$ with $h_1=\alpha+h_2$.
\item
\label{enit:Phiareequal}
If $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi_1$ and $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi_2$, then
$\Phi_1=\Phi_2$.
\end{enumerate}
\end{proposition}
\begin{proof}
For (1), fix $v_0\in\Omega$ and let $\alpha=h_1(v_0)-h_2(v_0)$,
so $\Phi(g)+h_1(v_0)=\alpha+\Phi(g)+h_2(v_0)$ for all $g\in G$.
Applying
\eqref{eqn:star} shows that $h_1=\alpha+h_2$ on $\Omega$, and thus on
$V$.
(2) is immediate from \eqref{eqn:star}.
\end{proof}
Let ${\mathscr A}_1(G)$ be the set of $\Phi\in {\mathscr A}(G)$ for which there exists
a rational function $h$ on $V$
with $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$.
We now identify the additive functions in ${\mathscr A}_1(G)$, analogous to
Proposition \ref{prop:x1is}.
\begin{proposition}
\label{prop:a1}
As vector spaces, ${\mathscr A}_1(G)\cong {\mathscr A}(H)$, where
$H=G/[G,G]\cdot G_{v_0}$ and $v_0\in\Omega$.
\end{proposition}
\begin{proof}
Let $\Phi\in{\mathscr A}_1(G)$ with $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$.
Evaluating \eqref{eqn:star} at $v_0\in\Omega$ and $g\in G_{v_0}$ shows
that $G_{v_0}\subseteq \ker(\Phi)$. Since ${\mathbb{G}_\mathrm{a}}$ is abelian,
$[G,G]\subseteq \ker(\Phi)$. Thus any $\Phi\in{\mathscr A}_1(G)$ factors
through
the quotient $\pi:G\to H$
to a unique $\overline{\Phi}\in{\mathscr A}(H)$,
with $\Phi=\overline{\Phi}\circ \pi$.
The map $\rho:{\mathscr A}_1(G)\to {\mathscr A}(H)$ defined by $\rho(\Phi)=\overline{\Phi}$
is ${\mathbb C}$-linear.
Conversely, define the ${\mathbb C}$-linear map $\sigma:{\mathscr A}(H)\to {\mathscr A}(G)$
by $\sigma(\overline{\Phi})=\overline{\Phi}\circ \pi$.
For $\overline{\Phi}\in {\mathscr A}(H)$,
we have $G_{v_0}\subseteq \ker(\overline{\Phi}\circ \pi)$, and hence
we may define a function
$\overline{h}:\Omega\to{\mathbb C}$ by
$\overline{h}(\rho(g)(v_0))=(\overline{\Phi}\circ \pi)(g)$.
By the argument%
\footnote{This uses the fact that ${\mathbb C}$ has characteristic $0$.}
of \cite[Proposition 2.11]{kimura},
$\overline{h}$ is
a regular function on $\Omega$
that may be extended to
a rational function $h$ on $V$.
By construction, $h(\rho(g)(v))-h(v)=(\overline{\Phi}\circ \pi)(g)$
for $v=v_0$ and all $g\in G$, but this implies that this equation holds for all $v\in\Omega$
and all $g\in G$.
Thus $h\stackrel{\mathrm{a}}{\longleftrightarrow} \overline{\Phi}\circ \pi$,
and $\sigma:{\mathscr A}(H)\to {\mathscr A}_1(G)$.
Finally, check that $\rho$ and $\sigma$ are inverses
using the uniqueness of the factorization through $\pi$.
\begin{comment}
For $\overline{\Phi}\in {\mathscr A}(H)$,
$\rho(\sigma(\overline{\Phi}))=\rho(\overline{\Phi}\circ \pi)=\overline{\Phi}$.
For $\Phi\in {\mathscr A}_1(G)$,
$\sigma(\rho(\Phi))=\sigma(\overline{\Phi})=\overline{\Phi}\circ \pi$,
for a $\overline{\Phi}$ with $\overline{\Phi}\circ \pi=\Phi$.
\end{comment}
\end{proof}
Additive relative invariants are the rational functions on $V$
having the following property.
\begin{proposition}
\label{prop:whichrationalfunctions}
Let $h:V\to{\mathbb C}$ be a rational function on $V$, holomorphic on $\Omega$.
Then $h$ is an additive relative invariant if, and only if,
for all $g\in G$,
$$h(\rho(g)(v))-h(v)\textrm{ is independent of $v\in\Omega$.}$$
\end{proposition}
\begin{proof}
By \eqref{eqn:star}, additive relative invariants have this property.
Conversely, define
$\Phi:G\to{\mathbb C}$ by
$\Phi(g)=h(\rho(g)(v))-h(v)$ for any $v\in \Omega$.
As a composition of regular functions, $\Phi$ is regular on $G$.
Let $g_1,g_2\in G$, and $v\in \Omega$. Then
\begin{eqnarray*}
\Phi(g_1 g_2^{-1})
&=&h\!\left(\rho(g_1 g_2^{-1})(v)\right)-h(v) \\
&=&h\!\left(\rho(g_1)(\rho(g_2^{-1})(v))\right)-h\!\left(\rho(g_2^{-1})(v)\right)
+h\!\left(\rho(g_2^{-1})(v)\right)-h(v) \\
&=&\Big(h(\rho(g_1)(v_1))-h(v_1)\Big)
+\Big(h(v_1)-h(\rho(g_2)(v_1))\Big),
\end{eqnarray*}
where $v_1=\rho(g_2^{-1})(v)\in \Omega$.
By our hypothesis,
$\Phi(g_1 g_2^{-1})=\Phi(g_1)-\Phi(g_2)$.
Thus $\Phi:G\to {\mathbb{G}_\mathrm{a}}$ is a homomorphism of algebraic groups, and
since $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$
we have
$\Phi\in {\mathscr A}_1(G)$.
\end{proof}
\subsection{Homogeneity}
Relative invariants are always homogeneous rational functions on $V$
because $\rho$ is a linear representation.
Similarly, additive relative invariants are
homogeneous of degree $0$.
\begin{proposition}
\label{prop:homogeneous}
If $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$ and $h$ is not the zero function,
then $h$ may be written as
$h=\frac{h_1}{f_1}$, where $h_1$ and $f_1$ are homogeneous polynomials
with $\deg(h_1)=\deg(f_1)$, $h_1$ and $f_1$ have no common factors,
and $f_1$ is a relative invariant.
\end{proposition}
\begin{proof}
First note that by Remark \ref{rem:omegac}, for any
$t\in{\mathbb C}^*$, we have
$v\in\Omega$ if and
only if $t\cdot v\in\Omega$; we use this fact implicitly in the
rest of the proof since Definition \ref{defn:addrelinv} describes the
behavior of $h$ only on $\Omega$.
Let $t\in{\mathbb C}^*$, and define the rational function $h_t$ on $V$ by
$h_t(v)=h(t\cdot v)$. Since $\rho$ is a linear representation,
\begin{equation}
\label{eqn:homog1}
h_t(\rho(g)(v))=h(t\cdot \rho(g)(v))=h(\rho(g)(t\cdot v)).
\end{equation}
Applying \eqref{eqn:star} to \eqref{eqn:homog1} gives
\begin{equation}
\label{eqn:homog2}
h_t(\rho(g)(v))=\Phi(g)+h(t\cdot v)=\Phi(g)+h_t(v).
\end{equation}
Since \eqref{eqn:homog2} holds for all $g\in G$ and $v\in \Omega$,
we have $h_t\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$.
By Proposition \ref{prop:basicproperties}\eqref{enit:hdifferbyalpha},
there exists a function
$\alpha:{\mathbb C}^*\to {\mathbb C}$ such that
\begin{equation}
\label{eqn:homog3}
h(t\cdot v)=h(v)+\alpha(t)
\end{equation}
for all $t\in{\mathbb C}^*$ and $v\in \Omega$.
If $s,t\in{\mathbb C}^*$ and $v\in\Omega$, then using \eqref{eqn:homog3}
repeatedly shows
\begin{equation*}
h(v)+\alpha(st)=h(s(t\cdot v))=h(t\cdot
v)+\alpha(s)=h(v)+\alpha(t)+\alpha(s),
\end{equation*}
or
$\alpha(st)=\alpha(s)+\alpha(t)$.
By \eqref{eqn:homog3}, we have $\alpha(1)=0$,
and hence $\alpha:({\mathbb C}^*,\cdot)\to({\mathbb C},+)$ is a group homomorphism.
Fixing some $v\in\Omega$ and instead using \eqref{eqn:homog3} to define
$\alpha$ shows that
$\alpha:{\mathbb{G}_\mathrm{m}}\to{\mathbb{G}_\mathrm{a}}$ is regular, hence
a homomorphism of complex linear
algebraic groups. By the Jordan decomposition,
$\alpha=0$.
Then $h(t\cdot v)=h(v)$ for all nonzero $t\in {\mathbb C}$ and $v\in\Omega$;
by density, $h$ is homogeneous
of degree $0$.
Thus we may write $h=\frac{h_1}{f_1}$, with $h_1$ and $f_1$ homogeneous polynomials
of equal degree, and without common factors. Since $h$ may only have
poles on $\Omega^c$, $f_1$ defines a hypersurface
in $\Omega^c$ or is a nonzero constant.
By Theorem \ref{thm:pvs1}, $f_1$ is a relative
invariant.
\end{proof}
This gives an apparently nontrivial result about the structure of
prehomogeneous vector spaces for which the exceptional orbit variety
has no hypersurface components.
\begin{corollary}
\label{cor:nontrivial}
Let $(G,\rho,V)$ be a prehomogeneous vector space, let
$v_0\in \Omega$,
and let $r$ be the number of irreducible hypersurface components of
$V\setminus \Omega$.
\begin{enumerate}
\item
\label{enit:X1is1}
If $r=0$,
then
${\mathscr A}_1(G)=\{0\}$ and $G=[G,G]\cdot G_{v_0}$.
\item
\label{enit:ddimleq1}
If $d:=\dim(G/[G,G]\cdot G_{v_0})\leq 1$, then
${\mathscr A}_1(G)=\{0\}$ and $r=d$.
\end{enumerate}
\end{corollary}
In particular,
$V\setminus \Omega$ has no hypersurface components if and only if
$G=[G,G]\cdot G_{v_0}$ for $v_0\in\Omega$.
\begin{proof}
Suppose $r=0$, so $X_1(G)=\{1\}$.
Let $\Phi\in {\mathscr A}_1(G)$, and choose $h$ with
$h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$.
Assume that $\Phi$ is nonzero and hence $h$ is not constant.
By Proposition \ref{prop:homogeneous},
write $h=\frac{h_1}{f_1}$ for polynomials $f_1$ and $h_1$ with
$\deg(h_1)=\deg(f_1)>0$, and $f_1$ a relative invariant.
If $f_1\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_1$, then by our hypothesis $\chi_1=1$.
By \eqref{eqn:relinv}, $f_1$ is a nonzero constant on $\Omega$,
contradicting the fact that $\deg(f_1)>0$.
Thus $\Phi=0$ and so ${\mathscr A}_1(G)=\{0\}$.
Since there are no nontrivial characters or additive functions,
by Proposition \ref{prop:ckcl} and Lemma \ref{lemma:rankdim}
we find that
$H=G/[G,G]\cdot G_{v_0}$ consists of a single point, hence
$G=[G,G]\cdot G_{v_0}$, proving \eqref{enit:X1is1}.
Now suppose $d\leq 1$. If $d=0$, then
by Corollary \ref{cor:numcomponents} and Proposition \ref{prop:a1},
$r=0-\dim({\mathscr A}_1(G))\geq 0$, implying the statement.
If $d=1$, then by the same reasoning either $r=0$ or $r=1$, but the
former is impossible by \eqref{enit:X1is1}.
Now apply Corollary \ref{cor:numcomponents} to show ${\mathscr A}_1(G)=\{0\}$.
\end{proof}
\subsection{Global equation}
Our definition of an additive relative invariant
only involves the behavior of the function on
$\Omega$.
We may describe the
behavior on all of $V$.
\begin{proposition}
\label{prop:globaleqn}
Let $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$ with $h$ nonzero.
As in Proposition \ref{prop:homogeneous}, write
$h=\frac{h_1}{f_1}$ for polynomials $h_1$ and $f_1$,
with $f_1\stackrel{\mathrm{m}}{\longleftrightarrow} \chi$.
Then for all $g\in G$ and $v\in V$,
\begin{equation}
\label{eqn:allgv}
h_1(\rho(g)(v))=\chi(g)\cdot
\left( f_1(v)\cdot \Phi(g)+h_1(v) \right).
\end{equation}
\end{proposition}
\begin{proof}
Define the regular functions
$L,R:G\times V\to {\mathbb C}$ using the left, respectively, right sides
of \eqref{eqn:allgv}.
By \eqref{eqn:star} and \eqref{eqn:relinv}, these functions agree on $G\times \Omega$, and
thus on $G\times V$.
\end{proof}
Of course, \eqref{eqn:allgv} also holds for
$h=0
\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi=0$ if $h_1=0$ and $f_1$ is a nonzero constant.
\subsection{Geometric interpretation}
We may now provide a
geometric interpretation of additive
relative invariants.
For polynomials $h_1,h_2,f$ on $V$ with $f$ irreducible and nonzero,
we will say that \emph{$h_1$ and $h_2$ agree to order $k$ on $V(f)$}
if $h_1-h_2\in (f)^{k+1}$, or equivalently, if in coordinates
$\frac{\partial^I}{\partial x^I}(h_1-h_2)(v)=0$ for all $v\in V(f)$
and all multi-indices $I$ with $|I|\leq k$.
\begin{comment}
\begin{proof}
If $h_1-h_2=f^{k+!}\cdot g$, then $\frac{\partial^I(\cdots)}{\partial
x^I}=f\cdot \text{stuff}$, so will be $0$.
For the other direction, suppose $h_1-h_2\in (f)^r$ for some $r\leq
k$.
Then $h_1-h_2=f^r\cdot g$.
When $|I|=r$, $v\in {\mathrm{smooth}}(V(f))$ and, say, $\frac{\partial
f}{\partial x_i}(v)\neq 0$,
\begin{align*}
\frac{\partial^r(h_1-h_2)}{\partial x_i^r}(v) &=
(\text{ones where $<r$ derivs fall on $f^r$})\cdot (\cdots)
+(\text{ones where $=r$ derivs fall on $f^r$})\cdot g(v) \\
&=0+(r!\cdot (\frac{\partial f}{\partial x_i}(v))^r+0)\cdot g(v)
\end{align*}
which is $0$ by assumption, so $g(v)=0$. By Nullstellensatz,
$h_1-h_2=f^{r+1}\cdot g'$, etc.
\end{proof}
\end{comment}
\begin{proposition}
\label{prop:funsagreeon}
Let $f_1,\ldots,f_r$ be the basic relative invariants, with
$f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i$.
Let $h=\frac{h_1}{f}$ for homogeneous polynomials $h_1$ and $f$ with
$\deg(h_1)=\deg(f)$,
$h$ written in lowest terms,
and $f=f_1^{k_1}\cdots f_r^{k_r}$ with $k_i\geq 0$. Let
$\chi=\chi_1^{k_1}\cdots \chi_r^{k_r}$.
Then the following are equivalent:
\begin{enumerate}
\item
\label{enit:haddrelinv}
$h$ is an additive relative invariant.
\item
\label{enit:hagreeorder}
For all $i=1,\ldots,r$ with $k_i>0$,
and all $g\in G$,
$h_1\circ \rho(g)$ and $\chi(g)\cdot h_1$ agree to order $k_i-1$ on
$V(f_i)$.
\item
\label{enit:hdivisible}
For all $g\in G$, $h_1\circ \rho(g)-\chi(g)\cdot h_1\in{\mathbb C}[V]$ is divisible by
$f$.
\end{enumerate}
In particular, when $f$ is reduced and we let
$V_\epsilon=\{x: f(x)=h_1(x)-\epsilon=0\}$
for all $\epsilon\in {\mathbb C}$,
then
$h$ is an additive relative invariant if and only if
$\rho(g)(V_\epsilon)=V_{\chi(g)\cdot \epsilon}$ for all $g\in G$ and
$\epsilon\in{\mathbb C}$.
\end{proposition}
\begin{proof}
If \eqref{enit:haddrelinv}, then by \eqref{eqn:allgv},
for all $g\in G$ and all $i=1,\ldots,r$ we have
$$(h_1\circ \rho(g))-\chi(g)\cdot h_1\in (f_i)^{k_i},$$
implying \eqref{enit:hagreeorder}.
\eqref{enit:hagreeorder} implies \eqref{enit:hdivisible} because
$f_1,\ldots,f_r$ all define
distinct irreducible hypersurfaces.
If \eqref{enit:hdivisible}, then for all $g,v$ we have
\begin{equation}
\label{eqn:h1rhogvminus}
h_1(\rho(g)(v))-\chi(g)\cdot h_1(v)=f(v)\cdot \beta(g,v)
\end{equation}
for some $\beta\in {\mathbb C}[G\times V]$.
For each $g\in G$, both sides of
\eqref{eqn:h1rhogvminus} should be zero or homogeneous of degree
$\deg(h_1)=\deg(f)$ as functions on $V$; hence,
$\beta(g,-)\in {\mathbb C}[V]$ is zero or
homogeneous of degree $0$, i.e., constant, and so $\beta\in {\mathbb C}[G]$.
Finally, rearrange \eqref{eqn:h1rhogvminus} and apply
\eqref{eqn:relinv} and Proposition \ref{prop:whichrationalfunctions}.
For the last part of the claim,
if $h$ is an additive relative invariant then
$\rho(g)(V_\epsilon)=V_{\chi(g)\cdot \epsilon}$ by \eqref{eqn:allgv}.
Conversely,
$\rho(g)(V_\epsilon)=V_{\chi(g)\cdot \epsilon}$
for all $\epsilon\in{\mathbb C}$
implies that
$h_1\circ \rho(g)-\chi(g)\cdot h_1$
vanishes on $V(f)$, and hence is divisible by $f$, i.e.,
\eqref{enit:hdivisible}.
\end{proof}
\begin{comment}
\begin{example}
In Example \ref{ex:denompowers} with $n=3$, consider the function
$h=\frac{h_1}{f}$ with $h_1=\frac{1}{2}x_2^2$, $f=f_1^2$, and
$f_1=x_1\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_1$.
Although $h$ is not an additive relative invariant,
$h_1\circ \rho(g)-\chi_1(g)^2\cdot h_1$ is divisible by $f_1$, but
\emph{not} $f=f_1^2$.
Hence the higher-order vanishing behavior is required.
\end{example}
\end{comment}
\subsection{Vanishing lemma}
\label{subsec:vanishing}
We now prove that certain
characters and additive functions
vanish on certain isotropy subgroups.
This is a key observation used in \S\ref{sec:lfds}.
\begin{lemma}
\label{lemma:newvanishingthm}
Let $v\in V$.
\begin{enumerate}[(i)]
\item
\label{enit:gvkerchi}
Let $f_1$ be a relative invariant, $f_1\stackrel{\mathrm{m}}{\longleftrightarrow} \chi$.
If $f_1$ is defined at $v$ and $f_1(v)\neq 0$, then $G_{v}\subseteq \ker(\chi)$.
\item
\label{enit:gvkerchiorphi}
Let $\frac{h_1}{f_1}$ be an additive relative invariant
written as in Proposition \ref{prop:homogeneous},
with $h_1$ nonzero, $\frac{h_1}{f_1}\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$, and
$f_1\stackrel{\mathrm{m}}{\longleftrightarrow} \chi$.
Then the following conditions are equivalent:
\begin{itemize}
\item $h_1(v)=0$ or $G_v\subseteq \ker(\chi)$,
\item $f_1(v)=0$ or $G_v\subseteq \ker(\Phi)$.
\end{itemize}
\end{enumerate}
\end{lemma}
\begin{proof}
Let $g\in G_v$.
For \eqref{enit:gvkerchi}, by
\eqref{eqn:relinv} we have
$$
f_1(v)=f_1(\rho(g)(v))=\chi(g)f_1(v),
$$
and \eqref{enit:gvkerchi} follows.
For \eqref{enit:gvkerchiorphi}, by
Proposition \ref{prop:globaleqn} we have
$$
h_1(v)=h_1(\rho(g)(v))=\chi(g) \left(
f_1(v) \Phi(g)+h_1(v) \right),
$$
or
\begin{equation}
\label{eqn:h1vthingy}
h_1(v)\cdot \left(1-\chi(g)\right)
= \chi(g)\cdot f_1(v) \cdot \Phi(g).
\end{equation}
Now consider the circumstances
when one side of \eqref{eqn:h1vthingy}
vanishes for all $g\in G_v$; the other side vanishes,
too.
\end{proof}
Lemma \ref{lemma:newvanishingthm} is best understood when
we consider generic points on the hypersurface components of
$\Omega^c$.
\begin{lemma}
\label{lemma:vanishingthm}
Let $f_1,\ldots,f_r$ be the basic relative invariants of $(G,\rho,V)$,
with $f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i$.
Let $v_i\in V(f_i)$, with $v_i\notin V(f_j)$ for $i\neq j$.
\begin{enumerate}
\item
\label{enit:figv}
If $i\neq j$, then $G_{v_i}\subseteq \ker(\chi_j)$.
\item
\label{enit:vanishing2}
Suppose $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$, with $h$ nonzero and $h=\frac{h_1}{g_1}$ as in
Proposition \ref{prop:homogeneous}, and
$g_1=f_1^{k_1}\cdots f_r^{k_r}$ the factorization of $g_1$.
Then:
\begin{enumerate}
\item
\label{enit:gvichi}
If $k_i>0$ and $h_1(v_i)\neq 0$, then
$G^0_{v_i}\subseteq \ker(\chi_i)$.
\item
\label{enit:gviphi}
If $k_i=0$, then
$G_{v_i}\subseteq \ker(\Phi)$.
\end{enumerate}
\end{enumerate}
\end{lemma}
\begin{proof}
\eqref{enit:figv} follows from Lemma
\ref{lemma:newvanishingthm}\eqref{enit:gvkerchi}.
For \eqref{enit:gvichi},
apply Lemma \ref{lemma:newvanishingthm}\eqref{enit:gvkerchiorphi}
to show that $G_{v_i}\subseteq \ker(\chi_1^{k_1}\cdots \chi_r^{k_r})$.
Then by \eqref{enit:figv}, $G_{v_i}\subseteq \ker(\chi_i^{k_i})$,
and this implies \eqref{enit:gvichi}.
In the situation of \eqref{enit:gviphi}, $g_1(v)\neq 0$ and
by \eqref{enit:figv} we have $G_{v_i}\subseteq \ker(\chi_1^{k_1}\cdots
\chi_r^{k_r})$.
Applying Lemma \ref{lemma:newvanishingthm}\eqref{enit:gvkerchiorphi}
then shows \eqref{enit:gviphi}.
\end{proof}
\begin{comment}
\begin{proof}
If $g\in G_{v_i}$, then by \eqref{eqn:relinv},
\begin{equation}
\label{lemmaeq:chiigv}
f_j(v_i)=f_j(\rho(g)(v_i))=\chi_j(g)f_j(v_i).
\end{equation}
Since $f_j(v_i)\neq 0$ for $i\neq j$, \eqref{lemmaeq:chiigv} implies that
$\chi_j(g)=1$, proving \eqref{enit:figv}.
In the situation of \eqref{enit:vanishing2},
Proposition \ref{prop:globaleqn} shows that
for all $g\in G$, $v\in V$,
\begin{equation}
\label{eqn:vanishingcomplicated}
h_1(\rho(g)(v))=\chi_1^{k_1}(g)\cdots \chi_r^{k_r}(g)\left(
f_1^{k_1}(v)\cdots f_r^{k_r}(v)\cdot \Phi(g)+h_1(v)\right).
\end{equation}
If $k_i>0$ and $g\in G_{v_i}$, then evaluating
\eqref{eqn:vanishingcomplicated} at $(g,v_i)$
and applying \eqref{enit:figv} gives
$$
h_1(v_i)=\chi_i^{k_i}(g)\cdot h_1(v_i),
$$
from which \eqref{enit:gvichi} follows.
If $k_i=0$ and $g\in G_{v_i}$, then evaluating
\eqref{eqn:vanishingcomplicated} at $(g,v_i)$ and
applying \eqref{enit:figv} gives
$$
h_1(v_i)=
f_1^{k_1}(v_i)\cdots f_r^{k_r}(v_i)\cdot \Phi(g)+h_1(v_i).
$$
As $f_1^{k_1}(v_i)\cdots f_r^{k_r}(v_i)\neq 0$, we have
$\Phi(g)=0$.
\end{proof}
\end{comment}
\begin{remark}
Since $h$ is written in lowest terms in
Lemma \ref{lemma:vanishingthm}\eqref{enit:vanishing2},
nearly any choice of $v_i$ in $V(f_i)$ has $h_1(v_i)\neq 0$.
Thus, if $v_i$ is generic and
$f_i$ appears nontrivially in the denominator of an additive relative
invariant, then
by Lemma \ref{lemma:vanishingthm}\eqref{enit:figv},
Lemma \ref{lemma:vanishingthm}\eqref{enit:gvichi},
and Theorem \ref{thm:pvs1},
the subgroup $G^0_{v_i}$
lies in the kernel of every $\chi\in X_1(G)$.
\end{remark}
\begin{remark}
\label{rem:fisquared}
\begin{comment}
In the situation of \eqref{enit:vanishing2},
suppose $k_i>0$.
Let
$S$ be the subset of points $v\in V(f_i)$ that are not
contained in any other hypersurface component and satisfy $h_1(v)\neq 0$.
Since $h$ is written in lowest terms,
$S$ is nonempty and Zariski open in $V(f_i)$.
Then by \eqref{enit:figv}, \eqref{enit:gvichi},
and Theorem \ref{thm:pvs1},
the subgroups $G^0_{v}$, $v\in S$,
lie in the kernel of every $\chi\in X_1(G)$.
\end{comment}
In the situation of Lemma \ref{lemma:vanishingthm}\eqref{enit:gvichi},
the results of
\cite[\S5]{pike-generalizebrion}
imply that
the ideal $I$ of Remark \ref{rem:omegac}
is contained in $(f_i)^2$.
In particular, if $I\nsubseteq (f_i)^2$ then there are
no additive relative invariants with $f_i$ appearing in the denominator.
\end{remark}
\begin{remark}
\label{rem:dimsorbits}
Lemma \ref{lemma:vanishingthm} says much about the
dimensions of the orbits of
certain algebraic subgroups $H$ of $G$.
In particular, if $G^0_v\subseteq H$ then
$H^0_v=G^0_v$
and it follows that
$\dim(H\cdot v)
=\dim(G\cdot v)-{\mathrm{codim}}(H)$.
\end{remark}
\subsection{Algebraic independence}
\label{subsec:algind}
\begin{comment}
The "Jacobian criterion" or "Jacobi's criterion" is any of the following:
Let K be a field of char 0, f_1,\ldots,f_m\in K[x_1,\ldots,x_n].
(ver1) Then f_1,\ldots,f_m are algebraically independent over K iff some m x m minor of the jacobian matrix is not identically zero.
This is stated for m=n in Humphreys, Reflection Groups and Coxeter Groups, p.63, \S 3.10
Another source for this, with proof, is "Two Themes in Commutative Algebra: Algebraic Dependence and Kähler Differentials ":
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.50.9377
and its cites, including a paper by Jacobi and pp.201-202 of Matsumura's Commutative Ring Theory. The paper by Jacobi is
"De Determinantibus functionalibus", available at:
http://gdz.sub.uni-goettingen.de/dms/load/pdf/?PPN=GDZPPN002142724
More generally,
(ver2) rank of the jacobian matrix over K(x_1,\ldots,x_n) = transcendence degree of {f_1,\ldots,f_m} over K
This can be found as Theorem 6 in http://arxiv.org/pdf/1102.2789v1.pdf or \S3 of http://arxiv.org/pdf/1202.4301v1.pdf
They give a few citations, and the latter gives a proof in an appendix.
\end{comment}
By Theorem \ref{thm:pvs1},
a set of basic relative
invariants are algebraically independent polynomials;
in fact, we may add to this set the numerators of
a basis of the additive relative invariants.
\begin{proposition}
\label{prop:algind}
Let $f_1,\ldots,f_r$ be a set of basic relative invariants, with
$f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i$.
Let $\Phi_1,\ldots,\Phi_s$ be a basis of ${\mathscr A}_1(G)$,
and for $1\leq i\leq s$
let $\frac{h_i}{g_i}$ be an additive relative invariant written in
lowest terms with
$\frac{h_i}{g_i}\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi_i$.
Let $v_i\in V(f_i)$ with $v_i\notin V(f_j)$ for $i\neq j$.
Then:
\begin{enumerate}
\item
\label{enit:fiinvariance}
Each $f_i$ is invariant under the action of $[G,G]\cdot G_{v_0}$,
and $G_{v_j}$ for $j\neq i$.
\item
\label{enit:hiinvariance}
Each $h_i$ is invariant under the action of $[G,G]\cdot G_{v_0}$,
and $G_{v_j}$ for $j$ with $g_i(v_j)\neq 0$.
\item
\label{enit:algebraicindependence}
The polynomials
$f_1,\ldots,f_r,h_1,\ldots,h_s\in {\mathbb C}[V]$ are algebraically independent
over ${\mathbb C}$.
\end{enumerate}
\end{proposition}
\begin{proof}
Claim \eqref{enit:fiinvariance} follows from \eqref{eqn:relinv},
Proposition \ref{prop:x1is}, and
Lemma \ref{lemma:vanishingthm}\eqref{enit:figv}.
Claim \eqref{enit:hiinvariance} follows from
Propositions \ref{prop:globaleqn} and
\ref{prop:a1}, and
Lemma \ref{lemma:vanishingthm}\eqref{enit:gviphi}.
A criterion of Jacobi (e.g., \cite[\S3.10]{humphreys-reflection_groups})
states that a set of $m$ polynomials in ${\mathbb C}[V]$, $m\leq \dim(V)$,
is algebraically independent over ${\mathbb C}$ if and only if
some $m\times m$ minor of their Jacobian matrix is not the zero
polynomial.
We shall equivalently show that
the wedge product of their exterior derivatives is not identically
zero.
Let ${\mathfrak{g}}$ be the Lie algebra of $G$, and for $X\in{\mathfrak{g}}$ let
$\xi_X$ be the vector field on $V$ defined by
$\xi_X(v)=\frac{d}{dt}\left(\rho(\exp(tX))(v)\right)|_{t=0}$ (see also
\S\ref{subsec:lfdsintro}).
If $f\stackrel{\mathrm{m}}{\longleftrightarrow} \chi$ and $X\in {\mathfrak{g}}$, then differentiating
\eqref{eqn:relinv} shows
\begin{equation}
\label{eqn:diffrelinv}
(\xi_X(f))(v)=df_{(v)}(\xi_X(v))=d\chi_{(e)}(X)\cdot f(v).
\end{equation}
Similarly, if
$X\in{\mathfrak{g}}$ and $g_i\stackrel{\mathrm{m}}{\longleftrightarrow}\chi'_i$ then differentiating \eqref{eqn:allgv} shows
\begin{equation}
\label{eqn:diffaddinv}
d(h_i)_{(v)}(\xi_X(v))=d(\chi'_i)_{(e)}(X)\cdot h_i(v)+g_i(v)\cdot
d(\Phi_i)_{(e)}(X).
\end{equation}
Since $(\chi_1,\ldots,\chi_r,\Phi_1,\ldots,\Phi_s)(G)=
({\mathbb{G}_\mathrm{m}})^r\times({\mathbb{G}_\mathrm{a}})^s$, choose
$X_1,\ldots,X_r,Y_1,\ldots,Y_s\in{\mathfrak{g}}$ such that
$d(\chi_i)_{(e)}(X_j)=\delta_{ij}$,
$d(\chi_i)_{(e)}(Y_j)=0$,
$d(\Phi_i)_{(e)}(X_j)=0$,
and
$d(\Phi_i)_{(e)}(Y_j)=\delta_{ij}$,
where $\delta_{ij}$ denotes the Kronecker delta function.
Because each $\chi'_i$ is some product of the $\chi_j$,
we have $d(\chi'_i)_{(e)}(Y_j)=0$.
Then by \eqref{eqn:diffrelinv} and \eqref{eqn:diffaddinv},
evaluating $df_1\wedge \cdots \wedge df_r\wedge dh_1\wedge\cdots\wedge
dh_s$ at $(\xi_{X_1},\cdots,\xi_{X_r},\xi_{Y_1},\cdots,\xi_{Y_s})(v)$
for any $v\in\Omega$
equals
\begin{equation*}
\begin{vmatrix}
\begin{smallmatrix}
f_1(v) & & \text{\large{0}} \\
& \ddots & \\
\text{\large{0}} & & f_r(v)
\end{smallmatrix}
&
\begin{smallmatrix}
\text{\Huge{0}}
\end{smallmatrix}
\\
\begin{smallmatrix}
\phantom{g_1(v)} \\
\text{\Huge{*}} \\
\phantom{0}
\end{smallmatrix}
&
\begin{smallmatrix}
g_1(v) & & \text{\large{0}} \\
& \ddots & \\
\text{\large{0}} & & g_r(v)
\end{smallmatrix}
\end{vmatrix}
\neq 0.\qedhere
\end{equation*}
\end{proof}
\begin{remark}
By Remark \ref{rem:dimsorbits} the dimension of the
$[G,G]\cdot G_{v_0}$--orbit of any $v\in\Omega$
is
$\dim(G\cdot v)-{\mathrm{codim}}([G,G]\cdot G_{v_0})
=\dim(V)-(r+s)$.
(Note that in $\Omega$ the orbits of
$[G,G]$ and $[G,G]\cdot G_{v_0}$ agree.)
\begin{comment}
\begin{proof}
Let $v\in \Omega$. Clearly $[G,G]\cdot v\subseteq [G,G]\cdot
G_{v_0}\cdot v$.
For the reverse, let $w\in [G,G]\cdot G_{v_0}\cdot v$, so $w=ghv$ for
$g\in[G,G]$, $h\in G_{v_0}$.
Say $v=k v_0$, so $w=gh(kv_0)$. Then
$$
w
=g(hkh^{-1}k^{-1})\cdot (khk^{-1})v
=g(hkh^{-1}k^{-1})v
\in [G,G]\cdot v.$$
\end{proof}
\end{comment}
Although
the level set of
$(f_1,\ldots,f_r,h_1,\ldots,h_s)$
containing $v$
and the orbit of such a $v$ will
thus
agree locally,
\begin{comment}
\begin{proof}
Let $W=V(f_i-f_i(v),h_j-h_j(v))$, an algebraic set.
As $df_1,\ldots,df_r,dh_1,\ldots,dh_s$ are linearly independent at
$v$ (by the proof above),
$v$ is a smooth point of $W$ and $W$ has codimension $r+s$.
Obviously, $([G,G]\cdot G_{v_0})\cdot v\subseteq {\mathrm{smooth}}(W)$,
with each side of the same dimension, so we must have equality, at
least locally.
The problem: why must $W$ be connected?
\end{proof}
\end{comment}
it is doubtful that
$f_1,\ldots,f_r,h_1,\ldots,h_s$ always generate the
subring of ${\mathbb C}[V]$ of all $[G,G]\cdot G_{v_0}$--invariant
polynomials,
as invariant subrings are not always
finitely generated.
\end{remark}
\subsection{Decomposition}
\label{subsec:decomp}
For prehomogeneous vector spaces, any nontrivial
factor of a relative invariant is itself a relative invariant;
a crucial fact used in the proof is the unique factorization of a
rational function on $V$.
An analogous statement for additive relative
invariants
might be that any additive relative invariant may be expressed as a
sum of additive relative invariants with `simpler' denominators.
Example \ref{ex:denompowers} shows that these denominators may be
powers of basic relative invariants, and hence we conjecture:
\begin{conjecture}
\label{conj:canbreakup}
There exists a basis $\Phi_1,\ldots,\Phi_s$ for ${\mathscr A}_1(G)$ such that
if we choose reduced rational functions $h_i$ with
$h_i\stackrel{\mathrm{a}}{\longleftrightarrow}\Phi_i$,
then for each $i$ the denominator of $h_i$ is a positive power of some
basic relative invariant.
\end{conjecture}
It would be natural to use a partial fraction expansion to prove
this,
but
such an expansion does not always
exist
for rational functions of several variables.
The question of the existence of a partial fraction decomposition
seems to be the main obstacle in proving this conjecture.
\begin{proposition}
\label{prop:canbreakup}
Let $\frac{h_1}{f_1f_2}$ be an additive relative invariant written in
lowest terms, with $f_1$ and $f_2$ polynomial relative invariants
having no common factors.
If
$$\frac{h_1}{f_1f_2}=\frac{\alpha}{f_1}+\frac{\beta}{f_2}$$
for $\alpha,\beta$ homogeneous polynomials of the same degree as
$f_1$ and $f_2$, respectively, then
$\frac{\alpha}{f_1}$ and $\frac{\beta}{f_2}$ are
additive relative invariants written in lowest terms.
\end{proposition}
\begin{proof}
The statement is true when $h_1=0$ and $f_1,f_2$ are constants, so
assume $h_1$ is nonzero.
Let $f_i\stackrel{\mathrm{m}}{\longleftrightarrow}\chi_i$ and $\frac{h_1}{f_1f_2}\stackrel{\mathrm{a}}{\longleftrightarrow}
\Phi$.
By algebra we have
\begin{equation}
\label{eqn:zstarstarstar}
h_1=\alpha\cdot f_2+\beta\cdot f_1
\end{equation}
and so by Proposition \ref{prop:globaleqn} we have for all $(g,v)\in
G\times V$,
\begin{multline}
\label{eqn:somehorrible}
\alpha(\rho(g)(v))\cdot f_2(\rho(g)(v))+\beta(\rho(g)(v))\cdot
f_1(\rho(g)(v)) \\
=\chi_1(g)\chi_2(g)\left(
f_1(v) f_2(v) \Phi(g)+\alpha(v)f_2(v)+\beta(v)f_1(v)\right).
\end{multline}
Rearranging \eqref{eqn:somehorrible} and using \eqref{eqn:relinv}
gives
\begin{multline}
\label{eqn:somehorrible2}
\left(\alpha(\rho(g)(v))-\chi_1(g)\alpha(v)\right)\cdot \chi_2(g) f_2(v) \\
=\chi_1(g)f_1(v) \left(\chi_2(g) f_2(v)
\Phi(g)+\chi_2(g)\beta(v)-\beta(\rho(g)(v))\right).
\end{multline}
For each
$g\in G$, the left side of \eqref{eqn:somehorrible2}
is divisible by $f_1$, and since $f_1$ and $f_2$ have no common
factors,
$\alpha\circ \rho(g)-\chi_1(g)\cdot \alpha$ is divisible by $f_1$.
By Proposition \ref{prop:funsagreeon},
$\frac{\alpha}{f_1}$ is an additive relative invariant,
and thus so
is
$\frac{\beta}{f_2}$.
If $\frac{\alpha}{f_1}$ or $\frac{\beta}{f_2}$ is not in lowest terms, then by
\eqref{eqn:zstarstarstar} neither is $\frac{h_1}{f_1f_2}$.
\end{proof}
\begin{remark}
If in Proposition \ref{prop:canbreakup} we omit the assumption that
$\alpha$ and $\beta$ are homogeneous, then we may take homogeneous
parts of \eqref{eqn:zstarstarstar}
to find
an $\alpha'$ and $\beta'$ that are homogeneous of the correct degree.
\end{remark}
\begin{comment}
\begin{proposition}
\label{prop:canbreakup}
Let $\frac{h_1}{f_1f_2}$ be an additive relative invariant
written in lowest terms, with $f_1$ and $f_2$ polynomial relative invariants,
$f_1$ irreducible, and
$f_1f_2$ reduced.
Let $f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i$.
If
$I=(f_1,f_2)$ is a radical ideal, and
every $v^*\in V(f_1,f_2)$ is in the closure of an orbit of
either
\begin{itemize}
\item
$\ker(\chi_1)$ acting on $V(f_1)\setminus V(f_2)$, or
\item
$\ker(\Phi)$ acting on $\Omega$
\end{itemize}
then there exist homogeneous polynomials $\alpha,\beta\in{\mathbb C}[V]$ such that
$$\frac{h_1}{f_1f_2}=\frac{\alpha}{f_1}+\frac{\beta}{f_2},$$
with $\frac{\alpha}{f_1}$ and $\frac{\beta}{f_2}$ additive relative
invariants written in lowest terms.
\end{proposition}
\begin{proof}
The statement holds for $h_1=0$, so assume $h_1$ is nonzero.
By Proposition \ref{prop:globaleqn}, for all $(g,v)$ we have
\eqref{eqn:zstar}.
Let $v^*\in V(f_1,f_2)$. By hypothesis, $v^*\in \overline{K\cdot
v_1}$ for some $v_1\in V(f_1)\setminus V(f_2)$, and hence there
is a sequence $(k_t)$ in $K$ with $\lim_{t\to \infty}
\rho(k_t)(v_1)=v^*$.
Evaluating \eqref{eqn:zstar} at $(k_t,v_1)$ gives
\begin{equation}
\label{eqn:zstarstar}
h_1(\rho(k_t)(v_1))=\chi_2(k_t)h_1(v_1),
\end{equation}
as $\chi_1(k_t)=1$ and $f_1(v_1)=0$.
Since by \eqref{eqn:relinv}
$$\lim_{t\to \infty} \chi_2(k_t)
=\lim_{t\to\infty} \frac{f_2(\rho(k_t)(v_1))}{f_2(v_1)}
=\frac{f_2(v^*)}{f_2(v_1)}=0$$
and $h_1$ is continuous,
\eqref{eqn:zstarstar} implies that
$h_1(v^*)=\lim_{t\to\infty} h_1(\rho(k_t)(v_1))=0$.
(The second item is similar NODO.)
Thus $h_1$ vanishes on $V(f_1,f_2)$, hence
$h_1\in \sqrt{I}=I$, and so
there exists $\alpha,\beta\in {\mathbb C}[V]$ satisfying
\eqref{eqn:zstarstarstar}.
Now take the appropriate homogeneous part of \eqref{eqn:zstarstarstar}
and apply Proposition \ref{prop:canbreakup}.
\end{proof}
\end{comment}
\section{Examples}
\label{sec:exs}
We now give some examples of additive relative invariants.
Our first two examples show there may be arbitrary numbers of
linearly independent additive relative invariants associated to a
single hypersurface, and arbitrary numbers of hypersurfaces.
\begin{comment}
Most of our examples have $\dim(G)=\dim(V)$; this is because
for a fixed $V$,
both $\dim(G_{v_0})$ and $\dim([G,G])$ generally increase with $\dim(G)$,
and
by Corollary \ref{cor:nontrivial}
our examples should have
$\dim(G/[G,G]\cdot G_{v_0})\geq 2$.
\end{comment}
\begin{comment}
\begin{example}
\label{ex:twobasic}
Let ${\mathrm{GL}}_2({\mathbb C})$ act on ${\mathbb C}^2$ by multiplication. Define the
connected algebraic subgroups
$$
G_1=\left\{ \begin{pmatrix} a & 0 \\ b & 1
\end{pmatrix}\in{\mathrm{GL}}_2({\mathbb C})\right\},
\qquad
G_2=\left\{ \begin{pmatrix} a & 0 \\ b & a
\end{pmatrix}\in{\mathrm{GL}}_2({\mathbb C})\right\}.$$
NODO
The two Saito matrices are, respectively,
(with columns as vfs.)
$$
\begin{pmatrix}
x & 0 \\
0 & x \end{pmatrix}
\qquad
\begin{pmatrix}
x & 0 \\
y & x \end{pmatrix},$$
showing that they both have $D=V(x)$ as the complement to the open
orbit, each defining $D$ with multiplicities.
Thus in each case, $f_1=x$ is the only relative invariant,
corresponding to the multiplicative character $\chi_1=a$ (using the
coordinates above NODO).
For $G_1$, $[{\mathfrak{g}}_1,{\mathfrak{g}}_1]$ has dimension 1, and so this is all of the
characters we may expect. For $G_2$, $[{\mathfrak{g}}_2,{\mathfrak{g}}_2]$ has dimension 0,
so that $G_2$ must have an additive function; indeed, it is
$$\Phi_1=\frac{b}{a},$$
and corresponds to the rational function
$$H=\frac{y}{x}.$$
(NODO: what are x,y?)
The isotropy subgroups at $v=(0,y)^T\neq 0$:
$$
G_1
\qquad
\left\{ \begin{pmatrix} 1 & 0 \\ b & 1
\end{pmatrix}\in{\mathrm{GL}}_2({\mathbb C})\right\}.
$$
Thus, $G_1$ has an isotropy subgroup of too large a dimension, and
$G_2$ has an abelian $1$-dimensional isotropy subgroup that is
isomorphic to $({\mathbb C},+)$.
\end{example}
\begin{example}
Let
$
G=\left\{ \left(\begin{smallmatrix} a & 0 \\ b & a
\end{smallmatrix}\right)\in{\mathrm{GL}}_2({\mathbb C})\right\}$
act on ${\mathbb C}\{x,y\}={\mathbb C}^2$ by multiplication.
Saito matrix $\begin{pmatrix} x & 0 \\ y & x \end{pmatrix}$.
$G$ has $D=V(x)$ as the exceptional orbit variety,
with $f_1=x\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_1=a$.
Since $G$ is abelian and the generic isotropy subgroup at $(x,y)=(1,0)$ is trivial,
$G$ has an additive function and thus an additive relative
invariant, namely,
$h_1=\frac{y}{x}\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi_1=\frac{b}{a}$.
The action of $g\in G$ sends
$x=y-\epsilon=0$ to $x=y-\chi_1(g)\cdot \epsilon=0$; in particular,
the origin is $G$--invariant.
\end{example}
\end{comment}
\begin{example}
\label{ex:2addinvs}
Let
$
G=\left\{ \left(\begin{smallmatrix}
a & 0 & 0\\
b & a & 0 \\
c & 0 & a
\end{smallmatrix}\right)\in{\mathrm{GL}}_3({\mathbb C})\right\}$
act on ${\mathbb C}\{x,y,z\}={\mathbb C}^3$ by multiplication, where this notation means
that $a,b,c$ may take any value that gives an invertible matrix.
Then $G$ has $D=V(x)$ as the exceptional orbit variety,
with $f_1=x\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_1=a$.
Since $G$ is abelian and the isotropy subgroup at
$(x,y,z)=(1,0,0)\in\Omega $ is trivial,
$G$ has $2$ linearly independent additive functions,
namely,
$h_1=\frac{y}{x}\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi_1=\frac{b}{a}$
and
$h_2=\frac{z}{x}\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi_2=\frac{c}{a}$.
On $V(x)$, radial subsets of the form $\alpha y+\beta z=0$
are invariant, and the $G$--action multiplies both coordinates by
$\chi_1=a$ as expected by Proposition \ref{prop:funsagreeon}.
\begin{comment}
\begin{verbatim}
2 add chars:
Trying subsets of size 3
For | 0 x 0 |
| x y 0 |
| 0 z x |
Everything seems good.
I=ideal(x) has 1 hypersurface components
generic isotropy has dim 0 and is image 0
bracket has dim 0 and is image 0
combined they have dim 0 and are image 0
* number of add. chars = 2
Has rank <1 on {monomialIdeal (x, y, z)}
Has rank <2 on {ideal(x)}
Has rank <3 on {ideal(x)}
On hypersurface component ideal(x):
Should have generic rank 1
At p=matrix {{0, 1, 0}},
isotropy subgp has dimension 2 and is image | 0 0 |
| x 0 |
| 0 x |
For | y 0 0 0 x 0 |
| 0 0 0 0 y 0 |
| 0 w y x z 0 |
| 0 0 0 0 w y |
Everything seems good.
I=ideal(y) has 1 hypersurface components
generic isotropy at p=| 0 1 0 0 | has dim 2 and is image | 0 0 |
| 0 0 |
| w x |
| 0 0 |
bracket has dim 1 and is image | 0 |
| 0 |
| y |
| 0 |
combined they have dim 3 and are image | 0 0 0 |
| 0 0 0 |
| w y x |
| 0 0 0 |
* number of add. chars = 2
* number of (mult and add) chars not including gen. isotropy = 2
Has rank <1 on {monomialIdeal (x, y, z, w)}
Has rank <2 on {ideal (x, y, w)}
Has rank <3 on {ideal(y)}
Has rank <4 on {ideal(y)}
On hypersurface component ideal(y):
Should have generic rank 2
At p=matrix {{1, 0, 0, 0}},
isotropy subgp has dimension 4 and is image | y 0 0 0 |
| 0 0 0 0 |
| 0 w y 0 |
| 0 0 0 y |
\end{verbatim}
\end{comment}
More generally, fix coordinates $x_1,\ldots,x_n$ on ${\mathbb C}^n$ and let
$G\subseteq {\mathrm{GL}}({\mathbb C}^n)$
consist of those matrices with all
diagonals equal, the first column unrestricted, and all other entries
equal to zero.
Then $G$ is abelian and has
$f_1=x_1$ and
$n-1$ linearly independent additive functions corresponding to
the additive relative invariants
$h_i=\frac{x_{i+1}}{x_1}$, $i=1,\ldots,n-1$.
\end{example}
\begin{example}
\label{ex:A}
Let
$
G=\left\{ \left(\begin{smallmatrix}
a & 0 & 0\\
b & a & 0 \\
0 & 0 & c
\end{smallmatrix}\right)\in{\mathrm{GL}}_3({\mathbb C})\right\}$
act on ${\mathbb C}\{x,y,z\}={\mathbb C}^3$ by multiplication.
Then $G$ has
$f_1=x\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_1=a$
and
$f_2=z\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_2=c$.
Since $G$ is abelian and the isotropy subgroup at
$(x,y,z)=(1,0,1)\in\Omega$ is trivial,
$\dim({\mathscr A}_1(G))=1$;
a generator is
$h_1=\frac{y}{x}\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi_1=\frac{b}{a}$.
On $V(x)$, $g\in G$ sends $(0,y,z)^T$ to $(0,\chi_1(g)
y,\chi_2(g)z)^T$; hence,
both
$V(x,y)$ and $V(x,z)$ are
$G$--invariant, but only the former has the behavior predicted by
Proposition \ref{prop:funsagreeon}.
Both $V(x)$ and $V(z)$ contain open orbits.
\begin{comment}
\begin{verbatim}
All are $V(x*y)$.
A
For | 0 x 0 |
| y 0 0 |
| 0 z x |
Everything seems good.
I=ideal(x*y) has 2 hypersurface components
generic isotropy has dim 0 and is image 0
bracket has dim 0 and is image 0
combined they have dim 0 and are image 0
* number of add. chars = 1
Has rank <1 on {monomialIdeal (x, y, z)}
Has rank <2 on {ideal (x, y), ideal (x, z)}
Has rank <3 on {ideal(x), ideal(y)}
On hypersurface component ideal(x):
Should have generic rank 2
At p=matrix {{0, 1, 1}},
isotropy subgp has dimension 1 and is image | 0 |
| 0 |
| x |
On hypersurface component ideal(y):
Should have generic rank 2
At p=matrix {{1, 0, 0}},
isotropy subgp has dimension 1 and is image | 0 |
| y |
| 0 |
B
For | 0 x 0 |
| y 0 0 |
| -z z x |
Everything seems good.
I=ideal(x*y) has 2 hypersurface components
generic isotropy has dim 0 and is image 0
bracket has dim 1 and is image | 0 |
| 0 |
| x |
combined they have dim 1 and are image | 0 |
| 0 |
| x |
* number of add. chars = 0
Has rank <1 on {ideal (x, y, z)}
Has rank <2 on {ideal (x, y), ideal (x, z)}
Has rank <3 on {ideal(x), ideal(y)}
On hypersurface component ideal(x):
Should have generic rank 2
At p=matrix {{0, 1, 1}},
isotropy subgp has dimension 1 and is image | 0 |
| 0 |
| x |
On hypersurface component ideal(y):
Should have generic rank 2
At p=matrix {{1, 0, 0}},
isotropy subgp has dimension 1 and is image | 0 |
| y |
| -z |
C
For | x 0 0 |
| 0 y 0 |
| 0 z y |
Everything seems good.
I=monomialIdeal(x*y) has 2 hypersurface components
generic isotropy has dim 0 and is image 0
bracket has dim 0 and is image 0
combined they have dim 0 and are image 0
* number of add. chars = 1
Has rank <1 on {monomialIdeal (x, y, z)}
Has rank <2 on {monomialIdeal (x, y), monomialIdeal (y, z)}
Has rank <3 on {monomialIdeal(x), monomialIdeal(y)}
On hypersurface component ideal(x):
Should have generic rank 2
At p=matrix {{0, 1, 0}},
isotropy subgp has dimension 1 and is image | x |
| 0 |
| 0 |
On hypersurface component ideal(y):
Should have generic rank 2
At p=matrix {{1, 0, 1}},
isotropy subgp has dimension 1 and is image | 0 |
| 0 |
| y |
\end{verbatim}
\end{comment}
More generally, we may take direct products of groups of the type in
Example \ref{ex:2addinvs}, realized as block diagonal matrices.
For $n\in{\mathbb N}$ and $k_1,\ldots,k_n\in {\mathbb Z}_{\geq 0}$,
this construction can produce a prehomogeneous vector space with
$\Omega^c$ consisting of
$n$ hyperplanes $H_1,\ldots,H_n$,
such that for each $i$ there are $k_i$ linearly independent additive
relative invariants having
poles on $H_i$.
\end{example}
Next, we observe that the existence of an additive relative invariant
does not depend only on the orbit structure of the action.
\begin{example}
Let
$
G=\left\{ \left(\begin{smallmatrix}
a & 0 & 0\\
b & \frac{a}{c} & 0 \\
0 & 0 & c
\end{smallmatrix}\right)\in{\mathrm{GL}}_3({\mathbb C})\right\}$
act on ${\mathbb C}\{x,y,z\}={\mathbb C}^3$ by multiplication.
Then $G$ has
$f_1=x\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_1=a$
and
$f_2=z\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_2=c$.
The generic isotropy subgroup is trivial and $\dim([G,G])=1$.
Thus $G/[G,G]\cdot G_{v_0}$ has no nontrivial additive functions,
and there are no nontrivial additive relative invariants.
The orbit structure of $G$ agrees with that of the group in
Example \ref{ex:A},
but here neither $V(x,y)$ nor $V(x,z)$ exhibit the behavior
described in Proposition \ref{prop:funsagreeon}.
\end{example}
\begin{comment}
\begin{example}
Let $G={\mathrm{GL}}_n({\mathbb C})$ act on $M(n,n,{\mathbb C})$, the space of $n\times n$ complex
matrices, by multiplication on the left.
The orbit of the identity matrix $I$ under $G$ is the set of invertible
matrices, the complement of the hypersurface defined by $f_1=\det:M(n,n,{\mathbb C})\to{\mathbb C}$.
$f_1$ is an irreducible relative invariant corresponding to
$\chi_1=\det:G\to{\mathbb C}$.
Since the isotropy group $G_I$ at $I$ is trivial,
$[G,G]={\mathrm{SL}}_n({\mathbb C})$, and $G/[G,G]$ has dimension $1$, $\chi_1$ will be
the only character.
\end{example}
\end{comment}
An additive relative invariant may unavoidably have a nontrivial power
of a basic relative invariant in its denominator.
\begin{example}
\label{ex:denompowers}
For $n\geq 2$, let $G\subseteq {\mathrm{GL}}_n({\mathbb C})$
consist of all invertible lower-triangular matrices $A$
such that each $(A)_{ij}$ depends only on $i-j$, that is,
$(A)_{i+1,j+1}=(A)_{i,j}$ whenever this makes sense.
Then $G$ is a connected abelian linear algebraic group, and its
action on ${\mathbb C}^n={\mathbb C}\{x_1,\ldots,x_n\}$ has an open orbit with
exceptional orbit variety
$V(x_1)$ and a trivial generic isotropy subgroup.
It follows that $G$ has $n-1$ linearly independent additive functions.
For $1\leq i\leq n-1$ define $\Phi_i$ by
\begin{equation*}
\Phi_i\left(
\begin{pmatrix}
a_1 & 0 & 0 & \cdots & 0
\\
a_2 & a_1 & 0 & \ddots & \vdots
\\
a_3 & a_2 & a_1 & \ddots & \vdots
\\
\vdots & \ddots & \ddots & \ddots & \vdots
\\
a_n & a_{n-1} & a_{n-2} & \cdots & a_1
\end{pmatrix}
\right)=
\frac{1}{(a_1)^i}
\begin{vmatrix}
\frac{1}{i}a_2 & a_1 & 0 & \cdots & 0
\\
\frac{2}{i}a_3 & a_2 & a_1 & \ddots & \vdots
\\
\frac{3}{i}a_4 & a_3 & a_2 & \ddots & \vdots
\\
\vdots & \ddots & \ddots & \ddots & \vdots
\\
\frac{i}{i}a_{i+1} & a_{i} & \cdots & \cdots & a_2
\end{vmatrix}.
\end{equation*}
For instance, when $n=4$ we have
\begin{align*}
\Phi_1&=\frac{a_2}{a_1}, &
\Phi_2&=\frac{1}{a_1^2}\left(\frac{1}{2} a_2^2 -a_1 a_3\right),
&
\text{and }
\Phi_3&=\frac{1}{a_1^3}\left(
\frac{1}{3}(a_2)^3-a_1a_2a_3+a_1^2 a_4 \right).
\end{align*}
For at least $n\leq 20$, computer calculations show that
each $\Phi_i:G\to {\mathbb{G}_\mathrm{a}}$ is a homomorphism, and hence
$\Phi_1,\ldots,\Phi_{n-1}$ is a basis of the space of additive
functions.
By \eqref{eqn:star},
the corresponding additive relative invariants are
of a form similar to $\Phi_i$ (e.g., substitute $x_i$ for $a_i$),
and for $i>1$ have nonreduced denominators.
\begin{comment}
NODO: A proof that this works in general?
To verify that these work:
\begin{verbatim}
n=20
phi[el_,A_]:=Det[
Table[ If[j==1,i/el,1]*A[[i+1,j]], {i,1,el},{j,1,el}] ]/(A[[1,1]])^el;
A=Table[ If[i-j>=0,a[i-j+1],0],{i,1,n},{j,1,n}];
B=Table[ If[i-j>=0,b[i-j+1],0],{i,1,n},{j,1,n}];
test[el_]:=Simplify[phi[el,A . B]-phi[el,A]-phi[el,B]];
Table[ test[el],{el,1,n-1}]
\end{verbatim}
To find the additive functions in the first place:
\begin{verbatim}
makeM[l_,n_]:=Table[If[i>=j,l[i-j+1],0],{i,1,n},{j,1,n}];
makeListOfVars[l_,n_]:=Table[l[i],{i,1,n}];
makeSumOfVars[l_,n_]:=Total[makeListOfVars[l,n]];
makegenericpoly[l_,n_,d_]:=
Total[
Map[
Function[a,Apply[ coeff,a[[1]] ]*FromCoefficientRules[{a[[1]]->1},makeListOfVars[l,n]] ],
CoefficientRules[
Expand[(makeSumOfVars[l,n])^d]
]
]];
makeListOfCoeffs[l_,n_,d_]:=
Map[
Function[a,Apply[ coeff,a[[1]] ] ],
CoefficientRules[
Expand[(makeSumOfVars[l,n])^d]
]
];
test[n_,d_]:=(
P=makegenericpoly[v,n,d];
Pv=makeListOfCoeffs[v,n,d];
phi[M_]:=(P)/(M[[1,1]])^d /. Table[v[i]->M[[i,1]],{i,1,n}];
M1=makeM[m1,n];
M2=makeM[m2,n];
X=phi[M1.M2]-phi[M1]-phi[M2];
Y=SolveAlways[(M1[[1,1]]*M2[[1,1]])^d*X==0,Join[Table[m1[i],{i,1,n}],Table[m2[i],{i,1,n}]] ];
Z=phi[M1] /. (Y[[1]]);
Collect[Z,Pv]
);
\end{verbatim}
\end{comment}
\end{example}
\begin{example}
Fix a positive integer $n$ and let $m\in \{n,n+1\}$.
Let $L\subseteq {\mathrm{GL}}_n({\mathbb C})$ (respectively, $U\subseteq {\mathrm{GL}}_m({\mathbb C})$)
consist of invertible lower triangular matrices
(resp., upper triangular unipotent matrices).
Let $G=L\times U$ act on the space $M(n,m,{\mathbb C})$ of $n\times m$ complex
matrices by $(A,B)\cdot M=AMB^{-1}$.
The classical \emph{LU factorization} of a complex
matrix asserts that this is a
prehomogeneous vector space (see \cite{DP-matrixsingI}).
For any matrix $M$, let $M^{(k)}$ denote the upper-leftmost $k\times
k$ submatrix of $M$.
By \cite[\S6]{DP-matrixsingI},
the basic relative invariants are of the form
$f_i(M)=\det(M^{(i)})$, $i=1,\ldots,n$,
and
$f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i(A,B)=\det(A^{(i)})$.
For $I$ the identity matrix, $v_0=I$ when $m=n$, or
$v_0=\begin{pmatrix} I & 0 \end{pmatrix}$ when $m=n+1$,
is in $\Omega$
and has a trivial isotropy subgroup $G_{v_0}$.
Since $\dim(G/[G,G]\cdot G_{v_0})=n+m-1$, the quotient has
$m-1$
linearly independent additive functions.
In fact, for $(A,B)\in G$ and $1\leq i\leq m-1$, let
$\Phi_i(A,B)=-(B)_{i,i+1}$.
A computation shows that these $\Phi_i$ are linearly independent
additive
functions of $G$,
with
$$h_i(M)=\frac{\det((\textrm{$M$ with column $i$
deleted})^{(i)})}{\det(M^{(i)})}\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi_i.$$
\begin{comment}
\begin{proof}
Want to show that
$$H_i(AMB^{-1})-H_i(M)=\Phi(A,B).$$
It suffices to prove this for $(A,e)$ and $(e,B)$.
For the $(A,I)$ elements, we find
$$
(AMI^{-1})^{(k\times (k+1))}=A^{(k)}M^{(k+1)}.
$$
As $A$ acts on the left, deleting column $k$ of the product
is easy to do by deleting column $k$ of $M^{(k+1)}$.
Thus,
$$
\det((\textrm{$AM$ with column $k$ deleted})^{(k)})
=\det(A^{(k)})\det((\textrm{$M$ with column $k$ deleted})^{(k)})
$$
and of course
$$\det((AM)^{(k)})=\det(A^{(k)})\det(M^{(k)});$$
since $\Phi(A,I)=0$, the claim is proven.
For the $(I,B)$ elements, we'll show
$$
H_i(MB)-H_i(M)=\Phi(I,B^{-1})=-\Phi(I,B).$$
Note that for a $k\times (k+1)$ matrix $N$,
$$
\det\begin{pmatrix} N \\ 0\cdots 010\end{pmatrix}
=-\det(\textrm{$N$ without column $k$}).$$
Let $M_{11}$ be the upper left $k\times (k+1)$ submatrix of $M$,
and $B_{11}$ the upper-left $(k+1)\times (k+1)$ submatrix of $B$.
Then the upper left $k\times (k+1)$ submatrix of $MB$ is
$M_{11}B_{11}$.
Also,
$$\begin{pmatrix} M_{11} \\ 0\cdots 010 \end{pmatrix}
\begin{pmatrix} B_{11} \end{pmatrix}
=\begin{pmatrix} M_{11}B_{11} \\ 0\cdots 0 1 \, (B)_{k,k+1}
\end{pmatrix}.$$
By multilinearity,
$$
\det\begin{pmatrix} M_{11} \\ 0\cdots 010 \end{pmatrix} \det(B_{11})
=\det\begin{pmatrix} M_{11}B_{11} \\ 0\cdots 0 1 \,0 \end{pmatrix}
+\det\begin{pmatrix} M_{11}B_{11} \\ 0\cdots 0 0 \, (B)_{k,k+1}
\end{pmatrix},
$$
and since $\det(B_{11})=1$,
$$
-\det((\textrm{$M$ without col $k$})^{(k)})
=-\det((\textrm{$MB$ without col $k$})^{(k)})
+(B)_{k,k+1}\det((MB)^{(k)}).
$$
Dividing by
$\det((MB)^{(k)})=\det((M)^{(k)})$
gives
$$
-H_k(M)=-H_k(MB)+(B)_{k,k+1},$$
or
$$H_k(MB)-H_k(M)=(B)_{k,k+1}=-\Phi(I,B)$$
as desired.
\end{proof}
\begin{verbatim}
with(LinearAlgebra):
n:=5; m:=6;
L:=Matrix(n,n,(i,j)-> `if`(i<j,0,l[i,j]));
R:=Matrix(m,m,(i,j)-> `if`(i=j,1,`if`(i>j,0,r[i,j])));
Rinv:=MatrixInverse(R);
M:=Matrix(n,m,(i,j)->`if`(i=j,1,0));
thing:=L.M.Rinv;
k:=1; simplify(-Determinant(thing[1..k,[1..(k-1),k+1]])/Determinant(thing[1..k,1..k]));
\end{verbatim}
\begin{verbatim}
Mathematica code:
n=3;
m=4;
M=Table[ms[i,j],{i,1,n},{j,1,m}];
L=Table[If[i<j,0,l[i,j]] ,{i,1,n},{j,1,n}];
U=Table[If[i<j,u[i,j],If[i==j,1,0]] ,{i,1,m},{j,1,m}];
H[i_,N_]:=Det[ Take[Drop[N,{},{i}],{1,i},{1,i}] ] / Det[ Take[N,{1,i},{1,i}] ];
Htest[i_]:=Simplify[H[i,L.M.Inverse[U]]-H[i,M]];
\end{verbatim}
\end{comment}
\end{example}
\begin{comment}
\begin{example}
(p.42 of Kimura's book)
\end{example}
\end{comment}
\section{Linear free divisors}
\label{sec:lfds}
We now consider prehomogeneous vector spaces
$(G,\rho,V)$ for which the complement of the open orbit
$\Omega$ is a type of hypersurface called a linear free divisor.
Our main theorem is that these have no nontrivial additive relative
invariants, but this has significant consequences for their structure.
\subsection{Introduction}
\label{subsec:lfdsintro}
Let ${\mathscr{O}_{\C^n,p}}$ denote the ring of germs of holomorphic functions on
${\mathbb C}^n$ at $p$,
and ${\mathrm{Der}}_{{\mathbb C}^n,p}$ the ${\mathscr{O}_{\C^n,p}}$--module of germs of holomorphic vector
fields on ${\mathbb C}^n$ at $p$.
Associated to a germ $(D,p)$ of a reduced analytic set in ${\mathbb C}^n$ is the
${\mathscr{O}_{\C^n,p}}$--module of \emph{logarithmic vector fields} defined by
$$
\Derlog{{\mathbb C}^n,p}{D}:=\{\eta\in {\mathrm{Der}}_{{\mathbb C}^n,p}: \eta(I(D))\subseteq I(D)\},$$
where $I(D)$ is the ideal of germs vanishing on $D$.
These are the vector fields tangent to $(D,p)$, and form
a Lie algebra using the Lie bracket of
vector fields.
Let $D$ be nonempty and $(D,p)\neq({\mathbb C}^n,p)$.
When $\Derlog{{\mathbb C}^n,p}{D}$ is a free ${\mathscr{O}_{\C^n,p}}$--module, necessarily of rank $n$,
then $(D,p)$ is called a \emph{free divisor}.
Free divisors are always pure hypersurface germs that are either smooth or
have singularities in codimension $1$,
and were first encountered as various types of discriminants.
Now let $(G,\rho,V)$ be a prehomogeneous vector space, let
${\mathfrak{g}}$ be the Lie algebra of $G$, and let $D:=V\setminus \Omega$
be the exceptional orbit variety.
Let $\Derlog{V}{D}_0$ denote the $\eta\in\Derlog{V}{D}$ that are
\emph{linear}, that is,
homogeneous of degree $0$ (e.g.,
$(x+y)\frac{\partial}{\partial x}-2z\frac{\partial}{\partial y}$).
Because $D$ is $G$--invariant, differentiating the action of $G$ gives a Lie algebra
\mbox{(anti-)}homomorphism
$\tau:{\mathfrak{g}}\to\Derlog{V}{D}_0$, where
$\xi_X:=\tau(X)$ is a vector field defined globally on $V$ by
$$\xi_X(v)=\tau(X)(v)=\frac{d}{dt}\!\left.\big(\rho(\exp(tX))(v)\big)\right|_{t=0}.$$
Thus, $\tau({\mathfrak{g}})\subseteq \Derlog{V}{D}_0$ are
finite-dimensional Lie subalgebras of the module $\Derlog{V}{D}$.
As $\tau({\mathfrak{g}})(v)=T_v(G\cdot v)$, the maximal minors of a matrix
containing
the coefficients of a basis of $\tau({\mathfrak{g}})$ are of degree $\dim(V)$ and
generate an ideal defining the set $D$.
When $\Derlog{V}{D}$ has a free basis of linear vector fields,
then $D$ is called a \emph{linear free divisor}.
By
\cite[Lemma 2.3]{gmns},
every linear free divisor $D$ is the exceptional orbit variety of
a prehomogeneous vector space
with the following properties:
\begin{definition}
\label{def:defineslfd}
Let $D$ be a linear free divisor in $V$.
If $(G,\rho,V)$ is a prehomogeneous vector space with
open orbit $V\setminus D$,
$G$ connected,
and $\dim(G)=\dim(V)$,
then say that
\emph{$G$ defines the linear free divisor $D$}.
\end{definition}
It follows that
$\ker(\rho)$ is finite, and
$\tau({\mathfrak{g}})=\Derlog{V}{D}_0$.
One such prehomogeneous vector space
may be constructed in the following way.
For a divisor $D\subset V$ in a finite-dimensional complex vector
space,
let ${\mathrm{GL}}(V)_D$ be the largest subgroup
of ${\mathrm{GL}}(V)$ that preserves $D$.
Note that ${\mathrm{GL}}(V)_D$ is algebraic.
Then $({\mathrm{GL}}(V)_D)^0$ is a connected complex linear algebraic group
with Lie algebra (anti-)isomorphic to
$\Derlog{V}{D}_0$.
For a linear free divisor $D\subset V$, the group
$({\mathrm{GL}}(V)_D)^0\subseteq {\mathrm{GL}}(V)$
with the inclusion representation is a prehomogeneous vector space that
defines $D$
(\cite[Lemma 2.3]{gmns}).
In fact,
if $(G,\rho,V)$ defines a
linear free divisor $D$, then $\rho(G)=({\mathrm{GL}}(V)_D)^0$.
\subsection{Brion's criterion}
\label{subsec:brion}
Brion gave the following useful criterion for $D$ to be a linear free
divisor.
\begin{theorem}[\cite{brion}, {\cite[Theorem 2.1]{freedivisorsinpvs}};
see also \cite{pike-generalizebrion}]
\label{thm:brion}
For $G=({\mathrm{GL}}(V)_D)^0$,
the following are equivalent:
\begin{enumerate}
\item
\label{cond:dlfd}
$D$ is a linear free divisor.
\item
\label{cond:bothopenisotrpy}
Both of these conditions hold:
\begin{enumerate}
\item
\label{cond:vmdopen}
$V\setminus D$ is a unique $G$-orbit, and the corresponding
isotropy groups are finite.
\item
\label{cond:isotropyrep}
Each irreducible component $D_i$ of $D$ contains an open
$G$-orbit $D_i^0$, and the corresponding isotropy groups are
extensions of finite groups by ${\mathbb{G}_\mathrm{m}}$.
\end{enumerate}
\end{enumerate}
When these hold,
$\tau({\mathfrak{g}})$
generates $\Derlog{V}{D}$, and each $D_i^0={\mathrm{smooth}}(D)\cap D_i$.
\end{theorem}
The proof of Theorem \ref{thm:brion} shows the following.
\begin{corollary}
\label{cor:tobrion}
Suppose that in the situation of Theorem \ref{thm:brion},
\eqref{cond:vmdopen} holds,
$D_i$ contains an open orbit $D_i^0$, and $v_i\in D_i^0$.
Then $G_{v_i}$ is an extension of a finite group by ${\mathbb{G}_\mathrm{m}}$
if and only if
the induced representation of $G^0_{v_i}$ on the normal line to $D_i$ at
$v_i$ is nontrivial.
\end{corollary}
\begin{comment}
DONE: We cannot strengthen this to isomorphism
\begin{proof}
The normal representation is $\rho_v:G^0_{v_i}\to {\mathrm{GL}}(N)$ for
$\dim(N)=1$.
If $\rho_v$ is nontrivial, then $G^0_{v_i}/\ker(\rho_v)\cong {\mathbb{G}_\mathrm{m}}$, and
by dimensional considerations, $\ker(\rho_v)$ is finite.
Moreover, $G_{v_i}$ is an extension of a finite group by $G^0_{v_i}$.
Conversely, if $G_{v_i}$ is an extension of a finite group by
${\mathbb{G}_\mathrm{m}}$, and $\rho_v$ is trivial, then we will obtain a contradiction.
$\rho_v$ trivial implies that $\rho_v$ fixes the points on the normal
line. As ${\mathbb{G}_\mathrm{m}}$ is reductive, we may split $\rho|_{{\mathbb{G}_\mathrm{m}}}$ into irreducibles,
one of which will be $\rho_v$, and we obtain a line fixed by ${\mathbb{G}_\mathrm{m}}$
which intersects $\Omega$; this contradicts \eqref{cond:vmdopen}.
\end{proof}
\end{comment}
The representation on the normal line is actually quite familiar.
\begin{lemma}
\label{lemma:interpnormal}
Let $(G,\rho,V)$ be a prehomogeneous vector space with $f$ as a basic
relative invariant, and let $v$ be a smooth point of $D:=V(f)$.
If $H\subseteq G_v$,
then the representation
$\rho_v:H\to{\mathrm{GL}}(L)$ on the
normal line $L=T_v V/T_vD$ to $D$ at $v$ is
\begin{equation*}
\rho_v(h)(\ell)=\chi(h)\ell,\qquad \textrm{for all }\ell\in L,
\end{equation*}
where $f\stackrel{\mathrm{m}}{\longleftrightarrow}\chi$.
\end{lemma}
Geometrically,
$\rho_v$ acts on a normal slice to $f=0$ at $v$, and such a slice
intersects all level sets of $f$.
At the same time, the action of $\rho(H)$ fixes $v$ and
translates between the level sets of $f$ according to $\chi$.
\begin{comment}
\begin{proof}
Since $H$ fixes $v$ and leaves invariant $D$,
$H$ leaves invariant $T_vD$.
Then the normal line is $L=T_vV/T_vD$, and $\rho|_H$ induces the
representation $\rho_v:H\to {\mathrm{GL}}(L)$, with
$\rho_v(h)(w+T_vD)=\rho(h)(w)+T_vD$.
Since $D$ is reduced and $v$ is a
smooth point, $T_vD=\ker(df_{(v)}:T_vV\to T_0{\mathbb C}\cong {\mathbb C})$.
In particular, $df_{(v)}$ induces a vector space isomorphism from
$L$ to ${\mathbb C}$.
By definition, then, we have the following
commutative diagram for all $h\in H$.
\begin{equation}
\label{eqn:whatisnormalline}
\xymatrix {
T_v V \ar[r]^{\rho(h)} \ar[d]_{df_{(v)}} & T_vV
\ar[d]^{df_{(v)}} \\
L \ar[r]^{\rho_v(h)} & L}
\end{equation}
If $w\in T_vV$ and $\lambda\in{\mathbb C}$, then
\begin{equation}
\label{eqn:fvlambda}
f(v+\lambda\cdot \rho(h)(w))=f(\rho(h)(v+\lambda w))=\chi(h)
f(v+\lambda w).
\end{equation}
Differentiating \eqref{eqn:fvlambda} with respect to $\lambda$ and
evaluating at $\lambda=0$, we have $df_{(v)}(\rho(h)(w))=\chi(h)\cdot
df_{(v)}(w)$, or by \eqref{eqn:whatisnormalline},
\begin{equation*}
\rho_v(h)(df_{(v)}(w))=\chi(h)\cdot df_{(v)}(w).
\end{equation*}
Thus $\rho_v(h)$ acts on $L$ by multiplication by $\chi(h)$.
\end{proof}
\end{comment}
\begin{proof}
The representation $\rho|_H$ fixes $v$ and hence
induces a representation $\rho'$ of $H$
on the tangent space $T_vV$;
by silently identifying $T_vV$ with $V$, we have $\rho'=\rho|_H$.
Since $\rho|_H$ leaves invariant $D$ and fixes the smooth point $v$ on
$D$,
the representation $\rho'$ leaves invariant $T_vD$.
Then the normal line is $L=T_vV/T_vD$, and $\rho'$ produces the quotient
representation $\rho_v:H\to {\mathrm{GL}}(L)$ defined by
$\rho_v(h)(w+T_vD)=\rho'(h)(w)+T_vD$.
Let $h\in H$ and $w\in T_vV$.
For $\lambda\in{\mathbb C}$ we have by \eqref{eqn:relinv},
\begin{equation}
\label{eqn:fvlambda}
f(v+\lambda\cdot \rho(h)(w))=f(\rho(h)(v+\lambda w))=\chi(h)
f(v+\lambda w).
\end{equation}
Differentiating \eqref{eqn:fvlambda} with respect to $\lambda$ and
evaluating at $\lambda=0$ gives
$df_{(v)}(\rho'(h)(w))=\chi(h) df_{(v)}(w)$,
or
$\rho'(h)(w)-\chi(h)w\in\ker(df_{(v)})$.
Since $f$ is reduced and $v$ is a
smooth point, $T_vD=\ker(df_{(v)}:T_vV\to T_0{\mathbb C})$.
The definition of $\rho_v$ then implies the result.
\end{proof}
\begin{remark}
If $(G,\rho,V)$ defines a linear free divisor $D$
in the sense of Definition \ref{def:defineslfd}, then
all of the results of this \S\ref{subsec:brion} hold for
$G$ as well as
$\rho(G)=({\mathrm{GL}}(V)_D)^0$.
\end{remark}
\subsection{The main theorems}
\label{subsec:mainthm}
Let $(G,\rho,V)$ define the linear free divisor
$D\subset V$ in the sense of Definition \ref{def:defineslfd}.
Let $f_1,\ldots,f_r$ be the basic relative invariants, so that
$\cup_{i=1}^r V(f_i)$ is the irreducible decomposition of
$D=V\setminus \Omega$.
Let $f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i$, and choose $v_0\in\Omega$.
For $1\leq i\leq r$ let $v_i$ be a generic point on $V(f_i)$,
an element of the open orbit of $G$ in $V(f_i)$.
\begin{theorem}
\label{thm:homomorphism}
Let
$G$ define the linear free divisor $D$, with the notation above.
Then:
\begin{enumerate}
\item
\label{enit:1}
The homomorphism $(\chi_1,\ldots,\chi_r):G\to ({\mathbb{G}_\mathrm{m}})^r$
is surjective and has kernel
$[G,G]\cdot G_{v_0}$.
\item
\label{enit:2}
$G/[G,G]\cdot G_{v_0}$ has only the trivial additive function and
$(G,\rho,V)$ has only constant additive relative invariants.
\item
\label{enit:3}
For $i\neq j$, we have $G^0_{v_j}\subseteq \ker(\chi_i)$.
\item
\label{enit:4}
The representation $\chi_i|_{G^0_{v_i}}:G^0_{v_i}\to{\mathbb{G}_\mathrm{m}}$ is surjective, with finite
kernel.
\item
\label{enit:5}
$\ker(\chi_i|_{G^0_{v_i}})=G^0_{v_i}\cap ([G,G]\cdot G_{v_0})$ is
finite.
\item
\label{enit:6}
For $i\neq j$, the subgroup $G^0_{v_i}\cap G^0_{v_j}$ is a finite subset of
$[G,G]\cdot G_{v_0}$.
\end{enumerate}
\end{theorem}
\begin{proof}
Let $G_1=[G,G]\cdot G_{v_0}$.
Claim
\eqref{enit:3} is just
Lemma \ref{lemma:vanishingthm}\eqref{enit:figv}.
By
Theorem \ref{thm:brion},
Corollary \ref{cor:tobrion}, and Lemma \ref{lemma:interpnormal},
each $\chi_i|_{G^0_{v_i}}$ is nontrivial, thus surjective.
As $\dim(G^0_{v_i})=1=\dim({\mathrm{GL}}({\mathbb C}))$, the kernel is finite, giving
\eqref{enit:4}.
If $\Phi\in{\mathscr A}(G/G_1)$ is nontrivial, then by Proposition \ref{prop:a1} there
exists a non-constant $h$ with $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$.
By Proposition \ref{prop:homogeneous}, write
$h=\frac{h_1}{g_1}$ as a reduced fraction, with
$g_1$ a polynomial relative invariant and
$g_1\stackrel{\mathrm{m}}{\longleftrightarrow} \chi$.
Since $\Phi$ is nontrivial, $g_1$ cannot be constant and so by
Theorem \ref{thm:pvs1} is a product of basic relative invariants;
let $f_i$ be an irreducible factor of $g_1$.
If $h_1(v_i)=0$, then by Proposition \ref{prop:globaleqn},
$h_1$ vanishes on $G\cdot v_i$ and hence on
$\overline{G\cdot v_i}=V(f_i)$; but then $f_i$ divides
$\gcd(h_1,g_1)$, a contradiction of how we wrote $h$, hence $h_1(v_i)\neq 0$.
By
Lemma \ref{lemma:vanishingthm}\eqref{enit:gvichi}
we have $G^0_{v_i}\subset \ker(\chi_i)$,
but this contradicts \eqref{enit:4}.
Thus $\Phi=0$ is trivial, proving
\eqref{enit:2}.
Since $G/G_1$ is connected and abelian,
by Proposition \ref{prop:ckcl},
$G/G_1\cong ({\mathbb{G}_\mathrm{m}})^k\times({\mathbb{G}_\mathrm{a}})^\ell$.
Choosing a free basis $\epsilon_1,\ldots,\epsilon_k$ of $X(G/G_1)$
and a basis $\Phi_1,\ldots,\Phi_\ell$ of ${\mathscr A}(G/G_1)$
gives an explicit isomorphism
$\theta=(\epsilon_1,\ldots,\epsilon_k,\Phi_1,\ldots,\Phi_\ell):G/G_1\to
({\mathbb{G}_\mathrm{m}})^k\times ({\mathbb{G}_\mathrm{a}})^\ell$.
Theorem \ref{thm:pvs1} and Proposition \ref{prop:x1is} show that
we may let each $\epsilon_i=\chi_i$.
Then, \eqref{enit:1} follows from \eqref{enit:2}.
For \eqref{enit:5},
such elements are contained in the kernel by \eqref{enit:1},
and conversely by \eqref{enit:3} and \eqref{enit:1}.
By \eqref{enit:4}, the kernel is finite.
For \eqref{enit:6},
observe that by \eqref{enit:3}, such elements are in the kernel of the
homomorphism of \eqref{enit:1}.
For finiteness, use \eqref{enit:5}.
\end{proof}
\begin{comment}
\begin{proof}
Let $H=[G,G]\cdot G_{v_0}$.
Since $G/H$ is connected and abelian,
by Proposition \ref{prop:ckcl},
$G/H\cong ({\mathbb{G}_\mathrm{m}})^k\times({\mathbb{G}_\mathrm{a}})^\ell$.
Choosing a free basis $\epsilon_1,\ldots,\epsilon_k$ of $X(G/H)$
and a basis $\Phi_1,\ldots,\Phi_\ell$ of ${\mathscr A}(G/H)$
gives an explicit isomorphism
$\theta=(\epsilon_1,\ldots,\epsilon_k,\Phi_1,\ldots,\Phi_\ell):G/H\to
({\mathbb{G}_\mathrm{m}})^k\times ({\mathbb{G}_\mathrm{a}})^\ell$.
Theorem \ref{thm:pvs1} and Proposition \ref{prop:x1is} show that
we may let $\epsilon_i=\chi_i$.
Then, \eqref{enit:1} will follow from \eqref{enit:2}.
If $\Phi\in{\mathscr A}(G/H)$ is nontrivial, then by Proposition \ref{prop:a1} there
exists a nonzero $h$ with $h\stackrel{\mathrm{a}}{\longleftrightarrow} \Phi$.
By Proposition \ref{prop:homogeneous}, write
$h=\frac{h_1}{g_1}$, with
$g_1$ a polynomial relative invariant,
$g_1\stackrel{\mathrm{m}}{\longleftrightarrow} \chi$, and $\gcd(h_1,g_1)=1$.
Since $\Phi$ is nontrivial, $g_1$ cannot be constant;
let $f_i$ be a factor of $g_1$.
If $h_1(v_i)=0$, then by Proposition \ref{prop:globaleqn},
$h_1$ vanishes on $G\cdot v_i$ and hence on
$\overline{G\cdot v_i}=V(f_i)$; but then $f_i$ divides
$\gcd(h_1,g_1)$, giving a contradiction, and hence $h_1(v_i)\neq 0$.
By
Lemma \ref{lemma:vanishingthm}\eqref{enit:gvichi}
we have $G^0_{v_i}\subset \ker(\chi_i)$.
Thus \eqref{enit:4} will provide the contradiction to prove
\eqref{enit:2}.
\eqref{enit:3} is just
Lemma \ref{lemma:vanishingthm}\eqref{enit:figv}.
By
Theorem \ref{thm:brion},
Corollary \ref{cor:tobrion}, and Lemma \ref{lemma:interpnormal},
each $\chi_i|_{G^0_{v_i}}$ is nontrivial, thus surjective.
As $\dim(G^0_{v_i})=1=\dim({\mathrm{GL}}({\mathbb C}))$, the kernel is finite, giving
\eqref{enit:4}, \eqref{enit:2}, and \eqref{enit:1}.
For \eqref{enit:5},
such elements are contained in the kernel by \eqref{enit:1},
and conversely by \eqref{enit:3} and \eqref{enit:1}.
By \eqref{enit:4}, the kernel is finite.
For \eqref{enit:6},
observe that by \eqref{enit:3}, such elements are in the kernel of the
homomorphism of \eqref{enit:1}.
For finiteness, use \eqref{enit:5}.
\end{proof}
\end{comment}
\begin{remark}
For each $i>0$, the group $G^0_{v_i}$ is a
$1$--dimensional connected complex linear algebraic group with
a surjective homomorphism to ${\mathbb{G}_\mathrm{m}}$,
and hence
$G^0_{v_i}\cong {\mathbb{G}_\mathrm{m}}$.
\end{remark}
Theorem \ref{thm:homomorphism} has a number of immediate consequences.
In particular,
we may easily compute the number of irreducible components.
\begin{theorem}
\label{thm:ofhomomorphism}
Let $G$ define the linear free divisor $D$, with the notation above.
Then:
\begin{enumerate}
\item
\label{enit:unique1}
The hypersurface $D$ has
$$r=\dim_{\mathbb C}(G/[G,G]\cdot
G_{v_0})=\dim_{\mathbb C}(G/[G,G])=\dim_{\mathbb C}({\mathfrak{g}}/[{\mathfrak{g}},{\mathfrak{g}}])$$
irreducible
components, where ${\mathfrak{g}}$ is the Lie algebra of $G$.
\item
\label{enit:unique2}
Every element of $G$ may be written in a finite number of ways
as a product of elements
from the subgroups $[G,G]\cdot G_{v_0},G^0_{v_1},\ldots,G^0_{v_r}$.
Each term of such a product is unique modulo
$[G,G]\cdot G_{v_0}$.
\item
\label{enit:unique3}
Let $S\subseteq \{1,\ldots,r\}$.
The subgroup of $G$ that leaves invariant all level sets of $f_i$
for $i\in S$ is normal in $G$, and is the product of the subgroups
$[G,G]\cdot G_{v_0}$,
and $G^0_{v_j}$ for $j\notin S$.
\item
\label{enit:unique5}
As algebraic groups, $G/[G,G]\cong ({\mathbb{G}_\mathrm{m}})^r$.
\item
\label{enit:unique4}
The subgroup $[G,G]$ contains all unipotent elements of $G$.
\end{enumerate}
\end{theorem}
\begin{proof}
For \eqref{enit:unique1}, combine
Theorem \ref{thm:homomorphism}\eqref{enit:2}
and Corollary \ref{cor:numcomponents}.
Note also that $G_{v_0}$ is finite by
Theorem \ref{thm:brion}.
Let $g\in G$.
By Theorem \ref{thm:homomorphism}\eqref{enit:4},
for $i>0$ there exists
$g_i\in G^0_{v_i}$ such that $\chi_i(g)=\chi_i(g_i)$.
By Theorem \ref{thm:homomorphism}\eqref{enit:3} and
\ref{thm:homomorphism}\eqref{enit:1},
$g(g_1\cdots g_r)^{-1}$ is in the kernel of each $\chi_i$, and
hence lies in $[G,G]\cdot G_{v_0}$.
This proves existence for \eqref{enit:unique2}.
To address uniqueness, let $g\in G$ and suppose
that for $j=1,2$ we have
$g=g_{0,j}\cdot g_{1,j}\cdots g_{r,j}$,
with
$g_{0,j}\in [G,G]\cdot G_{v_0}$
and
$g_{i,j}\in G^0_{v_i}$ for $i>0$.
By Theorem \ref{thm:homomorphism}\eqref{enit:1} and
\ref{thm:homomorphism}\eqref{enit:3},
for $i>0$ we have $\chi_i(g)=\chi_i(g_{i,1})=\chi_i(g_{i,2})$.
Then for $i>0$ we have
$g_{i,1}=g_{i,2}$ modulo $\ker(\chi_i|_{G^0_{v_i}})$,
and this kernel is finite and
contained in $[G,G]\cdot G_{v_0}$ by Theorem
\ref{thm:homomorphism}\eqref{enit:5}.
Since $g_{0,j}$ is uniquely determined by $g_{1,j},\ldots,g_{r,j}$,
there are precisely
$$
\prod_{i=1}^r \#(\ker(\chi_i|_{G^0_{v_i}}))$$
ways to write $g$ in this way. This proves \eqref{enit:unique2}.
By \eqref{eqn:relinv}, $g\in G$ leaves invariant all level sets of
$f_i$ if and only if $g\in\ker(\chi_i)$.
Hence, the subset $H$ leaving invariant all level sets of $f_i$ for
$i\in S$ is an intersection of kernels and thus a normal subgroup.
By Theorems \ref{thm:homomorphism}\eqref{enit:1} and
\ref{thm:homomorphism}\eqref{enit:3},
$[G,G]\cdot G_{v_0}$ and $G^0_{v_j}$ for $j\notin S$ are in $H$.
Conversely, by
\eqref{enit:unique2} and Theorems
\ref{thm:homomorphism}\eqref{enit:3} and
\ref{thm:homomorphism}\eqref{enit:5}, and the normality of
$[G,G]\cdot G_{v_0}$, any element of $H$ may be written as a product
of elements of these subgroups.
This proves \eqref{enit:unique3}.
To prove \eqref{enit:unique5},
consider the diagram
\begin{equation*}
\xymatrix@-1pc{
G/[G,G] \ar[r]^-\phi \ar[d]^{\kappa}
& {\mathbb{G}_\mathrm{m}}^k\times{\mathbb{G}_\mathrm{a}}^\ell \ar[d]^{\psi} \\
G/[G,G]\cdot G_{v_0} \ar[r]^-\chi
& {\mathbb{G}_\mathrm{m}}^r
}
\end{equation*}
where
$\phi$ is an isomorphism that exists by Proposition \ref{prop:ckcl},
$\kappa$ is the quotient map,
$\chi$ is the isomorphism induced from $(\chi_1,\ldots,\chi_r)$
by Theorem \ref{thm:homomorphism}\eqref{enit:1},
and $\psi$ makes the diagram commutative.
The Jordan decomposition implies that $\{1\}\times {\mathbb{G}_\mathrm{a}}^\ell\subseteq
\ker(\psi)$,
but this kernel is isomorphic to $\ker(\kappa)$, which is finite
because $G_{v_0}$ is finite.
Hence $\ell=0$ and $k\leq r$, and
the surjectivity of $\psi$ requires $k\geq r$,
proving \eqref{enit:unique5}.
Finally, \eqref{enit:unique4} is a consequence of \eqref{enit:unique5}
and the Jordan decomposition.
\end{proof}
\begin{example}[{\cite[Example 2.1]{mondbuchweitz}},{\cite[Example
5.1]{gmns}}]
\label{ex:aac}
On ${\mathbb C}^3$, fix coordinates $x,y,z$.
The linear free divisor
$x(xz-y^2)=0$ is defined by the solvable
group
$$G=\left\{\left(\begin{smallmatrix}
a & 0 & 0 \\
b & c & 0 \\
\frac{b^2}{a} & \frac{2bc}{a} & \frac{c^2}{a}
\end{smallmatrix}\right)\in
{\mathrm{GL}}({\mathbb C}^3)\right\}.$$
We have
$f_1=x\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_1=a$ and
$f_2=xz-y^2\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_2=c^2$.
At $v_0=(1,0,1)\in \Omega$,
$v_1=(0,1,0)\in V(f_1)$,
and $v_2=(1,0,0)\in V(f_2)$,
the (generic)
isotropy subgroups are defined by,
respectively,
$(a,b,c)=(1,0,\pm 1)$,
$(b,c)=(0,1)$,
and
$(a,b)=(1,0)$.
As $[G,G]$ is defined by $a=c=1$,
we see $[G,G]\cdot G_{v_0}$ is defined by $a=1$, $c=\pm 1$.
Theorems \ref{thm:homomorphism} and \ref{thm:ofhomomorphism}
are easy to verify.
\end{example}
Recall that $L(K)$ denotes the Lie algebra of an algebraic group $K$.
Let $\delta_{ij}$ denote the Kronecker delta function.
On the level of Lie algebras,
Theorem \ref{thm:homomorphism} implies
the following.
\begin{corollary}
\label{cor:homomorphismliealgebra}
As vector spaces,
$${\mathfrak{g}}=[{\mathfrak{g}},{\mathfrak{g}}]\oplus \bigoplus_{i=1}^r L(G_{v_i}).$$
For $i=1,\ldots,r$
there exist unique $X_i\in L(G_{v_i})$
such that $L(G_{v_i})={\mathbb C} X_i$,
and such that for all $j$ we have
$d(\chi_i)_{(e)}(X_j)=\delta_{ij}$
and
$\xi_{X_j}(f_i)=\delta_{ij}\cdot f_i$.
For $X\in [{\mathfrak{g}},{\mathfrak{g}}]$ and any $j$, $\xi_X(f_j)=0$.
\end{corollary}
\begin{proof}
Differentiating the homomorphism of Theorem
\ref{thm:homomorphism}\eqref{enit:1}
gives a homomorphism $\lambda:{\mathfrak{g}}\to L(({\mathbb{G}_\mathrm{m}})^r)=\oplus_{i=1}^r L({\mathbb{G}_\mathrm{m}})$ with kernel
$[{\mathfrak{g}},{\mathfrak{g}}]$.
By Theorems \ref{thm:homomorphism}\eqref{enit:3} and
\ref{thm:homomorphism}\eqref{enit:4},
under $\lambda$ each $L(G_{v_i})$ surjects onto the $i$th copy of
$L({\mathbb{G}_\mathrm{m}})$, and is zero on the rest.
This gives the vector space decomposition, and proves that for
$i\neq j$, $d(\chi_i)_{(e)}(L(G_{v_j}))=0$.
Now choose the unique $X_i\in L(G_{v_i})$ such that $d(\chi_i)_{(e)}(X_i)=1$.
The rest of the statement follows from
\eqref{eqn:diffrelinv}.
\end{proof}
Each $X_i$ depends on the choice of $v_i\in V(f_i)$,
but
any two choices will differ by an element of
$[{\mathfrak{g}},{\mathfrak{g}}]$.
\begin{remark}
\label{rem:worksmoregenerally}
More generally, let $(G,\rho,V)$ be a prehomogeneous vector space with
no nontrivial additive relative invariants.
(For instance, by Remark \ref{rem:fisquared}, this happens
if the ideal of Remark \ref{rem:omegac}
is not contained in $(f_i)^2$ for every basic relative invariant $f_i$.)
Then by the same argument in Theorems \ref{thm:homomorphism} and
\ref{thm:ofhomomorphism},
$G/[G,G]\cdot G_{v_0}$ is an
algebraic torus, of dimension equal to the number of irreducible hypersurface
components of the exceptional orbit variety $V\setminus \Omega$.
\end{remark}
\subsection{The structure of \texorpdfstring{$G$}{G}}
\label{subsec:structureg}
We now use Theorems \ref{thm:homomorphism} and
\ref{thm:ofhomomorphism} to study the structure of algebraic groups
defining linear free divisors.
Let $G$ be a connected complex algebraic group.
Let $\mathrm{Rad}(G)$ denote the \emph{radical} of $G$, the maximal
connected normal solvable subgroup.
\begin{comment}
\begin{verbatim}
http://www.encyclopediaofmath.org/index.php/Levi-Ma
Levi-Mal'tsev decomposition, Mostow's theorem?
or malcev?
p.285 of \cite{ov}.
\end{verbatim}
\end{comment}
The (algebraic) \emph{Levi decomposition} of $G$
writes
$$G=\mathrm{Rad}_u(G)\rtimes L,$$
where $\mathrm{Rad}_u(G)$ is the \emph{unipotent radical} of
$G$,
the largest connected unipotent normal subgroup of $G$, consisting of
all unipotent elements of $\mathrm{Rad}(G)$; and $L$ is
a \emph{Levi subgroup}, a maximal connected reductive algebraic subgroup of $G$,
unique up to conjugation
(\cite[11.22]{borel}).
Moreover, $L=\mathrm{Z}(L)^0\cdot [L,L]$
for
$\mathrm{Z}(L)$ the center of $L$,
$\mathrm{Z}(L)\cap [L,L]$ is finite,
$[L,L]$ is semisimple,
and $(\mathrm{Z}(L))^0=L\cap \mathrm{Rad}(G)$ is a maximal torus of
$\mathrm{Rad}(G)=\mathrm{Rad}_u(G)\rtimes \mathrm{Z}(L)^0$
\begin{comment}
Proof: Have $\supseteq$ by definition and the above.
For $g\in\rad(G)$, write $g=rl$, $r\in R$, $l\in L$. Then $l\in\mathrm{Rad}(G)$ implies that
$l\in \mathrm{Rad}(G)\cap L=(\mathrm{Z}(L))^0$.
And already have semidirect product.
\end{comment}
(\cite[14.2, 11.23]{borel}).
Groups defining linear free divisors have the following structure.
\begin{corollary}
\label{cor:ofhomomorphism2}
Let $G$ define the linear free divisor $D$ and have the
Levi decomposition
above. Then:
\begin{enumerate}
\item
\label{enit:gencomponents}
The number of irreducible components of $D$ equals
$\dim(\mathrm{Z}(L))$.
\item
\label{enit:genbracket}
$[G,G]=\mathrm{Rad}_u(G)\rtimes [L,L]$.
\item
\label{enit:geng}
$G=[G,G]\cdot (\mathrm{Z}(L))^0$, with $[G,G]\cap (\mathrm{Z}(L))^0$ finite.
\item
\label{enit:genisomorphism}
$(\chi_1,\ldots,\chi_r)|_{\mathrm{Z}(L)^0}:\mathrm{Z}(L)^0\to ({\mathbb{G}_\mathrm{m}})^r$ is
surjective, with a finite kernel.
\end{enumerate}
\end{corollary}
\begin{proof}
Let $R=\mathrm{Rad}_u(G)$.
Since $G$ is the semidirect product of $R$ and $L$,
a straightforward calculation shows that
$[G,G]=[R,R]\cdot [R,L]\cdot [L,L]$.
\begin{comment}
\begin{proof}
First, we show the RHS is a group.
Use \cite[17.2]{humphreys}.
By (b), since $R\mathrel{\unlhd} G$,
$[R,R]$ is a connected, algebraic, normal subgroup of $G$ (and $R$).
By (a), $[R,L]$ and $[L,L]$ are connected algebraic subgroups of $G$.
Now we show it is an abstract group...
By \cite[I.3, p.17]{langsalgebra},
since $[R,L]\subseteq R$ and $[R,R]\mathrel{\unlhd} R$,
$N_{[R,R]}=R\supseteq [R,L]$, making
$[R,L]\cdot [R,R]=[R,R]\cdot [R,L]$ a subgroup of $R$.
By the same,
since $[R,L]\cdot [R,R]\subseteq R$,
and $[L,L]\subseteq L$,
we have for $b\in [L,L]$, $h\in R$, $k\in L$,
\begin{align*}
b[h_1,k_1][h_2,h_3]b^{-1}
&=
b[h_1,k_1]b^{-1}b[h_2,h_3]b^{-1} \\
&=[bh_1b^{-1},bk_1b^{-1}][bh_2b^{-1},bh_3b^{-1}] \\
&\in [R,L]\cdot [R,R];
\end{align*}
thus, $N_{[R,L]\cdot [R,R]}\supseteq [L,L]$;
hence, the RHS is a group.
$[G,G]$ certainly contains all such elements.
Conversely, if $g_1=h_1k_1$, $g_2=h_2k_2$ for $h_i\in R$, $k_i\in L$,
we write the bracket as a product of an element of $R$ and an element
of $L$, and then as a product of elements from $[R,R]$, $[R,L]=[L,R]$,
and $[L,L]$:
\begin{align*}
[g_1,g_2]
&=
h_1(k_1h_2k_1^{-1})
\left( (k_1k_2k_1^{-1})h_1^{-1}(k_1k_2^{-1})\right)
\cdot
\left( (k_1k_2k_1^{-1}k_2^{-1}) h_2^{-1}
(k_2k_1k_2^{-1}k_1^{-1})\right)
\cdot (k_1k_2k_1^{-1}k_2^{-1}) \\
&=
[h_1,h_2]
[h_2h_1h_2^{-1},k_1]
[k_1,h_2h_1]
[h_2h_1,k_1k_2k_1^{-1}]
[k_1k_2k_1^{-1},h_2]
[h_2,[k_1,k_2]]
[k_1,k_2].
\end{align*}
\end{proof}
\end{comment}
By Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique4} and connectedness,
$R\subseteq [G,G]$.
Since $R\mathrel{\unlhd} G$, we have $[R,R],[R,L]\subseteq R$ and hence
$$
[G,G]
= R\cdot [G,G]
= R\cdot [R,R]\cdot [R,L]\cdot [L,L]
= R\cdot [L,L].$$
As $R$ consists of unipotent elements and
$\mathrm{Rad}(G)\cap L$ consists of semisimple elements,
$R\cap L=\{e\}$ and hence $R\cap [L,L]=\{e\}$.
Since $R\subseteq [G,G]$ are normal subgroups of $G$,
we have $R\mathrel{\unlhd} [G,G]$.
This proves the decomposition \eqref{enit:genbracket}
as abstract groups, and hence as complex algebraic groups
(\cite[\S3.3--3.4]{ov}).
From this, we conclude
\begin{equation*}
\label{eqn:gstructure}
G=R\cdot L=R\cdot [L,L]\cdot (\mathrm{Z}(L))^0=[G,G]\cdot (\mathrm{Z}(L))^0.
\end{equation*}
Clearly, $[L,L]\subseteq [G,G]\cap L$.
If $g\in[G,G]\cap L$, then by \eqref{enit:genbracket} we have
$g=k\cdot \ell$ for $k\in R$ and $\ell\in [L,L]\subseteq L$, hence
$k=g\ell^{-1}\in L\cap R=\{e\}$ and so $g=\ell\in[L,L]$; thus $[G,G]\cap L=[L,L]$.
In particular, $[G,G]\cap (\mathrm{Z}(L))^0=
[L,L]\cap (\mathrm{Z}(L))^0$, which is finite.
This proves \eqref{enit:geng}.
By \eqref{enit:geng} and an isomorphism theorem,
$$
G/[G,G]
\cong (\mathrm{Z}(L))^0/([G,G]\cap \mathrm{Z}(L)^0),$$
and hence $\dim(\mathrm{Z}(L))=\dim(G/[G,G])$.
Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique1}
then proves \eqref{enit:gencomponents}.
Finally, \eqref{enit:genisomorphism}
may be checked on the level of Lie algebras.
By \eqref{enit:geng}, we have
${\mathfrak{g}}=[{\mathfrak{g}},{\mathfrak{g}}]\oplus {\mathfrak{z}}({\mathfrak{l}})$, where
${\mathfrak{g}}$ and ${\mathfrak{l}}$ are the Lie algebras of $G$ and $L$,
and ${\mathfrak{z}}({\mathfrak{l}})$ is the center of ${\mathfrak{l}}$.
For $\chi=(\chi_1,\ldots,\chi_r)$,
we have
$\ker(d\chi_{(e)})=[{\mathfrak{g}},{\mathfrak{g}}]$
by Theorem \ref{thm:homomorphism}\eqref{enit:1},
and hence $d\chi_{(e)}|_{{\mathfrak{z}}({\mathfrak{l}})}$ is
an isomorphism.
In fact, by Theorem \ref{thm:homomorphism}\eqref{enit:1}
the kernel in \eqref{enit:genisomorphism} is
$(\mathrm{Z}(L))^0\cap ([G,G]\cdot G_{v_0})$.
\end{proof}
\begin{remark}
An arbitrary connected complex linear algebraic group with Levi
decomposition $G=\mathrm{Rad}_u(G)\rtimes L$ has
$$[G,G]=
[\mathrm{Rad}_u(G),\mathrm{Rad}_u(G)]\cdot [\mathrm{Rad}_u(G),L]\cdot [L,L]
\subseteq \mathrm{Rad}_u(G)\rtimes [L,L],$$
and hence
$[\mathrm{Rad}_u(G),\mathrm{Rad}_u(G)]\cdot [\mathrm{Rad}_u(G),L]\subseteq \mathrm{Rad}_u(G)$;
if $G$ defines a linear free divisor, then
by
Corollary \ref{cor:ofhomomorphism2}\eqref{enit:genbracket}
these are both equalities.
\end{remark}
\begin{remark}
\label{rem:boreltori}
If $B$ is a \emph{Borel subgroup} of $[L,L]$, a maximal connected
solvable subgroup, and $T\subseteq B$ is a maximal
torus of $[L,L]$, then it follows from \cite[11.14]{borel}
that $\mathrm{Rad}_u(G)\cdot B\cdot \mathrm{Z}(L)^0$ is a Borel subgroup of $G$
and $T\cdot \mathrm{Z}(L)^0$ is a maximal torus of $G$.
In particular, the number of irreducible components is at most
the dimension of the maximal torus.
\begin{comment}
For instance, $\mathrm{Z}(L)^0$ is not a maximal torus for
\cite[Theorem 2.11(2)]{gms} or \ref{ex:complicated}.
\begin{lemma}
NODO not used.
Let $\phi:G\to H$ be a surjective morphism of algebraic groups with
$\ker(\phi)$ contained in $\mathrm{Rad}(G)$, i.e., in every Borel subgroup.
If $B_G$ is a Borel subgroup of $G$, then $\phi(B_G)$ is a Borel
subgroup of $H$;
the same holds for maximal tori.
If $B_H$ is a Borel subgroup of $H$, then $\phi^{-1}(B_H)$ is a Borel
subgroup of $G$;
if $\ker(\phi)$ is a torus in $\mathrm{Z}(G)$
then the same is true of maximal tori.
\end{lemma}
\begin{proof}
The first statement (for Borel and for tori)
is essentially \cite[11.14(1)]{borel}, although that also states that
every Borel subgroup of $H$ arises in this way.
Thus for the second statement, there exists some Borel $B_G$ in $G$
with $\phi(B_G)=B_H$. Since $\ker(\phi)\subseteq B_G$,
it follows that $B_G=\phi^{-1}(B_H)$.
We now prove that every maximal torus in $H$ is the image of a maximal
torus in $G$.
Now let $T_H$ be a maximal torus of $H$. Find a $B_H$ with maximal
torus $T_H$. Then $B_G=\phi^{-1}(B_H)$ is a Borel in $G$ with a
maximal torus $T_G$. By \cite[11.14(1)]{borel}, $\phi(T_G)$ is a
maximal torus in $B_H$, and hence is conjugate in $B_H$ to $T_H$,
i.e., there exists a $\lambda \in B_H$ with
$\lambda\phi(T_G)\lambda^{-1}=T_H$.
Choose $\tau\in B_H$ with $\phi(\tau)=\lambda$. Then
$\phi(\tau T_G \tau^{-1})=T_H$, and $\tau T_G \tau^{-1}$ is a maximal
torus in $B_G$ and $G$.
Let $T_H$ be a maximal torus in $H$ and find a maximal torus $T_G$ in
$G$ with $\phi(T_G)=T_H$.
Then $\phi^{-1}(T_H)=T_G\cdot \ker(\phi)$.
Since a torus in $\mathrm{Z}(G)$ is contained in every maximal torus,
$\ker(\phi)\subseteq T_G$ and hence $\phi^{-1}(T_H)=T_G$.
\end{proof}
\begin{lemma}
Let $B\subseteq [L,L]$ be a Borel subgroup and let $T$ be a maximal
torus in $B$.
Then
$\mathrm{Rad}_u(G)\cdot B\cdot \mathrm{Z}(L)^0$ is a Borel subgroup in $G$ and
$T\cdot \mathrm{Z}(L)^0$ is a maximal torus in $G$.
\end{lemma}
\begin{proof}
By \cite[11.14(2)]{borel} there exists a Borel subgroup $B_L$ and
maximal torus $T_L$ of $L$
such that $B=(B_L\cap [L,L])^0$ and $T=(T_L\cap [L,L])^0$.
Since $T_L\cdot \mathrm{Z}(L)^0$ is a torus,
$\mathrm{Z}(L)^0\subseteq T_L\subseteq B_L$.
and hence
\begin{equation}
\label{eqn:gpcenter}
\mathrm{Z}(L)^0\cdot T\subseteq T_L\qquad \text{and}\qquad
\mathrm{Z}(L)^0\cdot B\subseteq B_L.
\end{equation}
Note that $\mathrm{Z}(L)^0\cap T\subseteq \mathrm{Z}(L)^0\cap B\subseteq
\mathrm{Z}(L)^0\cap [L,L]$ is finite.
Applying \cite[11.14(1)]{borel} to
the maps $L\to L/\mathrm{Z}(L)^0\cong [L,L]/(\mathrm{Z}(L)^0\cap [L,L])$
and $[L,L]\to [L,L]/(\mathrm{Z}(L)^0\cap [L,L])$, we see that
$\dim(B_L)=\dim(B)+\dim(\mathrm{Z}(L)^0)$
and $\dim(T_L)=\dim(T)+\dim(\mathrm{Z}(L)^0)$.
By connectedness, the containments of \eqref{eqn:gpcenter} are equalities.
By \cite[11.14(2)]{borel} there exists a Borel subgroup $B_G$ and a
maximal torus $T_G$ of $G$ such that
$B_L=(B_G\cap L)^0$ and $T_L=(T_G\cap L)^0$.
Since $\mathrm{Rad}_u(G)$ is contained in every Borel subgroup,
$\mathrm{Rad}_u(G)\cdot B_L\subseteq B_G$.
By \cite[11.14(1)]{borel} these are of the same dimension and hence
$B_G=\mathrm{Rad}_u(G)\cdot B_L$.
If $\dim(T_L)<\dim(T_G)$, then
by applying \cite[11.14(1)]{borel} to $G\to G/\mathrm{Rad}_u(G)\cong L$
we see that $T_G\cap \mathrm{Rad}_u(G)$ is nontrivial, but this contradicts
the Jordan decomposition. Hence $T_L=T_G$.
\end{proof}
\end{comment}
\end{remark}
Consider the following examples of linear free divisors.
\begin{example}
\label{ex:aacpart2}
We continue Example \ref{ex:aac}.
Since $G$ is solvable, $[L,L]$ is trivial and hence
$\mathrm{Rad}_u(G)$ is defined by $a=c=1$,
and a maximal torus $L=\mathrm{Z}(L)$ is defined by $b=0$.
Corollary \ref{cor:ofhomomorphism2} is easy to check;
in particular, the $2$-element subgroup
$L\cap G_{v_0}$ lies in the kernel of
$(\chi_1,\chi_2):G\to{\mathbb{G}_\mathrm{m}}^2$ restricted to $(\mathrm{Z}(L))^0$.
\end{example}
The following example is neither reductive nor solvable.
\begin{example}[{\cite[Example 9.4]{DP-matrixsingI}}]
\label{ex:complicated}
Define the algebraic group
$$
G=\left\{\left(\begin{smallmatrix}
a & 0 & 0 & 0 \\
0 & b & c & 0 \\
0 & d & e & 0 \\
f & g & h & i\end{smallmatrix}\right)\in{\mathrm{GL}}({\mathbb C}^4)\right\}.$$
Let $S$ be the space of $4\times 4$ symmetric matrices with
the usual coordinates
$x_{ij}$, $1\leq i\leq j\leq 4$.
Let $V\subset S$ be the subspace where $x_{11}=0$.
Let $\rho:G\to{\mathrm{GL}}(V)$ be defined by $\rho(A)(M)=AMA^T$.
Note that $\ker(\rho)=\{\pm I\}$.
A Levi decomposition of $G$ has
$L$ defined by $f=g=h=0$
and $\mathrm{Rad}(G)$ defined by $c=d=b-e=0$
(the Borel subgroup of lower-triangular $g\in G$ is not normal in $G$).
Then
$\mathrm{Rad}_u(G)$ is defined by $a=b=e=i=1$ and $c=d=0$,
$[L,L]$ is defined by $f=g=h=0$ and $be-cd=a=i=1$,
and
$L\cap \mathrm{Rad}(G)=\mathrm{Z}(L)^0$ is defined by $b=e$ and $c=d=f=g=h=0$.
Finally,
$[G,G]$ is defined by
$a=be-cd=i=1$.
The exceptional orbit variety is the linear free divisor
defined by $f_1\cdot f_2\cdot f_3$,
where
$$
f_1=\begin{vmatrix}
x_{22} & x_{23} \\
x_{23} & x_{33}
\end{vmatrix},
\qquad
f_2=
\begin{vmatrix}
0 & x_{12} & x_{13} \\
x_{12} & x_{22} & x_{23} \\
x_{13} & x_{23} & x_{33} \end{vmatrix},
\qquad
f_3=
\begin{vmatrix}
0 & x_{12} & x_{13} & x_{14} \\
x_{12} & x_{22} & x_{23} & x_{24} \\
x_{13} & x_{23} & x_{33} & x_{34} \\
x_{14} & x_{24} & x_{34} & x_{44} \end{vmatrix},
$$
corresponding to the characters
$\chi_1=(cd-be)^2$,
$\chi_2=a^2(cd-be)^2$,
$\chi_3=a^2(cd-be)^2i^2$, respectively.
Let
$$
v_0=
\left(\begin{smallmatrix}
0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{smallmatrix}\right),
\qquad
v_1=
\left(\begin{smallmatrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{smallmatrix}\right),
\qquad
v_2=
\left(\begin{smallmatrix}
0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 \\
1 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \end{smallmatrix}\right),
\qquad
v_3=
\left(\begin{smallmatrix}
0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \end{smallmatrix}\right)
\in V
$$
be generic points in $\Omega$ and on each $V(f_i)$.
\begin{comment}
\begin{verbatim}
v={{0,1,0,0},{1,1,0,0},{0,0,1,0},{0,0,0,1}}; mvm=m . v . Transpose[m] - v; Solve[mvm[[1,1]]==0 && mvm[[1,2]]==0 && mvm[[1,3]]==0 && mvm[[1,4]]==0 && mvm[[2,2]] ==0 && mvm[[2,3]]==0 && mvm[[2,4]]==0 && mvm[[3,3]]==0 && mvm[[3,4]]==0 && mvm[[4,4]]==0,{a,b,c,d,e,f,g,h,i},Complexes]
\end{verbatim}
\end{comment}
Then $G_{v_0}$, $G_{v_1}$, $G_{v_2}$, $G_{v_3}$ respectively
consist of all elements of $G$ of the form
$$
\left(
\begin{smallmatrix}
\gamma & 0 & 0 & 0 \\
0 & \gamma & 0 & 0 \\
0 & 0 & \delta & 0 \\
0 & 0 & 0 & \epsilon
\end{smallmatrix}\right),
\qquad
\left(
\begin{smallmatrix}
\frac{1}{a} & 0 & 0 & 0 \\
0 & a & 0 & 0 \\
0 & 0 & \delta & 0 \\
0 & 0 & 0 & \epsilon
\end{smallmatrix}\right),
\qquad
\left(
\begin{smallmatrix}
\frac{1}{a} & 0 & 0 & 0 \\
0 & \frac{1}{a} & 0 & 0 \\
0 & 0 & a & 0 \\
0 & 0 & 0 & a
\end{smallmatrix}\right),
\qquad
\left(
\begin{smallmatrix}
\epsilon & 0 & 0 & 0 \\
0 & \epsilon & 0 & 0 \\
0 & 0 & \delta & 0 \\
0 & 0 & 0 & a
\end{smallmatrix}\right),
$$
where $\gamma^2=\delta^2=\epsilon^2=1$ and $a\in{\mathbb C}$.
\begin{comment}
$$\begin{pmatrix}
a & 0 & 0 & 0 \\
0 & b & c & 0 \\
0 & d & e & 0 \\
f & g & h & i \end{pmatrix}$$
Then $G_{v_0}$
is 8 points, defined by
$a=b=\pm 1$,
$e=\pm 1$,
$i=\pm 1$,
$c=d=f=g=h=0$;
$G_{v_1}$ is defined by
$a=\frac{1}{b}$,
$c=d=f=g=h=0$,
$e=\pm 1$,
$i=\pm 1$;
$G_{v_2}$ by
$a=b=\frac{1}{e}$,
$c=d=f=g=h=0$,
$i=e$;
and
$G_{v_3}$ by
$a=b=\pm 1$,
$e=\pm 1$,
$c=d=f=g=h=0$,
$i\neq 0$.
\end{comment}
With these calculations,
it is
straightforward to check
the conclusions of
Theorems \ref{thm:homomorphism}
and \ref{thm:ofhomomorphism} and Corollary
\ref{cor:ofhomomorphism2}.
\end{example}
\subsection{The structure of the Lie algebra}
\label{subsec:liealgebra}
We now summarize our results for the structure of the Lie algebra ${\mathfrak{g}}$ of $G$.
In contrast to the terminology for groups,
the usual \emph{Levi decomposition} of ${\mathfrak{g}}$
expresses ${\mathfrak{g}}$ as the semidirect sum of the \emph{radical}
${\mathfrak{rad}\,\g}$,
defined as
the maximal solvable ideal,
and a semisimple \emph{Levi subalgebra};
these correspond to the Lie algebras
of $\mathrm{Rad}(G)$ and $[L,L]$, respectively.
\begin{proposition}
Let $(G,\rho,V)$ define a linear free divisor $D\subset V$.
Let $G$ have a Levi decomposition as above, and let
${\mathfrak{g}}$, ${\mathfrak{r}}$, and ${\mathfrak{l}}$ be the Lie algebras of $G$, $\mathrm{Rad}_u(G)$, and
$L$, respectively, and let ${\mathfrak{z}}({\mathfrak{l}})$ denote the center of ${\mathfrak{l}}$. Then
as vector spaces
${\mathfrak{g}}={\mathfrak{r}}\oplus {\mathfrak{z}}({\mathfrak{l}})\oplus [{\mathfrak{l}},{\mathfrak{l}}]$
where
the ideal ${\mathfrak{r}}$ consists of nilpotent elements,
${\mathfrak{z}}({\mathfrak{l}})$ is abelian and consists of semisimple elements,
$[{\mathfrak{l}},{\mathfrak{l}}]$ is semisimple,
the ideal $[{\mathfrak{g}},{\mathfrak{g}}]={\mathfrak{r}}\oplus [{\mathfrak{l}},{\mathfrak{l}}]$,
the ideal ${\mathfrak{r}}\oplus {\mathfrak{z}}({\mathfrak{l}})$ equals the radical of ${\mathfrak{g}}$,
and
${\mathfrak{l}}={\mathfrak{z}}({\mathfrak{l}})\oplus [{\mathfrak{l}},{\mathfrak{l}}]$.
We thus have:
$$
\begin{matrix}
[{\mathfrak{r}},{\mathfrak{r}}]\subseteq {\mathfrak{r}}
& [{\mathfrak{r}},[{\mathfrak{l}},{\mathfrak{l}}]]\subseteq {\mathfrak{r}}
& [{\mathfrak{r}},{\mathfrak{z}}({\mathfrak{l}})]\subseteq {\mathfrak{r}} \\
& [[{\mathfrak{l}},{\mathfrak{l}}],[{\mathfrak{l}},{\mathfrak{l}}]]=[{\mathfrak{l}},{\mathfrak{l}}]
& [[{\mathfrak{l}},{\mathfrak{l}}],{\mathfrak{z}}({\mathfrak{l}})]=0 \\
& & [{\mathfrak{z}}({\mathfrak{l}}),{\mathfrak{z}}({\mathfrak{l}})]=0.
\end{matrix}$$
\end{proposition}
\begin{proof}
Most of this follows from the algebraic Levi decomposition of $G$.
Since
$[L,L]$ is semisimple and $\mathrm{Rad}_u(G)$ is unipotent,
their Lie algebras are semisimple and nilpotent, respectively.
Since ${\mathfrak{r}}$ is nilpotent, all of its elements are nilpotent.
Then apply Corollary \ref{cor:ofhomomorphism2}\eqref{enit:genbracket}.
\end{proof}
\begin{remark}
Theorem 6.1 of \cite{gmns} describes a
normal form for a basis of the logarithmic vector
fields of a linear free divisor.
A key ingredient is a maximal subspace of simultaneously
diagonalizable linear logarithmic vector fields,
i.e., the vector fields corresponding to the
Lie algebra ${\mathfrak{t}}$ of a maximal torus.
By Remark \ref{rem:boreltori}, ${\mathfrak{t}}$ may always be chosen to
contain ${\mathfrak{z}}({\mathfrak{l}})$.
\begin{comment}
Much of this may be recovered in the following way.
Let $B$ be a Borel subgroup of $L$ and let
$B'$ be its opposite Borel subgroup
(see \cite[14.1 Cor. 1]{borel}).
Then $T=B\cap B'$ is a maximal torus of $L$, and hence of $G$.
Let $B_u$ and $B_u'$ be the unipotent subgroup of $G$.
Note that
${\mathfrak{g}}={\mathfrak{r}}\oplus {\mathfrak{b}}_u\oplus {\mathfrak{t}}\oplus {{\mathfrak{b}}}_u'$, with
${\mathrm{ad}}({\mathfrak{t}})$ leaving invariant ${\mathfrak{r}}$, ${\mathfrak{b}}_u$, and ${\mathfrak{b}}_u'$.
The adjoint representation of $G$ on ${\mathfrak{g}}$, restricted to $T$,
is a torus action and hence diagonalizes.
Then ${\mathfrak{r}}$, ${\mathfrak{b}}_u$, and ${\mathfrak{b}}_u'$ are direct sums of lines invariant
under $T$, and a corresponding basis
shows that there is a basis
$\sigma_0,\sigma_1,\ldots,\sigma_s,\nu_1,\ldots,\nu_{n-s-1}$
of ${\mathfrak{g}}$ with $\sigma_i$ semisimple, $\nu_i$ nilpotent,
$[\sigma_i,\nu_j]\in {\mathbb C}\cdot \nu_j$,
and $\sigma_0,\ldots,\sigma_s$ a basis for ${\mathfrak{t}}$.
NODO how $\mathbb{Q}$, and how that formula?
\end{comment}
\end{remark}
\subsection{Some special cases}
\label{subsec:specialcases}
We now apply our results to several special types of linear free divisors.
\subsubsection{Abelian groups}
The \emph{normal crossings divisor} in a vector space $V$ is
given by the union of all coordinate hyperplanes for some choice of
vector space coordinates.
It is a linear free divisor,
and is the only linear free divisor defined by an abelian group.
\begin{corollary}[{\cite[Theorem 2.12]{freedivisorsinpvs}}]
Let $V$ be a finite-dimensional complex vector space and suppose that
a connected complex linear algebraic group $G\subseteq {\mathrm{GL}}(V)$ defines
a linear free divisor $D$ in $V$.
Then $G$ is abelian if and only if $D$ is
equivalent, under a change of coordinates in $V$, to the normal crossings divisor.
\end{corollary}
\begin{proof}
Let $n=\dim(G)=\dim(V)$.
If $D$ is the normal crossings divisor, then after choosing a basis of
$V$, $G$ is the diagonal group in ${\mathrm{GL}}(V)$, isomorphic to
$({\mathbb{G}_\mathrm{m}})^{n}$. Thus $G$ is abelian.
If $G$ is abelian,
then by Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique1}, $D$ has
$\dim(G)=n$
irreducible components.
By Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique5},
$G$ is isomorphic to $({\mathbb{G}_\mathrm{m}})^n$, and hence is a maximal torus in
${\mathrm{GL}}(V)$.
As all such tori are conjugate in ${\mathrm{GL}}(V)$,
we may choose coordinates on $V$ so that $G$ is the group of diagonal
matrices.
A calculation then shows that $D$ is the normal crossings divisor.
\end{proof}
\begin{comment}
(By Corollary \ref{cor:ofhomomorphism2}\eqref{enit:geng},
$G$ is an algebraic torus.)
Since the irreducible components $D_1,\ldots,D_n$ of $D$ are defined by homogeneous polynomials of
positive degree, each $D_i$ is a hyperplane through $0$ defined by a linear
form.
By the following Lemma, $D$ must be equivalent to the normal crossings
divisor.
\end{proof}
\begin{lemma}
If a central hyperplane arrangement $H=\cup_{i=1}^n H_i$ in $V$, $n=\dim(V)$,
is a linear free divisor, then $\cap_{i=1}^n H_i=\{0\}$ and $H$ is
equivalent to the normal crossings divisor.
\end{lemma}
\begin{proof}
Let $I=\cap_{i=1}^n H_i$, and let each $\alpha_i$ be a
linear form defining $H_i$.
Since $I$ is the kernel of
$\alpha=(\alpha_1,\ldots,\alpha_n):V\to{\mathbb C}^n$,
$\alpha$ is invertible if and only if $I=\{0\}$.
Suppose that $I\neq\{0\}$,
so that $I$ contains a line $L=\{\lambda v:\lambda\in{\mathbb C}\}$ for some nonzero
$v\in V$. Define the vector field $\xi(x)=v$ (identifying
$T_x V$ with $V$) on $V$. Since each hyperplane $D_i$ contains the
line $L$, $\xi$ is tangent to each $D_i$
(since, e.g., $\alpha_i(p+\lambda
v)=\alpha_i(p)+\lambda\alpha_i(v)=\alpha_i(p)+\lambda\cdot 0$).
Thus $\xi$ is tangent to $D=\cup_{i=1}^n D_i$, so
$\xi\in\Derlog{V}{D}$. Since $\xi$ has degree $-1$, $\xi$ cannot
be expressed in terms of linear vector fields,
contradicting the assumption that $D$ is a linear free divisor.
Thus $I=\{0\}$, $\alpha$ is invertible, and
in using $\alpha$ to identify $V$ with ${\mathbb C}^n$ we see that $D$
is the normal crossings divisor.
\end{proof}
\end{comment}
\subsubsection{Irreducible linear free divisors}
We now examine the groups that
produce irreducible
linear free divisors.
Observe that if $G\subseteq {\mathrm{GL}}(V)$ defines a linear free divisor
$D\subset V$, then $\lambda\cdot I\in ({\mathrm{GL}}(V)_D)^0=G$
for $I$ the identity element and
$\lambda\in{\mathbb C}^*$.
Recall that an algebraic group $H$ is called \emph{perfect} if
$[H,H]=H$. For instance, semisimple groups are perfect.
\begin{corollary}
\label{cor:irreducible}
Let $D\subset V$ be a linear free divisor defined by the group
$G\subseteq {\mathrm{GL}}(V)$.
Let $H=G\cap {\mathrm{SL}}(V)$, and let $K=({\mathbb C}^*)\cdot I\subseteq G$.
The following are equivalent:
\begin{enumerate}
\item
\label{en:1component}
$D$ has $1$ irreducible component.
\item
\label{en:h0gg}
$H^0=[G,G]$.
\item
\label{en:gkgg}
$G=K\cdot [G,G]$.
\item
\label{en:h0perfect}
$H^0$ is perfect.
\item
\label{en:isperfect}
There exists a perfect connected codimension $1$ algebraic subgroup
$J$ of $G$.
\end{enumerate}
When these hold, $J=[G,G]=H^0$.
\end{corollary}
\begin{proof}
Clearly $G=K\cdot H$, and there are a finite number of ways to write
$g\in G$ as a product of elements of $K$ and $H$.
It follows that $\dim(H)=n-1$.
The multiplication morphism $K\times H^0\to G$ has a connected image
of dimension $\dim(G)$, and hence
$G=K\cdot H^0$.
If
\eqref{en:1component}, then
by Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique1},
$\dim([G,G])=n-1$.
Since $[G,G]\subseteq H$ are of the same dimension
and $[G,G]$ is connected, we have $[G,G]=H^0$ and \eqref{en:h0gg}.
If \eqref{en:h0gg}, then by the above work, $G=K\cdot [G,G]$, giving
\eqref{en:gkgg}.
Suppose \eqref{en:gkgg}.
Since $[G,G]\subseteq H$ we have
$\dim([G,G])\leq n-1$.
Also,
$\dim(G)\leq \dim(K)+\dim([G,G])$
shows $\dim([G,G])\geq n-1$, and hence
$\dim([G,G])=n-1$.
Since they are connected and of the same dimension, $[G,G]=H^0$.
Since $K$ is in the center of $G$,
we have $[K\cdot N,K\cdot M]=[N,M]$ for any subgroups $N$ and $M$ of $G$.
In particular,
\begin{equation*}
H^0=[G,G]=[K\cdot [G,G],K\cdot [G,G]]=[[G,G],[G,G]]=[H^0,H^0],
\end{equation*}
giving \eqref{en:h0perfect}.
If \eqref{en:h0perfect}, then since $\dim(H^0)=n-1$,
we have \eqref{en:isperfect}.
If \eqref{en:isperfect}, then
since
$J=[J,J]\subseteq [G,G]\subseteq H$ and
$\dim(H)=n-1$,
we have $\dim([G,G])=n-1$.
By Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique1},
$D$ has $1$ irreducible component, proving \eqref{en:1component}.
Finally, note that $J\subseteq [G,G]\subseteq H^0$ are all
connected algebraic groups of the same dimension, hence equal.
\end{proof}
\begin{remark}
The case when $H$ is semisimple was thoroughly explored
in \cite{freedivisorsinpvs};
by the Levi decomposition of $G$, this is equivalent to
$\dim(\mathrm{Z}(L))=1$ and $\mathrm{Rad}_u(G)=\{e\}$.
Are there other irreducible linear free divisors, with $H^0$ perfect and
$\mathrm{Rad}_u(G)\neq \{e\}$?
Since an irreducible representation that is a
prehomogeneous vector space has
$H$ semisimple
by \cite[Theorem 7.21]{kimura},
in such an example the representation must be reducible.
\end{remark}
\begin{comment}
The Lie algebra version of Corollary \ref{cor:irreducible} is that
${\mathfrak{g}}$ is the direct sum of a 1--dimensional
ideal and a perfect ideal ${\mathfrak{h}}$, or
that ${\mathfrak{g}}$ contains a codimension $1$ perfect subalgebra.
\end{comment}
\subsubsection{Reductive groups}
For reductive groups, we have the following.
\begin{corollary}[{\cite[Lemma 2.6]{freedivisorsinpvs}}]
\label{cor:reductive}
For a linear free divisor $D$ defined by a reductive group $G$, the number of
irreducible components of $D$ equals the dimension of the center of
$G$.
Let $H$ be the subgroup of $G$ leaving invariant the level sets of
the product
$f_1\cdots f_r$.
Then
$D$ is irreducible if and only if $H^0$ is semisimple.
\end{corollary}
\begin{proof}
Let $G$ have a Levi decomposition.
Since $\mathrm{Rad}_u(G)$ is trivial and $G$ is itself a Levi subgroup,
apply Corollary \ref{cor:ofhomomorphism2}\eqref{enit:gencomponents}
to get the first statement.
If $D$ is irreducible, then $r=1$ and
so by Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique3},
$H=[G,G]\cdot G_{v_0}$. In particular, $H^0=[G,G]$,
and this is semisimple by the structure theory.
Conversely, suppose $H^0$ is semisimple.
Since $H=\ker(\chi_1\cdots \chi_r)$,
by Theorem \ref{thm:homomorphism}\eqref{enit:1},
$H^0$ has codimension $1$ in $G$.
Since $H^0$ is perfect, by Corollary \ref{cor:irreducible},
$D$ is irreducible.
\end{proof}
Granger--Mond--Schulze also show (\cite[Theorem
2.7]{freedivisorsinpvs})
that for a linear free divisor $D\subset V$ defined by a reductive
group $G$,
the number of irreducible hypersurface components of $D$ equals
the number of irreducible $G$--modules in $V$.
\begin{comment}
NODO: follows from Schur's lemma?
No. They use Schur's lemma in one direction but don't say what
they're using.
\end{comment}
\begin{example}
Let $D\subset V$ be a linear free divisor constructed from a
\emph{quiver} $Q$,
a finite connected oriented graph with vertex set $Q_0$,
edge set $Q_1$, and a dimension vector $d:Q_0\to {\mathbb N}$
(see \cite{gmns,mondbuchweitz,freedivisorsinpvs}).
\begin{comment}
A vector space of dimension $d(v)$ is attached to
each vertex $v\in Q_0$.
The \emph{representation space} $\textrm{Rep}(Q,d)$ is the product, over $e\in
Q_1$, of the
${\mathrm{Hom}}_{{\mathbb C}}(--,--)$'s between the vector spaces corresponding to
the source and the target of $e$.
The \emph{quiver group} $G$ is the product, over $v\in Q_0$, of
${\mathrm{GL}}(d(v),{\mathbb C})$, and
$G$ acts on $\textrm{Rep}(Q,d)$ by a change of variables in the appropriate
source or target spaces of the ${\mathrm{Hom}}$'s.
When $d$ is a \emph{real Schur root}, then
there is an open orbit by \cite[Theorem 4.1]{gmns}
\end{comment}
Here, the group $G$ is a product over $Q_0$ of general linear
groups,
so $G$ is reductive with $\dim(\mathrm{Z}(G))=|Q_0|$,
and $V$ is the space of representations of $(Q,d)$.
When $d$ is a real Schur root and $Q$ has no oriented cycles, then
$G$ has an open orbit and
a theorem of Kac states that
the complement has
$|Q_0|-1$ irreducible hypersurface components
(\cite[\S4]{gmns}).
The apparent disagreement with
Corollary \ref{cor:reductive} is resolved by
observing that $G$ does not define $D$ in the sense of
Definition \ref{def:defineslfd} as
the representation $\rho$ of $G$ has a
$1$-dimensional kernel contained in $\mathrm{Z}(G)$,
whereas $\rho(G)$
defines $D$ and
is
reductive with center of dimension $|Q_0|-1$.
\end{example}
\subsubsection{Solvable groups}
\begin{comment}
Notes to myself:
If $G\subset L_n$, then $G_u=G\cap U_n$.
BUT, not necessarily true that $G\cap D_n$ is a maximal torus (though I don't have a counterexample.) Must be more careful of the basis.
TODO: idea: Z(G) is reductive, hence $\rho|_{Z(G)}$ is reducible. Use
this to prove all solvable lfds come from block representations?
\end{comment}
Recall that by
the Lie--Kolchin Theorem,
any solvable linear algebraic group $G\subset {\mathrm{GL}}(V)$
has a basis of $V$
in which
$G$ is lower triangular.
\begin{corollary}
\label{cor:solvable}
Let $D\subset V$ be a linear free divisor defined by a solvable group
$G\subseteq {\mathrm{GL}}(V)$.
Fix any basis which makes $G$ lower triangular, and
let $\phi:G\to ({\mathbb{G}_\mathrm{m}})^{\dim(V)}$ send $g$ to the diagonal entries of
$g$.
Then:
\begin{enumerate}
\item
\label{enit:solvablenum}
The number of components of $D$ equals the dimension of the maximal torus
of $G$, and also $\dim(\phi(G))$.
\item
\label{enit:solvablehyp}
$D$ has a hyperplane component.
\item
\label{enit:solvableunip}
Let $G_u$ be the subgroup consisting of unipotent elements of $G$.
Then $[G,G]=G_u=\ker(\phi)$.
\item
\label{enit:solvablefactor}
Every $\chi\in X_1(G)$ factors through $\phi$.
\end{enumerate}
\end{corollary}
\begin{proof}
At first we proceed without the hypothesis that $G$ defines a linear
free divisor.
Since $G$ is connected and solvable, the Levi decomposition of $G$ has
$\mathrm{Rad}(G)=G$, $\mathrm{Rad}_u(G)=G_u$, $L$ is a
maximal torus of $G$, and
in this case $[G,G]\subseteq G_u$.
Note that $\phi$ is a homomorphism of linear algebraic groups, and by
the definition of unipotent, $G_u=\ker(\phi)$.
Then since $\dim(G)=\dim(G_u)+\dim(L)=\dim(\ker(\phi))+\dim(L)$,
we have
\begin{equation}
\label{eqn:imphistar}
\dim(L)=\dim(\phi(G)).
\end{equation}
Now let $\chi\in X(G)$. By the Jordan decomposition, $G_u\subseteq
\ker(\chi)$, and hence $\ker(\phi)\subseteq \ker(\chi)$.
If $\overline{\chi}$ and $\overline{\phi}$ are the induced
homomorphisms on $G/G_u$, then the homomorphism
$\lambda=\overline{\chi}\circ (\overline{\phi})^{-1}:\phi(G)\to {\mathbb{G}_\mathrm{m}}$
satisfies $\chi=\lambda\circ \phi$.
Since
$\lambda$
is a character on a subtorus of $({\mathbb{G}_\mathrm{m}})^{\dim(V)}$,
by \cite[8.2]{borel}
$\lambda$ extends to a character $\psi:({\mathbb{G}_\mathrm{m}})^{\dim(V)}\to {\mathbb{G}_\mathrm{m}}$
with $\chi=\psi\circ \phi$.
This proves \eqref{enit:solvablefactor}.
Now assume that $G$ defines a linear free divisor $D$.
By Corollary \ref{cor:ofhomomorphism2}\eqref{enit:gencomponents},
the number of components of $D$ equals $\dim(\mathrm{Z}(L))$.
Then the observation that $L=\mathrm{Z}(L)$ and \eqref{eqn:imphistar} implies
\eqref{enit:solvablenum}.
By Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique4} we have
$G_u\subseteq [G,G]$, and hence $[G,G]=G_u=\ker(\phi)$, proving
\eqref{enit:solvableunip}.
Finally,
the Lie--Kolchin Theorem guarantees an invariant complete flag in $V$,
hence an invariant hyperplane $H$.
As $H$ cannot
intersect $\Omega$, $H$ is a component of $D$, proving
\eqref{enit:solvablehyp}.
In these lower-triangular coordinates, this $H$ is defined by the
``first coordinate''.
\end{proof}
\begin{remark}
By Corollary \ref{cor:solvable}\eqref{enit:solvablefactor}, the characters
corresponding to the basic relative invariants
are functions of the
diagonal entries.
\end{remark}
\begin{comment}
NODO: big comment here
Q: When doing classification, can we send elts not in bracket to
diagonal?
In fact, we can say more.
NODO: is this useful?
\begin{lemma}
\label{lemma:niceform}
Let $G\subset {\mathrm{GL}}(V)$ be a connected, solvable, algebraic subgroup.
Fix a basis of $V$, let $L$ be the group of invertible
lower-triangular matrices, and let $D$ (resp., $U$) be the subgroup
of $L$ of diagonal (resp., unipotent) matrices.
There exists an $g\in {\mathrm{GL}}(V)$ such that
$H=gGg^{-1}\subseteq L$,
$H\cap U$ is the subgroup of unipotent elements of $H$,
and $H\cap D$ is a maximal torus of $H$.
\end{lemma}
\begin{proof}
By the Lie--Kolchin Theorem, there exists a $a\in {\mathrm{GL}}(V)$ such that
$aGa^{-1}\subseteq L$.
Let $T_G$ be a maximal torus of $aGa^{-1}$.
Since $T_G$ is contained in a maximal torus $T_L$ of $L$,
and $T_L$ must be conjugate to the maximal torus $D$ in $L$,
there exists a
$b\in L$ such that $bT_Gb^{-1}\subseteq D$ is a maximal torus of
$baG(ba)^{-1}\subseteq L$.
Let $g=ba$.
Finally, for any connected algebraic $K\subseteq L$,
the set of unipotent elements of $K$ is
$K\cap U$.
\end{proof}
With this, then generic isotropy group for NODO is ???
\begin{lemma}
\label{lemma:flags}
Let $V_0\subset V_1\subset \cdots \subset V_n$ be a complete flag with
$\dim(V_i)=i$, let $W$ be a subspace of $V_n$, and
let $k_i=\dim(W\cap V_i)$. Then
$k_{i+1}-1\leq k_i\leq k_{i+1}$; that is,
at each step in
$W\cap V_0\subseteq W\cap V_1\subseteq \cdots \subseteq W\cap V_n=W$,
the dimension increases, by at most $1$.
\end{lemma}
\begin{proof}
Clearly $k_i\leq k_{i+1}$.
Moreover,
\begin{align*}
k_{i+1}-k_i
&=
(\dim(W)+i+1-\dim(W+V_{i+1}))-(\dim(W)+i-\dim(W+V_i)) \\
&=1+(\dim(W+V_i)-\dim(W+V_{i+1})\leq 1,
\end{align*}
since $W+V_i\subseteq W+V_{i+1}$.
\end{proof}
NODO: if $G_{v_0}$ is trivial, then this would be much easier to
explain.
\begin{corollary}
If $G$ is solvable and produces a linear free divisor (NODO: clean
this up), then the Lie algebra ${\mathfrak{g}}$ of $G$ is a semidirect sum
of the ideal $[{\mathfrak{g}},{\mathfrak{g}}]$ and an abelian subalgebra $\mathfrak{a}$.
\end{corollary}
\begin{proof}
$G$ is the semidirect product of $G_u$ and a torus $G_s$.
Since by NODO the identity component of $G_u$ is the same as the
identity component of $[G,G]$, the Lie algebra of $G_u$ is $[{\mathfrak{g}},{\mathfrak{g}}]$.
Since $G_s$ is a torus, its Lie algebra $\mathfrak{a}$ is abelian.
\end{proof}
In particular, the representation of $\mathfrak{a}$ on automorphisms
of $[{\mathfrak{g}},{\mathfrak{g}}]$ which is used to define the semidirect sum must be (in
some sense) surjective.
\end{comment}
\subsection{Degrees of the components}
\label{subsec:whatelse}
Let $G$ define a linear free divisor $D$
with basic relative invariants $f_1,\ldots,f_r$.
We have seen that $r$
may be computed from
the Lie algebra ${\mathfrak{g}}$ of $G$;
may the degrees of the $f_i$ be computed from ${\mathfrak{g}}$?
If we only use the abstract Lie algebra structure of ${\mathfrak{g}}$, then the
answer is no:
\begin{example}
Consider the following two linear free divisors in ${\mathbb C}^5$:
\begin{align*}
D_1:\quad & (x_3 x_5-x_4^2)\begin{vmatrix} 0 & x_1 & x_2 \\ x_1 & x_3 & x_4
\\ x_2 & x_4 & x_5 \end{vmatrix}=0, \\
\text{and }D_2:\quad &
(x_2^2x_3^2-4x_1x_3^3-4x_2^3x_4+18x_1x_2x_3x_4-27x_4^2x_1^2)x_5=0.
\end{align*}
This $D_1$ is \cite[Example 9.4]{DP-matrixsingI},
while $D_2$ is the product-union of a hyperplane with
\cite[Theorem 2.11(2)]{freedivisorsinpvs}.
The degrees of the polynomials defining the
irreducible components of $D_1$ and $D_2$
differ.
However,
the groups defining $D_1$ and $D_2$ have the same abstract Lie algebra
structure:
${\mathfrak{gl}}_2({\mathbb C})\oplus {\mathfrak{gl}}_1({\mathbb C})$.
Thus, $D_1$ and $D_2$ are constructed from inequivalent
representations of the same abstract Lie algebra.
\begin{comment}
\begin{verbatim}
needsPackage "VectorFields";
R=QQ[x1,x2,x3,x4,x5];
f=(x3*x5-x4^2)*det(matrix {{0,x1,x2},{x1,x3,x4},{x2,x4,x5}});
D=derlogV(ideal (f));
LV1n1=matrix {{x1}, {0}, {x3}, {0}, {-x5}};
LV1n2=matrix {{x2}, {0}, {2*x4}, {x5}, {0}};
LV2n1=matrix {{0}, {x1}, {0}, {x3}, {2*x4}};
LV2n2=matrix {{0}, {x2}, {-x3}, {0}, {x5}};
vR=matrix {{0}, {0}, {x3}, {x4}, {x5}};
assert(image matrix mutableMatrix D==image (LV1n1|LV1n2|LV2n1|LV2n2|vR));
-- check the brackets
assert(bracket(LV1n1,LV1n2)==matrix mutableMatrix (-LV1n2));
assert(bracket(LV1n1,LV2n1)==matrix mutableMatrix LV2n1);
assert(bracket(LV1n1,LV2n2)==0);
assert(bracket(LV1n2,LV2n1)==matrix mutableMatrix (LV2n2-LV1n1));
assert(bracket(LV1n2,LV2n2)==matrix mutableMatrix (-LV1n2));
assert(bracket(LV2n1,LV2n2)==matrix mutableMatrix LV2n1);
assert(bracket(LV1n1,vR)==0);
assert(bracket(LV1n2,vR)==0);
assert(bracket(LV2n1,vR)==0);
assert(bracket(LV2n2,vR)==0);
\end{verbatim}
and
\begin{verbatim}
needsPackage "VectorFields";
R=QQ[x1,x2,x3,x4,x5];
f=(x2^2*x3^2-4*x1*x3^3-4*x2^3*x4+18*x1*x2*x3*x4-27*x4^2*x1^2)*x5;
D=derlogV(ideal (f));
LV1n1=matrix {{2*x1}, {x2}, {0}, {-x4}, {0}};
LV1n2=matrix {{x2}, {2*x3}, {3*x4}, {0}, {0}};
LV2n1=matrix {{0}, {3*x1}, {2*x2}, {x3}, {0}};
LV2n2=matrix {{-x1}, {0}, {x3}, {2*x4}, {0}};
vR=matrix {{0}, {0}, {0}, {0}, {x5}};
assert(image matrix mutableMatrix D==image (LV1n1|LV1n2|LV2n1|LV2n2|vR));
-- check the brackets
assert(bracket(LV1n1,LV1n2)==matrix mutableMatrix (-LV1n2));
assert(bracket(LV1n1,LV2n1)==matrix mutableMatrix LV2n1);
assert(bracket(LV1n1,LV2n2)==0);
assert(bracket(LV1n2,LV2n1)==matrix mutableMatrix (LV2n2-LV1n1));
assert(bracket(LV1n2,LV2n2)==matrix mutableMatrix (-LV1n2));
assert(bracket(LV2n1,LV2n2)==matrix mutableMatrix LV2n1);
assert(bracket(LV1n1,vR)==0);
assert(bracket(LV1n2,vR)==0);
assert(bracket(LV2n1,vR)==0);
assert(bracket(LV2n2,vR)==0);
\end{verbatim}
\end{comment}
\end{example}
Of course, since
the representation of the Lie algebra determines the divisor,
the representation undoubtedly
contains the information necessary to compute these degrees.
How may we do so effectively?
\begin{lemma}
\label{lemma:degrees}
Let $G\subseteq {\mathrm{GL}}(V)$ define the linear free divisor $D$ with
basic relative invariants $f_1,\ldots,f_r$, with $f_i\stackrel{\mathrm{m}}{\longleftrightarrow}
\chi_i$.
Choose $X_i\in{\mathfrak{g}}\subseteq {\mathfrak{gl}}(V)$ such that $d(\chi_i)_{(e)}(X_j)=\delta_{ij}$, or
equivalently, $\xi_{X_i}(f_j)=\delta_{ij} f_j$.
Let $I$ denote the identity endomorphism in
both $G\subseteq {\mathrm{GL}}(V)$ and
${\mathfrak{g}}\subseteq{\mathfrak{gl}}(V)$, with $\xi_I$ the Euler
vector field.
Let the module $A\subseteq {\mathrm{Der}}_V$ consist of the vector fields
annihilating all $f_i$.
Then:
\begin{align}
\xi_I(f_i)&=\deg(f_i) f_i,
\label{eqn:degree3} \\
d(\chi_i)_{(e)}(I)&=\deg(f_i),
\label{eqn:degree0} \\
\chi_i(\lambda\cdot I)&=\lambda^{\deg(f_i)}\quad
\text{for all $\lambda\in{\mathbb C}^*$},
\label{eqn:degree4} \\
I&=\sum_{j=1}^r \deg(f_j) X_j\bmod{[{\mathfrak{g}},{\mathfrak{g}}]},
\label{eqn:degree1} \\
\text{and }
\xi_I&=\sum_{j=1}^r \deg(f_j) \xi_{X_j}\bmod{A}.
\label{eqn:degree2}
\end{align}
\end{lemma}
\begin{proof}
Corollary \ref{cor:homomorphismliealgebra} shows that such $X_i$
exist.
By \eqref{eqn:diffrelinv}, the $\xi_{X_i}$ have the claimed effects on $f_j$ for
all $i,j$.
Then \eqref{eqn:degree3} is just the Euler relation,
and applying \eqref{eqn:diffrelinv} shows
\eqref{eqn:degree0}.
Integrating \eqref{eqn:degree0} shows \eqref{eqn:degree4}.
By \eqref{eqn:degree0}, for each $i$ we have
$$d(\chi_i)_{(e)}\Big(I-\sum_{j=1}^r \deg(f_j)
X_j\Big)=\deg(f_i)-\deg(f_i)=0,$$
and since $\cap_{i=1}^r \ker(d(\chi_i)_{(e)})=[{\mathfrak{g}},{\mathfrak{g}}]$ by
Theorem \ref{thm:homomorphism}\eqref{enit:1},
we have \eqref{eqn:degree1}.
A similar argument using \eqref{eqn:degree3}
shows that $\xi_I-\sum_{j=1}^r \deg(f_j) \xi_{X_j}$ annihilates each $f_i$,
giving \eqref{eqn:degree2}.
\end{proof}
It is unclear whether the embedding ${\mathfrak{g}}\subseteq {\mathfrak{gl}}(V)$ may be used
to find the degrees without first finding either the $\chi_i$,
$d(\chi_i)_{(e)}$, or $f_i$.
\begin{remark}
For $d\in{\mathbb N}$ let $R_d\subseteq {\mathbb{G}_\mathrm{m}}$ denote the group of $d$th roots of
unity. Then in the situation of Lemma \ref{lemma:degrees}
with $G\subseteq {\mathrm{GL}}(V)$,
\eqref{eqn:degree4} implies that
$$
\ker(\chi_i|_{{\mathbb{G}_\mathrm{m}}\cdot I})=\ker(\chi_i)\cap ({\mathbb{G}_\mathrm{m}}\cdot I)
= R_{\deg(f_i)}\cdot I.$$
Then by Theorem \ref{thm:homomorphism}\eqref{enit:1} and number
theory,
$([G,G]\cdot G_{v_0})\cap ({\mathbb{G}_\mathrm{m}}\cdot I)=R_{d}\cdot I$ for
$d$ the greatest common divisor of
$\{\deg(f_1),\ldots,\deg(f_r)\}$.
\begin{comment}
When instead $\rho:G\to{\mathrm{GL}}(V)$ defines $D$ with $\rho$ not the
inclusion map, we must account for $\ker(\rho)$.
For instance, in Example \ref{ex:complicated},
$[G,G]\cdot G_{v_0}=\ker(\rho)$ and hence the GCD of the
degrees of the basic relative invariants is $1$.
\end{comment}
\end{remark}
\subsection{Homotopy groups of \texorpdfstring{$V\setminus D$}{V-D}}
\label{subsec:homotopy}
The results above
can
give some insight into the topology of the complement of a linear free
divisor $D$
and two types of (global) Milnor fiber associated to $D$.
\begin{proposition}
\label{prop:homotopy}
Let $G\subseteq {\mathrm{GL}}(V)$ define a linear free divisor $D\subset V$ with
irreducible components defined by $f_1,\ldots,f_r$,
and $f_i\stackrel{\mathrm{m}}{\longleftrightarrow} \chi_i$.
Let $L$ be a Levi factor of $G$.
Let $v_0\in\Omega$, let $F=(f_1,\ldots,f_r):V\to{\mathbb C}^r$, let
$K$ be the fiber of $f_1\cdots f_r$ containing $v_0$,
and let $P\subset K$ be the fiber of $F$ containing $v_0$.
Use $v_0$ or $e\in G$ as the base point for all homotopy groups.
Then
for $n>1$,
$$
\pi_n(L)\cong
\pi_n([L,L])\cong
\pi_n(G)\cong
\pi_n([G,G])\cong
\pi_n(P)\cong
\pi_n(\Omega)\cong
\pi_n(K),$$
and for $n=1$ we have the exact
diagrams in
Figure \ref{fig:pi1}.
\end{proposition}
\begin{figure}[ht]
\caption{The exact diagrams describing the fundamental groups
of the spaces in Proposition \ref{prop:homotopy}.
For a map $\phi$, let $\phi_*$ denote the associated map of
fundamental groups. Each $i$ is constructed from an inclusion map.
We do not claim commutativity of these diagrams.}
\label{fig:pi1}
\begin{gather*}
\xymatrix@-1pc{
& & & 0 \ar[d] & 0 \ar[dl] \\
& & & \pi_1(P) \ar[dl]_{i_*} \ar[d]^{i_*} \\
& 0 \ar[r] & \pi_1(K)\ar[dl]^{{p_8}_*} \ar[r]^{i_*} & \pi_1(\Omega)
\ar[d]^{{p_7}_*} \ar[rr]^-{(m\circ p_7)_*} & & {\mathbb Z}
\ar@{=}[d] \ar[r] & 0 \\
0\ar[r] & {\mathbb Z}^{r-1} \ar[dl] \ar[rr] & & {\mathbb Z}^r \ar[rr]^{m_*} \ar[d] & & {\mathbb Z} \ar[r] & 0 \\
0 & & & 0 & &
}
\\
\xymatrix@-1pc{
& 0 \ar[d] & 0 \ar[d] \\
0 \ar[r] & \pi_1([G,G]) \ar[r]^-{{p_2}_*} \ar[d]_{i_*} & \pi_1(P)
\ar[r]
\ar[d]^{i_*} &
[G,G]\cap G_{v_0} \ar[r] & 0 \\
0 \ar[r] & \pi_1(G) \ar[r]^{{p_1}_*} \ar[d]_{{p_5}_*} & \pi_1(\Omega) \ar[r] \ar[d]^{{p_7}_*} & G_{v_0} \ar[r] & 0 \\
& {\mathbb Z}^r \ar[d] & {\mathbb Z}^r \ar[d] \\
& 0 & 0
}
\\
\xymatrix@-1pc{
& 0 \ar[d]& 0\ar[d] & \\
0 \ar[r] & \pi_1([G,G]) \ar[r]^-{i_*} \ar[d]_{{p_4}_*} & \pi_1(G)
\ar[r]^-{{p_5}_*} \ar[d]^{{p_3}_*} & {\mathbb Z}^r
\ar@{=}[d] \ar[r] & 0 \\
0 \ar[r] & \pi_1([L,L]) \ar[r]_-{i_*} \ar[d] & \pi_1(L) \ar[r]_-{{p_6}_*} \ar[d]& {\mathbb Z}^r \ar[r] & 0 \\
& 0 & 0
}
\end{gather*}
\end{figure}
\begin{proof}
If $G_1\subset G_2\subset G_3$ are Lie groups, then
by \cite[\S7.4--7.5]{steenrod}
the map of cosets
$G_3/G_1\to G_3/G_2$
is a fiber bundle with fiber $G_2/G_1$.
Thus, we have the following fiber bundles:
\begin{align*}
p_1:G &\to G/G_{v_0}\cong \Omega
&
p_2:[G,G]&\to [G,G]/([G,G]\cap G_{v_0})
\\
p_3:G &\to G/\mathrm{Rad}_u(G)\cong L
&
p_4:[G,G]&\to [G,G]/\mathrm{Rad}_u(G)\cong [L,L]
\\
p_5:G&\to G/[G,G]\cong {\mathbb{G}_\mathrm{m}}^r
&
p_6:L&\to L/[L,L]\cong {\mathbb{G}_\mathrm{m}}^r \\
q:G/G_{v_0}&\to G/[G,G]\cdot G_{v_0}
\end{align*}
The identifications of the codomains
of $p_1$, $p_3$, $p_4$, $p_5$, and $p_6$
are by, respectively,
the orbit map $\alpha_{v_0}$ of $v_0$,
the definition of the Levi decomposition,
Corollary \ref{cor:ofhomomorphism2}\eqref{enit:genbracket},
Theorem \ref{thm:ofhomomorphism}\eqref{enit:unique5},
and
the isomorphism
$$L/[L,L]\cong (G/\mathrm{Rad}_u(G))/([G,G]/\mathrm{Rad}_u(G))\cong G/[G,G]$$
that follows from the Levi decomposition and
Corollary \ref{cor:ofhomomorphism2}\eqref{enit:genbracket}.
Since $G_{v_0}$ and $[G,G]\cap G_{v_0}$ are finite,
$p_1$ and $p_2$ are covering spaces with deck transformation groups
isomorphic to $G_{v_0}$ and $[G,G]\cap G_{v_0}$, respectively
(\cite[Proposition 1.40]{hatcher}).
For $q$, we use the orbit map to identify
$G/G_{v_0}$ with $\Omega$ and
then by Theorem \ref{thm:homomorphism}\eqref{enit:1}
compose with the isomorphism $G/[G,G]\cdot G_{v_0}\to ({\mathbb{G}_\mathrm{m}})^r$
induced by
$(\chi_1,\ldots,\chi_r):G\to ({\mathbb{G}_\mathrm{m}})^r$.
This gives a fiber bundle
$$p_7:\Omega\to ({\mathbb{G}_\mathrm{m}})^r$$
defined by
$p_7(\rho(g)(v_0))=(\chi_1,\ldots,\chi_r)(g)$
with fiber homeomorphic to $\ker(\chi_1,\ldots,\chi_r)/G_{v_0}=[G,G]\cdot
G_{v_0}/G_{v_0}$.
As the action of $[G,G]$ on $[G,G]\cdot G_{v_0}/G_{v_0}$
is
smooth and transitive, and the isotropy subgroup at
$e G_{v_0}$ is $[G,G]\cap G_{v_0}$,
the fiber of $p_7$ is
isomorphic to $[G,G]/([G,G]\cap G_{v_0})$.
Let $m:({\mathbb{G}_\mathrm{m}})^r\to {\mathbb{G}_\mathrm{m}}$ and $n:({\mathbb C}^*)^r\to {\mathbb C}^*$
both be defined by $(a_1,\ldots,a_r)\mapsto a_1\cdots a_r$.
By \eqref{eqn:relinv} we have a commutative diagram
\begin{equation}
\label{eqn:somediagramofmult}
\xymatrix@-0.5pc{
&
({\mathbb{G}_\mathrm{m}})^r \ar[r]^m \ar[d]^{\beta} & {\mathbb{G}_\mathrm{m}} \ar[d]^{\gamma} \\
\Omega \ar[ur]^{p_7} \ar[r]^F \ar@/_1pc/[rr]_{f_1\cdots f_r}
& ({\mathbb C}^*)^r \ar[r]^{n} & {\mathbb C}^*}
\end{equation}
where
$\beta(a_1,\ldots,a_r)=(a_1 f_1(v_0),\ldots,a_r f_r(v_0))$
and $\gamma(a)=a (f_1\cdots f_r)(v_0)$ are homeomorphisms.
By \eqref{eqn:somediagramofmult}, $P$ is
homeomorphic to the fiber of $p_7$, that is,
$P\cong [G,G]/([G,G]\cap G_{v_0})$,
and hence $P$ is the codomain of $p_2$.
Also, $K$ is homeomorphic
to
$(m\circ p_7)^{-1}(1)$;
restricting $p_7$ gives a fiber bundle
$$p_8:K\to \ker(m)\cong ({\mathbb{G}_\mathrm{m}})^{r-1}$$
with fiber $P$.
By \cite[Proposition 4.48]{hatcher},
a fiber bundle is a Serre fibration, that is,
it possesses the homotopy lifting property for CW complexes,
and these have the usual homotopy long exact sequence of a fibration
(\cite[Theorem 4.41]{hatcher});
apply this sequence
to all $p_i$ and $m\circ p_7$.
Note that the connected unipotent group $\mathrm{Rad}_u(G)$
is diffeomorphic to some ${\mathbb C}^p$, and hence is contractible
(\cite[\S3.3.6]{ov}).
Also,
$G_{v_0}$ and $[G,G]\cap G_{v_0}$ are finite,
$P\cong [G,G]/[G,G]\cap G_{v_0}$ is connected,
$\pi_1(({\mathbb{G}_\mathrm{m}})^k)\cong{\mathbb Z}^k$, and $\pi_i(({\mathbb{G}_\mathrm{m}})^k)\cong 0$ for $i> 1$.
The long exact sequences show that for $n>1$
there is a diagram, not necessarily commutative,
\begin{comment}
\begin{align*}
\pi_n(G)&\cong \pi_n(\Omega) &
\pi_n([G,G])&\cong \pi_n([G,G]/[G,G]\cap G_{v_0}) \\
\pi_n(G)&\cong \pi_n(L) &
\pi_n([G,G])&\cong \pi_n([L,L]) \\
\pi_n([G,G]/[G,G]\cap G_{v_0})&\cong \pi_n(\Omega) &
\pi_n(\ker(m\circ p_7))&\cong \pi_n(P).
\end{align*}
\end{comment}
$$
\xymatrix@-1pc{
\pi_n([L,L])\ar[d]_{p_6}
&
\ar[l]_{p_4} \pi_n([G,G]) \ar[r]^-{p_2} \ar[d]_{p_5}
&
\pi_n(P) \ar[r]^{p_8} \ar[d]_{p_7}
&
\pi_n(K) \ar[dl]^{m\circ p_7}
\\
\pi_n(L)
&
\ar[l]_{p_3} \pi_n(G) \ar[r]^{p_1}
&
\pi_n(\Omega)
&
}$$
where an arrow labeled by $\phi$ represents
an isomorphism
occurring
in the long exact sequence of the fibration $\phi$,
either $\phi_*$ or $i_*$ for $i$ the inclusion of the
fiber.
The sequences also give
the exact
diagrams in Figure~\ref{fig:pi1}.
\end{proof}
Thus, the homotopy groups may largely be computed from the homotopy
groups of the semisimple part $[L,L]$ of $G$.
For instance, if $G$ is solvable then $[L,L]=\{e\}$ and hence
by
Proposition \ref{prop:homotopy},
$\Omega$, $P$, and $K$ are $K(\pi,1)$ spaces, as shown in \cite{kpi1}.
\begin{comment}
\begin{remark}
NODO
As in \cite{kpi1},
we can pull back the form $\frac{dz}{z}$ on ${\mathbb C}$
along $f_i$ to get a
meromorphic form
$\frac{df_i}{f_i}$ on $V$ which is holomorphic on $\Omega$.
\end{remark}
\begin{remark}
topology of semisimple gps
NODO
\cite[\S5.2.1, 5.2.5, 5.3.2]{ov}
\end{remark}
Applying the Serre spectral sequence to
the fiber bundle $p_1$ might compute the cohomology
of $\Omega$, but
the nontrivial action of $\pi_1(\Omega)$ on
the cohomology of $G_{v_0}$ makes this difficult.
A more productive approach seems to be the following.
Let $k$ be a field of characteristic $0$ or a prime not dividing
$|G_{v_0}|$.
Since $p_1$ is a covering space,
the cohomology of $G$ and $\Omega$
with coefficients in $k$
are related in a simple way
(e.g., \cite[Proposition 3G.1]{hatcher}).
To compute the cohomology of $G$, it is useful to
replace $G$ by $\widetilde{G}=\mathrm{Rad}_u(G)\rtimes(\mathrm{Z}(L)^0\times [L,L])$,
a $|\mathrm{Z}(L)^0\cap [L,L]|$-sheeted covering space over $G$.
Then topologically $\widetilde{G}$ is a product, and we may
apply the
K\"unneth formula.
NOTE: I believe this is completely correct, just a distraction.
\begin{remark}
For the cohomology $H^k(\Omega;F)$,
consider the covering space
$$
\xymatrix{
\widetilde{G}=\mathrm{Rad}_u(G)\rtimes (\mathrm{Z}(L)^0\times [L,L])
\ar[r]^-{\phi} & G \ar[r]^-{\alpha} &
G/G_{v_0}\cong \Omega},$$
where $\phi$ is an $n=|\mathrm{Z}(L)^0\cap [L,L]|$-fold covering
given by multiplication, and $\alpha$ is an
$m=|G_{v_0}|$-fold covering.
If $F$ is a field of characteristic
$0$ or a prime not dividing $nm$, then
by \cite[Proposition 3G.1]{hatcher},
each $H^k(\Omega;F)$ is isomorphic to
the $\phi^{-1}(G_{v_0})$-invariant
classes in $H^k(\widetilde{G};F)$.
Note that $\widetilde{G}$ is a product topologically.
See also
\cite[Theorem 1.6]{gmns}
and
\cite{cohomology-of-complement}.
\end{remark}
\end{comment}
\begin{comment}
\begin{lemma}
Let $\rho:G\to{\mathrm{GL}}(V)$ be a representation of a connected complex
linear algebraic group, with $\ker(\rho)$ finite.
Let $G$ have a Levi decomposition as above.
For $g\in G$, let $c_g:\mathrm{Rad}_u(G)\to \mathrm{Rad}_u(G)$ be defined by
$c_g(r)=grg^{-1}$.
Then the algebraic group
$$\tilde{G}=\mathrm{Rad}_u(G)\rtimes(\mathrm{Z}(L)^0\times [L,L])$$
with group operation
$$(r_1,z_1,l_1)\cdot (r_2,z_2,l_2)=(r_1\cdot c_{(z_1l_1)}(r_2),z_1z_2,l_1l_2)$$
is connected and has a surjective homomorphism $\phi:\tilde{G}\to G$ with
finite kernel $\{(e,z,z^{-1}): z\in\mathrm{Z}(L)^0\cap [L,L]\}$.
\end{lemma}
\begin{proof}
By the Levi decomposition, there exists an isomorphism
$\alpha:\mathrm{Rad}_u(G)\rtimes L\to G$.
Since $\mathrm{Z}(L)^0$ and $[L,L]$ are algebraic normal subgroups of $L$ with
finite intersection and $L=\mathrm{Z}(L)^0\cdot [L,L]$,
there is a surjective homomorphism $\beta:\mathrm{Z}(L)^0\times [L,L]\to L$
given by multiplication in $L$.
Finally, let $\phi=\alpha\circ (\mathrm{id}_{\mathrm{Rad}_u(G)} \times \beta)$
and observe that $\phi$ is surjective, and
the kernel is as claimed and thus finite.
\end{proof}
To do cohomology:
Let $G\subseteq {\mathrm{GL}}(V)$ define the linear free divisor $D\subset V$.
Let $G$ have a Levi decomposition as above.
Let $n=|\mathrm{Z}(L)^0\cap [L,L]|\cdot |G_{v_0}|$.
Let $F$ be a field of characteristic $0$ or a prime not dividing
$n$.
Then let
$$\tilde{G}=\mathrm{Rad}_u(G)\rtimes(\mathrm{Z}(L)^0\times [L,L]).$$
Have a surjective homomorphism
$\phi:\tilde{G}\to G$ with finite kernel.
This produces a $\tilde{G}$-action on $V$, and the isotropy subgroup
of $v_0$ is $H=\phi^{-1}(G_{v_0})$, which is finite.
Then have the orbit map
$q:\tilde{G}\to \tilde{G}/H\cong G/G_{v_0}\cong \Omega$.
By \cite[Proposition 3G.1]{hatcher},
$q^*:H^k(\Omega;F)\to H^k(\tilde{G};F)$
is injective, with image $H^k(\tilde{G};F)^{H}$
(i.e., the $H$-invariant classes).
By \cite[Theorem 3.16]{hatcher},
$H^k(\tilde{G};F)\cong H^k(\mathrm{Rad}_u(G);F)\otimes_F
H^k((\mathrm{Z}(L)^0;F)\otimes_F H^k([L,L];F)$.
Since $\mathrm{Rad}_u(G)$ is contractible, $\mathrm{Z}(L)^0\cong {\mathbb{G}_\mathrm{m}}^r$,
and
${\mathbb{G}_\mathrm{m}}^r$ is diffeomorphic to $(S^1)^r\times (0,\infty)^r$,
we have
$H^k(\tilde{G};F)\cong
H^k((S^1)^r;F)\otimes_F H^k([L,L];F)$.
homotopic maps give same map on cohomology?
and p-c gp, so ?
\end{comment}
\subsection{(Non-linear) free divisors}
Let $(D,p)$ be an arbitrary free divisor in $({\mathbb C}^n,p)$.
May the number of components of
$(D,p)$ be computed from the structure of $M=\Derlog{{\mathbb C}^n,p}{D}$?
One natural guess,
$\dim_{\mathbb C}(M/\mathscr{O}_{{\mathbb C}^n,p}\cdot [M,M])$,
does not work.
For example, if $D$ is a hyperplane in ${\mathbb C}^2$, then the number
computed is $0$; in fact, $M=\mathscr{O}_{{\mathbb C}^n,p}\cdot [M,M]$ whenever $\Derlog{{\mathbb C}^n,p}{D}\nsubseteq
\mathscr{M}_p\cdot {\mathrm{Der}}_{{\mathbb C}^n,p}$ for $\mathscr{M}_p$ the maximal
ideal.
Other examples give answers too large:
for the plane curve $D=V((a^2-b^3)(a^7-b^{13}))$,
the number computed is $35$.
\begin{comment}
\begin{verbatim}
needsPackage "VectorFields";
-- Use the local ordering
R=QQ[a,b,MonomialOrder=>RevLex,Global=>false];
tryit = (f) -> (
M=derlogV(ideal (f));
stdio<<M<<endl;
MM=computeCommutator(M);
stdio<<MM<<endl;
N=subquotient(matrix mutableMatrix M,matrix mutableMatrix MM);
stdio<<"ZZZ"<<endl;
B=basis(N);
stdio<<"M/[M,M] has dimension "<<numColumns(B)<<endl;
);
-- Gives 0
tryit(a);
-- Gives 1, not the correct answer
tryit(a*b*(a-b));
\end{verbatim}
Also check this last number using Singular:
\begin{verbatim}
ring R=0,(a,b),ds;
// Use M2 to compute derlogV((a^2-b^3)*(a^7-b^13)).
module D=[13*a*b^3-13*a^3, 7*b^4-7*a^2*b],
[65*b^12-338*a^8+273*a^5*b^2,-182*a^7*b+147*a^4*b^3+35*a^6];
// Also use M2 to compute [D,D], i.e., computeCommutator(D)
module cD=
[4615*b^15-2925*a^2*b^12-23660*a^8*b^3+21970*a^10+12285*a^5*b^5-12285*a^7*b^2, 910*a*b^13-12740*a^7*b^4+11830*a^9*b+6615*a^4*b^6-4130*a^6*b^3-2485*a^8],
[169*a^2*b^6-338*a^4*b^3+169*a^6, 91*a*b^7-182*a^3*b^4+91*a^5*b],
[845*a*b^15-845*a^3*b^12-4394*a^9*b^3+4394*a^11+3549*a^6*b^5-3549*a^8*b^2, -2366*a^8*b^4+2366*a^10*b+1911*a^5*b^6-1456*a^7*b^3-455*a^9],
[91*a*b^7-182*a^3*b^4+91*a^5*b, 49*b^8-98*a^2*b^5+49*a^4*b^2],
[455*b^16-455*a^2*b^13-2366*a^8*b^4+2366*a^10*b+1911*a^5*b^6-1911*a^7*b^3, -1274*a^7*b^5+1274*a^9*b^2+1029*a^4*b^7-784*a^6*b^4-245*a^8*b],
[845*a*b^15-845*a^3*b^12-4394*a^9*b^3+4394*a^11+3549*a^6*b^5-3549*a^8*b^2, 455*b^16-455*a^2*b^13-2366*a^8*b^4+2366*a^10*b+1911*a^5*b^6-1911*a^7*b^3],
[4225*b^24-43940*a^8*b^12+35490*a^5*b^14+114244*a^16-184548*a^13*b^2+74529*a^10*b^4, -11830*a^7*b^13+9555*a^4*b^15+2275*a^6*b^12+61516*a^15*b-99372*a^12*b^3-11830*a^14+40131*a^9*b^5+9555*a^11*b^2],
[-2366*a^8*b^4+2366*a^10*b+1911*a^5*b^6-1456*a^7*b^3-455*a^9, -1274*a^7*b^5+1274*a^9*b^2+1029*a^4*b^7-784*a^6*b^4-245*a^8*b],
[-11830*a^7*b^13+9555*a^4*b^15+2275*a^6*b^12+61516*a^15*b-99372*a^12*b^3-11830*a^14+40131*a^9*b^5+9555*a^11*b^2, 33124*a^14*b^2-53508*a^11*b^4-12740*a^13*b+21609*a^8*b^6+10290*a^10*b^3+1225*a^12];
module q1=modulo(D,cD);
module q2=std(q1);
vdim(q2);
\end{verbatim}
NODO: Mention Hauser--M\"uller
\end{comment}
\bibliographystyle{amsalpha}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 8,360
|
Shops will be holding tournaments to coincide with the release of COLOSSAL WARFARE!
Join in to win awesome prizes!
Each card has a special version!
Are you going to hit the jackpot?
Players who place in 1st and 2nd place get even more prizes!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 761
|
\section{Introduction}
\citet{Bass12} identified the soft gamma-ray source \object{IGR~J12319$-$0749} found by the {\em INTEGRAL} satellite as a radio- and X-ray-emitting object. Based on its optical spectrum, the source is a quasar at $z$=3.12 \citep{Mass12}. From the broad emission lines, the mass of the central black hole is estimated to be 2.8$\times$10$^9$~M$_{\odot}$. \citet{Bass12} collected several other pieces of evidence that suggest the source is an extreme blazar, a flat-spectrum radio quasar with powerful jets. If this is the case, the source is the second most distant blazar detected by {\em INTEGRAL} so far. The X-ray source located within the {\em INTEGRAL} error circle has been recently followed up with the {\em Swift} satellite's X-ray telescope (XRT), and UV-optical telescope (UVOT). The X-ray image confirms that the object coincides with a National Radio Astronomy Observatory (NRAO) Very Large Array (VLA) Sky Survey (NVSS) radio source \citep[VLA D configuration, flux density 60.4 mJy at 1.4 GHz]{Cond98}. The radio source is also listed as an unresolved object ($<$5$\arcsec$) in the VLA Faint Images of the Radio Sky at Twenty-cm (FIRST) survey catalogue \citep[][VLA B configuration, 62.9 mJy flux density and 61.0 mJy/beam peak brightness at 1.4 GHz]{Whit97}. No other radio measurements (at frequencies other than 1.4~GHz, and with better resolution) are known. The X-ray flux seems variable over a period of a few months \citep{Bass12}. From the sparsely available, non-simultaneous data and upper limits, \citet{Bass12} tried to reconstruct the spectral energy distribution of IGR~J12319$-$074, and concluded that it is similar to that of another high-redshift blazar, IGR~J22517+2218 \citep[at $z$=3.668;][]{Falc98}.
Here we report on our high-resolution radio interferometric observation of the radio counterpart of IGR~J12319$-$0749, using the technique of very long baseline interferometry (VLBI) with the European VLBI Network (EVN). This is an excellent tool for confirming with imaging that a source is indeed a blazar with compact radio emission on milli-arcsecond (mas) angular scale; VLBI is also capable of determining the accurate astrometric position of a compact radio-emitting object.
\section{EVN observations and data reduction}
\begin{figure}[!h]
\centering
\includegraphics[bb=68 169 507 626, height=80mm, angle=270, clip=]{20778-fig1.ps}
\caption{
Naturally-weighted 5-GHz e-EVN image of the quasar J1231$-$0747 (IGR~J12319$-$0749). The lowest contours are drawn at $\pm$0.27~mJy/beam, corresponding to $\sim$3$\sigma$ image noise. The positive contour levels increase by a factor of 2. The peak brightness is 84.1~mJy/beam. The Gaussian restoring beam is 8.8~mas $\times$ 2.0~mas with major axis position angle $75\degr$. The restoring beam (FWHM) is indicated with an ellipse in the lower-left corner. The image is centered on the brightness peak at right ascension $12^{\rm h}31^{\rm m}57\fs68547$ and declination $-7\degr47\arcmin18\farcs0901$.}
\label{image}
\end{figure}
The EVN observation of IGR~J12319$-$0749 (or J1231$-$0747, the name derived from the more accurate coordinates of the radio counterpart, and used in this paper hereafter) took place on 2012 June 19 at 5~GHz frequency. We utilized the electronic VLBI (e-VLBI) mode \citep{Szom08} where, unlike the conventional VLBI, the signals are not recorded at the telescope sites, but are transmitted to the correlator via wide-band optical fibre networks. The real-time correlation of the data took place at the EVN Data Processor at the Joint Institute for VLBI in Europe (JIVE), Dwingeloo, The Netherlands. At the maximum recording rate of 1024~Mbit~s$^{-1}$, eight antennas of the radio telescope network participated in the experiment: Effelsberg (Germany), Jodrell Bank Mk2 telescope (UK), Medicina, Noto (Italy), Toru\'n (Poland), Onsala (Sweden), Hartebeesthoek (South Africa), and the phased array of the Westerbork Synthesis Radio Telescope (WSRT, The Netherlands). Inter-continental baselines up to the length of $\sim$8100~km were provided by the Hartebeesthoek antenna. Our short exploratory e-EVN experiment (project code RSF06) lasted for 2~h. Eight intermediate frequency channels (IFs) were used in both left and right circular polarisations. The total bandwidth was 128~MHz per polarisation.
The target source J1231$-$0747 was observed in phase-reference mode \citep[e.g.][]{Beas95}. We did not have prior information about the expected correlated flux density of J1231$-$0747. For a successful detection, we planned sufficiently long coherent integration time to be spent on the source to improve the sensitivity of the observations. Phase-referencing involves regular nodding of the radio telescopes between the target and a nearby bright and compact reference source. We used J1233$-$1025 as the phase-reference calibrator, at $2\fdg65$ separation from our target. The target--reference cycles of $\sim$6~min allowed us to spend $\sim$4.5~min on the target source in each cycle, leading to $\sim$80~min accumulated integration time on J1231$-$0747. The absolute astrometric position of the reference source is listed in the catalogue of the current realisation of the International Celestial Reference Frame \citep[ICRF2,][]{Fey09}. Phase-referencing is suitable for accurately determining the position of the target source with respect to the reference source.
The US National Radio Astronomy Observatory (NRAO) Astronomical Image Processing System ({\sc AIPS}) was used for the data calibration in a standard way \citep[e.g.][]{Diam95}. The visibility amplitudes were calibrated using the antenna gains, and the system temperatures measured at the antennas during the experiment. Fringe-fitting \citep{Schw83} was performed for the phase-reference calibrator (J1233$-$1025), and the fringe-finder sources (J1125+2610, J1159+2914). The data were then exported to the Caltech {\sc Difmap} package \citep{shep94} for imaging. The conventional mapping procedure involving several iterations of {\sc CLEAN}ing \citep{Hogb74} and phase (then amplitude) self-calibration resulted in the images and brightness distribution models for the calibrators. Overall antenna gain correction factors (typically $\sim$10\% or less) were determined and applied to the visibility amplitudes in {\sc AIPS}. Then fringe-fitting was repeated in {\sc AIPS}, now taking the {\sc CLEAN} component model of the phase-reference calibrator into account. The residual phase corrections resulting from the non-pointlike structure of the phase-reference calibrator were considered this way. The solutions obtained were interpolated and applied to the target source data. Then the visibility data of J1231$-$0747 were also exported to {\sc Difmap} for imaging. The phase-referenced image obtained was used for determining the position of the brightness peak with the {\sc AIPS} task {\sc MAXFIT}. The right ascension $12^{\rm h}31^{\rm m}57\fs68547$ and the declination $-7\degr47\arcmin18\farcs0901$ have accuracies of 0.7~mas and 1~mas, respectively, estimated from the phase-reference calibrator source position accuracy, the target--calibrator separation, the angular resolution of the interferometer array, and the signal-to-noise ratio.
From the phase-referenced data, it turned out that the target source itself is sufficiently bright and compact for fringe-fitting. Therefore, we applied the {\sc AIPS} task {\sc FRING} for J1231$-$0747, as was done earlier for the phase-reference source and the fringe-finders. This way the absolute position information is lost for the target. However, the final naturally-weighted image (Fig.~\ref{image}) obtained in {\sc Difmap} with the standard hybrid-mapping cycles of {\sc CLEAN}ing, and phase and amplitude self-calibration is somewhat more sensitive than the phase-referenced image.
\section{Results and discussion}
Our 5-GHz VLBI image (Fig.~\ref{image}) shows J1231$-$0747 (IGR~J12319$-$0749) as a compact radio source without any extended feature above the brightness limit of $\sim$0.5~mJy/beam (5$\sigma$) at mas or 10-mas angular scales. The radio source appears practically unresolved with the e-EVN on baselines up to $\sim$136 million wavelengths (M$\lambda$), i.e. from the European antennas to Hartebeesthoek. To physically characterise the source, we used {\sc Difmap} to fit a circular Gaussian brightness distribution model directly to the self-calibrated visibility data. The best-fit model component has 84.6$\pm$4.2~mJy flux density and 0.24$\pm$0.01~mas diameter (full width at half maximum, FWHM). This size can be compared with the minimum resolvable angular size \citep[e.g.][]{Kova05,Loba05} that can be obtained in this VLBI experiment, assuming natural weighting
\begin{equation}
\vartheta_{\rm lim}= b_{\psi} \sqrt{\frac{4 \ln 2}{\pi} \ln{\left(\frac{\rm SNR}{\rm SNR-1}\right)}}.
\end{equation}
Here $b_{\psi}$ is the half-power beam size along a given position angle $\psi$, and SNR the signal-to-noise ratio, i.e. the ratio between the peak brightness and the image noise.
A source is considered unresolved if its measured visibility amplitude at the longest baseline differs from that of a point source of the same flux density by not more than the 1-$\sigma$ uncertainty.
This method is often applied to determine the minimum resolvable size of compact VLBI-detected components \citep[e.g.][]{Lee08,Savo08,Reyn09,Abdo11,OSul11}
In our case, SNR=829, and the beam size varies between 2.0~mas and 8.8~mas, corresponding to the FWHM of the elongated elliptical Gaussian restoring beam in the direction of its minor and major axes, respectively (Fig.~\ref{image}). The resulting minimum resolvable angular size is the largest in the direction of the major axis, $\vartheta_{\rm lim, maj}$=0.29~mas. Our fitted model component size (0.24 mas) is similar but somewhat smaller than this value.
With a more conservative approach, we adopt the 3-$\sigma$ image noise level which reduces the signal-to-noise ratio to 276. In this case, the minimum resolvable angular size in the major axis direction becomes $\vartheta_{\rm lim, maj}$=0.50~mas. We can use our modelfit result as the size estimate in the minor axis direction ($\vartheta_{\rm min}$=0.24~mas) since it still exceeds the minimum resolvable size along this position angle.
The measured parameters allow us to estimate the apparent brightness temperature ($T_{\rm b}$) of the compact radio-emitting region \citep{Cond82} in the rest frame of the source
\begin{equation}
T_{\rm b} = 1.22\times 10^{12} (1+z) \frac{S}{\vartheta_{\rm lim, maj} \vartheta_{\rm min} \nu^2},
\end{equation}
where $S$ is the flux density (Jy) and $\nu$ the observing frequency (GHz). For J1231$-$0747, we obtain $T_{\rm b}$=(1.42$\pm$0.09) $\times$ $10^{11}$~K. Because of the unresolved nature of the radio source, this value can be considered a lower limit to the brightness temperature.
A reasonable assumption about the intrinsic brightness temperature ($T_{\rm b,int}$) of the source would lead to an estimate of the Doppler-boosting factor $\delta$=$T_{\rm b}/T_{\rm b,int}$ \citep{Read94} in the jet. It is customary to assume the equipartition value, $T_{\rm b,int} = T_{\rm eq} \simeq 5\times 10^{10}$~K \citep{Read94,Laht99} as a good approximation of the intrinsic brightness temperature of compact AGN \citep[e.g.][]{Jors06,Hova09,Vere10,Wu12}. This is valid if the energy in the radiating particles is equal to the energy stored in the magnetic field. Departure from equipartition is certainly possible for sources in their maximum brightness state, which could increase $T_{\rm b,int}$ \citep{Homa06}. On the other hand, VLBI studies of large samples found that for the majority of objects the typical $T_{\rm b,int}$ measurements are $\sim$$2-3\times 10^{10}$~K, very close to but somewhat below the canonical equipartition value \citep[e.g.][]{Kell04,Homa06}.
Bearing in mind that there is always an uncertainty in the determination of Doppler factors from single-epoch brightness temperature measurements, we follow the general practice and adopt $T_{\rm b,int} = 5\times10^{10}$~K for J1231$-$0747. In this case $\delta \ga 2.8$. It can naturally be explained in the framework of the standard orientation-based unified model for radio-loud active galactic nuclei \citep{Urry95}, i.e. with synchrotron emission of the plasma in an approaching relativistic jet pointing close to our line of sight. A rough estimate of the jet inclination angle with respect to the line of sight $\theta$ can be given by assuming a typical bulk Lorentz factor found for quasar jets, $\Gamma$=10 \citep[e.g.][]{Kell04}. Using
\begin{equation}
\cos \theta = \frac{\Gamma - \delta^{-1}}{\sqrt{\Gamma^2 - 1}},
\end{equation}
the jet inclination is $\theta$$\approx$14$\degr$, which becomes smaller if the Doppler factor and/or the Lorentz factor are higher than assumed here. The compact high-resolution radio structure of J1231$-$0747 and its measured brightness temperature are consistent with the VLBI imaging data for most radio-loud quasars at around $z$=3 \citep[e.g.][]{Gurv92,Gurv94,Frey97,Para99,OSul11}.
Our VLBI experiment provides the first spectral data point available for the source at 5~GHz. The measured $S_{\rm 5}$=84.6$\pm$4.2~mJy is a lower limit to the total flux density of J1231$-$0747 if there is any extended emission around the central compact source that is resolved out by the EVN. This value is higher than the 1.4-GHz flux densities from the NVSS and FIRST surveys (consistently $\sim$60~mJy), which may indicate an inverted power-law spectrum of this quasar in the observed GHz frequency range. Note that the emitted (rest-frame) frequencies are higher by a factor of (1+$z$) because of the expansion of the Universe. In our case, $\nu_{\rm obs}$=5~GHz corresponds to $\nu_{\rm em}$=20.6~GHz. Since the different flux density measurements are made at widely separated epochs, it is also possible that J1231$-$0747 shows significant flux density variations. Both the flat or slightly inverted radio spectrum and the variability are characteristic to compact blazars.
\section{Conclusions}
Our e-EVN observation of the suspected radio counterpart of IGR~J12319$-$0749 (J1231$-$0747), the second-highest redshift soft gamma-ray blazar detected with the {\em INTEGRAL} satellite, provided strong additional support for its blazar nature suggested by \citet{Bass12}. The derived position of the compact radio source is consistent with the less accurate a-priori coordinates taken from the FIRST survey \citep{Whit97} within the errors, thus strengthening the identification of the gamma-ray source with the radio source, the only compact source within the FIRST beam. We found that J1231$-$0747 is a $\sim$85-mJy radio source with sub-mas angular size. The slightly inverted radio spectrum or the flux density variability implied by our measurement at 5~GHz are expected from a blazar.
With the usual assumption of energy equipartition between the magnetic field and the relativistic particles in the jet \citep{Read94}, we estimated the lower limit to the brightness temperature of J1231$-$0747. It indicates a Doppler-boosted radio jet inclined within $\sim$14$\degr$ to the line of sight. The size of the radio source and the parameters of the jet could be better constrained with future higher-resolution VLBI imaging observations, in particular with longer interferometric baselines in the east--west direction. Simultaneous multi-frequency observations with large radio telescopes or interferometers would reveal the radio part of the SED and determine the turnover frequency in the radio spectrum. Flux density monitoring would shed light on the variability properties of this blazar, possibly providing additional constraints on the physical parameters of its jet.
From an observational point of view, its sufficiently high brightness and compactness make J1231$-$0747 an ideal new phase-reference calibrator object for any future VLBI experiment studying nearby weak radio sources.
\begin{acknowledgements}
We are grateful to the chair of the EVN Program Committee, Tom Muxlow, for granting us short exploratory e-VLBI observing time. We thank the two anonymous referees for their suggestions. The EVN is a joint facility of European, Chinese, South African, and other radio astronomy institutes funded by their national research councils. e-VLBI research infrastructure in Europe is supported by the European Community's Seventh Framework Programme (FP7/2007-2013) under grant agreement RI-261525 NEXPReS. The research leading to these results has received funding from FP7 under grant agreement No. 283393 (RadioNet3), the Hungarian Scientific Research Fund (OTKA, grant no.\ K104539), and the China--Hungary Collaboration and Exchange Programme by the International Cooperation Bureau of the Chinese Academy of Sciences. T. An is grateful for the financial support from the National Natural Science Foundation of Science and Technology of China (2013CB837901).
\end{acknowledgements}
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module.exports = require('./dist/backbone.module.js');
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"redpajama_set_name": "RedPajamaGithub"
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Yesterday I played the remake of Castle of Illusion Starring Mickey Mouse on PS3. I was always a big fan of the original game on the Megadrive, although I have heard that this is one of those rare games where the Master System port is actually superior. However, no matter which system you play it on, Castle of Illusion is a wonderful platformer that captures the spirit of the Magic Kingdom brilliantly.
Once you've (re-)learned the basics, you'll be throwing apples at enemies and collecting gems in no time. The game doesn't seem to have added much innovation to the genre (admittedly I've not played it all yet), but what it does, it does very well. The platforming is particularly fun. Mickey's jumps can be controlled with great precision, and even in the early stages there are times when it gets a bit hairy as you hop from falling leaf to falling leaf in an attempt to reach more stable ground. I particularly enjoyed the part where Mickey has to run towards the screen in an attempt to outrun a giant apple rolling downhill, in an amusing nod to Raiders of The Lost Ark.
I'm looking forward to playing more, and it's got me in the mood to revisit the original too. I'm not looking forward to the toy levels though – I could never get past the annoying part when the screen flips upside down and all the controls are inverted!
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"redpajama_set_name": "RedPajamaC4"
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Haywire Magazine
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Browse: Home » Columns » Art Tickles: Swings and Roundabouts
Art Tickles: Swings and Roundabouts
July 7, 2014 · by Taylor Hidalgo · in Art Tickles, Columns
Taylor Hidalgo has always been the sort of child who got hurt finding alternative uses for playground equipment.
In a very simplified sense, games are about choices.
The more complex a game's mechanics, or the more intricate its systems, the more choices players have at their disposal. Action games have more direct interactions, giving players options for how to combat opponents and how best to avoid obstacles. Adventure games present problems that must be overcome with planning, cunning, or quick-witted problem solving. Open world games present even more opportunity, letting the players not only choose how to resolve conflicts, but also whether or not to engage in them at all.
These choices are a function of not just a player's prerogative, but also the developer's intention. For a mechanic to exist, a developer must first have considered the mechanic, and coded it into the game. While seemingly obvious, the depth of that can be misleading. Each encounter, each mechanic, each enemy behavior, each scene, they all come from a designed and deliberate place. Players' decisions may come about independent of a developer's direct influence, but the coded mechanics and available options all inform the framework from which the player builds their choices.
Deeper than that, though, how a game is mechanically assembled affects how the game is able to communicate itself. Not just from a conflict or narrative standpoint, but also from a standpoint of player agency. A player is only as free to resolve conflicts as the scale of a game allows. A title like Fallout: New Vegas, for instance, offers players a wide host of options. The map is fluid and freeform, allowing the player total freedom on which areas they explore, whether or not to engage in fights or events they see coming, or whose side to take in ongoing disputes. These options are seemingly infinite, with the potential for multiple undiscovered quests, approaches, and tactics even after hundreds of hours or repeat playthroughs and exploration. It is an experience that cannot really be replicated in less open worlds.
Conversely, a game like Mirror's Edge is much more straightforward. There is a defined, structured sequence of stages and story beats, whose progression is strictly linear. The narrative cannot, by its nature, be in any other order. The world is closed off, structured, and free from player intervention. While this limits the player's agency from a narrative experience standpoint, the mechanical results play out rather differently. While each encounter is pre-built and more heavily scripted, the player's ability to navigate therein reveals a much deeper agency than Fallout's.
Fights are less determined by statistics, more determined by input and behaviors. Players can breeze by guards and police, vaulting obstacles and climbing past dangers. Or they can come into direct, aggressive contact with the police. Or, sometimes, by creeping up in a sniper's perch, pick enemies off one by one. Each of these options come from how the player chooses to address a problem. While the story may not be personalized, the individual play style is decidedly unique. Players still have agency despite the narrative limitations, and it comes from the choice of how to interact with the mechanics.
Therefore, the depth of pre-coded mechanics isn't just about what tools a player is given access to, but also the scale on which these tools are built. The more a player can move through each individual environment, the more choices they have with controlling their pace or their conflict. In Mirror's Edge, this means more ways to take cover, more places to lead police, more walls to climb or rails to vault. Each of these options gives players better control over their environment. Despite heavy scripting, players have more choice over how these situations play out than casual observation might imply.
Which is very different to Fallout: New Vegas, whose interactions with most environments are very static. Distant hills and mountains can be climbed, different buildings entered, but once conflicts genuinely begin, the player has a much more limited range of ways to interact. Speech checks must be completed, combat must be resolved, and positioning and tactics are more defined by environment than player agency.
Further, a lot of experiences are tethered to location. Players who have a massive world to explore also have a massive world to traverse. Travel from point to point is an excellent time to build atmosphere, tone, and tension, which Fallout does quite well, but also limits the pace with which players can arrive at game experiences. Any fetch quest, escort mission, or backtracking hike builds onto an existing pace limiter, preventing players from truly getting the same instant feedback that some other games can provide. Quests aren't simply about the experience of the fight or puzzle, but a culmination of events or behaviors. Many of them must ultimately be turned in to an NPC, a limitation that not even fast travel can truly eliminate.
Which isn't to say that Fallout is worse than Mirror's Edge, or vice versa, but that scale is often a limiting and defining factor in how a player is allowed to experience a game. Open world games define their player agency by freedom of selection and exploration, but these freedoms often come at the cost of player ability, more tightly restricting how a player can explore a world due to more limiting factors. Conflicts are often pared down to simpler mechanics, exploration and player indulgence are less instantly gratifying and more a result of larger scopes. They do, however, tend to result in more patient and indulgent experiences.
So even in a complicated sense, games are still about choices. Not just the decisions players make, but also the design decisions developers make. The scope and size of the game worlds also define how a player is able to choose their experience. What options are made available also restrict those that can emerge organically. In that way, it's up to players not just how they want to experience the games they choose, but also to pick games based on what sort of choices they'll want to be able to make. Players should always have choice about which playgrounds to play in, even if that means getting fewer choices.
Swings and roundabouts, I suppose.
Taylor Hidalgo is a writer by hobby, grasping at the edges of professional work. He's a fan of the sound of language, the sounds of games and the sound of deadlines looming nearby. He sometimes says things on Twitter and his blog.
Tags: Fallout New Vegas, Mirror's Edge
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"redpajama_set_name": "RedPajamaCommonCrawl"
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Q: ORA-00923 dot net core devart provider FirstOrDefault() query error I'm developing a .net core web API which use database oracle and data provider devart when I try to get some data on login operation I have a this problem
{"ORA-00923: FROM keyword not found where expected"}
my code drop catch on this line throw exception
var result = db.tables.Where(x => x.code== userParams.code&& x.password == userParams.Password).FirstOrDefault();
I changed after the where funtion
I tried
.First()
.SingleOrDefault()
.FirstOrDefault()
Also, I tried then that worked but I think this functions illogical
.SingleOrDefault()
.Take(1)
My versions
.net core 3.1
OracleDB version 11g
EFCore Devart
EF 3.1
Anyone can know anything about this problem?
A: Thank you @Devart I updated my Devart.Data.Oracle.EFCore version ith this version. It worked now :)
https://www.nuget.org/packages/Devart.Data.Oracle.EFCore/9.10.909
A: If SingleOrDefault is working i would redifne your query like this
var result = db.tables.SingleOrDefault(x => x.code == userParams.code && x.password == userParams.Password);
By doing so, i also think you are skipping som unnecessarily sub querying.
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Q: Why is Lebesgue integral first defined for step function and then for larger classes of functions? Most books on Lebesgue integration define the concept first for step functions (and simple functions) and later on extend the definition to Lebesgue measurable functions. Why is this approach preferred over the original definition of Lebesgue which develops measure theory first and then defines the integral as a limit of sum (somewhat like the usual definition of Riemann integral)? Is there any pedagogic advantage? To me this approach looks quite artificial compared to the limit of a sum definition.
EDIT: In response to the comments from "Did" I am trying to elaborate further. I find two approaches available to study Lebesgue Integration:
1) Integral is defined for step functions using a simple sum. Then we consider increasing sequences of such step functions whose integrals converge to some value $I$. In this case the sequence of step functions converges a.e to some function $f$ and we say that $I$ is integral of $f$. Such functions as $f$ are called upper functions and difference of two such upper functions is called a Lebesgue Integrable function. This is the approach in Apostol's Mathematical Analysis. Concept of measure is defined later as integral of the indicator function. Royden's Real Analysis also follows the same approach.
2) Lebesgue's original definition where he defines the concept of measure first and then partitions the range of functions into multiple subintervals and forms a sum analogous to a Riemann sum and the limit of this sum as we make finer and finer partitions of the range is defined as the integral of the function.
I find almost all books to be using the first approach and not second. And Lebesgue's definition seems to be used only for historical context. To me the second approach looks intuitive. I wonder if there is any pedagogic advantage of the first approach. I am a beginner in these topics and still trying to come to terms with the step functions approach which is so desperately based on the convergence of sequence of functions.
A: The problem ist how to define the sum without step functions. The only way I could think of would use the density Function
$$ d_f(t) := \mu(\{s\in D| f(s) > t\})$$
and the decreasing rearrangement. When you define the integral for step funcitons first however, you can prove lebesgue's convergence theorem and define the integral without such constructs. What limit would you use to define the integral of an arbitrary function without using an approximation with step functions?
A: It is often helpful to appeal to intuition when first introducing a subject, and step functions and other simple to integrate function aid in the creation of an intuitional basis for the more formalized concepts that will be presented.
In addition (speaking about Riemann integration) the pedagogical methods I have seen that introduce integration for step functions first (e.g Tom M. Apostle's Calculus Vol. 1) use set theory as a basis to create a simple definition of "measurable functions" (obviously different from the Lebesguian measure theory, but in the same spirit) and then proceed to use step functions to appeal to intuition while developing the idea of the integral as a sum, then introduce more complicated functions (adding progressively more complicated functions into our definition of measurable and by extension creating a more generalized definition) and introducing the limiting process.
Considering the nature of Lebesgue integration it would seem useful to understand first intuitively what you are doing with step functions then to formalize it and develop a theory of measurable functions for which you can apply Lebesgue integration. I am not a teacher (nor have I ever analysis), so I can only speak from my readings of analysis and my experience as a student.
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Fraenkel Gallery
DIANE ARBUS: curated by CARRIE MAE WEEMS
June 3 - August 19, 2021 Fraenkel Gallery is pleased to present an exhibition of 45 photographs by Diane Arbus, curated by acclaimed contemporary artist Carrie Mae Weems. A long-time admirer of Arbus's work, Weems has selected images spanning Arbus's fifteen-year career, from 1956 until her death in 1971. The exhibition will be on view at 49 Geary Street from June 3 to August 13, 2021, and will be followed by an exhibition devoted to Weems's own work in September. Launch Viewing Room
ELISHEVA BIERNOFF: Starting From Wrong
April 1 - May 28, 2021 Fraenkel Gallery is pleased to present its second exhibition of new work by Elisheva Biernoff. Titled Starting from Wrong, the exhibition features twelve meticulously detailed paintings measuring no larger than 4 x 5 inches each. All completed since 2017, Biernoff's recent paintings are carefully observed, two-sided works based on found and anonymous photographs. The exhibition will be on view from April 1 to May 28, 2021, and will be accompanied by a catalogue published by Fraenkel Gallery, to be released on April 7. Please join us on April 22 for the premiere of a conversation between the artist and Fraenkel Gallery President Frish Brandt. Read more
21 JANUARY-25 MARCH 2021 Fraenkel Gallery is pleased to present new work by Christian Marclay, incorporating collage, video animation, and photography. The exhibition continues Marclay's investigation into the relationship between sound and image through sampling elements from art and popular culture, and reflects the anxiety and frustration of the current global pandemic and political crises. The exhibition will be on view in the gallery from January 21 to March 26, 2021. Read more
Wardell Milan: Death, Wine, Revolt
ONLINE VIEWING ROOM 29 October 2020 - 15 January 2021 Fraenkel Gallery is pleased to present new work by Wardell Milan. The gallery's second solo show of the New York-based artist will be on view from October 29 to December 22, 2020. The exhibition features Milan's ongoing series "Death, Wine, Revolt," which combines photography, drawing, painting, collage, and sculpture to explore themes of over-indulgence, destruction, and revolution. While earlier series such as "Parisian Landscapes" looked inward, to personal questions of freedom and desire, Milan made the works on view in response to the turmoil of the global moment. Read more
ONLINE VIEWING ROOM 10 Sep - 23 Oct 2020 Read more
ONLINE VIEWING ROOM 30 Jul - 7 Sep 2020 On The Road is a celebration of the American road and the mythic role it has played in photography. Sheltered in place and longing for road trips, we are inspired by our artists' ongoing interest in roads, both as visual phenomena and metaphors. In a time of unprecedented immobility, these photographs from the past 60 years explore a range of approaches to the American landscape and its relationship to the car. We hope they evoke a sense of adventure, independence, and camaraderie. Read more
ART BASEL ONLINE 2020
ONLINE VIEWING ROOM 17 - 26 Jun 2020 Fraenkel Gallery is pleased to participate in Art Basel Online and present this extended selection in our online viewing room with work by Diane Arbus, Bernd & Hilla Becher, Nan Goldin, Lee Friedlander, Peter Hujar and others. Featuring Richard Avedon's portraits of Andy Warhol and The Factory; Sophie Calle's pairing of photography and text; and Hiroshi Sugimoto's Opticks. Read more
HIROSHI SUGIMOTO: BEGINNINGS
ONLINE VIEWING ROOM 20 May - 6 Jul 2020 Fraenkel Gallery is pleased to present HIROSHI SUGIMOTO: Beginnings, the third installation in our series of online exhibitions, 'For the Pleasure of Looking', which highlight our philosophy of living with art. On view simultaneously with Hiroshi Sugimoto: Opticks, the exhibition offers a look into three series that form the foundation of Sugimoto's work. The presentation will be on view through June 14, 2020. Read more
ONLINE VIEWING ROOM 30 Apr - 14 May 2020 DANCING IN THE STREETS is our second presentation in a new series of online exhibitions. With this selection, we look forward to the moment when we can all celebrate together. Featuring the work of Lee Friedlander, Helen Levitt, Garry Winogrand, Peter Hujar, Alec Soth, and more. Read more
ONLINE VIEWING ROOM 16 - 29 Apr 2020 Fraenkel Gallery is pleased to inaugurate For the Pleasure of Looking, a new series of online exhibitions that highlight our philosophy of living with art. Unusual times call for unusual measures, and our current situation certainly calls for MAKE MINE A DOUBLE, Fraenkel Gallery's first in a series of idiosyncratic digital exhibitions. With unexpected works by artists central to the gallery's mission – Diane Arbus, Bernd & Hilla Becher, Christian Marclay, and Anonymous among others – MAKE MINE A DOUBLE suggests several perspectives from which to consider that timely request. Feel free to interpret the title any way that fits; we offer the sixteen possibilities below. Read more
HIROSHI SUGIMOTO: OPTICKS
ONLINE VIEWING ROOM 2 Apr - 15 Aug 2020 Over more than four decades, Hiroshi Sugimoto has explored the ways photography can be used to record traces of invisible but elemental forces. Opticks is a new series of large-scale color photographs depicting the color of light Sugimoto observed through a prism in his Tokyo studio. Using Polaroid film, he recorded sections of the rainbow spectrum projected into a darkened chamber, paying particular attention to the spaces and gaps between hues. Opticks presents unique enlargements of these Polaroids. Read more
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We've teamed up with Aviation Art to produce a limited edition collection of World War Two RAF prints.
Produced from the original paintings of Lincolnshire artist Eddie Ash, who is well known for his Red Arrows and Battle of Britain paintings for the RAF, these are accurate reproductions of specific wartime aircraft in flight, on 250 gsm lumi silk art paper, measuring 280 x 210 mm and are ideal for framing.
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{
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\section{Introduction}
The Bernstein-Gelfand-Gelfand category~${\mathcal O}$ associated to a triangular decomposition of a finite dimensional contragredient Lie (super)algebra is an important and intensively studied object in modern representation theory, see e.g. \cite{MR2428237}. Category~${\mathcal O}$ for basic classical Lie superalgebras ${\mathfrak g}$ is not yet as well understood as for semisimple Lie algebras and exhibits many novel features. However, for the particular case of the general linear superalgebra $\mathfrak{gl}(m|n)$, category~${\mathcal O}$ has a Kazhdan-Lusztig (KL) type theory, introduced by Brundan in~\cite{Brundan} and proved to be correct by Cheng, Lam and Wang in~\cite{CLW}. This determines the characters of simple modules algorithmically. Moreover, in this case, category~${\mathcal O}$ is Koszul, as proved by Brundan, Losev and Webster in~\cite{Brundan3, BLW}. In the current paper, we note that this theory is also an `abstract KL theory' in the sense of~\cite{CPS1, CPS2}. Observe that also for $\mathfrak{osp}(2m+1|2n)$, a KL type theory has been introduced and established by Bao and Wang in~\cite{BW}.
Our main focus in the current paper is the study of three homological invariants in category~${\mathcal O}$, specifically for $\mathfrak{gl}(m|n)$, and their applications to the open question concerning the classification of non-equivalent blocks.
The {\em associated variety} of a module $M\in{\mathcal O}$ is the set of self-commuting odd elements $x$ of the Lie superalgebras for which $M$ has non-trivial ${\mathbb C}[x]$-homology, see \cite{Duflo, Vera}. Since the associated variety of a module in ${\mathcal O}$ consists of orbits of the even Borel subgroup, we classify such orbits. Then we investigate the associated variety of Verma modules for $\mathfrak{gl}(m|n)$ with distinguished Borel subalgebra, leading in particular to a complete description for the cases $\mathfrak{gl}(1|n)$, $\mathfrak{gl}(m|1)$ and $\mathfrak{gl}(2|2)$.
The {\em projective dimension} of an object in an abelian category with enough projective objects is the length of a minimal projective resolution. Contrary to category~${\mathcal O}$ for semisimple Lie algebras, category~${\mathcal O}$ for ${\mathfrak g}$ contains modules with infinite projective dimension. We obtain two characterisations for the abelian subcategory of modules having finite projective dimension. The first one (valid for $\mathfrak{gl}(m|n)$) is as the category of modules having trivial associated variety. The second one (valid for arbitrary~${\mathfrak g}$) is as an abelian category generated by the modules induced from the underling Lie algebra. Then we determine the projective dimension of injective modules for $\mathfrak{gl}(m|n)$ and use this to obtain the finitistic global homological dimension of the blocks, which builds on and extends some results of Mazorchuk in~\cite{CouMaz2, MR2366357, preprint}. Concretely, we show that this global categorical invariant of the blocks is determined by the singularity of the core of the central character. The results also provide the means to describe the level of `dominance' of a simple module in terms of the projective dimension of its injective envelope. The relevance of this categorical description of the dominance lies in the fact that, contrary to the Lie algebra case, one cannot use the projective dimension of the simple module itself for this, as the latter dimension will be infinite as soon as the module is atypical.
To deal with modules with infinite projective dimension we define the {\em complexity} of a module as the polynomial growth rate of a minimal projective resolution. The concrete motivation is to obtain a tool to homologically and categorically describe atypicality, similar to the description of singularity and dominance by projective dimension in the previous paragraph. We prove that our notion of complexity is well defined on category~${\mathcal O}$ for any basic classical Lie superalgebra, meaning that complexity of all modules is finite. Then we study the complexity of Verma (for the distinguished Borel subalgebra) and simple modules for $\mathfrak{gl}(m|n)$ and the relation with the degree of atypicality. Similar results for the category of finite dimensional weight modules of $\mathfrak{gl}(m|n)$ have been obtained by Boe, Kujawa and Nakano in~\cite{BKN1, BKN2}.
Integral blocks in category~${\mathcal O}$ for a Lie algebra are equivalent if they have the same singularity, see \cite{SoergelD}. Similarly, the blocks of the category of finite dimensional modules of a basic classical Lie superalgebra depend (almost) only on the degree of atypicality, see \cite{MR2734963, Lilit}. The {\em classification of non-equivalent blocks} in category~${\mathcal O}$ for a Lie superalgebra is an open question. Our main result here is the fact that a combination of the two aforementioned global categorical characteristics does not suffice to separate between non-equivalent blocks. More precisely, we use our results on projective dimensions to materialise subtle local differences in blocks with similar global properties into categorical invariants. We do this explicitly for regular atypical blocks for $\mathfrak{sl}(3|1)$, resulting in the fact that {\em all} such blocks are non-equivalent (while they have the same singularity and degree of atypicality). In particular this implies that, contrary to category~${\mathcal O}$ for Lie algebras and the category of finite dimensional weight modules for Lie superalgebras, category~${\mathcal O}$ for Lie superalgebras can contain infinitely many non-equivalent blocks. Furthermore the results demonstrate that, in general, equivalences between integral blocks will be very rare. This is summarised in Figure \ref{tabb}, where~${\mathcal F}$ represents the category of finite dimensional weight modules and ${\mathfrak g}_{\bar{0}}$ a reductive Lie algebra.
\begin{figure}\label{tabb}
\caption{Equivalence of blocks}
\[
\xymatrix{
*+[F]\txt{All blocks in ${\mathcal F}({\mathfrak g}_{\bar{0}})$ \\ equivalent} \ar[d]\ar[rr]&& *+[F]\txt{Blocks in ${\mathcal F}({\mathfrak g})$ \\ determined by atypicality} \ar[d]\\
*+[F]\txt{Blocks in ${\mathcal O}({\mathfrak g}_{\bar{0}})$\\determined by singularity}\ar[rr]&&*+[F]\txt{Blocks in ${\mathcal O}({\mathfrak g})$\\ {\bf not just} determined by combination\\ of atypicality and singularity}
}
\]
\end{figure}
We hope that our results can be applied in the quest to obtain a full classification of non-equivalent blocks in category~${\mathcal O}$ for $\mathfrak{gl}(m|n)$. Furthermore, in further work we aim to strengthen the equivalence between trivial associative variety and zero complexity for a module to a more general link between complexity and the associated variety and in particular to determine the complexity of simple modules.
The paper is structured as follows. In Section~\ref{secpre} we recall some preliminary results. In Section~\ref{secext} we study extensions between Verma and simple modules, in connection with Brundan's Kazhdan-Lusztig theory. In Section~\ref{secfpd} we obtain the characterisations for the modules with finite projective dimension. In Section~\ref{secassvar} we study the self-commuting cone and associated varieties in relation with category~${\mathcal O}$. In Section~\ref{secblocks} we study projective dimensions and obtain the result on the non-equivalence of blocks. In Section~\ref{seccomp} we introduce and study the notion of complexity. Finally, in the appendices we illustrate certain results for the example of $\mathfrak{gl}(2|1)$ and carry out some technicalities.
\section{Preliminaries}
\label{secpre}
In this paper we work with basic classical Lie superalgebras. We refer to Chapters 1-4 in~\cite{bookMusson} for concrete definitions. We denote a basic classical Lie superalgebra by ${\mathfrak g}$ and its even and odd part by ${\mathfrak g}_{\bar{0}}\oplus{\mathfrak g}_{\bar{1}}={\mathfrak g}$. A Borel subalgebra will be denoted by ${\mathfrak b}$, a Cartan subalgebra by ${\mathfrak h}$ (which is the same as a Cartan subalgebra of ${\mathfrak g}_{\bar{0}}$) and the set of positive roots by $\Delta^+$. The even and odd positive roots are then denoted by $\Delta^+_{\bar{0}}$ and $\Delta^+_{\bar{1}}$. The set of integral weights is denoted by $P_0\subset{\mathfrak h}^\ast$. The Weyl group $W=W({\mathfrak g}:{\mathfrak h})$ is the same as the Weyl group $W({\mathfrak g}_{\bar{0}}:{\mathfrak h})$. We fix a $W$-invariant form $(\cdot,\cdot)$ on ${\mathfrak h}^\ast$ as in Theorem 5.4.1 of~\cite{bookMusson}. We define $\rho=\frac{1}{2}(\sum_{\alpha\in\Delta^+_{\bar{0}}}\alpha)-\frac{1}{2}(\sum_{\gamma\in\Delta_{\bar{1}}^+}\gamma)$. The dot action is then given by
$$w\cdot\lambda=w(\lambda+\rho)-\rho.$$
\subsection{Basic classical Lie superalgebras of type~$A$}
Mostly ${\mathfrak g}$ will be one of the following
\begin{equation}\label{listA}{\mathfrak g}=\begin{cases}\mathfrak{gl}(m|n)\\ \mathfrak{sl}(m|n)\,\mbox{ with }\, m\not=n\\\mathfrak{pgl}(n|n)\simeq \mathfrak{gl}(n|n)/\mathfrak{z}(\mathfrak{gl}(n|n)).\end{cases}\end{equation}
Then we use the standard ${\mathbb Z}$-grading ${\mathfrak g}={\mathfrak g}_{-1}\oplus{\mathfrak g}_0\oplus{\mathfrak g}_1,$
with ${\mathfrak g}_{\bar{0}}={\mathfrak g}_0$ and ${\mathfrak g}_{\bar{1}}={\mathfrak g}_{-1}\oplus {\mathfrak g}_{1}$. We fix an element in the centre of ${\mathfrak g}_0$ which we denote by $z\in\mathfrak{z}({\mathfrak g}_0)$ and which satisfies
\begin{equation}\label{gradel}
[z,X]=X,\quad\forall \,X\in{\mathfrak g}_1\quad\mbox{and}\qquad [z,Y]=-Y,\quad\forall\, Y\in {\mathfrak g}_{-1}.
\end{equation}
The necessity of such an element is the reason we can not always include the other classical Lie superalgebras of type~$A$, {\it viz.} $\mathfrak{sl}(n|n)$ and $\mathfrak{psl}(n|n)$. Note that technically $\mathfrak{sl}(n|n)$ and $\mathfrak{pgl}(n|n)$ are not basic. In Remark \ref{error} we provide an example of properties concerning the associated variety that fail for $\mathfrak{sl}(n|n)$.
We choose the standard basis $\{\varepsilon_i,i=1,\cdots,m\}\cup \{\delta_j,j=1,\cdots,n\}$ of ${\mathfrak h}^\ast$ for $\mathfrak{gl}(m|n)$. Unless stated otherwise, the positive roots of any algebra in~\eqref{listA} are chosen to be
\begin{equation}\label{roots}
\Delta^+=\left\{
\begin{array}{ll}
\varepsilon_{i} - \varepsilon_{j} &\hbox{for $1\le i< j\le m$,}\\
\varepsilon_{i} -\delta_j &\hbox{for
$1\le i\le m$
and $1\le j\le n$,}\\
\delta_{i} - \delta_{j} &\hbox{for $1\le i< j\le n$.}
\end{array}
\right.
\end{equation}
This choice corresponds to the so-called distinguished system of positive roots, or the distinguished Borel subalgebra.
The Lie superalgebra generated by the positive root vectors is denoted by ${\mathfrak n}$, so we have ${\mathfrak b}={\mathfrak h}\oplus{\mathfrak n}$, ${\mathfrak g}_1={\mathfrak n}\cap {\mathfrak g}_{\bar{1}}$ and set ${\mathfrak n}_0:={\mathfrak n}\cap{\mathfrak g}_0$.
We define $\rho_0=\frac{1}{2}(\sum_{\alpha\in\Delta^+_0}\alpha)$. Note that $\rho_1=\rho-\rho_0$ is orthogonal to all even roots, so $w\cdot\lambda=w(\lambda+\rho_0)-\rho_0$ for $\lambda\in{\mathfrak h}^\ast$ and $w\in W$.
For the remainder of this subsection we restrict to ${\mathfrak g}=\mathfrak{gl}(m|n)$. We will use
\[\delta=-\sum_{i=1}^m i\varepsilon_{i} +\sum_{i=1}^n (m-i+1)\delta_i\;\,\in {\mathfrak h}^\ast, \]
rather than $\rho$, because the coefficients of $\delta$ are integers. The difference $\rho-\delta$ is orthogonal to all roots, so
$$w\cdot\lambda=w(\lambda+\delta)-\delta\qquad\mbox{and}\qquad (\lambda+\rho,\gamma)=(\lambda+\delta,\gamma),$$
for $w\in W$ and $\gamma\in \Delta_{\bar{1}}^+$. We fix a bijection between integral weights $P_0\subset{\mathfrak h}^\ast$ and ${\mathbb Z}^{m|n}$, by
\begin{equation} \label{hat}P_0\leftrightarrow {\mathbb Z}^{m|n},\quad\lambda\mapsto \mu^\lambda\quad \mbox{with}\quad \mu^\lambda_i=(\lambda+\delta,\varepsilon_i)\quad\mbox{and}\quad \mu^\lambda_{m+j}=(\lambda+\delta,\delta_j).\end{equation}
As in~\cite{BLW} we use the notation $\Lambda:={\mathbb Z}^{m|n}$. Note that, by the above $\mu^{w\cdot\lambda}=w(\mu^\lambda)$, where the latter term uses the regular action of $W\simeq S_m\times S_n$ on $\Lambda={\mathbb Z}^{m|n}$.
For any $\mu=(\alpha_1,\alpha_2,\cdots\alpha_m|\beta_1,\cdots\beta_n)\in \Lambda$, we refer to the integers $\alpha_1,\cdots\alpha_m$ as the labels of $\mu$ on the left side and to the integers $\beta_1,\cdots\beta_n$ as the labels of~$\mu$ on the right side.
The set of integral regular dominant weights is then described by
$$P_0^{++}=\{\lambda\in P_0\,|\, w\cdot\lambda <\lambda,\,\,\forall\, w\in W \},$$
where~$\mu\leq\lambda$ if $\lambda-\mu$ is a sum of positive roots.
We denote by $\Lambda^{++}$ the corresponding subset in $\Lambda$.
This set has the structure of a poset, which describes the highest weight structure
of the category of finite dimensional weight modules ${\mathcal F}$, see e.g. \cite{Brundan, MR1443186}.
Another convenient choice of Borel subalgebra is the anti-distinguished Borel subalgebra ${\mathfrak b}={\mathfrak b}_{\bar{0}}\oplus {\mathfrak g}_{-1}$. Note that the anti-distinguished Borel subalgebra of $\mathfrak{gl}(m|n)$ is mapped to the distinguished Borel subalgebra of $\mathfrak{gl}(n|m)$ under the isomorphism $\mathfrak{gl}(m|n)\simeq \mathfrak{gl}(n|m)$.
\subsection{BGG category~${\mathcal O}$}
\label{prelO}
Category~$s{\mathcal O}$ for a basic classical Lie superalgebra ${\mathfrak g}$ with Borel subalgebra ${\mathfrak b}$ is defined as the full subcategory of all ${\mathfrak g}$-modules, where the objects are finitely generated; ${\mathfrak h}$-semisimple and locally $U({\mathfrak b})$-finite. Note that this definition does not depend on the actual choice of Borel subalgebra, only the even part ${\mathfrak b}_{\bar{0}}:={\mathfrak b}\cap{\mathfrak g}_{\bar{0}}.$ For each Borel subalgebra ${\mathfrak b}$ (with ${\mathfrak b}\cap {\mathfrak g}_{\bar{0}}={\mathfrak b}_{\bar{0}}$), this category has a (different) structure of a highest weight category. An alternative definition of $s{\mathcal O}$ is as the full subcategory of all ${\mathfrak g}$-modules, where the objects $M$ satisfy~${\rm Res}^{\fg}_{\fg_{\bar{0}}} M\in{\mathcal O}^0$, after neglecting parity, where~${\mathcal O}^0$ is the corresponding category for ${\mathfrak g}_{\bar{0}}$. In what follows we will work with the full Serre subcategory~${\mathcal O}$ of $s{\mathcal O}$, defined similarly as in Section 2 of \cite{Brundan3}. This means that to each weight one attaches a parity (in a consistent way) and only modules in which the corresponding weight spaces appear in said parity are allowed. Then one has $s{\mathcal O}\simeq {\mathcal O}\oplus{\mathcal O}$, abstractly as categories, see Lemma 2.2 of \cite{Brundan3}. We denote the Serre subcategory of ${\mathcal O}$ generated by modules admitting the central character $\chi$ by ${\mathcal O}_{\chi}$. We denote the central character corresponding to $L(\lambda)$ by~$\chi_\lambda$.
Now we turn to this category for $\mathfrak{gl}(m|n)$. For an overview of the current knowledge we refer to the survey \cite{Brundan3}. The indecomposable blocks of category~${\mathcal O}$ for $\mathfrak{gl}(m|n)$ have been determined by Cheng, Mazorchuk and Wang in Theorem 3.12 of \cite{CMW}. They also proved, in Theorem 3.10 of
\cite{CMW}, that every non-integral block is equivalent to an integral block in the category ${\mathcal O}$ for a direct sum of several general linear superalgebras.
Therefore, we can restrict to integral blocks for several of our purposes.
The Serre subcategory of ${\mathcal O}$ of modules with integral weight spaces is denoted by ${\mathcal O}_{{\mathbb Z}}$. The blocks of category~${\mathcal O}_{\mathbb Z}$ can be described by linkage classes. The linkage class $\xi$ generated by $\mu\in P_0$ is
$$\xi=[\mu]=\{\lambda\in P_0\,|\,\chi_{\mu}=\chi_{\lambda}\}.$$
The indecomposable block ${\mathcal O}_\xi$ is then defined as the full Serre subcategory of ${\mathcal O}$ generated by the set of simple modules $\{L(\lambda)\,|\,\lambda\in\xi\}.$ The degree of atypicality of the weights in the linkage class is denoted by $\sharp \xi$. Furthermore we denote the central character corresponding to the block by $\chi_\xi$. The Bruhat order $\preceq$ of \cite{Brundan, BLW, CMW} on $P_0$ is the minimal partial order satisfying
\begin{itemize}
\item if $s\cdot\lambda \le \lambda $ for a reflection $s\in W$ and $\lambda\in P_0$, we have $s\cdot\lambda \preceq \lambda $;
\item if $(\lambda+\rho,\gamma)=0$ for $\lambda\in P_0$ and $\gamma\in \Delta^+_1$, we have $\lambda-\gamma\preceq \lambda$.
\end{itemize}
The set $\Lambda\simeq P_0$ equipped with this Bruhat order is the poset for ${\mathcal O}_{{\mathbb Z}}$ as a highest weight category.
Note that the connected components of the Bruhat order are precisely the linkage classes.
As in~\cite{MR2734963} we define the core of $\chi_\xi$, which we denote by $\chi'_\xi$. This is the typical central character of $\mathfrak{gl}(m-\sharp\xi|n-\sharp\xi)$, corresponding to a weight in $\xi$ where~$\sharp\xi$ labels on each side are removed in order to create a typical weight. We also fix $w_0^\xi\in W$ as the longest element in the subgroup of the Weyl group of $\mathfrak{gl}(m-k|n-k)$ which stabilises a dominant weight corresponding to~$\chi'_\xi$.
We will use the translation functors on ${\mathcal O}$ introduced in~\cite{Brundan} and studied further in~\cite{BLW, Ku}. Denote by $U={\mathbb C}^{m|n}$ the tautological module and let $F$ (resp. $E$) be the exact endofunctor of ${\mathcal O}_{{\mathbb Z}}$ defined by tensoring with $U$ (resp. $U^\ast$). Then $\{F_i\,|\,i\in{\mathbb Z}\}$ and $\{E_i\,|\,i\in{\mathbb Z}\}$ are the subfunctors of $F$ and $E$ corresponding to projection on certain blocks. According to Theorem 3.10 of~\cite{BLW}, this defines an $\mathfrak{sl}(\infty)$ tensor product categorification on ${\mathcal O}_{{\mathbb Z}}$.
Finally we introduce notation for some structural modules in category~${\mathcal O}$. For each~$\lambda\in{\mathfrak h}^\ast$ we denote the Verma module (the ${\mathfrak g}$-module induced from the one dimensional ${\mathfrak b}$-module on which ${\mathfrak h}$ acts through $\lambda$) by $M(\lambda)$. Its simple top is denoted by $L(\lambda)$. The indecomposable projective cover of $L(\lambda)$ is denoted by $P(\lambda)$ and the indecomposable injective hull by~$I(\lambda)$. The corresponding modules in ${\mathcal O}^0$ are denoted by $L_0(\lambda),M_0(\lambda),P_0(\lambda), I_0(\lambda)$. We denote the Kac modules by $K(\lambda)=U({\mathfrak g})\otimes_{U({\mathfrak g}_0\oplus{\mathfrak g}_1)}L_0(\lambda)$ and the dual Kac modules by $\overline{K}(\lambda)=U({\mathfrak g})\otimes_{U({\mathfrak g}_0\oplus{\mathfrak g}_{-1})}L_0(\lambda)$. These modules are also co-induced, e.g.
\begin{equation}\label{indcoind}U({\mathfrak g})\otimes_{U({\mathfrak g}_0\oplus{\mathfrak g}_{1})}L_0(\lambda)\simeq{\rm Hom}_{U({\mathfrak g}_0+{\mathfrak g}_1)}(U({\mathfrak g}),L_0(\lambda-2\rho_1)).\end{equation}
Note that for structural modules we will sometimes write $L(\mu^\lambda)$ to denote $L(\lambda)$ when~$\lambda\in P_0$, with slight abuse of notation.
By Theorem 25$(i)$ and Corollary 14 of~\cite{CouMaz2}, for any $\lambda\in{\mathfrak h}^\ast$ and $N\in{\mathcal O}$ we have
\begin{equation}
\label{VerH}
{\rm Ext}^i_{{\mathcal O}}(M(\lambda),N)\simeq {\rm Hom}_{{\mathfrak h}}({\mathbb C}_{\lambda},H^i({\mathfrak n},N)),
\end{equation}
with $H^i({\mathfrak n},-)\simeq {\rm Ext}^i_{{\mathfrak n}}({\mathbb C},-)$, the Lie superalgebra cohomology of ${\mathfrak n}$.
\subsection{Brundan-Kazhdan-Lusztig theory} We review a few items of~\cite{Brundan, BLW, CLW}, to which we refer for details, see also the survey \cite{Brundan3}.
Let $\mathbb V$ be the natural $\mathfrak{sl}(\infty)$ module and $\mathbb W$ its restricted dual. The Lie algebra $\mathfrak{sl}(\infty)$ is generated by the Chevalley generators $\{e_i,f_i\,|\,i\in{\mathbb Z}\}$. Then $\Lambda={\mathbb Z}^{m|n}$ naturally parametrises a monomial basis of $\mathbb V^{\otimes m}\otimes \mathbb W^{\otimes n}$. We denote such a monomial basis by $\{v_\lambda \,|\,\lambda\in \Lambda\}$. Identifying $[M(\lambda)]\in K({\mathcal O}_{{\mathbb Z}})$ with $v_\lambda$ then leads to an isomorphism of vector spaces \begin{equation}K({\mathcal O}^{\Delta}_{{\mathbb Z}})\leftrightarrow \mathbb V^{\otimes m}\otimes \mathbb W^{\otimes n},\label{GrothTens}
\end{equation}
with $K({\mathcal O}^{\Delta}_{{\mathbb Z}})$ the Grothendieck group of the exact full subcategory~${\mathcal O}^{\Delta}_{{\mathbb Z}}$ of ${\mathcal O}_{{\mathbb Z}}$, whose objects are
the modules admitting a Verma flag. It follows immediately from the standard filtration of $M(\lambda)\otimes U$ that we can define an $\mathfrak{sl}(\infty)$-action on $K({\mathcal O}_{{\mathbb Z}})$ by $e_i [M]:= [E_i M]$ and $f_i[M]=[F_i M]$. The isomorphism \eqref{GrothTens} then becomes an $\mathfrak{sl}(\infty)$-module isomorphism.
For the quantised enveloping algebra $U_q(\mathfrak{sl}(\infty))$ we denote the corresponding module by $\dot{\mathbb V}^{\otimes m}\otimes \dot{\mathbb W}^{\otimes n}$. It turns out that $\dot{\mathbb V}^{\otimes m}\otimes \dot{\mathbb W}^{\otimes n}$ admits the Lusztig canonical basis. This basis is denoted by $\{\dot{b}_\mu,\, \mu\in \Lambda\}$ and the monomial basis by $\{\dot{v}_\lambda,\, \lambda\in\Lambda\}$. Then the polynomials $d_{\mu,\lambda}(q)$ and the inverse matrix, the KL polynomials $p_{\lambda,\nu}(-q)$, are defined as
$$\dot{b}_\mu=\sum_{\lambda\in\Lambda}d_{\mu,\lambda}(q)\dot{v}_{\lambda}\quad\mbox{and}\quad\dot{v}_{\lambda}=\sum_{\nu\in\Lambda}p_{\lambda,\nu}(-q)\dot{b}_\nu.$$
In \cite{Brundan}, Brundan conjectured and in~\cite{CLW} Cheng, Lam and Wang proved that
\begin{equation}\label{resCheng}(P(\mu): M(\lambda))=d_{\mu,\lambda}(1)=[M(\lambda):L(\mu)].\end{equation}
Furthermore Brundan, Losev and Webster proved in Theorem A of~\cite{BLW} that ${\mathcal O}$ (has a graded lift which) is standard Koszul. This implies that the KL polynomials can be interpreted as a minimal projective resolution of the Verma module, or
\begin{equation}\label{defp}p_{\lambda,\nu}(q)=\sum_{k\ge 0}q^k\dim{\rm Ext}^k_{{\mathcal O}}(M(\lambda), L(\nu)),\end{equation}
see also Section 5.9 of~\cite{BLW}. For $q=-1$, equation~\eqref{defp} is a direct consequence of equation~\eqref{resCheng} and the Euler-Poincar\'e principle.
Take an interval $I\subset {\mathbb Z}$. Let $\mathfrak{sl}(I)\simeq \mathfrak{sl}(|I|+1)$ denote the Lie subalgebra of $\mathfrak{sl}(\infty)$ generated by the
Chevalley generators $\{e_i,f_i\,|\,i\in I\}$. We set $I_+:=I\cup (I+1)$. Then $\Lambda_I$ is the sub-poset of $\Lambda$, consisiting of all vectors in
$\mathbb Z^{m|n}$ with labels in the interval $I_+$. It is clear that $\Lambda_I$ is in bijection with the monomial basis of the $\mathfrak{sl}(I)$-module
${\mathbb V_I}^{\otimes m}\otimes {\mathbb W_I}^{\otimes n}.$
Since $\mathbb W_I$ is isomorphic to $\Lambda^{|I|}\mathbb V_I$ as a $\mathfrak{sl}(I)$-module, $\Lambda_I$ also corresponds to a poset of another highest weight
category.
The relevant category is a subcategory of the parabolic category~${\mathcal O}$ for $\mathfrak{gl}(m+|I|n)$, where the parabolic subalgebra has Levi part
$\mathfrak{gl}(1)^{\oplus m}\oplus\mathfrak{gl}(|I|)^{\oplus n}$, generated by simple modules with suitable restrictions on highest weights,
see Definition~3.13 in~\cite{LW}.
We denote this category by ${\mathcal O}_I'$ and the bijection of the weights in $\Lambda_I$ with the set $\Lambda_I'$ of highest weights of the simple modules in
${\mathcal O}'_I$ by $\phi_I:\Lambda_I\to\Lambda_{I}'$.
In Section 2.8 in~\cite{BLW}, two ideals $\Lambda_{\le I}$ and $\Lambda_{< I}$ in the poset are constructed with the property
$\Lambda_I=\Lambda_{\le I}\backslash \Lambda_{< I}$. Then we have Serre (highest weight) subcategories ${\mathcal O}_{< I}$ and ${\mathcal O}_{\le I}$ in
${\mathcal O}_{{\mathbb Z}}$ and the quotient category~${\mathcal O}_I={\mathcal O}_{\le I}/{\mathcal O}_{< I}$. This is a highest weight category with poset $\Lambda_I$. By the general theory of such
subquotients of highest weight categories, it follows that
\begin{equation}\label{extsubquo}
{\rm Ext}^\bullet_{{\mathcal O}}(M(\mu), L(\lambda))\simeq {\rm Ext}^\bullet_{{\mathcal O}_I}(M(\mu), L(\lambda))
\end{equation} if $\lambda,\mu\in\Lambda_I$, see Section 2.5 in~\cite{BLW}.
It is proved in~\cite{BLW} via uniqueness of tensor product categorifications in~\cite{LW}, that the categories ${\mathcal O}_I$ and ${\mathcal O}_I'$ are equivalent.
\section{Extensions and Kazhdan-Lusztig theory}
\label{secext}
In this entire section we consider ${\mathfrak g}=\mathfrak{gl}(m|n)$ with, unless specified otherwise, distinguished Borel subalgebra. The results extend to $\mathfrak{sl}(m|n)$ if $m\not=n$.
\subsection{Length function and abstract Kazhdan-Lusztig theories}
\label{seclen}
We define a length function $l:\Lambda\times \Lambda \to{\mathbb N}$ on a poset $(\Lambda,\preceq)$ to be a function with domain $\{(\lambda,\mu)\,|\,\mu \preceq \lambda\}$ which satisfies $l(\lambda,\mu)=l(\lambda,\kappa)+l(\kappa,\mu)$ if $\mu\preceq\kappa\preceq\lambda$, with $l(\lambda,\mu)=0$ if and only if $\lambda=\mu$. Note that, in principal, a length function should be a function $l':\Lambda\to{\mathbb Z}$ such that $l(\lambda,\mu):=l'(\lambda)-l'(\mu)$ satisfies the above properties. However, in our case, it is possible to construct such an $l'$ from our $l$ by the procedure in Section 3-g in~\cite{Brundan}. As we will only need the difference in length between two comparable weights, we ignore this technicality.
Before going to $\mathfrak{gl}(m|n)$, we review the length function for any integral (possibly singular) block in (possibly) parabolic category~${\mathcal O}$ for some $\mathfrak{gl}(d)$ with $d\in{\mathbb N}$. In this case we can assume $\lambda$ and $\mu$ to belong to the same (integral) Weyl group orbit. For an integral block of category~${\mathcal O}$ we set $l(\lambda,\mu)=l(\lambda)-l(\mu)$, with $l(\lambda),l(\mu)$ as defined in Theorem 3.8.1 in~\cite{CPS1}. There is a unique dominant element in the orbit of $\lambda$ and $\mu$, which we denote by $\kappa$. Then there are unique elements of the Weyl group $w_1,w_2\in W$ such that these are the longest element satisfying $\mu=w_1\cdot\kappa$ and $\lambda=w_2\cdot\kappa$. Then we have $l(\lambda,\mu)=l(w_1)-l(w_2)$, where~$l:W\to{\mathbb N}$ is the length function on $W$ as a Coxeter group. For an integral block in parabolic category~${\mathcal O}$, we just have the same length function as for the corresponding block in the original category~${\mathcal O}$, but restricted to the poset of weights dominant for the Levi subalgebra. Also the Bruhat order is the restriction of the Bruhat order in the non-parabolic case.
Now we can define a length function for category~${\mathcal O}_{{\mathbb Z}}$ for $\mathfrak{gl}(m|n)$ and distinguished Borel subalgebra. For the cases ${\mathfrak g}=\mathfrak{gl}(2|1)$ and ${\mathfrak g}=\mathfrak{gl}(1|2)$, this will be made explicit in Appendix \ref{gl21}.
\begin{lemma}\label{defL}
For any two $\lambda,\mu\in \Lambda$ with $\mu \preceq\lambda $ and any interval $I$ such that $\lambda,\mu\in\Lambda_I$, set $l_I(\lambda,\mu)=l(\phi_I(\lambda),\phi_I(\mu))$.
This value $l_I(\lambda,\mu)$ does not depend on the particular interval $I$. This leads to a well-defined length function $\mathtt{l}$ on $\Lambda\times \Lambda$
defined by
$$\mathtt{l}(\lambda,\mu)=l_I(\lambda,\mu)\qquad\mbox{for any $I$ such that }\lambda,\mu\in \Lambda_I.$$
\end{lemma}
\begin{proof}
Take $\mu=(\alpha_1,\cdots,\alpha_m|\beta_1,\cdots,\beta_n)\in \Lambda$. We choose some $b\in{\mathbb Z}$ greater than the maximal value of the labels of $\mu$ and an
$a\in{\mathbb Z}$ smaller than the minimum of the labels of~$\mu$. Take the interval $I=[a,b]$. By the description in Section 2.2 in~\cite{BLW} and Section~3.5
in~\cite{LW} we have
\begin{eqnarray*}&&\phi_{[a,b]}(\alpha_1,\cdots,\alpha_m|\beta_1,\cdots,\beta_n)\\
&=&(\alpha_1,\cdots,\alpha_m,b+1,b,\cdots^{\hat{\beta_1}},a,b+1,b,\cdots^{\hat{\beta_2}},a,\cdots,b+1,b,\cdots^{\hat{\beta_n}},a),
\end{eqnarray*}
where~$\hat{c}$ implies that the value $c$ is left out in the sequence of numbers which otherwise descend by $1$ in each step.
We have to prove that $l_{[a,b]}(\lambda,\mu)=l_{[a',b']}(\lambda,\mu)$ for $\lambda,\mu\in \Lambda_{[a,b]}$ and any $a'\le a$ and $b'\ge b$. By construction, $\phi_{[a,b]}(\lambda)$ and $\phi_{[a,b]}(\mu)$ are in the same orbit of $S_{m+n(b-a+1)}$. Denote the unique dominant weight in the orbit by $D_{a,b}$. Assume now that $w_\mu^{a,b}$ is the longest element in $S_{m+n(b-a+1)}$ such that
$$\phi_{[a,b]}(\mu)= w_\mu^{a,b} D_{a,b}.$$
We now embed $S_{m+n(b-a+1)}$ into $S_{m+n(b-a+2)}$ by identifying it with $S_{m+n(b-a+1)}\times S_1^n$. It follows that
$$w_{\mu}^{a-1,b} = x \, w_\mu^{a,b} y\, \quad \in S_{m+n(b-a+2)},$$
where~$y$ is the longest element of the subgroup $S_1^{m+n(b-a+1)}\times S_n$ and $x$ is given by
$$(s_{m+1}s_{m+2}\cdots s_{m+nd})(s_{m+d+2}s_{m+n+2}\cdots s_{m+nd+1})\cdots (s_{m+(n-1)(d+1)+1}\cdots s_{p-1}),$$
where we set $d=b-a+1$ and $p=m+n(b-a+2)=m+nd+n$. As $x$ and $y$ clearly do not depend on $\mu$, we find that indeed
$$l(w_\mu^{a,b})-l(w_\lambda^{a,b})=l(w_\mu^{a-1,b})-l(w_\lambda^{a-1,b}).$$
A similar reasoning for $b\mapsto b+1$ concludes the proof.
\end{proof}
This definition has the following immediate consequence.
\begin{theorem}
\label{KLtheory}
For $\lambda,\mu\in\Lambda$, we have
$${\rm Ext}^i_{{\mathcal O}}(M(\mu),L(\lambda))\not=0\quad\Rightarrow \quad\mu\preceq\lambda\,\mbox{ with }\,i\le \mathtt{l} (\lambda,\mu)\,\mbox{ and }\,i\equiv \mathtt{l} (\lambda,\mu)\,({\rm mod} 2).$$
\end{theorem}
\begin{proof}
Through equations \eqref{defp}, \eqref{extsubquo} and the fact that KL polynomials of ${\mathcal O}_I$ correspond to those in ${\mathcal O}_I'$, this can be reduced to the corresponding statement on extensions between standard and simple modules in singular blocks of parabolic category~${\mathcal O}$ for Lie algebras, by Lemma~\ref{defL}.
This result is known in these categories. We give a sketch of a proof for completeness. For blocks in non-parabolic category~${\mathcal O}$ this is Theorem~3.8.1 in~\cite{CPS1}. To prove it in parabolic category~${\mathcal O}$, we consider the Koszul dual statement, concerning the radical filtration of standard modules, see \cite{Backelin}. Then, by the above, the statement is correct for regular blocks in parabolic category~${\mathcal O}$ for Lie algebras.
The full (dual) result follows immediately from graded translation to the wall, see \cite{Stroppel}.
\end{proof}
By applying the work of Cline, Parshall and Scott this leads to the following corollary.
\begin{corollary}
\label{corsim}
The highest weight category~${\mathcal O}_{{\mathbb Z}}$ with length function $\mathtt{l}$ has an abstract Kazhdan-Lusztig theory, according to Definition 2.1 in~\cite{CPS2}. Consequently, we have
$$\sum_{k\ge 0} q^k \dim{\rm Ext}_{{\mathcal O}}^k(L(\lambda),L(\mu))=\sum_{\nu\in\Lambda} p_{\nu,\lambda}(q)p_{\nu,\mu}(q).$$
\end{corollary}
\begin{proof}
This is a special case of Corollary 3.9 in~\cite{CPS2}, using the duality on ${\mathcal O}$.
\end{proof}
Comparison with equation~\eqref{defp} yields
\begin{equation}
\label{extsimple}
\dim {\rm Ext}^j_{{\mathcal O}}(L(\lambda),L(\mu))=\sum_{i=0}^j\sum_{\nu\in\Lambda}\dim{\rm Ext}^i_{{\mathcal O}}(M(\nu),L(\lambda))\,\dim{\rm Ext}_{{\mathcal O}}^{j-i}(M(\nu),L(\mu)).
\end{equation}
This formula follows also by standard methods from the observation that category~${\mathcal O}$ for~$\mathfrak{gl}(m|n)$ is standard Koszul, see \cite{BLW}, the fact that there are no extensions from Verma modules to dual Verma modules and use of the duality functor.
\begin{remark}
The analogue of equation~\eqref{extsubquo} does not hold for
$${\rm Ext}^\bullet_{{\mathcal O}}(L(\mu), L(\lambda))\leftrightarrow {\rm Ext}^\bullet_{{\mathcal O}_I}(L(\mu), L(\lambda)),$$
which is confirmed by equation~\eqref{extsimple}, as the summation over $\nu$ goes out of $\Lambda_I$. However, using the subsequent Lemma~\ref{vancentre} it is possible to show that for each~$\lambda,\mu\in \Lambda_I$ and a fixed degree $j$, there is an interval $J_{\lambda,\mu,j}$, such that
$$\dim{\rm Ext}^j_{{\mathcal O}}(L(\mu), L(\lambda))\simeq\dim {\rm Ext}^j_{{\mathcal O}_{J_{\lambda,\mu,j}}}(L(\mu), L(\lambda)).$$
\end{remark}
\begin{remark}
\label{lengthnotsame}
\begin{enumerate}
\item The length function $\mathtt{l}$ of Lemma~\ref{defL} does not reduce to the length function for $\mathfrak{gl}(m)\oplus\mathfrak{gl}(n)$, when restricted to one Weyl group orbit.
\item The restriction of $\mathtt{l}$ to $\Lambda^{++}$ does not correspond to the known length function, as defined by Brundan in Section $3$-$g.$ of~\cite{Brundan}. It is impossible to find a length function on $\Lambda$ with such a restriction.
\end{enumerate}
Both properties are illustrated in Appendix \ref{gl21}.
\end{remark}
The Brundan KL theory for ${\mathcal F}$ is also an abstract KL theory, as proved in Theorem~4.51(i) of \cite{Brundan}. There are connections between the KL polynomials for both categories.
\begin{lemma}
\label{lemOF}
The restriction of the polynomials $p_{\lambda,\nu}(q)$ in equation~\eqref{defp} to $\lambda,\nu\in \Lambda^{++}$ gives the KL polynomials of~\cite{Brundan, MR1443186} for the category~${\mathcal F}$. Namely,
$$\dim{\rm Ext}_{{\mathcal O}}^j(M(\lambda),L(\nu))=\dim{\rm Ext}^j_{{\mathcal F}}(K(\lambda),L(\nu)).$$
\end{lemma}
In the proof we will use the Hochschild-Serre (HS) spectral sequences, see e.g. Section 16.6 in~\cite{bookMusson}. Concretely we will apply the HS
spectral sequence for the ideal ${\mathfrak g}_1\subset {\mathfrak n}$:
\begin{equation}
\label{HSss}
H^p({\mathfrak n}_0,H^q({\mathfrak g}_1,L(\mu)))\,\Rightarrow\, H^{p+q}({\mathfrak n},L(\mu)).
\end{equation}
\begin{proof}
Equation~\eqref{VerH} allows us to use the spectral sequence \eqref{HSss}. The fact that $H^q({\mathfrak g}_1,L(\nu))$ is a finite dimensional ${\mathfrak g}_0$-module and Kostant
cohomology (see Theorem 5.14 of \cite{Kostant}) imply that for any $\lambda\in\Lambda^{++}$
$${\rm Hom}_{{\mathfrak h}}( {\mathbb C}_\lambda,H^p({\mathfrak n}_0,H^q({\mathfrak g}_1,L(\mu)))=0 \quad\text{for}\quad p>0.$$
Hence the spectral sequence collapses and we obtain
$${\rm Hom}_{{\mathfrak h}}( {\mathbb C}_\lambda,H^j({\mathfrak n}, L(\nu)))={\rm Hom}_{{\mathfrak b}_0}({\mathbb C}_\lambda,H^j({\mathfrak g}_1,L(\nu)))={\rm Hom}_{{\mathfrak g}_0}(L_0(\lambda),H^j({\mathfrak g}_1,L(\nu))),$$
since $H^j({\mathfrak g}_1,L(\nu))$ is a semisimple ${\mathfrak g}_0$-module.
By the analogue of equation~\eqref{VerH} in the category~${\mathcal F}$ we have
$$\dim{\rm Ext}_{{\mathcal O}}^j(K(\lambda),L(\nu))=\dim{\rm Hom}_{{\mathfrak g}_0}(\L_0(\lambda),H^j({\mathfrak g}_1,L(\nu))).$$
Hence the statement follows.
\end{proof}
\subsection{Further vanishing properties of extensions}
In this subsection we continue to consider ${\mathfrak g}=\mathfrak{gl}(m|n)$. We assume that the Borel subalgebra is the distinguished or anti-distinguished one. The essential (and characterising) property of these choices is that every positive root which is simple in $\Delta^+_{\bar{0}}$ is also simple in $\Delta^+$.
\begin{lemma}
\label{lemtwist}
Consider $\lambda,\mu\in{\mathfrak h}^\ast$ and a simple reflection $s\in W$.
\begin{enumerate}[$($i$)$]
\item If $s\cdot\lambda=\lambda$ and $s\cdot\mu <\mu$, we have ${\rm Ext}_{{\mathcal O}}^{\bullet}(M(\lambda),L(\mu))=0$.
\item If $s\cdot\lambda<\lambda$ and $s\cdot\mu<\mu$, we have ${\rm Ext}^j_{{\mathcal O}}(M(s\cdot\lambda),L(\mu))\simeq {\rm Ext}^{j-1}_{{\mathcal O}}(M(\lambda),L(\mu))$.
\end{enumerate}
\end{lemma}
\begin{proof}
We prove these results using the right exact twisting functors $T_s$ and the left derived functors ${\mathcal L}_iT_s$ on category~${\mathcal O}$, as studied in~\cite{CMW, CouMaz}. For both $(i)$ and $(ii)$ we have $${\mathcal L}_iT_s\left(M(\lambda)\right)\simeq\delta_{i,0}M(s\cdot\lambda)\qquad\mbox{and}\qquad {\mathcal L}_iT_s( L(\mu))\simeq\delta_{i,1}L(\mu),$$ see Lemmata 5.4 and 5.7 and Theorem 5.12$(i)$ in~\cite{CouMaz}. The combination of the displayed equations with Proposition 5.11 in~\cite{CouMaz}, leads to
$${\rm Ext}^j_{{\mathcal O}}(M(s\cdot\lambda),L(\mu))\simeq {\rm Ext}^{j-1}_{{\mathcal O}}(M(\lambda),L(\mu)).$$
A step-by-step explanation of this procedure can be found in the proof of Proposition 3 in~\cite{MR2366357}. This yields $(ii)$. In case $s\cdot\lambda=\lambda$, we obtain by iteration that $${\rm Ext}^j_{{\mathcal O}}(M(\lambda),L(\mu))={\rm Hom}_{{\mathcal O}}(M(\lambda),L(\mu))=0,$$ proving $(i)$.
\end{proof}
An alternative proof of Lemma~\ref{lemtwist} follows from equation~\eqref{VerH} and a HS spectral sequence reducing the statement to an $\mathfrak{sl}(2)$ property.
\begin{remark}
The combination of Lemmata \ref{lemtwist} and \ref{lemOF} completely determines the KL polynomials $p_{\lambda,\nu}$ in case
$\nu\in\Lambda^{++}$ in terms of the KL polynomials for ${\mathcal F}$.
\end{remark}
Next we prove that some KL polynomials can be obtained from the ones for the underlying Lie algebra.
\begin{lemma}\label{KLorbit}
For $\mu,\lambda\in {\mathfrak h}^\ast$ in the same orbit of the Weyl group we have
\begin{eqnarray*}\dim{\rm Ext}^j_{{\mathcal O}}(M(\mu),L(\lambda))&=&\dim{\rm Ext}^j_{{\mathcal O}^0}(M_0(\mu),L_0(\lambda))\qquad\mbox{and}\\
\dim{\rm Ext}^1_{{\mathcal O}}(L(\mu),L(\lambda))&=&\dim{\rm Ext}^1_{{\mathcal O}^0}(L_0(\mu),L_0(\lambda)).
\end{eqnarray*}
\end{lemma}
\begin{proof}
The second equality follows from the first and equation~\eqref{extsimple}.
Now we prove the first equality. We use the element $z\in {\mathfrak h}$, defined in equation~\eqref{gradel}. Note that ${\mathfrak h}$ and therefore $z$ act on
$H^{p+q}({\mathfrak n},L(\lambda)$ as well as on the spectral sequence term $H^p({\mathfrak n}_0,H^q({\mathfrak g}_1,L(\lambda)))$. Recall that $H^q({\mathfrak g}_1,L(\lambda)))$ is a
${\mathfrak g}_0$-subquotient of $S^q{\mathfrak g}_{-1}\otimes\Lambda{\mathfrak g}_{-1}\otimes L_0(\lambda)$. Therefore, if $q>0$, then
any weight $\nu$ of $H^q({\mathfrak g}_1,L(\lambda)))$ satisfies
the condition $\nu(z)<\lambda(z)$.
Next we notice that $\lambda(z)=\mu(z)$. Therefore,
$${\rm Hom}_{{\mathfrak h}}({\mathbb C}_\mu,H^p({\mathfrak n}_0,H^q({\mathfrak g}_1,L(\lambda))))=0\qquad\mbox{if}\quad q>0.$$
This yields
$${\rm Hom}_{{\mathfrak h}}({\mathbb C}_\mu,H^j({\mathfrak n},L(\lambda)))\simeq{\rm Hom}_{{\mathfrak h}}({\mathbb C}_\mu,H^j({\mathfrak n}_0,L(\lambda)^{{\mathfrak g}_1}))\simeq {\rm Ext}_{{\mathcal O}^0}^j(M_0(\mu),L_0(\lambda)), $$
since $L(\lambda)^{{\mathfrak g}_1}\simeq L_0(\lambda)$, concluding the proof. An alternative proof of the first equality follows from the algorithm to calculate the canonical basis in Section 3 of~\cite{Brundan3}.
\end{proof}
The following vanishing lemma will be useful in Section \ref{secfpd}.
\begin{lemma}
\label{vancentre}
Consider $\lambda,\mu\in{\mathfrak h}^\ast$, then we have
$${\rm Ext}^j_{{\mathcal O}}(M(\lambda),L(\mu))\not=0\quad\Rightarrow \quad \max\{j-l(w_0),0\}\, \le \,\mu(z)-\lambda(z)\, \le\, j+\dim{\mathfrak g}_1,$$
with $z\in\mathfrak{z}({\mathfrak g}_0)\subset{\mathfrak h}$ as in equation~\eqref{gradel} and $w_0$ standing for the longest element in $W$.
\end{lemma}
\begin{proof}
We use the reformulation~\eqref{VerH} and the spectral sequence \eqref{HSss}. This implies that the extension vanishes unless $\lambda=w\cdot\nu$,
for some $w\in W$ and a highest weight $\nu$ of a simple ${\mathfrak g}_0$-subquotient of $H^q({\mathfrak g}_1,L(\mu))$. Moreover,
$j-q\le l(w_0)=\dim {\mathfrak n}_0$.
In particular we have
$$[S^q({\mathfrak g}_{-1})\otimes \Lambda({\mathfrak g}_{-1})\otimes L_0(\mu):L_0(w^{-1}\cdot\lambda)]\not=0.$$
Hence $q\leq \mu(z)-\lambda(z)$. This implies
$$j-l(w_0)\, \le\, q\,\le \,\mu(z)-\lambda(z)$$
and
$$j+\dim{\mathfrak g}_1\geq q+\dim{\mathfrak g}_1\geq \mu(z)-\lambda(z).$$
The proposed inequalities thus follow.
\end{proof}
\subsection{Socle of the tensor space}
\begin{theorem}
The socle of the $\mathfrak{sl}(\infty)$-module $\mathbb V^{\otimes m}\otimes \mathbb W^{\otimes n}$ contains $\{b_{\mu}\,|\,\mu\in \Lambda\}$.
Furthermore, under the $\mathfrak{sl}(\infty)$-module morphism $\mathbb V^{\otimes m}\otimes \mathbb W^{\otimes n}\simeq K({\mathcal O}^{\Delta}_{{\mathbb Z}})$, this socle corresponds to the subgroup of $K({\mathcal O}^{\Delta}_{{\mathbb Z}})$ generated by the projective modules.
\end{theorem}
\begin{proof}
We denote the socle by $\mathbb V^{\{m,n\}}$. It is proved in Theorem 2.2 of~\cite{Penkov} that $\mathbb V^{\{m,n\}}$ corresponds to the
intersection of the kernels of all contraction maps $$ \mathbb V^{\otimes m}\otimes \mathbb W^{\otimes n}\twoheadrightarrow \mathbb V^{\otimes m-1}\otimes \mathbb W^{\otimes n-1}.$$
So ${v}_\lambda$ is in $\mathbb V^{\{m,n\}}$ if and only if $\lambda$ is typical.
According to the finiteness discussion of the algorithm to compute Lusztig's canonical basis in Section 3 of~\cite{Brundan3} it follows that for any $\nu\in\Lambda$,
$$\dot{b}_\nu=\sum_{\lambda} f_\lambda(q) A_\lambda \dot{v}_\lambda,$$
for a finite sum of typical (dominant) $\lambda\in\Lambda$, certain $A_\lambda\in U_q(\mathfrak{sl}(\infty))$ and $f_{\lambda}(q)\in{\mathbb Z}[q,q^{-1}]$. Evaluating this in $q=1$ yields the first part.
To prove the second statement we consider the description of the socle in Theorem~2.1 of~\cite{Penkov}. This implies that the socle is the direct sum of highest weight modules (with respect to the system of positive roots introduced there), where the highest weight vectors are linear combinations of $v_\lambda$ for typical $\lambda\in\Lambda$. This implies that the socle is inside the submodule generated by the basis $\{b_{\mu}\,|\,\mu\in \Lambda\}$.
\end{proof}
\section{Finiteness of homological dimension and the associated variety}
\label{secfpd}
In this section (except Lemma~\ref{propind}) ${\mathfrak g}$ is in the list \eqref{listA}. In Section 4.4 of~\cite{preprint}, Mazorchuk proved that the finitistic global dimension of (parabolic) category~${\mathcal O}$ is finite for classical Lie superalgebras. In this section, we relate the finiteness of projective dimensions with the associated variety of \cite{Duflo, Vera}. For any $M\in{\mathfrak g}$-mod we define its associated variety $X_M$ as
\begin{equation}\label{defasva}
\begin{aligned}
&X=\{x\in{\mathfrak g}_{\bar{1}}\,\,\mbox{with}\,\, [x,x]=0 \},\\
&X_M=\{x\in X\,\,\mbox{with}\,\, xM\not=\ker_xM\}.
\end{aligned}\end{equation}
Let ${\mathcal A}_0$ denote the additive and Karoubian closure of the category of modules of the form ${\rm Ind}^{\fg}_{\fg_{\bar{0}}} N_0$ with $N\in {\mathcal O}^0$.
This category does not need to be abelian. By iteration we define ${\mathcal A}_j$ as the full subcategory of ${\mathcal O}$, containing the modules in ${\mathcal A}_{j-1}$ as
well as modules in ${\mathcal O}$ which can be written as a kernel or cokernel of an injective or surjective morphism between two modules in ${\mathcal A}_{j-1}$ or an
extension of two modules in ${\mathcal A}_{j-1}$. By taking the full subcategory of ${\mathcal O}$ consisting of modules in some ${\mathcal A}_j$ for $j\in{\mathbb N}$, we obtain an abelian
full Serre subcategory which we denote by~${}^{({\mathfrak g},{\mathfrak g}_{\bar{0}})}{\mathcal O}$. The equivalence between $(iii)$ and $(iv)$ in the following theorem
actually implies firstly that ${}^{({\mathfrak g},{\mathfrak g}_{\bar{0}})}{\mathcal O}\simeq {\mathcal A}_{2l(w_0)}$ and secondly that this category can also be obtained by a similar procedure using only cokernels.
\begin{theorem}
\label{finpdind}
Let ${\mathfrak g}$ be a Lie superalgebra from the list ~\eqref{listA} and $M\in{\mathcal O}$. Denote by ${\rm pd}_{{\mathcal O}}$ the projective dimension in ${\mathcal O}$. The following conditions are
equivalent:
\begin{eqnarray*}
(i)\quad {\rm pd}_{{\mathcal O}}M<\infty;\quad&(iii)& M\in {}^{({\mathfrak g},{\mathfrak g}_{\bar{0}})}{\mathcal O};\\
(ii)\;\,\quad X_M=\{0\};\quad&(iv)& {\rm pd}_{{\mathcal O}}M\le 2l(w_0).
\end{eqnarray*}
Consequently we have $\rm fin.dim{\mathcal O}=2l(w_0)$.
\end{theorem}
The statement on the finitistic global dimension will be improved upon, by determining it each for each indecomposable block individually in Theorem \ref{fdblock}.
\begin{remark}
As the proof of Theorem \ref{finpdind} reveals, for arbitrary basic classical Lie superalgebra we still have the property $$(i)\Leftrightarrow (iii)\Leftrightarrow (iv) \Rightarrow (ii).$$
However, already for $\mathfrak{sl}(1|1)$ we have $(ii)\not\Rightarrow (i)$, by Remark \ref{error}.
\end{remark}
The remainder of this section is devoted to proving this theorem.
\begin{lemma}
\label{propind}
Let ${\mathfrak g}$ be a classical Lie superalgebra, $M\in{\mathcal O}$ and $N_0\in {\mathcal O}^0$. We have
\begin{enumerate}[$($i$)$]
\item ${\rm pd}_{{\mathcal O}}M\ge {\rm pd}_{{\mathcal O}^0}{\rm Res}^{\fg}_{\fg_{\bar{0}}} M$;
\item If $M$ is a direct summand of ${\rm Ind}^{\fg}_{\fg_{\bar{0}}} N_0$, then ${\rm pd}_{{\mathcal O}}M\le {\rm pd}_{{\mathcal O}^0}N_0$;
\item ${\rm pd}_{{\mathcal O}}{\rm Ind}^{\fg}_{\fg_{\bar{0}}} N_0={\rm pd}_{{\mathcal O}^0}N_0$.
\end{enumerate}
Consequently we have $\rm fin.dim{\mathcal O}=2l(w_0)$.
\end{lemma}
\begin{proof}
The functors ${\rm Res}^{\fg}_{\fg_{\bar{0}}}$ and ${\rm Ind}^{\fg}_{\fg_{\bar{0}}}$ are exact and map projective modules to projective modules, this implies $(i)$ and $(ii)$.
Claim $(iii)$ follows from combining $(i)$ and $(ii)$ since the ${\mathfrak g}_{\bar{0}}$-module ${\mathbb C}$ is a direct summand of $\Lambda{\mathfrak g}_{\bar{1}}$, so $N_0$ is a direct summand ${\rm Res}^{\fg}_{\fg_{\bar{0}}} {\rm Ind}^{\fg}_{\fg_{\bar{0}}} N_0\simeq \Lambda{\mathfrak g}_{\bar{1}}\otimes N_0$.
Property $(iii)$ implies $\rm fin.dim{\mathcal O}\ge 2l(w_0)$ and the reversed inequality is Theorem~3 of~\cite{preprint}.
\end{proof}
\begin{proposition}
\label{propflag}
Take ${\mathfrak g}$ to be in~\eqref{listA}. Assume that $M\in{\mathcal O}$, is ${\mathfrak g}_{-1}$-free (resp. ${\mathfrak g}_{1}$-free), then $M$ has a Kac flag (resp. dual Kac flag).
\end{proposition}
\begin{proof}
Take $M$ to be ${\mathfrak g}_{-1}$-free. The ${\mathfrak g}_0$-module $N:=M/{\mathfrak g}_{-1}M$ decomposes according to the eigenvalues of $z$, in equation~\eqref{gradel}, as
$N=\bigoplus_{\alpha\in{\mathbb R}}N_\alpha.$ Since $M$ is finitely generated there is only a finite amount of $\alpha$ for which $N_\alpha\not=0$. We take $\alpha_0$ to be the highest of these. Then $N_{\alpha_{0}}$ is isomorphic to a ${\mathfrak g}_0\oplus{\mathfrak g}_1$-submodule of ${\rm Res}^{{\mathfrak g}}_{{\mathfrak g}_0\oplus{\mathfrak g}_1}M$. Since $M$ is ${\mathfrak g}_{-1}$-free, we find
$$U({\mathfrak g})\otimes_{U({\mathfrak g}_0\oplus {\mathfrak g}_{1})}N_{\alpha_0}\hookrightarrow M.$$
Since $U({\mathfrak g})\otimes_{U({\mathfrak g}_0\oplus {\mathfrak g}_{1})}N_{\alpha_0}$ clearly has a filtration by Kac modules, the proof can be completed iteratively by considering the cokernel of
the above morphism.
The proof for a ${\mathfrak g}_1$-free $M$ is identical.
\end{proof}
\begin{remark}\label{error}
For ${\mathfrak g}=\mathfrak{sl}(n|n)$, the element $z\in\mathfrak{z}({\mathfrak g}_0)$ does not exist, which leads to counterexamples of Proposition \ref{propflag} and Theorem \ref{finpdind}. Consider $\mathfrak{sl}(1|1)=\langle x,y,e\rangle$ with
$$[x,x]=0=[y,y],\,\quad [x,y]=e,\,\quad [e,x]=0=[e,y].$$ The two dimensional module $V=\langle v_1,v_2\rangle$ with action $xv_1=v_2=yv_1$
and with trivial action of $e$ satisfies $X_V=\{0\}$ but has no (dual) Kac flag and is not projective in ${\mathcal F}={\mathcal O}$.
\end{remark}
\begin{lemma}
\label{finpdflag}
If $M\in{\mathcal O}$ admits a Kac flag and a dual Kac flag, then ${\rm pd}_{{\mathcal O}}M< \infty$.
\end{lemma}
\begin{proof}
We will prove ${\rm id}_{{\mathcal O}}M<\infty$, for the injective dimension, which is equivalent by Section 3 in~\cite{preprint}.
The proof could also be done immediately for projective dimension using an unconventional definition of the Kac modules. For any $\lambda,\mu\in{\mathfrak h}^\ast$, we prove
$${\rm Ext}^j_{{\mathcal O}}(K(\mu),\overline{K}(\lambda))=0={\rm Ext}^j_{{\mathcal O}}(\overline{K}(\lambda), K(\mu))\qquad\mbox{for}\quad j>2l(w_0).$$
Indeed, applying Frobenius reciprocity twice (using Theorem~25$(i)$ in~\cite{CouMaz2} and equation~\eqref{indcoind}) yields
$${\rm Ext}^j_{{\mathcal O}}(\overline{K}(\lambda), K(\mu))={\rm Ext}^j_{{\mathcal O}^0}(L_0(\lambda),L_0(\mu-2\rho_1)),$$
and a similar argument holds for the other equality.
In particular this implies that for any $\lambda,\mu\in{\mathfrak h}^\ast$ we have
\begin{equation}\label{vanishM}{\rm Ext}^j_{{\mathcal O}}(K(\mu),M)=0={\rm Ext}^j_{{\mathcal O}}(\overline{K}(\lambda), M)\qquad\mbox{for}\quad j>2l(w_0).\end{equation}
Since $M$ is finitely generated, the element $z\in\mathfrak{z}({\mathfrak g}_0)$ in equation~\eqref{gradel} has eigenvalues in an interval of finite length $p$. We prove that for any $\alpha\in{\mathfrak h}^\ast$,
$${\rm Ext}^j_{{\mathcal O}}(L(\alpha),M)=0\qquad \mbox{if}\qquad j>p+\dim{\mathfrak g}_1+4l(w_0).$$
Assume that the above extension would not be zero. The short exact sequence $N\hookrightarrow K(\alpha)\twoheadrightarrow L(\alpha)$ and the vanishing properties in~\eqref{vanishM} imply that there must be a $\beta\in{\mathfrak h}^\ast$ for which $L(\beta)$ is a subquotient of $N$ (and therefore satisfies $\beta(z)\le \alpha(z)-1$) such that
$${\rm Ext}^{j-1}_{{\mathcal O}}(L(\beta),M)\not=0.$$
This procedure and the dual one using dual Kac modules can be repeated so that we come to the conclusion that there must exist $\kappa,\nu\in {\mathfrak h}^\ast$ with $\kappa(z)\ge\alpha(z)+j-2l(w_0)$ and $\nu(z)\le \alpha(z)-j+2l(w_0)$ such that
\begin{equation}\label{ext2l}{\rm Ext}^{2l(w_0)}_{{\mathcal O}}(L(\kappa),M)\not=0\qquad\mbox{and}\qquad {\rm Ext}^{2l(w_0)}_{{\mathcal O}}(L(\nu),M)\not=0.\end{equation}
The combination of equation~\eqref{extsimple} and Lemma~\ref{vancentre} yields that for any two $\mu,\mu'$ we have
\begin{equation*}{\rm Ext}^i_{{\mathcal O}}(L(\mu),L(\mu'))=0\qquad \mbox{unless}\qquad |\mu(z)-\mu'(z)|\le \dim{\mathfrak g}_1+i.\end{equation*}
Equation~\eqref{ext2l} therefore implies that both $\kappa(z)$ and $\nu(z)$ must lie in an interval of length $p+2(\dim{\mathfrak g}_1+2l(w_0))$. However, the construction above implies that
$$\kappa(z)-\nu(z)\ge 2j-4l(w_0),\quad \mbox{with} \quad j>p+\dim{\mathfrak g}_1+4l(w_0).$$
This means we have proved that
$${\rm id}_{{\mathcal O}}M < p+\dim{\mathfrak g}_1+4l(w_0),$$
for some finite $p\in{\mathbb N}$.
\end{proof}
\begin{lemma}\label{auxass1} Let $\mathfrak l={\mathfrak l}_{\bar{1}}$ be a finite dimensional abelian Lie superalgebra with trivial even part, equipped with a ${\mathbb Z}$-grading
$\mathfrak l=\mathfrak l_0\oplus\mathfrak l_1$. Let $M$ be a graded $\mathfrak l$-module (may be infinite dimensional) and assume there exists $k\in{\mathbb Z}$ such that $M_j=0$ for
$j\geq k$. If $M$ is free over $\mathfrak l_0$ and $\mathfrak l_1$, then it is also free over $\mathfrak l$.
\end{lemma}
\begin{proof} Let $\{v_1,\dots,v_p\}$ be a basis of $\mathfrak l_0$ and $\{u_1,\dots,u_q\}$ a basis of $\mathfrak l_1$, set
$v=v_1\dots v_p$ and $u=u_1\dots u_q$, both are elements in $U(\mathfrak l)=\Lambda(\mathfrak l)$.
Let $M'\subset M$ be a maximal graded subspace such that $M'\xrightarrow{vu}vu M$ is an isomorphism and $N$ be the submodule generated by $M'$. Then $N$ is
free over $\mathfrak l$, therefore $N$ is both projective and injective. Let $M''=M/N$, then $M\simeq M''\oplus N$. We claim that $M''=0$.
Assume the opposite. Note that $M''$ satisfies all the assumptions of the lemma and $vuM''=0$.
Pick up the maximal $j$ such that
$M''_j\neq 0$. Then $M''_j$ is a free $\mathfrak l_0$-module and one can find $m\in M''_j$ such that $vm\neq 0$. On the other hand, $\mathfrak l_1 m=0$. Since
$M''$ is free over $\mathfrak l_1$ we obtain that $m=u m'$ for some $m'\in M''$. This implies that $v u m'\neq 0$, which is a contradiction.
\end{proof}
\begin{lemma}\label{auxass2} Consider $M\in {\mathcal O}$ which satisfies $X\cap {\mathfrak g}_{\pm1}=\{0\}$. Then $M$ is ${\mathfrak g}_{\pm1}$ free.
\end{lemma}
\begin{proof}
We will prove the statement if $X\cap {\mathfrak g}_{1}=\{0\}$. The case $X\cap {\mathfrak g}_{-1}=\{0\}$ is similar. Consider ${\mathfrak g}=\mathfrak{gl}(m|n)$ with $m>1$.
The proof is by induction on $m$, using the previous
lemma.
Define the grading on ${\mathfrak g}_1$ and on $M$ by setting the degree of the root space ${\mathfrak g}_{\alpha}$ to be $(\alpha,\varepsilon_1)$ and the degree of the weight space
$M_\lambda$ to be $(\lambda,\varepsilon_1)$. Then $\mathfrak l:={\mathfrak g}_1$ and $M$ satisfy all the conditions of Lemma \ref{auxass1}.
Now ${\mathfrak l}_0\subset\mathfrak{gl}(m-1|n)$, with ${\mathfrak l}_0=\mathfrak{gl}(m-1|n)_1$. As a $\mathfrak{gl}(m-1|n)$-module, every graded component of $M$, with respect to
the above grading, is a direct summand of $M$ and an object in category ${\mathcal O}$ for $\mathfrak{gl}(m-1|n)$. By the induction hypothesis, $M$ is thus free over ${\mathfrak l}_0$.
Similarly, ${\mathfrak l}_1=\mathfrak{gl}(1|n)_1$. Again we find that $M$, now regarded as a $\mathfrak{gl}(1|n)$-module,
decomposes into modules belonging to category ${\mathcal O}$.
For this one could consider the graded components with respect to the grading given by
$\lambda\mapsto -\sum_{i=2}^m(\lambda,i\varepsilon_i)$.
Again by the induction assumption $M$ is free over ${\mathfrak g}_1$.
Lemma \ref{auxass1} then implies that $M$ is free over ${\mathfrak g}_1$. The base case $m=1$ can be covered by similar induction on $n$.
\end{proof}
\begin{proof}[Proof of Theorem \ref{finpdind}]
The equivalence of $(i)\Leftrightarrow (iv)$ follows from Lemma~\ref{propind}.
Assume that $M\in{\mathcal O}$ is a direct summand of a module induced from one in~${\mathcal O}^0$. Lemma~\ref{propind}$(ii)$ therefore yields ${\rm pd}_{{\mathcal O}}M<\infty$. Recall the categories ${\mathcal A}_j$ from the beginning of this section. Lemma~6.9 in~\cite{MR2428237} or Section 2.3 in~\cite{CouMaz2} shows that if every module in ${\mathcal A}_{j-1}$ has finite projective dimension in ${\mathcal O}$, then so has every module in ${\mathcal A}_j$. Hence we find $(iii)\Rightarrow (i)$.
Now assume that ${\rm pd}_{{\mathcal O}}M<\infty$ holds. Since $M$ has a finite resolution by projective modules and projective modules are direct summands of modules induced from projective modules in ${\mathcal O}^0$ we obtain $M\in{}^{({\mathfrak g},{\mathfrak g}_0)}{\mathcal O}$. This proves $(i)\Rightarrow (iii)$.
By Lemma \ref{auxass2}, $X_M\cap{\mathfrak g}_{\pm 1}=\{0\}$ implies that $M$ is ${\mathfrak g}_{ \pm 1}$-free, property $(ii)\Rightarrow (i)$ follows from Proposition \ref{propflag} and Lemma~\ref{finpdflag}. Since every projective module is a direct summand in a module induced from ${\mathfrak g}_0$, it has trivial associated variety, see Lemma 2.2(1) in~\cite{Duflo}. If $M$ has a finite resolution by projective modules, it has finite projective dimension as a ${\mathbb C} [x]$-module for any $x\in{\mathfrak g}_{\bar{1}}$ with $[x,x]=0$, so it is ${\mathbb C} [x]$-free. Thus $X_M=\{0\}$ and we obtain $(i)\Rightarrow (ii)$.
The last statement is a special case of Lemma~\ref{propind}.
\end{proof}
\section{$B_{\bar{0}}$-orbits in the self-commuting cone and some results on associated variety in category~${\mathcal O}$}
\label{secassvar}
The associated variety $X_M$ in \eqref{defasva} of an object $M$ in ${\mathcal O}$ is intrinsically not a categorical invariant, contrary to projective dimension and complexity, although results as Theorem~\ref{finpdind} indicate interesting links with categorical invariants. However, it is thus not possible to use the equivalences of categories in~\cite{CMW} to reduce to the integral case. Therefore, throughout the entire section, we consider weights in ${\mathfrak h}^\ast$, not just in $P_0$.
In the first two subsections we will consider the associated variety of modules in category~${\mathcal O}$ for ${\mathfrak g}$ a basic classical Lie superalgebra in the list
\begin{equation}\label{list}
\mathfrak{sl}(m|n), m\neq n;\,\,\mathfrak{gl}(m|n);\,\, \mathfrak{osp}(m|2n);\,\, D(2,1,\alpha);\,\, G(3);\,\, F(4),\end{equation}
with arbitrary Borel subalgebra. In the last three subsections we will focus on ${\mathfrak g}=\mathfrak{gl}(m|n)$ with the distinguished Borel subalgebra.
\subsection{$B_{\bar{0}}$-orbits} Let $B_{\bar{0}}$ be the Borel subgroup of the algebraic group $G_{\bar{0}}$
with Lie algebra~${\mathfrak b}_{\bar{0}}$. The group $B_{\bar{0}}$ acts on $X$ of \eqref{defasva} by adjoint action. If $M$ is in category~${\mathcal O}$, then the simply connected cover of
$B_{\bar{0}}$ acts on $M$. Thus the associated variety $X_M$, as defined in equation~\eqref{defasva}, is a $B_{\bar{0}}$-invariant subvariety of $X$. Therefore it is
important to study $B_{\bar{0}}$-orbits in~$X$. It is proven in~\cite{Duflo} that $X$ has finitely many $G_{\bar{0}}$-orbits. We will show in this subsection that the same is true for $B_{\bar{0}}$-orbits.
Let $S=\{\alpha_1,\dots,\alpha_k\}$ be a set of mutually orthogonal linearly independent isotropic roots and $x_1,\dots, x_k$ be some non-zero elements in the
root subspaces ${\mathfrak g}_{\alpha_1},\dots,{\mathfrak g}_{\alpha_k}$ respectively. Then $x_S:=x_1+\dots+x_k\in X$. For such an $x_S\in X$, we say that its rank is $k=|S|$. Let ${\mathcal S}$ denote the set of all subsets $S$ of mutually orthogonal
linearly independent isotropic roots and $X/B_{\bar{0}}$ denote the set of $B_{\bar{0}}$-orbits in~$X$. In Theorem 4.2 of \cite{Duflo} it was proved that $X/G_{\bar{0}}\,\simeq\, {\mathcal S}/{W}$, now we derive an analogous description of $X/B_{\bar{0}}$.
Define the map
$$\Phi:{\mathcal S}\to X/B_{\bar{0}};\,\quad \Phi(S):=B_{\bar{0}} x_S\\,\, \mbox{ for all }S\in{\mathcal S}.$$
We assume that $\Phi(\emptyset)=0$. Note that $\Phi$ does not depend on a choice of $x_1,\dots,x_k$, since any two such elements are conjugate under the action of a maximal torus in $G_{\bar{0}}$.
\begin{theorem}\label{mainass}
Consider ${\mathfrak g}$ in the list \eqref{list}, the map $\Phi$ is a bijection, so $ X/B_{\bar{0}}\,\simeq\,{\mathcal S}$.
\end{theorem}
\begin{proof} First, we will prove that $\Phi$ is surjective, {\it i.e.} that every~$B_{\bar{0}}$-orbit contains an $x_S$ for some $S\in {\mathcal S}$.
Recall that Theorem 4.2 in~\cite{Duflo} implies that every~$G_{\bar{0}}$-orbit contains an $x_S$ for some $S\in{\mathcal S}$. Due to the Bruhat decomposition
$$G_{\bar{0}}=\bigsqcup_{w\in W}B_{\bar{0}}wB_{\bar{0}},$$
it suffices to prove that for every~$S\in{\mathcal S}$ and $w\in W$, $B_{\bar{0}}wB_{\bar{0}}x_S$ is a union of $B_{\bar{0}}x_{S'}$ for some $S'\in{\mathcal S}$.
Moreover, using induction on the length of $w$ (and $B_{\bar{0}}swB_{\bar{0}}x_S\subset B_{\bar{0}}s B_{\bar{0}} w B_{\bar{0}}x_S$ for a simple reflection $s$), it is sufficient to prove the latter statement only in the case when~$w=r_{\alpha}$ is a simple reflection.
Let $G_\alpha$ be the $SL_2$-subgroup in $G_{\bar{0}}$ associated with the root $\alpha$ and $B_\alpha:=G_\alpha\cap B_{\bar{0}}$. Since
$$B_{\bar{0}}r_{\alpha}B_{\bar{0}}\subset B_{\bar{0}}G_\alpha,$$
we have to show that $G_\alpha x_S$ lies in a union of $B_{\bar{0}}x_{S'}$ for some $S'\in{\mathcal S}$.
We need the following well-known facts about root system of the superalgebras in \eqref{list}. By odd $\alpha$-chain we mean the maximal subset of odd roots of the form $\beta+p\alpha$ with $p\in{\mathbb Z}$ for some $\beta\in\Delta$.
\begin{enumerate}
\item The length of any odd $\alpha$-chain consisting of odd roots is at most $3$;
\item If $\beta$ is an isotropic root then either $\beta+\alpha$ is not a root or $\beta-\alpha$ is not a root;
\item If $S\in {\mathcal S}$, then at most $2$ roots of $S$ are not preserved by $r_{\alpha}$.
\end{enumerate}
We consider the three cases allowed by statement $(3)$ individually. If all roots of $S$ are preserved by $r_\alpha$, then $G_\alpha x_S=x_S$ and the statement is trivial.
Assume that there exists exactly one root $\alpha_i\in S$, which is not preserved by $r_{\alpha}$. Consider the $G_{\alpha}$-submodule $V_i\subset {\mathfrak g}_{\bar{1}}$ generated
by $x_i$.
By (2) $x_i$ and $r_{\alpha}(x_i)$ are the lowest and the highest weight vectors in $V_i$ and from representation theory of $SL_2$ we have
$$G_{\alpha}x_i=B_{\alpha}x_i\cup B_{\alpha}r_{\alpha}(x_i).$$ Since for any $\alpha_j\in S$ with $j\neq i$ we have $G_{\alpha}x_j=x_j$, we obtain
$$G_{\alpha}x_S=B_{\alpha}x_S\cup B_{\alpha}x_{r_{\alpha}(S)}\subset B_{\bar{0}} x_S\cup B_{\bar{0}} x_{r_\alpha(S)}.$$ Hence the statement is proved in this case.
Finally, assume that there are two distinct roots $\alpha_i,\alpha_j\in S$ which are not preserved by $r_{\alpha}$. Since this case is only possible for
Lie superalgebras of defect greater than $1$, we may assume that ${\mathfrak g}$ is either general linear or orthosymplectic. In this case
$\alpha_i=\pm\varepsilon_a\pm\delta_b$ and $\alpha_j=\pm\varepsilon_c\pm\delta_d$ for some $a\neq c$ and $b\neq d$. That implies that $\alpha_i,\alpha_j$ and
$\alpha$ are roots of some root subalgebra ${\mathfrak g}'$ isomorphic to $\mathfrak{sl}(2|2)$. Furthermore, the corresponding subgroup $G'_{\bar{0}}$ preserves $x_l$ for
all $l\neq i,j$. Therefore it suffices to check the analogous statement for $G'$. It can be done by direct
computation and we leave it to the reader.
Now we will prove that $\Phi$ is injective, {\it i.e.} $x_{S'}\in B_{\bar{0}}x_S$ implies $S=S'$. Assume that $x_{S'}=\operatorname{Ad}_g x_S$ for some $g\in B_{\bar{0}}$.
Note $|S|=|S'|$ by the property $X/G_{\bar{0}}\simeq{\mathcal S}/{W} $ of~\cite{Duflo}. Let
$$\mathfrak m=\bigoplus_{\alpha\in S}{\mathfrak g}_\alpha,\quad \mathfrak m'=\bigoplus_{\alpha\in S'}{\mathfrak g}_\alpha,\quad {\mathfrak h}'=\operatorname{Ad}_g({\mathfrak h}).$$
Then $\mathfrak m$ (resp. $\mathfrak m'$) is the minimal ${\mathfrak h}$-submodule of ${\mathfrak g}$ containing $x_S$ (resp. $x_{S'}$). On the other hand, $\mathfrak m'$ is also the minimal
${\mathfrak h}'$-submodule containing $x_{S'}$. Furthermore $\operatorname{Ad}_gh-h\in {\mathfrak n}^+_{\bar{0}}$ for all $h\in {\mathfrak h}$, which implies that the spectra of $h$ and of
$\operatorname{Ad}_gh$ in $\mathfrak m'$ coincide. Since the former is $\{\alpha(h)\,|\,\alpha\in S\}$ and the latter is $\{\alpha(h)\,|\,\alpha\in S'\}$,
we obtain $S=S'$.
\end{proof}
For any ${\mathfrak g}$-module $M\in{\mathcal O}$, we introduce the notation
$${\mathcal S}(M)=\{S\in{\mathcal S} | \Phi(S)\subset X_M\}.$$
In particular, Theorem \ref{mainass} implies that for $M,N\in{\mathcal O}$ we have ${\mathcal S}(M)={\mathcal S}(N)$ if and only if $X_M=X_N$.
\subsection {General properties of the associated variety for modules in category~${\mathcal O}$} Recall from Lemma~6.2 of \cite{Duflo}
that if $x\in X$ and $M$ is a ${\mathfrak g}$-module, then
$M_x:=\ker x/{\rm im}\, x$ is ${\mathfrak g}_x$-module, where~${\mathfrak g}_x:=\ker{\rm ad} x/{\rm im}\,{\rm ad} x$. If ${\mathfrak g}$ is a basic classical superalgebra, then
${\mathfrak g}_x$ is also a basic classical superalgebra.
For example, if ${\mathfrak g}=\mathfrak{gl}(m|n)$ and the rank of $x$ is $k$,
then ${\mathfrak g}_x$ is isomorphic to $\mathfrak{gl}(m-k|n-k)$. If $x=x_S$ we identify~${\mathfrak g}_x$ with the root subalgebra in ${\mathfrak g}$, whose roots are orthogonal
but not proportional to the roots from $S$.
For a Lie superalgebra ${\mathfrak l}$ we denote by $Z({\mathfrak l})$ the center of the universal enveloping algebra~$U({\mathfrak l})$ and by
$\check{Z}({\mathfrak l})$ the set of central characters. In Section 6 of \cite{Duflo}, a map $\eta:Z({\mathfrak g})\to Z({\mathfrak g}_x)$ was introduced. In Theorem 6.11 of {\it op. cit.}, it was proved that all fibres of the
dual map $\check\eta:\check{Z}({\mathfrak g}_x)\to \check{Z}({\mathfrak g})$ are finite, more precisely any fibre consists of at most two points. If $M$ admits a
generalised central character $\chi$ then $M_x$ is a direct sum of ${\mathfrak g}_x$-modules admitting generalised central characters from $\check\eta^{-1}(\chi)$. The degree of atypicality of the central characters in $\check\eta^{-1}(\chi)$ is equal to the degree of atypicality of $\chi$ minus the rank of $x$, see e.g. equation (3) in~\cite{Vera}. In the case ${\mathfrak g}=\mathfrak{gl}(m|n)$, the map $\check\eta$ is injective and it maps a central character $\chi'$ of ${\mathfrak g}_x$ to the central character $\chi$ of
${\mathfrak g}$ with the same core.
Below we summarise the general properties of the functor ${\mathfrak g}$-mod to ${\mathfrak g}_x$-mod, which sends $M$ to $M_x$.
\begin{lemma}\label{general}Consider ${\mathfrak g}$ in the list \eqref{list}.
\begin{enumerate}
\item For any ${\mathfrak g}$-modules $M$ and $N$ we have $(M\otimes N)_x\simeq M_x\otimes N_x$.
\item Let $M$ be a ${\mathfrak g}$-module from the category~${\mathcal O}$ which admits a generalised central character with atypicality degree $k$ and $S\in{\mathcal S}$. If $S\in {\mathcal S}(M)$, then
$|S|\leq k$.
\item Let $M$ be from the category~${\mathcal O}$, $x=x_S$ for some $S\in{\mathcal S}$ and ${\mathfrak b}_x:={\mathfrak b}\cap{\mathfrak g}_x$, then ${\mathfrak b}_x$ acts locally finitely on $M_x$ and $M_x$ is a weight module.
\end{enumerate}
\end{lemma}
\begin{proof} To prove (1) note that we have the obvious homomorphism $M_x\otimes N_x\to (M\otimes N)_x$ of ${\mathfrak g}_x$-modules. To check that it is an isomorphism
consider $M$ and $N$ as $\mathbb C[x]$-modules. Then
$$M\simeq M^f\oplus M_x,\quad N\simeq N^f\oplus N_x,$$
where~$M^f$ and $N^f$ are free $\mathbb C[x]$-modules. Since $M^f\otimes N$ and $M\otimes N^f$ are free we obtain an isomorphism
$M_x\otimes N_x\simeq (M\otimes N)_x$.
To show (2) use the map $\check\eta$. If $|S|>k$, then $\check\eta^{-1}(\chi)$ is empty and therefore $M_x=0$.
Finally, (3) is trivial since $M_x$ is a subquotient of $M$, and ${\mathfrak b}_x$ acts locally finitely and ${\mathfrak h}\cap {\mathfrak g}_x$ diagonally on $M$.
\end{proof}
\begin{remark} We believe that (3) can be strengthened. Namely, if $M$ lies in the category~${\mathcal O}$, then $M_x$ belongs to the category~${\mathcal O}$ for ${\mathfrak g}_x$, {\it i.e.} $M_x$
is finitely generated. But we do not have a proof of this at the moment.
\end{remark}
\subsection{On associated variety of Verma modules for $\mathfrak{gl}(m|n)$}
In this subsection we assume ${\mathfrak g}=\mathfrak{gl}(m|n)$ and consider the distinguished Borel subalgebra
${\mathfrak b}={\mathfrak b}_{\bar{0}}\oplus {\mathfrak g}_1$. We set ${\mathcal S}(\lambda):={\mathcal S}(M(\lambda))$.
We start with the following technical lemma.
\begin{lemma}\label{technical} Let ${\mathfrak s}$ be $(1|2)$ dimensional superalgebra with odd generators $\xi,\eta$ and even generator $u$ satisfying
$[\xi,\eta]=0$, $[u,\xi]=\xi$ and $[u,\eta]=-\eta$. Assume that $M$ is an ${\mathfrak s}$-module semisimple over ${\mathbb C} u$ and such that the spectrum of $u$ in $M$ is bounded
from above, {\it i.e.} there exists $\gamma_0$ such that $\operatorname{Re}\gamma<\gamma_0$ for any eigenvalue $\gamma$ of $u$. Then
$M_{\eta}=0$ implies $M_{\eta+\xi}=0$.
\end{lemma}
\begin{proof} Note that any $u$-eigenvector $v\in M$ such that $\xi\eta v\neq 0$ generates a projective ${\mathfrak s}$-submodule
(in the category of ${\mathfrak s}$-modules semisimple over $u$). Hence we have a decomposition $M=P\oplus L$ for some projective $P$ and $L$ such that
$\xi\eta L=0$. Obviously, $P_{\xi+\eta}=0$ and we have to check only that $L_{\xi+\eta}=0$. Since $L_\eta=0$, it follows that $L$ is free over $\eta$, hence
we have a $u$-invariant decomposition $L=L'\oplus L''$ such that $\xi L'\subset L''$, $\eta L'=L''$,
$\xi L''=\eta L''=0$ and $\eta:L'\to L''$ is an isomorphism. Let $\zeta: L''\to L'$ be the inverse of $\eta$. Then there are $n'\in {\rm End}(L')$ and $n''\in{\rm End}(L'')$ such that
$$\zeta(\xi+\eta)=\operatorname{Id}_{L'}+n',\quad (\xi+\eta)\zeta=\operatorname{Id}_{L"}+n'',$$
where~$[u,n']=2n'$ and $[u,n'']=2n''$. The latter condition and the assumption on the spectrum of $u$ imply that
$n'$ and $n''$ are locally nilpotent and hence $\operatorname{Id}_{L'}+n'$ and $\operatorname{Id}_{L"}+n''$ are both invertible.
Hence we find $\operatorname{Ker}(\xi+\eta)=L''$ and $\operatorname{Im}(\xi+\eta)=L''$. That implies $M_{\eta+\xi}=0$.
\end{proof}
\begin{lemma}\label{free} If $M\in{\mathcal O}$ is free over $U({\mathfrak g}_{-1})$, then $X_M\subset{\mathfrak g}_1$ and therefore $S\in{\mathcal S}(M)$ implies $S\subset \Delta_{\bar{1}}^+$.
\end{lemma}
\begin{proof} If $x\in {\mathfrak g}_{-1}$, then $M_x=0$ since $M$ is free over $\mathbb C[x]$. Let $x\in X$. Then $x$ can be written uniquely as $x^++x^-$ with
$x^{\pm}\in{\mathfrak g}_{\pm 1}$. We claim that if $x^-\neq 0$, then $M_x=0$. Indeed, we apply Lemma~\ref{technical} with $u=z$, where~$z$ is introduced in equation~\eqref{gradel}, $\xi=x^+$ and $\eta=x^-$ and use the fact
that $M_{\eta}=0$.
\end{proof}
If $\alpha$ is a root of ${\mathfrak g}$, we denote by $X_{\alpha}$ some non-zero element from the root space ${\mathfrak g}_\alpha$ and set $H_{\alpha}:=[X_\alpha,X_{-\alpha}]$.
\begin{lemma}\label{hereditary} Let $M$ be from the category~${\mathcal O}$, and $S\in{\mathcal S}(M)$. Let $S$ be a disjoint union of two subsets $S_{1}$ and $S_{-1}$ and
$h\in{\mathfrak h}^*$ be an element of the Cartan subalgebra, non-negative on all even positive roots. Assume that $\alpha(h)=i$ for all $\alpha\in S_i$ where~$i=\pm 1$.
Then $S_{-1}\in{\mathcal S}(M)$.
\end{lemma}
\begin{proof} Follows again from Lemma~\ref{technical}. We write $X=X^++X^-$, where~$X^\pm =\sum_{\alpha\in S_{\pm 1}}X_\alpha$, set $\xi=X^+,\eta=X^-, u=h$.
\end{proof}
\begin{remark} More generally, it seems plausible that if $S\in{\mathcal S}(M)$, then any subset $S'\subset S$ is also in ${\mathcal S}(M)$.
\end{remark}
\begin{lemma}
\label{reducegl}
Let $S=\{\varepsilon_{i_s}-\delta_{j_s}\,|\, s\in[1,k]\}$ be a set of $k$ mutually orthogonal positive odd roots. Set $a:=\min \{i_s\,|\, s\in[1,k]\}$ and $b:=\max\{ j_s\,|\, s\in [1,k]\}$ and let ${\mathfrak g}'\simeq\mathfrak{gl}(m-a+1|b)$ be the subalgebra of ${\mathfrak g}\simeq\mathfrak{gl}(m|n)$ generated by
$X_{\pm(\varepsilon_i-\delta_j)}$ with $i\ge a$ and $j\le b$. Let $\lambda'$ be the restriction of $\lambda$ to the Cartan subalgebra of ${\mathfrak g}'$ and ${\mathcal S}'$ be the set of subsets of mutually orthogonal linearly independent odd roots in ${\mathfrak g}'$ (clearly, ${\mathcal S}'\subset{\mathcal S}$). Then we have
${\mathcal S}(\lambda')={\mathcal S}(\lambda)\cap {\mathcal S}'$.
\end{lemma}
\begin{proof} Let $S\in{\mathcal S}'$ and $x=x_S$.
The Verma module $M(\lambda)$ is isomorphic to $M(\lambda')\otimes S({\mathfrak g}/({\mathfrak g}'+{\mathfrak b}))$ as a ${\mathfrak g}'$-module and therefore as a $\mathbb C[x]$-module, where we considered adjoint
action on ${\mathfrak g}/({\mathfrak g}'+{\mathfrak b})$. In particular $M(\lambda')$ is a direct summand in $M(\lambda)$. Therefore $M(\lambda')_x\neq 0$ implies $M(\lambda)_x\neq 0$.
On the other hand, if $M(\lambda')_x=0$, then $M(\lambda)_x= 0$ by Lemma~\ref{general}(1).
\end{proof}
\begin{corollary}\label{reductyp}
Consider the set $\{\varepsilon_{i_s}-\delta_{j_s}\}$ of all positive atypical roots for $\lambda\in{\mathfrak h}^\ast$.
For a set $S$ of mutually orthogonal odd positive roots of the form $\varepsilon_i-\delta_j$ with $i> i_s$ and $j< j_s$ for every~$s$, we have $S\not\in {\mathcal S}(\lambda)$.
\end{corollary}
\begin{proof}
This follows from Lemma~\ref{reducegl}, since $\lambda'$ is a typical weight.
\end{proof}
\begin{lemma}
\label{nointeg}
Consider $S=\{\alpha_1,\cdots,\alpha_k\}\in {\mathcal S}$. If $S\in {\mathcal S}(\lambda)$, then $( \lambda,\alpha_i)\in\mathbb Z$ for all $i=1,\dots,k$.
\end{lemma}
\begin{proof}
Let $h_i\in [{\mathfrak g}_{\alpha_i},{\mathfrak g}_{-\alpha_i}]$, such that $\beta(h_i)=(\beta,\alpha_i)$ for any weight $\beta$. For any $t_1,\dots,t_k\in \mathbb C\setminus 0$ consider
the $\mathfrak{sl}(1|1)$-triple $\{x_S,y,h\}$, where~$h=t_1h_1+\dots+t_kh_k$. By assumtion, $S\in {\mathcal S}(\lambda)$, so $M(\lambda)$ cannot be a typical module for the $\mathfrak{sl}(1|1)$-triple and hence there exists a weight $\mu$
such that $\mu(h)=0$. Since $\lambda-\mu$ is an integral linear combination of roots, we have
$$\lambda(h)=\sum_{i=1}^kt_i(\lambda,\alpha_i)\in\sum_{i=1}^k\mathbb Z t_i.$$
For generic choice of $t_1,\dots,t_k$ this implies $(\lambda,\alpha_i)\in\mathbb Z$ for all $i=1,\dots,k$.
\end{proof}
\begin{lemma}\label{dominant} Let ${\mathfrak g}=\mathfrak{gl}(n|n)$ and $\lambda\in P_0^{++}$ of degree of atypicality $n$.
Then $X_{M(\lambda)}={\mathfrak g}_1$.
\end{lemma}
\begin{proof} We consider the ${\mathbb Z}$-grading ${\mathfrak g}={\mathfrak g}_{-1}\oplus{\mathfrak g}_0\oplus{\mathfrak g}_1$. By Lemma~\ref{free} it suffices to prove ${\mathfrak g}_1\subset X_{M(\lambda)}$.
Note that every $M\in {\mathcal O}$ can be equipped with the $\mathbb Z$-grading induced by the action of $z$ in equation \eqref{gradel}, let $M^+$ denote the highest degree component.
As for any $x\in{\mathfrak g}_1$ we have $M(\lambda)^+\subseteq \ker x$, it would suffice to prove $M(\lambda)^+\not\subseteq xM(\lambda)$. We will prove this
by using the property $L(\lambda)^+\not\subseteq xL(\lambda)$ for any $x\in {\mathfrak g}_1$, which is known to be true by Section 10 in~\cite{Duflo}.
Consider the exact sequence $M(\lambda)\xrightarrow{\pi} L(\lambda)\to 0$ of graded ${\mathfrak g}$-modules. Now assume that $M(\lambda)^+\subseteq xM(\lambda)$, so any
$a\in M(\lambda)^+$ can be written as $a=xb$ for some $b\in M(\lambda)$. Then clearly $\pi(a)=x\pi(b)$ and, as the graded map $\pi$ is surjective, we find a
contradiction with the fact that $L(\lambda)^+\subseteq xL(\lambda)$.
\end{proof}
Now we focus on the case $|S|=1$.
\begin{lemma}\label{oneroot} Let $\alpha=\varepsilon_i-\delta_j$ be a positive odd root. If $(\lambda+\rho,\alpha)=0$, then $\{\alpha\}\in{\mathcal S}(\lambda)$.
\end{lemma}
\begin{proof} Consider the subset $\Gamma$ of odd positive roots defined by
$$\Gamma=\{\varepsilon_p-\delta_q | p>i,q\leq j,\,\,\text{or}\,\, p\geq i,q<j\}.$$
We set $\Sigma_\Gamma=\sum_{\gamma\in\Gamma}\gamma$, then we have $(\Sigma_\Gamma,\alpha)=-(\rho,\alpha)$.
Let $v\in M(\lambda)$ be a highest weight vector and
$$w:=\prod_{\gamma\in\Gamma} X_{-\gamma}v.$$
Then a quick check yields $X_{\alpha}w=0$.
We claim that $w\notin{\rm im} X_{\alpha}$. Indeed, $w$ is a weight vector of weight $\displaystyle\mu= \lambda-\Sigma_{\Gamma}$.
Assume $w=X_{\alpha}w'$. Without loss of generality we may assume that $w'$ is a weight vector of weight $\mu-\alpha$. Note that the weight space
$M(\lambda)_{\mu-\alpha}$ is one-dimensional. Therefore we may assume that $w'$ is proportional to $X_{-\alpha}w$. On the other hand
$$X_{\alpha}X_{-\alpha}w=H_{\alpha}w=0$$
since $(\mu,\alpha)=(\lambda-\Sigma_\Gamma,\alpha)=(\lambda+\rho,\alpha)=0$.
Therefore $X_\alpha\in X_{M(\lambda)}$ and the lemma is proven.
\end{proof}
\begin{lemma}
\label{oneroot2} Let $\gamma=\varepsilon_p-\delta_q$ be a positive odd root with $(\lambda+\rho,\gamma)=0$. Then for $\alpha=\varepsilon_i-\delta_j$ with $i\le p$, $j\ge q$, $(\lambda+\rho,\varepsilon_i-\varepsilon_p)\in{\mathbb Z}_{\ge 0}$ and $(\lambda+\rho,\delta_q-\delta_j)\in{\mathbb Z}_{\le 0}$ we have $\{\alpha\}\in{\mathcal S}(\lambda)$.
\end{lemma}
\begin{proof}
If $i=p$ and $j=q$, this is Lemma~\ref{oneroot}. Otherwise we set $\beta=\varepsilon_i-\varepsilon_p$, $\beta'=\delta_q-\delta_j$ and $r=r_\beta r_\beta'$ (or $r=r_\beta$ if $\beta'=0$ or $r=r_{\beta'}$ if $\beta=0$). By assumption and application of Verma's theorem in Theorem~4.6 in~\cite{MR2428237}, we have an embedding $M(r\cdot\lambda)\subset M(\lambda)$.
Since $(r\cdot\lambda+\rho,\varepsilon_i-\delta_j)=0$ we can repeat the construction in the proof of Lemma~\ref{oneroot} of a vector $w$ for the highest weight vector~$v$ of $M(r\cdot\lambda)$. We again have $X_\alpha w=0$. Suppose that $w=X_\alpha w'$. We may assume that
$w'$ has weight $\nu=\mu-\alpha$, where~$\mu$ is the weight of $w$. Although the corresponding weight space $M(\lambda)_{\nu}$ is not one dimensional, any vector
in this space is of form $$X_{-\alpha}\prod_{\gamma\in\Gamma}X_{-\gamma} u$$ for some $u\in U({\mathfrak g}_{\bar{0}})v$ of weight $r_{\beta}\cdot\lambda$. Therefore we have
$$X_{\alpha}X_{-\alpha}\prod_{\gamma\in\Gamma}X_{-\gamma} u\,\,\in\,\, H_\alpha \prod_{\gamma\in\Gamma}X_{-\gamma} u+{\rm im} X_{-\alpha}.$$
But $(\mu,\alpha)=0$, hence $\displaystyle H_\alpha \prod_{\gamma\in\Gamma}X_{-\gamma} u=0$. Since $w\notin {\rm im} X_{-\alpha}$, we obtain that
$X_{\alpha}w'$ is never $w$. Thus, $M(\lambda)_{X_\alpha}\neq 0$.
\end{proof}
\begin{proposition}\label{typicalverma} Let $M$ be a Verma or simple module. Then $X_M=0$ if and only if the highest weight of $M$ is typical.
\end{proposition}
\begin{proof}
The case where~$M$ is a Verma module is an immediate consequence of Lemma~\ref{oneroot}, so we consider $M\simeq L(\lambda)$ simple for $\lambda\in{\mathfrak h}^\ast$. If $\lambda$ is typical, the claim follows from Theorem~\ref{finpdind}, so we are left with $\lambda$ atypical.
Consider $(\lambda+\rho,\gamma)=0$ for $\gamma$ an odd root. If $\gamma$ is simple the result follows immediately since $X_{-\gamma}v=0$ for a highest weight vector $v$ (since $U({\mathfrak n}^+)X_{-\gamma}v=0$) while $v\not\in{\rm im} X_{-\gamma}$. If $\gamma$ is not simple we can consider a sequence of odd reflections (see Section 3.5 in~\cite{bookMusson}) to obtain a system of positive roots in which $\gamma$ is simple. If one of these odd reflections is atypical for the simple module we can take the corresponding root as $\gamma$, so we consider the situation where each odd reflection is typical. In the new system of roots (with corresponding half-sum $\widetilde{\rho}$) the simple module will have highest weight $\widetilde{\lambda}=\lambda+\rho-\widetilde{\rho}$, which thus satisfies $( \widetilde{\lambda}+\widetilde{\rho},\gamma)=0$, so we end up in the setting where~$\gamma$ is simple.
\end{proof}
\subsection{The cases ${\mathfrak g}=\mathfrak{gl}(1|n)$ and ${\mathfrak g}=\mathfrak{gl}(m|1)$}
For these cases we determine ${\mathcal S}(\lambda)$, or equivalently $X_{M(\lambda)}$, for any $\lambda\in{\mathfrak h}^\ast$.
\begin{theorem}\label{verma} Let ${\mathfrak g}=\mathfrak{gl}(1|n)$ and $\lambda$ be some atypical weight. Let $p\leq n$ be such that
$(\lambda+\rho, \varepsilon_1-\delta_p)=0$ and $(\lambda+\rho,\varepsilon_1-\delta_j)\neq 0$ for all $j<p$. Then $\{\varepsilon_1-\delta_i\}\in {\mathcal S}(\lambda)$
if and only if $i\geq p$ and $(\lambda+\rho,\varepsilon_1-\delta_i)\in\mathbb Z_{\leq 0}$.
\end{theorem}
\begin{proof} If $\alpha=\varepsilon_1-\delta_i$ for some $i<p$, then $\{\alpha\}\notin{\mathcal S}(\lambda)$, by Corollary~\ref{reductyp}. If $\alpha=\varepsilon_1-\delta_i$ satisfies $i\geq p$ and $(\lambda+\rho,\varepsilon_1-\delta_i)\in\mathbb Z_{\leq 0}$, then $\{\alpha\}\in{\mathcal S}(\lambda)$, by Lemma~\ref{oneroot2}.
Now let us assume that $\alpha=\varepsilon_1-\delta_i$ for some $i>p$, but $(\lambda+\rho,\alpha)\notin\mathbb Z_{\leq 0}$. Then we have either
$(\lambda,\alpha)\notin\mathbb Z$ or $(\lambda,\alpha)\in{\mathbb Z}_{\ge i}$. The first case is covered by Lemma~\ref{nointeg}. For the second case, we use Lemma~\ref{reducegl} with $S=\{\varepsilon_1-\delta_p\}$ and hence ${\mathfrak g}'\simeq \mathfrak{gl}(1|p)$. It thus suffices to consider the Verma module $M(\lambda')$ of ${\mathfrak g}'$ and to show that
$M(\lambda')_{X_\alpha}=0$. Note that $(\beta,\alpha)\geq 0$ for any negative even root $\beta$ of ${\mathfrak g}'$ and $(\beta,\alpha)=-1$
for any negative odd root $\beta\neq -\alpha$ of ${\mathfrak g}'$. Therefore
$(\mu,\alpha)\neq 0$ for any weight $\mu$ of $M(\lambda')$. Therefore $M(\lambda')$ as a module over the $\mathfrak{sl}(1|1)$-subalgebra, generated by $X_{\pm \alpha}$,
is a direct sum of typical modules. Thus, $M(\lambda')_{X_\alpha}=0$.
\end{proof}
\begin{remark} The above theorem implies that in contrast with the finite dimensional case, see Lemma 2.1 in \cite{Duflo}, there are $M\in{\mathcal O}$ for which the associated variety $X_M$ is not closed.
For example, if ${\mathfrak g}=\mathfrak{gl}(1|2)$ and $\lambda=3\delta_2$, then $X_{M(\lambda)}={\mathfrak g}_1\setminus \mathbb C (\varepsilon_1-\delta_2)$ is not closed.
\end{remark}
The isomorphism $\mathfrak{gl}(1|n)\simeq \mathfrak{gl}(n|1)$ links the highest weight structure of category~${\mathcal O}$ with distinguished system of positive roots for $\mathfrak{gl}(1|n)$ to category~${\mathcal O}$ with anti-distinguished system of positive roots for $\mathfrak{gl}(n|1)$. The following result is therefore not identical to Theorem \ref{verma}, but can be proved similarly.
\begin{theorem}\label{verma2}
Let ${\mathfrak g}=\mathfrak{gl}(m|1)$ and $\lambda$ be some atypical weight. Let $p\leq m$ be such that
$(\lambda+\rho, \varepsilon_p-\delta_1)=0$ and $(\lambda+\rho,\varepsilon_j-\delta_1)\neq 0$ for all $j>p$. Then $\{\varepsilon_i-\delta_1\}\in {\mathcal S}(\lambda)$
if and only if $j\le p$ and $(\lambda+\rho,\varepsilon_i-\delta_1)\in\mathbb Z_{\ge 0}$.
\end{theorem}
\subsection{The case ${\mathfrak g}=\mathfrak{gl}(2|2)$} In this case we have four positive odd roots
$$\alpha=\varepsilon_2-\delta_1, \beta=\varepsilon_1-\delta_1,\gamma=\varepsilon_2-\delta_2, \delta=\varepsilon_1-\delta_2.$$
We represent weights by using the bijection ${\mathfrak h}^\ast\simeq {\mathbb C}^{2|2}$, as a natural extension of equation~\eqref{hat} and determine all ${\mathcal S}(\lambda)$ for $\lambda\in{\mathfrak h}^\ast$.
\begin{lemma}\label{atypicality1} Let $\lambda$ be a weight with degree of atypicality $1$. Then up to the shift by the weight $(t,t|t,t)$, for any $t\in{\mathbb C}$,
we have the following options.
\begin{enumerate}
\item $\mu^\lambda=(0,a|b,0)$ with $a,b\in{\mathbb C}$ such that $a\not=b$ and $ab\neq 0$. Then
${\mathcal S}(\lambda)=\{\{\delta\},\emptyset\}$.
\item $\mu^\lambda=(0,a|0,b)$ with $a,b\in{\mathbb C}$ such that $a\not=b$ and $a\neq 0$. If
$b\notin \mathbb Z_{\geq 0}$, then ${\mathcal S}(\lambda)=\{\{\beta\},\emptyset\}$. If
$b\in \mathbb Z_{\geq 0}$, then ${\mathcal S}(\lambda)=\{\{\beta\},\{\delta\},\emptyset\}$.
\item $\mu^\lambda=(a,0|b,0)$ with $a,b\in{\mathbb C}$ such that $a\not=b$ and $b\geq 0$. If
$a\notin \mathbb Z_{\geq 0}$, then ${\mathcal S}(\lambda)=\{\{\gamma\},\emptyset\}$. If
$a\in \mathbb Z_{\geq 0}$, then ${\mathcal S}(\lambda)=\{\{\gamma\},\{\delta\},\emptyset\}$.
\item $\mu^\lambda=(a,0|0,b)$ with $a,b\in{\mathbb C}$ such that $a\not=b$. If
$b\notin \mathbb Z_{\geq 0}$ and $a\notin \mathbb Z_{\geq 0}$, then ${\mathcal S}(\lambda)=\{\{\alpha\},\emptyset\}$. If
$b\in \mathbb Z_{\geq 0}$ but $a\notin \mathbb Z_{\geq 0}$, then ${\mathcal S}(\lambda)=\{\{\gamma\},\{\alpha\},\emptyset\}$.
If $a\in \mathbb Z_{\geq 0}$ but
$b\notin \mathbb Z_{\geq 0}$, then ${\mathcal S}(\lambda)=\{\{\beta\},\{\alpha\},\emptyset\}$. Finally, if $a\in \mathbb Z_{\geq 0}$ and $b\in \mathbb Z_{\geq 0}$, then
${\mathcal S}(\lambda)=\{\{\delta\},\{\gamma\},\{\beta\},\{\alpha\},\emptyset\}$.
\end{enumerate}
\end{lemma}
\begin{proof} The proof of this lemma is a straightforward application of Lemma~\ref{oneroot2} and reduction to the case of $\mathfrak{gl}(1|2)$. We leave it
as an exercise to the reader.
\end{proof}
\begin{lemma}\label{atypicality2} Let $\lambda$ be a weight with degree of atypicality $2$ and $A(\lambda)$ denote the set of all odd positive roots atypical to
$\lambda$.
\begin{enumerate}
\item If $\lambda$ is regular dominant integral, then ${\mathcal S}(\lambda)$ is the set of all subsets of mutually orthogonal roots in $\Delta_1^+$.
\item If $\lambda$ is regular integral and neither dominant nor anti-dominant, then
${\mathcal S}(\lambda)=\{\{\delta\},\{\beta,\gamma\},\{\beta\},\{\gamma\},\emptyset\}$.
\item If $\lambda=-\rho$, then ${\mathcal S}(\lambda)$ is the set of all subsets of mutually orthogonal roots in $\Delta_1^+=A(-\rho)$.
\item If $\lambda$ is regular, non-integral or anti-dominant integral, then ${\mathcal S}(\lambda)$ is the set of all subsets of $A(\lambda)$.
\end{enumerate}
\end{lemma}
\begin{proof} We first observe that (1) is a particular case of Lemma~\ref{dominant}.
Next, we prove (2). We assume that $\mu^\lambda=(a,0|a,0)$ with positive integral $a$, the case $(0,a|0,a)$ being similar. Then ${\mathcal S}(\lambda)$ contains $\{\beta\}$
and and $\{\gamma\}$ by Lemma~\ref{oneroot} and $\{\delta\}$ by Lemma~\ref{oneroot2}. On the other hand, $\{\alpha\}\notin {\mathcal S}(\lambda)$ by Lemma~\ref{reducegl}.
Also we can apply Lemma~\ref{hereditary} to $S=\{\alpha,\delta\}$ with $h=\varepsilon_1-\varepsilon_2$ and conclude that
$\{\alpha,\delta\}\notin {\mathcal S}(\lambda)$. It remains to show that $\{\beta,\gamma\}\in {\mathcal S}(\lambda)$.
For this we take $w=X_{-\alpha}v$, where~$v$ denotes the highest weight vector, and let
$x=X_{\beta}+X_{\gamma}$. Then $xw=0$ and we claim that $w\notin{\rm im} x$ by the same argument in the proof of Lemma~\ref{oneroot2} since we have
$$X_{\beta}X_{-\beta}X_{-\alpha}v=(\lambda-\alpha,\beta)X_{-\alpha}v=0,\quad X_{\gamma}X_{-\gamma}X_{-\alpha}v=(\lambda-\alpha,\gamma)X_{-\alpha}v=0.$$
Let us prove (3) now. As $\mu^\lambda=(0,0|0,0)$, Lemma~\ref{oneroot} implies that ${\mathcal S}(\lambda)$ contains all singletons.
Furthermore, $\{\beta,\gamma\}\in {\mathcal S}(\lambda)$ by the same
argument as above. To prove that $\{\alpha,\delta\}\in {\mathcal S}(\lambda)$ set $x=X_{\alpha}+X_{\delta}$. Let $M$ denote the projection of
$M(0,-1|0,0)\otimes U$ on the most atypical block. Lemma~\ref{general} (1),(2) implies that $M_x=0$. On the other hand, $M$ has a filtration by three Verma modules,
$M(0,0|0,0)$, $M(0,-1|-1,0)$ and $M(0,-1|0,-1)$. From the previous cases we have $M(0,-1|-1,0)_x\neq 0$ and $M(0,-1|0,-1)_x=0$. Thus, we must have
$M(0,0|0,0)_x\neq 0$.
Finally, let us deal with (4).
Here we have several subcases to consider.
If $A(\lambda)=\{\beta,\gamma\}$, then we may assume $\lambda=(a,0|a,0)$ with non-integral $a$.
Any subset of $A(\lambda)$ is in ${\mathcal S}(\lambda)$ by the same argument as
in the previous case. Furthemore, $\{\alpha\},\{\delta\}\notin {\mathcal S}(\lambda)$ by Lemma~\ref{nointeg}, and
$\{\alpha,\delta\}\notin {\mathcal S}(\lambda)$ by Lemma~\ref{hereditary}.
If $A(\lambda)=\{\alpha,\delta\}$, then we may assume
$\mu^\lambda=(-a,0|0,-a)$ with $a\in \mathbb Z_{> 0}$. Then $\{\beta\},\{\gamma\}\notin {\mathcal S}(\lambda)$ by Lemma
\ref{reductyp}. Moreover, Lemma~\ref{hereditary} implies that $\{\beta,\gamma\}\notin {\mathcal S}(\lambda)$.
On the other hand, $\{\alpha\},\{\delta\}\in{\mathcal S}(\lambda)$ by Lemma~\ref{oneroot}. Finally, to prove that $\{\alpha,\delta\}\in {\mathcal S}(\lambda)$ we use the same trick
with translation functor as in (3). More precisely, we set again $x=X_{\alpha}+X_{\delta}$ and consider the projection $M$ of $M(-a-1,0|0,-a)\otimes U$
on the most atypical block. Now $M$ is filtred by two Verma modules: $M(-a-1,0|0,-a-1)$and $M(-a,0|0,-a)$. Using the result for $a=0$ we obtain by induction
in $a$ that $M(-a,0|0,-a)_x\neq 0$.
\end{proof}
\section{Projective dimensions and blocks of category~${\mathcal O}$}
\label{secblocks}
In this entire section we consider ${\mathfrak g}=\mathfrak{gl}(m|n)$ or ${\mathfrak g}=\mathfrak{sl}(m|n)$ with distinguished Borel subalgebra. We denote by $\mathtt{a}:W\to\mathbb{N}$ Lusztig's $\mathtt{a}$-function, see \cite{Lu}.
\subsection{Projective dimensions of structural modules}
\begin{theorem}
\label{pdstruct}
We have the following connection between projective dimensions of structural modules in the categories ${\mathcal O}$ and ${\mathcal O}^0$, for $\lambda\in{\mathfrak h}^\ast$:
\begin{enumerate}[$($i$)$]
\item ${\rm pd}_{{\mathcal O}}L(\lambda)={\rm pd}_{{\mathcal O}^0}L_0(\lambda)$ if $\lambda$ is typical, otherwise ${\rm pd}_{{\mathcal O}}L(\lambda)=\infty$;
\item ${\rm pd}_{{\mathcal O}}M(\lambda)={\rm pd}_{{\mathcal O}^0}M_0(\lambda)$ if $\lambda$ is typical, otherwise ${\rm pd}_{{\mathcal O}}M(\lambda)=\infty$;
\item ${\rm pd}_{{\mathcal O}}I(\lambda)={\rm pd}_{{\mathcal O}^0}I_0(\lambda)$.
\end{enumerate}
This implies that for $\lambda\in P_0$, we have
$${\rm pd}_{{\mathcal O}}I(\lambda)=\mathtt{a}(w_0x_\lambda)$$
with $x_\lambda$ the longest Weyl group element such that $x^{-1}_\lambda\cdot\lambda$ is dominant.
\end{theorem}
\begin{proof}
Properties $(i)$ and $(ii)$ for typical $\lambda$ follow from the fact that Brundan's KL polynomials for typical weights correspond to those for $\mathfrak{g}_0$, see e.g. Lemma~\ref{KLorbit}. Properties $(i)$ and $(ii)$ for atypical~$\lambda$ follow from the combination of Proposition~\ref{typicalverma} and Theorem \ref{finpdind} $(i)\leftrightarrow (ii)$.
According to Lemma~\ref{propind}$(i)$ and $(ii)$, to prove $(iii)$ it suffices to prove that $I_0(\lambda)$ is a direct summand of ${\rm Res}^{{\mathfrak g}}_{{\mathfrak g}_0} I(\widetilde\lambda)$, with $\widetilde\lambda:=\lambda+2\rho_1$ (as ${\rm pd} I_0(\widetilde\lambda)={\rm pd} I_0(\lambda)$) and that $I(\lambda)$ is a direct summand of ${\rm Ind}^{{\mathfrak g}}_{{\mathfrak g}_0}I_0(\lambda)$. Both the induced and restricted module are injective and ${\rm Ind}^{{\mathfrak g}}_{{\mathfrak g}_0}\simeq {\rm Coind}^{{\mathfrak g}}_{{\mathfrak g}_0}$, so the properties
$${\rm Hom}_{{\mathfrak g}_0}(L_0(\lambda), {\rm Res}^{{\mathfrak g}}_{{\mathfrak g}_0} I(\widetilde\lambda))\simeq {\rm Hom}_{{\mathfrak g}}({\rm Ind}^{{\mathfrak g}}_{{\mathfrak g}_0} L_0(\lambda), I(\widetilde\lambda))=[{\rm Ind}^{{\mathfrak g}}_{{\mathfrak g}_0} L_0(\lambda): L(\widetilde\lambda)]\not=0\,\,\,\mbox{and} $$
$${\rm Hom}_{{\mathfrak g}}(L(\lambda), {\rm Ind}^{{\mathfrak g}}_{{\mathfrak g}_0} I_0(\lambda))\simeq {\rm Hom}_{{\mathfrak g}_0}({\rm Res}^{{\mathfrak g}}_{{\mathfrak g}_0} L(\lambda), I_0(\lambda))=[{\rm Res}^{{\mathfrak g}}_{{\mathfrak g}_0} L(\lambda): L_0(\lambda)]\not=0 $$
conclude the proof of $(iii)$.
Finally, the projective dimension of $I(\lambda)$ follows from $(iii)$ and Theorem 16 of~\cite{MR2366357}, see also Theorem 44(ii) in~\cite{SHPO4}.
\end{proof}
\begin{remark}
The theorem reduces the question concerning projective dimension of simple and Verma modules to the corresponding questions for Lie algebras. For integral regular weights these are well-known, see \cite{MR2366357}. For singular blocks only estimates and special cases are known at the moment, see \cite{SHPO4, CouMaz2}.
\end{remark}
Theorem 44(ii) in \cite{SHPO4} and Theorem \ref{pdstruct} lead to the following immediate consequences.
\begin{corollary}
\label{pdextr}
Consider $\lambda\in{\mathfrak h}^\ast$, then
\begin{enumerate}[$($i$)$]
\item $\dim L(\lambda) <\infty \quad\Leftrightarrow\quad {\rm pd}_{{\mathcal O}} I(\lambda)=2l(w_0);$
\item $\lambda$ is anti-dominant $\quad\Leftrightarrow\quad {\rm pd}_{{\mathcal O}} I(\lambda)=0.$
\end{enumerate}
\end{corollary}
Property $(ii)$ was first obtained through other methods in Theorem 2.22 of~\cite{BLW}.
\subsection{Finitistic global dimension of blocks}
\begin{theorem}
\label{fdblock}
The finitistic global homological dimension of the block ${\mathcal O}_\xi$ for an integral linkage class $\xi$ is given by
$$\rm fin.dim{\mathcal O}_\xi=2\mathtt{a}(w_0w_0^\xi).$$
\end{theorem}
\begin{proof}
The proof of Theorem 3 in~\cite{preprint} implies that $\rm fin.dim{\mathcal O}_\xi$ for a classical Lie superalgebra is equal to the highest projective dimension of an injective module. The result therefore follows immediately from Theorem 26(ii) in \cite{CouMaz2} and Theorem \ref{pdstruct}.
\end{proof}
\subsection{Blocks in category~${\mathcal O}$}
In this subsection we demonstrate a principle which is responsible for the fact that different integral blocks in category~${\mathcal O}$ will almost never be equivalent, even when they have the same degree of atypicality and singularity of core.
The origin of this new phenomenon is that each atypical integral block contains simple objects which are more regular or more singular. The category behaves differently `around' these objects. The `distance' between these objects in the category is determined by how far the core is distanced from the walls of the Weyl chamber. We make this explicit for blocks for $\mathfrak{sl}(3|1)$ with regular core, by using our results on projective dimensions.
In order to avoid the obvious equivalences of blocks coming from the centre~$\mathfrak{z}({\mathfrak g})$, see Lemma 3.5 in~\cite{CMW}, we consider ${\mathfrak g}=\mathfrak{sl}(3|1)$ rather than $\mathfrak{gl}(3|1)$. We use the fact that an equivalence of abelian categories is always given in terms of exact functors.
\begin{theorem}
\label{thmsl31}
Consider ${\mathfrak g}=\mathfrak{sl}(3|1)$. No two atypical integral blocks ${\mathcal O}_{\xi}$ with regular core $\chi'_\xi$ are equivalent.
\end{theorem}
\begin{proof}
We use the notation $\Lambda={\mathbb Z}^{3|1}$ of $\mathfrak{gl}(3|1)$-weights, silently making the relevant identification.
The integral linkage classes with regular core are given by $\xi^p=[(p,1,0|0)]$, parametrised by $p\in{\mathbb N}$ with $p>1$. We label the set $\Lambda^{++}\cap\xi^p$ as $\{\lambda^p_i\,|\, i\in{\mathbb Z}\}$ with
\begin{itemize}
\item $\lambda^p_{i}= (p,1,i|i)$ for $i\le 0$;
\item $\lambda^p_i=(p,i+1,1|i+1)$ for $0< i <p-1$;
\item $\lambda^p_i=(i+2,p,1|i+2)$ for $i\ge p-1$.
\end{itemize}
From Corollary \ref{pdextr} we know that finite dimensional and anti-dominant simple modules are categorically defined. Any equivalence of categories between ${\mathcal O}_{\xi^p}$ and ${\mathcal O}_{\xi^{p'}}$ must therefore preserve these two types of simple modules.
We will construct a categorical invariant in the form of a graph. This graph is given by the ${\rm Ext}^1$-quiver of the subcategory of finite dimensional modules in ${\mathcal O}_\xi$, where in each node $\lambda\in \Lambda^{++}$ we write the number of anti-dominant simple subquotients in $P(\lambda)$. This number is denoted by $\flat\lambda$.
The subsequent Lemma~\ref{Vermaanti} implies that the number of anti-dominant simple subquotients in $P(\lambda)$ is equal to twice the number of Verma modules in its standard filtration. By equation~\eqref{resCheng}, this means
$$\flat\lambda \;=\; 2\sum_{\nu\in\Lambda}d_{\lambda,\nu}(1).$$
Computing $d_{\lambda,\nu}$ is a direct application of the bumping procedure, see Example 3.3 in~\cite{Brundan3}. It follows that $\sum_{\nu\in\Lambda}d_{\lambda,\nu}(1)-1$ is equal to the minimal strictly positive number of times the (unique) atypical positive root of $\lambda$ can be added to $\lambda$ such that the result is another regular weight.
The ${\rm Ext}^1$-quiver of the category of finite dimensional weight modules of $\mathfrak{sl}(3|1)$ is well-known to be of Dynkin type~$A_\infty$, which follows e.g. from the penultimate paragraph in Appendix~\ref{gl21} and Theorem 2 of \cite{MR2734963}. The combination of these results yields the following graphs: for $p\ge 3$ we have
\begin{displaymath}
\xymatrix{
\cdots \bullet_{\flat\lambda^p_{-2}=4}\ar[r] &\bullet_{\flat\lambda^p_{-1}=4}\ar[r]\ar[l] &\bullet_{\flat\lambda^p_0=6}\ar[r]\ar[l] & \bullet_{\flat\lambda^p_{1}=4} \ar[r]\ar[l]& \bullet_{\flat\lambda^p_{2}=4}\ar[r]\ar[l] & \cdots\ar[l] }
\end{displaymath}
\begin{displaymath}
\xymatrix{
\cdots \bullet_{\flat\lambda^p_{p-4}=4}\ar[r] &\bullet_{\flat\lambda^p_{p-3}=4}\ar[r]\ar[l] &\bullet_{\flat\lambda^p_{p-2}=6}\ar[r]\ar[l] & \bullet_{\flat\lambda^p_{p-1}=4} \ar[r]\ar[l]& \bullet_{\flat\lambda^p_{p}=4}\ar[r]\ar[l] & \cdots\ar[l] },
\end{displaymath}
meaning $p-3$ nodes between the two exceptional nodes and if $p=2$ we have
\begin{displaymath}
\xymatrix{
\cdots\bullet_{\flat\lambda^2_{-2}=4}\ar[r] &\bullet_{\flat\lambda^2_{-1}=4}\ar[r]\ar[l] & \bullet_{\flat\lambda^2_{0}=8} \ar[r]\ar[l]& \bullet_{\flat \lambda_{1}^2=4}\ar[r]\ar[l] & \bullet_{\flat\lambda^2_{2}=4}\ar[r]\ar[l] & \bullet\ar[l]\cdots }.
\end{displaymath}
Since each diagram is different from the others, the result follows.
\end{proof}
\begin{corollary}
Category~${\mathcal O}_{{\mathbb Z}}$ for a basic classical Lie superalgebra can contain infinitely many nonequivalent blocks.
\end{corollary}
\begin{remark}
An alternative categorical invariant to $\flat \lambda$ would be to take $[P(\lambda):L(\lambda)]$. By BGG reciprocity and the fact that in the $\mathfrak{sl}(3|1)$ case the standard filtration of $P(\lambda)$ is multiplicity free (which follows from computation), $[P(\lambda):L(\lambda)]$ corresponds to the number of Verma modules in the standard filtration of $P(\lambda)$, so $\flat\lambda=2[P(\lambda):L(\lambda)]$.
\end{remark}
\begin{lemma}
\label{Vermaanti}
Any atypical integral Verma module of $\mathfrak{sl}(3|1)$ contains exactly two simple subquotients which have an anti-dominant highest weight.
\end{lemma}
\begin{proof}
Take $\mu\in P_0$. Any Verma module $M(\mu)\simeq U({\mathfrak g})\otimes_{U({\mathfrak g}_0+{\mathfrak g}_1)} M_0(\mu)$ has a Kac filtration. Moreover, the number of times $K(\lambda)$ appears in this filtration is equal to $[M_0(\mu): L_0(\lambda)]$. Since $\Lambda{\mathfrak g}_{-1}$ is a finite dimensional ${\mathfrak g}_0$-module, the only possibility for $K(\lambda)$ to have an anti-dominant simple subquotient is if $\lambda$ is anti-dominant itself. Now every Verma module $M_0(\mu)$ has exactly one anti-dominant simple subquotient, it hence suffices to prove the lemma for anti-dominant Verma modules $M(\mu)=K(\mu)$.
By construction $K(\mu)$ admits $4$ different eigenvalues of $z$ in equation~\eqref{gradel}. Lemma~6.10(i) in~\cite{CouMus} implies that any anti-dominant atypical simple module for $\mathfrak{sl}(3|1)$ admits $3$ different such eigenvalues. This implies that any anti-dominant simple subquotient $L(\nu)$ of $K(\mu)$ with $\nu\not=\mu$ must satisfy~$\nu\in W\cdot(\mu-\gamma)$ with $\gamma\in\Delta^+_1$ atypical for $\mu$. Note that $W\cdot(\mu-\gamma)$ does not depend on the atypical root $\gamma$, in case there is more than one. This hence leaves only one possibility besides $\mu$, {\it viz.} the unique anti-dominant weight in $W\cdot(\mu-\gamma)$, which we denote by $\nu$. We claim that $[K(\mu):L(\nu)]\le 1$. This follows from looking at weight spaces corresponding to the weight $\mu-2\rho_1$. In $K(\mu)$, this has dimension one, in $L(\mu)$ dimension zero and in $L(\nu)$ dimension one. The fact that there appears at least two anti-dominant simple modules follows from the fact that both the socle and top of $K(\mu)$ must be anti-dominant.
\end{proof}
\begin{remark}
Classically the equivalences between regular integral blocks are often given by translation functors, see Section 7.8 in~\cite{MR2428237}. The same holds for finite dimensional modules for Lie superalgebras. It is interesting to note how this fails for category~${\mathcal O}$ for superalgebras. The functor $F_p$ maps the block corresponding to $(p,1,0|0)$ to the one corresponding to $(p+1,1,0|0)$. According to Theorem 2.4 in~\cite{Ku}, this maps every simple module to a simple module with only three exceptions: $F_p$ maps the singular modules $L(p,p,1|p)$, $L(p,1,p|p)$ and $L(1,p,p|p)$ to indecomposable modules which are not simple.
The principal used in the proof of Theorem \ref{thmsl31}, that the singular objects appear at different positions in the two blocks, is also responsible for the problematic behaviour of the translation functor. This principle namely causes translation onto the walls for some modules and translation out of the wall for other.
\end{remark}
\section{Complexity in category~${\mathcal O}$}
\label{seccomp}
In this section we introduce the notion of complexity in category~${\mathcal O}$ for basic classical Lie superalgebras, as the rate of polynomial growth of a minimal projective resolution of a module. We prove that this is well-defined, {\it i.e.} it is finite for every module. Then we study the relation between degree of atypicality and complexity of Verma and simple modules for $\mathfrak{gl}(m|n)$. Similar results for the category~${\mathcal F}$ have been obtained by Boe, Kujawa and Nakano in~\cite{BKN1, BKN2}.
\subsection{Definition and basic properties}
The usual notion of complexity, as introduced by Alperin, measures the rate of growth of the dimension in a minimal projective resolution. Since the projective objects in category~${\mathcal O}$ are infinite dimensional we need to consider instead the number of indecomposable projective objects. This variation has also been studied for the category of finite dimensional modules of $\mathfrak{gl}(m|n)$ in Section 9 in~\cite{BKN2} and is (contrary to the original approach) a categorical invariant.
\begin{definition}\label{defC}
For $M\in{\mathcal O}$ we define $c_{{\mathcal O}}(M)$, the complexity of $M$ in category~${\mathcal O}$, as
$$c_{{\mathcal O}}(M)=r\left(\sum_{\mu\in{\mathfrak h}^\ast}\dim{\rm Ext}^\bullet_{{\mathcal O}}(M,L(\mu))\right).$$
The rate of growth $r(c^\bullet)$ of a sequence of numbers $c^\bullet$ is defined as the smallest non-negative integer $k$ such that there is a constant $C>0$ for which $ c^j\le C j^{k-1}$ for all $j>0$. In case the $c^j$ are not finite or no such integer exists, we set $r(c^\bullet)=\infty$.
\end{definition}
By definition, the complexity of a module is zero if and only if it has finite projective dimension. Immediate from the definition we have the following properties.
\begin{lemma}
\label{lemcompseq}
Consider a short exact sequence $A_1\hookrightarrow A_2\twoheadrightarrow A_3$ in category~${\mathcal O}$, then
$$c_{{\mathcal O}}(A_i)\le\max\{c_{{\mathcal O}}(A_j),c_{{\mathcal O}}(A_k)\}$$
for any permutation $\{i,j,k\}$ of $\{1,2,3\}$.
\end{lemma}
As main results of this subsection we prove that this notion of complexity is well-defined for category~${\mathcal O}$ for basic classical Lie superalgebras and that translation functors cannot increase complexity.
\begin{proposition}
For any $M\in{\mathcal O}$, the value $c_{{\mathcal O}}(M)$ is finite dimensional, more precisely
$$c_{{\mathcal O}}(M)\le\dim{\mathfrak g}_{\bar{1}}.$$
\end{proposition}
\begin{proof}
We prove this by induction on the (finite) projective dimension of ${\rm Res}^{\fg}_{\fg_{\bar{0}}} M$ in category~${\mathcal O}^0$. Assume that the property holds for any $K\in{\mathcal O}$ with ${\rm pd}_{{\mathcal O}^0}{\rm Res}^{\fg}_{\fg_{\bar{0}}} K< p$. Denote the projective cover of an $M\in{\mathcal O}$, with ${\rm pd}_{{\mathcal O}^0}{\rm Res}^{\fg}_{\fg_{\bar{0}}} M= p$, by $P$ and the kernel of the morphism $P\twoheadrightarrow M$ by $N$. Since ${\rm pd}_{{\mathcal O}^0}{\rm Res}^{\fg}_{\fg_{\bar{0}}} N< p$ and $c_{{\mathcal O}}(P)=0$, the induction step and Lemma~\ref{lemcompseq} imply that $$c_{{\mathcal O}}(M)\le c_{{\mathcal O}}(N)\le \dim{\mathfrak g}_{\bar{1}}.$$
It remains to be proved that $c_{{\mathcal O}}(M)\le\dim{\mathfrak g}_{\bar{1}}$ in case ${\rm Res}^{\fg}_{\fg_{\bar{0}}} M$ is projective in ${\mathcal O}^0$. We consider the Chevalley-Eilenberg resolution of ${\mathbb C}$ for $({\mathfrak g},{\mathfrak g}_{\bar{0}})$-relative homological algebra. It was proved explicitly in Proposition 2.4.1 of~\cite{BKN1} that this is a $({\mathfrak g},{\mathfrak g}_{\bar{0}})$-projective resolution of ${\mathbb C}$. Tensoring this resolution with $M$ yields an exact complex
$$\cdots\to U({\mathfrak g})\otimes_{U({\mathfrak g}_{\bar{0}})}(S^j({\mathfrak g}_{\bar{1}})\otimes {\rm Res}^{\fg}_{\fg_{\bar{0}}} M) \to\cdots \to U({\mathfrak g})\otimes_{{\mathfrak g}_{\bar{0}}}({\mathfrak g}_{\bar{1}}\otimes{\rm Res}^{\fg}_{\fg_{\bar{0}}} M)\to M\to 0.$$
Since ${\rm Res}^{\fg}_{\fg_{\bar{0}}} M$ is projective in ${\mathcal O}^0$, this is a projective resolution in ${\mathcal O}$ of $M$. Applying Frobenius reciprocity then implies
$$\sum_{\mu\in{\mathfrak h}^\ast}\dim {\rm Ext}^j_{{\mathcal O}}(M,L(\mu))\le \sum_{\mu\in{\mathfrak h}^\ast}\dim{\rm Hom}_{{\mathfrak g}_{\bar{0}}}(S^j({\mathfrak g}_{\bar{1}})\otimes {\rm Res}^{\fg}_{\fg_{\bar{0}}} M, {\rm Res}^{\fg}_{\fg_{\bar{0}}} L(\mu)).$$
Lemma~\ref{bound2} applied to ${\mathfrak g}_{\bar{0}}$ and Lemma~\ref{bound3} then allow to conclude
$$\sum_{\mu\in{\mathfrak h}^\ast}\dim {\rm Ext}^j_{{\mathcal O}}(M,L(\mu))\le C_{{\mathfrak g}_{\bar{0}}}\,\widetilde{C}_{{\mathfrak g}}\,q\, \dim S^j({\mathfrak g}_{\bar{1}}) .$$
with $q$ the number of indecomposable projective modules of ${\mathcal O}^0$ in ${\rm Res}^{\fg}_{\fg_{\bar{0}}} M$. The result thus follows from the fact that the polynomial grow rate of $\dim S^j({\mathfrak g}_{\bar{1}})$ is $\dim{\mathfrak g}_{\bar{1}}$.
\end{proof}
\begin{proposition}
\label{propT}
Consider any finite dimensional module $V$ and translation functor
$$T^{\chi,\chi'}_V:{\mathcal O}_{\chi}\to {\mathcal O}_{\chi'}\,\,:\quad M\in{\mathcal O}_{\chi}\mapsto T^{\chi,\chi'}_V(M)= (M\otimes V)_{\chi'}\in{\mathcal O}_{\chi'}.$$
Then we have $c_{{\mathcal O}}(T^{\chi,\chi'}_V(M))\le c_{{\mathcal O}}(M)$.
\end{proposition}
\begin{proof}
Consider a minimal projective resolution of $M$. Its rate of polynomial growth is~$c_{{\mathcal O}}(M)$. This projective resolution is mapped by the exact functor $T^{\chi,\chi'}_V$ to a (not necessarily minimal) projective resolution of $T^{\chi,\chi'}_V(M)$. The polynomial rate of that resolution is smaller or equal to $c_{{\mathcal O}}(M)$, by Lemma~\ref{bound2}.
\end{proof}
\begin{corollary}
Consider a translation functor $T=T^{\chi,\chi'}_V$ with adjoint $\widetilde T=T^{\chi',\chi}_{V^\ast}$ and $M\in{\mathcal O}_{\chi}$. If for $M':=TM$ we have $\widetilde T M'\simeq M$, then $c_{{\mathcal O}}(M)=c_{{\mathcal O}}(M').$
\end{corollary}
\subsection{Complexity of Verma modules for $\mathfrak{gl}(m|n)$}
By the equivalences of categories in \cite{CMW}, it suffices to compute the complexity for Verma modules in integral blocks. Their complexity is in principle determined by Brundan's KL polynomials. Using equation~\eqref{defp}, for any $\lambda\in\Lambda$ we introduce the notation
\begin{equation}\label{KLVcompl}p_{\lambda}^j=\sum_{\nu\in\Lambda}\dim{\rm Ext}_{{\mathcal O}}^j(M(\lambda),L(\nu))=\frac{1}{j!} \sum_{\nu\in\Lambda} \left(\frac{\partial^j}{\partial q^j}p_{\lambda,\nu}(q)\right)_{q=0}.\end{equation}
\begin{theorem}
\label{complVerma}
There are constants $C_{k}$, such that for any $\lambda\in\Lambda$ with $\sharp[\lambda]=k$ we have
\begin{equation}\label{eqthm}p_{\lambda}^j\;\le\;C_{k}\, j^{k-1},\qquad \forall j>0.
\end{equation}
The complexity of a Verma module satisfies $c_{{\mathcal O}}(M(\lambda))=\sharp[\lambda]$ if $\lambda\in\Lambda$ is regular and $c_{{\mathcal O}}(M(\lambda))\le\sharp[\lambda]$ if $\lambda\in\Lambda$ is singular.
\end{theorem}
\begin{theorem}
\label{complK}
Any module $M\in {\mathcal O}_\xi$ which is either ${\mathfrak g}_1$-free or ${\mathfrak g}_{-1}$-free has $c_{{\mathcal O}}(M)\le \sharp\xi$.
\end{theorem}
The remainder of this subsection is devoted to the proof of these theorems, but first we observe that the corresponding property for category~${\mathcal F}$ as derived in~\cite{BKN2} can be made even more precise.
\begin{lemma}
\label{lemF}
For any $\lambda\in\Lambda^{++}$ with $\sharp[\lambda]=k$, we have
$$\sum_{\nu\in\Lambda^{++}}\dim{\rm Ext}^j_{{\mathcal F}}(K(\lambda),L(\nu))\,=\,\binom{k+j-1}{k-1}\,=\,\frac{1}{(k-1)!}(j^{k-1}+\frac{1}{2}k(k-1)j^{k-2}+\cdots).$$
\end{lemma}
\begin{proof}
Theorem 4.51 and Corollary 3.39(ii) in~\cite{Brundan} imply that
$$\sum_{\nu\in\Lambda^{++}}\dim{\rm Ext}^j_{{\mathcal F}}(K(\lambda),L(\nu))\,=\,\sharp\{\theta\in {\mathbb N}^{k}\,|\,\, |\theta|=j \},$$
which proves the statement.
\end{proof}
For the subsequent proofs, we divide the atypical weights into four mutually exclusive types. For $\lambda\in\Lambda$, we set $a_\lambda$ equal to the highest label which appears on both sides.
\begin{enumerate}[(a)]
\item There is no label in $\lambda$ higher than $a_\lambda$.
\item There is a label in $\lambda$ higher than $a_\lambda$, but no label equal to $a_\lambda+1$.
\item There is a label equal to $a_\lambda+1$, but only one occurrence of $a_\lambda$ on each side.
\item There is a label equal to $a_\lambda+1$, as well as multiple occurrences of $a_\lambda$ on some side.
\end{enumerate}
We also set $v(\lambda)$ equal to the number of labels in $\lambda$ strictly higher than $a_\lambda$. So $v(\lambda)=0$ iff $\lambda$ satisfies (a).
Furthermore, we denote by {\bf P}$[k]$ for $0\le k\le\min(m,n)$ the property that there is a constant $C_{k}$, such that \eqref{eqthm} is true for all $\lambda\in\Lambda$ with $\sharp[\lambda]=k$. We will freely use the constant $C:=C_{\mathfrak g}$ from Lemma~\ref{bound2}.
\begin{lemma}
\label{newlem1}
Assume that property {\bf P}$[k-1]$ holds and consider $\lambda\in\Lambda$ with $\sharp[\lambda]=k$.
\begin{enumerate}
\item If $\lambda$ satisfies $($a$)$, we have
$$p^j_\lambda\le (m+n)^2CC_{k-1}\sum_{l=0}^j (j-l)^{k-2}\le (m+n)^2CC_{k-1}j^{k-1}.$$
\item If $\lambda$ satisfies $($b$)$, denote the lowest label in $\lambda$ strictly higher than $a_\lambda$ by $b_\lambda$, set $d:=b_\lambda-a_\lambda-1>0$. Then we have
$$p^j_\lambda\le (m+n)^2CC_{k-1}j^{k-1}+\sum_{i=1}^yp^{j-d}_{\lambda_{(i)}},$$
for some $y<m+n$ with $\lambda_{(i)}\in\Lambda$ satisfying $($c$)$, $v(\lambda_{(i)})=v(\lambda)$ and $\sharp[\lambda_{(i)}]=k$.
\end{enumerate}
\end{lemma}
\begin{proof}
We consider $\lambda\in\Lambda$ with $\sharp [\lambda]=k$ and assume it satisfies either $($a$)$ or $($b$)$. Set $a:=a_\lambda$. We refer to the side with strictly most occurrences of $a$ as the big side and the other as the small side. If there is an equal number of $a$ on each side, we choose the big and small side randomly. Denote the number of $a$'s appearing on the big side by~$y$.
Fix one occurrence of $a$ on the small side. We create $\lambda'\in\Lambda$ by replacing that label by $a+1$. As $\lambda$ satisfies $(a)$ or $(b)$, by construction $\lambda'$ has degree of atypicality $k-1$. If the small side is the left-hand side we set $T=E_a$, otherwise $T=F_a$. Then $TM(\lambda')$ has a standard filtration of length $y+1$, where~$\lambda$ is the lowest weight appearing. The $y$ other highest weights, which we denote by $\{\lambda^{(i)}\,|\, i=1,\cdots, y\}$, are obtained from $\lambda$ by raising one of the occurrences of $a$ on the big side and the fixed occurrence of $a$ on the small side. Thus we can define a module $\widetilde M\in{\mathcal O}$ by the short exact sequence
\begin{equation}\label{sesproof}0\to \widetilde M \to T M(\lambda')\to M(\lambda)\to 0.\end{equation}
By considering the long exact sequence obtained by applying $\oplus_{\nu}{\rm Hom}_{{\mathcal O}}(-,L(\nu))$ to \eqref{sesproof} in combination with {\bf P}$[k-1]$ and Lemma~\ref{bound2}, we find
\begin{equation}\label{neweqnew}p^j_\lambda\le \sum_{i=1}^yp^{j-1}_{\lambda^{(i)}}+(m+n)CC_{k-1} j^{k-2},\end{equation}
where~$m+n$ is the dimension of the tautological module for $\mathfrak{gl}(m|n)$.
If $\lambda$ satisfies (a), so do the $\lambda^{(i)}$. So we can apply the above procedure on each of the $\lambda^{(i)}$, with the added simplification that the highest label in $\lambda^{(i)}$ appears only once on each side. Hence the analogue of $y$ is equal to $1$ in the following steps. Applying this $j-1$ times and using the estimate $y\le m+n$ proves $(1)$.
Now assume that $\lambda$ satisfies (b) and recall the constant $d$ introduced in the statement of part (2) of the lemma. If $d=1$, then the $\lambda^{(i)}$ satisfy~$($c$)$ and the claim in part (2) follows immediately from equation~\eqref{neweqnew}. If $d>1$, then the $\lambda^{(i)}$ satisfy~$($b$)$ and we can apply the procedure again on them. Hence we can repeat the procedure $d$ times, where again only the first time we will need a constant~$y$ bigger than $1$. This yields
$$p^j_\lambda\le (m+n)^2CC_{k-1}\sum_{l=0}^{d-1} (j-l)^{k-2}+ \sum_{i=1}^yp^{j-d}_{\lambda_{(i)}}$$
with $\lambda_{(i)}\in\Lambda$ obtained from $\lambda$ by adding $d$ to our fixed occurrence of $a$ on the small side and adding $d$ to the $i$th occurrence of $a$ on the big side. By construction, the $\lambda_{(i)}$ satisfy~(c). This completes the proof of part $(2)$.
\end{proof}
\begin{lemma}
\label{newlem2}
Assume that $\lambda\in\Lambda$ satisfies $(c)$ and $\sharp[\lambda]=k$, then we have
$$p^j_\lambda\le (m+n)C \left(p_{\lambda'}^j+p_{\lambda''}^{j-1}\right),$$
with $\lambda',\lambda''\in\Lambda$ satisfying $\sharp[\lambda']=\sharp[\lambda'']=k$, $v(\lambda')< v(\lambda)$ and $v(\lambda'')<v(\lambda)$.
\end{lemma}
\begin{proof}
We define $\lambda'$ as obtained from $\lambda$ by raising the occurrence of $a_\lambda$ on the side where no $a_\lambda+1$ appears by $1$ and $\lambda''$ as obtained from $\lambda$ by raising both occurrences of $a_\lambda$ by one. By definition there is a short exact sequence
$$0\to M(\lambda'')\to T M(\lambda')\to M(\lambda)\to 0$$
with $T=E_a$ if $a_\lambda+1$ appears on the right and $T=F_a$ otherwise. The corresponding long exact sequence and Lemma~\ref{bound2} then imply
$$p^j_\lambda\le (m+n)C p_{\lambda'}^j+p_{\lambda''}^{j-1}.$$
The properties of $\lambda',\lambda''$ follow from construction.
\end{proof}
\begin{lemma}
\label{newlem3}
Assume that $\lambda\in\Lambda$ with $\sharp[\lambda]=k$ satisfies $(d)$. Then we have
$$p_\lambda^j\le (m+n)^{m+n}C^{m+n} p^j_{\lambda_{+}}, $$
for some $\lambda_+\in\Lambda$ satisfying $($b$)$, $\sharp[\lambda_+]=k$ and $v(\lambda_+)=v(\lambda)$.
\end{lemma}
\begin{proof}
We define $\lambda_+$ as obtained from $\lambda$ by adding $1$ to every label strictly bigger than~$a_\lambda$. By composing the appropriate $E_i$ and $F_i$ ($v(\lambda)$ in total) into a translation functor $T$ we have $M(\lambda)^{\oplus d}=T M(\lambda_+)$ for some number $d$ with $1\le d\le v(\lambda)!$. Lemma~\ref{bound2} then implies
$$p_\lambda^j\le (m+n)^{v(\lambda)}C^{v(\lambda)} p^j_{\lambda_{+}}, $$
which proves the lemma.
\end{proof}
\begin{lemma}
\label{lemcVerma}
Assume that {\bf P}$[k-1]$ holds, with $1\le k\le \min\{m,n\}$, then also {\bf P}$[k]$ holds.
\end{lemma}
\begin{proof}
We will prove by induction on $v\in [0,m+n-2k]$ that there is a constant $C^{(v)}_{k}$ such that if $\lambda$ is of atypicality degree $k$ and $v(\lambda)\le v$, then $p_\lambda^j\le C_{k}^{(v)} j^{k-1}$ holds for all $j>0$. This proves the lemma for $C_{k}=C_{k}^{(m+n-2k)}$.
If $v(\lambda)=0$, then $\lambda$ satisfies (a), so Lemma~\ref{newlem1}(1) implies this result with $C^{(0)}_{k}=(m+n)^2CC_{k-1}$. Now assume that the property holds for all $v$ up to $\widehat{v}$ and consider $\lambda\in\Lambda$ with $\sharp[\lambda]=k$ and $v(\lambda)=\widehat{v}+1$.
\begin{enumerate}[(i)]
\item If $\lambda$ satisfies (c), then the induction hypothesis and Lemma~\ref{newlem2} imply
$$p^j_\lambda\le D_1 j^{k-1}\qquad\mbox{with}\quad D_1:=2(m+n)CC^{(\widehat v)}_{k}.$$
\item If $\lambda$ satisfies (b), then (i) and Lemma~\ref{newlem1}(2) imply
$$p^j_\lambda\le (m+n)^2CC_{k-1}j^{k-1}+(m+n)D_1 (j-d)^{k-1}\le D_2 j^{k-1}$$
for $D_2:=(m+n)D_1+(m+n)^2CC_{k-1} $.
\item If $\lambda$ satisfies (d), then (ii) and Lemma~\ref{newlem3} imply
$$p^j_\lambda\le D_3 j^{k-1}\qquad\mbox{with}\quad D_3:=(m+n)^{m+n}C^{m+n}D_2.$$
\end{enumerate}
Thus we can take $C_{k}^{(\widehat v+1)}=D_3$, which concludes the proof.
\end{proof}
\begin{proof}[Proof of Theorem \ref{complVerma}]
First we note that there is a constant $C_{0}$, such that property {\bf P}[0] holds. This follows from the equivalence of this question to the one in blocks of ${\mathcal O}_{{\mathbb Z}}^0$, see e.g. Lemma \ref{KLorbit}, since there are finitely many non-equivalent blocks each containing finitely many Verma modules. Lemma~\ref{lemcVerma} then iteratively proves the first statement in the theorem and thus also $c_{{\mathcal O}}(M(\lambda))\le \sharp[\lambda]$.
Now we establish the equality for regular weights. First we take $\kappa\in\Lambda^{++}$ and use Lemma~\ref{lemOF} to obtain
\begin{eqnarray*}
\sum_{\nu\in\Lambda^{++}}\dim{\rm Ext}_{{\mathcal O}}^j(M(\kappa),L(\nu))=\sum_{\nu\in\Lambda^{++}}{\rm Ext}^j_{{\mathcal F}}(K(\kappa), L(\nu)).
\end{eqnarray*}
This has polynomial growth rate $\sharp[\kappa]$ by Lemma~\ref{lemF}. Now for any $\kappa\in\Lambda^{++}$ and $w\in W$ we consider the subsequence of \eqref{KLVcompl}
$$\sum_{\nu\in\Lambda^{++}}\dim{\rm Ext}^j_{{\mathcal O}}(M(w\kappa), L(\nu))= \sum_{\nu\in\Lambda^{++}}\dim{\rm Ext}^{j-l(w)}_{{\mathcal O}}(M(\kappa), L(\nu)),$$
where the equality follows from Lemma~\ref{lemtwist}$(ii)$. This proves that $p^j_{w\kappa}$ has polynomial growth rate at least $\sharp [w\kappa]$.
\end{proof}
\begin{proof}[Proof of Theorem \ref{complK}]
First we prove that $c_{{\mathcal O}}(K(\lambda))\le \sharp[\lambda]$ for any $\lambda\in \Lambda$. For an anti-dominant $\mu$, we have $M(\mu)=K(\mu)$, so the result follows from Theorem \ref{complVerma}. Then we use (finite) induction by considering the Bruhat order $\prec_0$ for $\mathfrak{g}_0$ on $P_0$. Assume that $c_{{\mathcal O}}(K(\nu))\le k$ for all $\nu\prec_0\lambda$ with $\sharp[\lambda]=k$. Then the module $N$ defined by the exact sequence
$$0\to N\to M(\lambda)\to K(\lambda)\to 0,$$
has a filtration by $K(\nu)$ with $\nu\prec_0\lambda$. By the induction hypothesis and Lemma~\ref{lemcompseq} we have $c_{{\mathcal O}}(N)\le \sharp[\lambda]$. Lemma~\ref{lemcompseq} and Theorem \ref{complVerma} then imply $c_{{\mathcal O}}(K(\lambda))\le \sharp[\lambda]$.
Now take an arbitrary module which is ${\mathfrak g}_{-1}$-free. By Proposition \ref{propflag} this module has a filtration by Kac modules, the result thus follows from Lemma~\ref{lemcompseq}. All results are also valid for the anti-distinguished system of positive roots, which proves the claim for~${\mathfrak g}_1$.
\end{proof}
\subsection{Complexity of simple modules for $\mathfrak{gl}(m|n)$} Also the complexity of simple modules is in principle determined by Brundan's KL polynomials, see Corollary \ref{corsim}.
We investigate a relation between the complexity of a simple module and its ${\mathfrak n}$-cohomology. Therefore we introduce
$$c_{{\mathfrak n}}(M):=r(\dim H^\bullet({\mathfrak n},M)),$$
for $M\in{\mathcal O}$, with $r$ as introduced in Definition \ref{defC}.
\begin{proposition}
\label{compKost}
For any $\lambda\in\Lambda$, we have
$$\max\{c_{{\mathcal O}}(M(\lambda))\,,\,c_{{\mathfrak n}}(L(\lambda))\}\,\,\le\,\, c_{{\mathcal O}}(L(\lambda))\,\,\le\,\,\sharp[\lambda]+c_{{\mathfrak n}}(L(\lambda)).$$
\end{proposition}
\begin{proof}
Set $\sharp[\lambda]=k$, by equation~\eqref{extsimple} and Theorem \ref{complVerma} we have
$$\sum_{\mu\in\Lambda}{\rm Ext}^j_{{\mathcal O}}(L(\lambda),L(\mu))\le\sum_{i=0}^j C_{k}(j-i)^{k-1}\sum_{\kappa}\dim{\rm Ext}^i_{{\mathcal O}}(M(\kappa),L(\lambda)),$$
By equation~\eqref{VerH} and by setting $p=c_{{\mathfrak n}}(L(\lambda))$, there exists some constant $C$ for which
$$\sum_{\kappa\in\Lambda}\dim {\rm Ext}^i_{{\mathcal O}}(M(\kappa),L(\lambda))\le C i^{p-1},\quad\qquad\forall i\in{\mathbb N}.$$
Combining the two above equations leads to
$$\sum_{\mu\in\Lambda}{\rm Ext}^j_{{\mathcal O}}(L(\lambda),L(\mu))\le\sum_{i=0}^j C_{k}(j-i)^{k-1}Ci^{p-1}\le C C_{k} j^{k+p-1},$$
which implies the second inequality.
By considering only the extremal terms in the summation~\eqref{extsimple}, we find
$$\dim{\rm Ext}^j_{{\mathcal O}}(L(\lambda),L(\mu))\;\ge \;\max\left\{\dim {\rm Ext}^j_{{\mathcal O}}(M(\lambda),L(\mu)),\dim {\rm Ext}^j_{{\mathcal O}}(M(\mu),L(\lambda)) \right\}.$$
The first inequality in the claim then follows from equation~\eqref{VerH}.
\end{proof}
For finite dimensional simple modules we can improve the estimates.
\begin{proposition}
If $\kappa\in\Lambda^{++}$, we have
$$2\sharp[\kappa]\,\le\, c_{{\mathcal O}}(L(\kappa))\,\le\,\sharp[\kappa]+r\left(\sum_{\nu\in\Lambda^{++}}\dim{\rm Ext}^\bullet_{{\mathcal F}}(K(\nu),L(\kappa))\right).$$
\end{proposition}
\begin{proof}
Equation~\eqref{extsimple} gives the following lower bound for $\sum_{\mu\in\Lambda}\dim{\rm Ext}^j_{{\mathcal O}}(L(\kappa),L(\mu))$:
\begin{equation}\label{subseq}\sum_{\lambda,\nu\in\Lambda^{++}}\sum_{i=0}^j\dim{\rm Ext}^{i}_{{\mathcal O}}(M(\lambda),L(\kappa))\dim{\rm Ext}^{j-i}_{{\mathcal O}}(M(\lambda),L(\nu)).\end{equation}
By Lemma~\ref{lemOF} and the abstract KL theory of ${\mathcal F}$, see Theorem 4.51 and Corollary 4.52 in~\cite{Brundan}, we then find that the summation in \eqref{subseq} is equal to $\sum_{\nu\in\Lambda^{++}}{\rm Ext}^j_{{\mathcal F}}(L(\kappa),L(\nu)).$
This has polynomial growth rate $2\sharp[\kappa]$ by Theorem 9.1.1 in~\cite{BKN2}, proving the first inequality.
By Lemma~\ref{lemtwist}$(i)$ and $(ii)$ and Lemma~\ref{lemOF} we have
$$\dim H^j({\mathfrak n},L(\kappa))=\sum_{i=0}^{l(w_0)}\left(\sharp W(i) \right)\sum_{\lambda\in\Lambda^{++}} \dim{\rm Ext}^{j-i}_{{\mathcal F}}(K(\lambda),L(\kappa)),$$
with $\sharp W(i)$ the number of elements in $W$ of length $i$. This proves the second inequality by Proposition \ref{compKost}.
\end{proof}
We end this subsection with a conjecture.
\begin{conjecture}
\label{consimple}
For any $\lambda\in\Lambda$ we have $c_{{\mathcal O}}(L(\lambda))=2\sharp[\lambda]$ and $c_{{\mathcal O}}(M(\lambda))=\sharp[\lambda]$.
\end{conjecture}
If this conjecture is true we obtain in particular that for an integral block ${\mathcal O}_{\xi}$
\begin{itemize}
\item a categorical interpretation of the singularity, by the finitistic global dimension $2\mathtt{a}(w_0w_0^{\xi})$, see Theorem \ref{fdblock}.
\item a categorical interpretation of the atypicality, by the global complexity $2\sharp\xi$, see Conjecture \ref{consimple}.
\end{itemize}
\subsection{Link between complexity and associated variety}
We note two explicit connections between complexity in category~${\mathcal O}$ for $\mathfrak{gl}(m|n)$ and the associated variety, which follow from Theorem \ref{finpdind}, Lemma~\ref{general}(2) and Theorem \ref{complK}.
\begin{proposition}
\begin{enumerate}
\item For any $M\in{\mathcal O}$, we have $c_{{\mathcal O}}(M)=0\Leftrightarrow X_M=\{0\}$.
\item If $M\in{\mathcal O}$ is ${\mathfrak g}_{-1}$-free and admits a generalised central character of atypicality degree $k$, we have both $c_{{\mathcal O}}(M)\le k$ and $|S|\le k$ for any $S\in {\mathcal S}(M)$.
\end{enumerate}
\end{proposition}
This results seem to suggest that there must be some deeper connection between complexity and the associated variety. Similar connections appear in~\cite{BKN2}.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,973
|
package com.swfarm.biz.chain.dao.impl;
import com.swfarm.biz.chain.bo.PurchaseTaskFeedback;
import com.swfarm.biz.chain.dao.PurchaseTaskFeedbackDao;
import com.swfarm.pub.framework.dao.GenericDaoHibernateImpl;
public class PurchaseTaskFeedbackDaoImpl extends
GenericDaoHibernateImpl<PurchaseTaskFeedback, Long> implements
PurchaseTaskFeedbackDao {
public PurchaseTaskFeedbackDaoImpl(Class<PurchaseTaskFeedback> type) {
super(type);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,085
|
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 5,991
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Whenever Agrus Kos, Wojek Veteran attacks, attacking red creatures get +2/+0 and attacking white creatures get +0/+2 until end of turn.
Whenever Akki Lavarunner deals damage to an opponent, flip it.
: Akroma, Angel of Fury gets +1/+0 until end of turn.
At the beginning of your upkeep, if you have more cards in hand than each opponent, you may sacrifice a Swamp. If you do, return Akuta, Born of Ash from your graveyard to the battlefield.
As Alhammarret, High Arbiter enters the battlefield, each opponent reveals their hand. You choose the name of a nonland card revealed this way.
Your opponents can't cast spells with the chosen name (as long as this creature is on the battlefield).
: Target player puts the top three cards of their library into their graveyard.
If a nontoken creature an opponent owns would die or a creature card not on the battlefield would be put into an opponent's graveyard, exile that card instead.
, : Prevent all combat damage that would be dealt this turn. Activate this ability only before the combat damage step.
Anya, Merciless Angel gets +3/+3 for each opponent whose life total is less than half their starting life total.
As long as an opponent's life total is less than half their starting life total, Anya has indestructible.
Eminence — At the beginning of combat on your turn, if Arahbo, Roar of the World is in the command zone or on the battlefield, another target Cat you control gets +3/+3 until end of turn.
Whenever another Cat you control attacks, you may pay . If you do, it gains trample and gets +X/+X until end of turn, where X is its power.
At the beginning of your upkeep, sacrifice Arcades Sabboth unless you pay .
: Arcades Sabboth gets +0/+1 until end of turn.
: Target artifact creature's controller sacrifices it. That player may search their library for a noncreature artifact card, put it onto the battlefield, then shuffle their library.
Whenever you cast a spell, put the cards in your hand on the bottom of your library in any order, then draw that many cards.
Whenever Ashling, the Extinguisher deals combat damage to a player, choose target creature that player controls. The player sacrifices that creature.
: Until end of turn, Autumn Willow can be the target of spells and abilities controlled by target player as though it didn't have shroud.
: Attach target Aura attached to a creature to another creature.
, Sacrifice another creature: Exile target nonland permanent. Activate this ability only if you have at least 10 life more than your starting life total.
Tap an untapped Wizard you control: Draw a card.
You may play two additional lands on each of your turns.
, Exile Balthor the Defiled: Each player returns all black and all red creature cards from their graveyard to the battlefield.
Other Barbarian creatures get +1/+1.
: Another target Barbarian creature gets +1/+0 until end of turn.
: Regenerate another target Vampire.
Whenever a Forest enters the battlefield, green creatures you control get +1/+1 and gain trample until end of turn.
Grandeur — Discard another card named Baru, Fist of Krosa: Create an X/X green Wurm creature token, where X is the number of lands you control.
: Ben-Ben, Akki Hermit deals damage to target attacking creature equal to the number of untapped Mountains you control.
: Target spell or permanent becomes the color of your choice until end of turn.
Bontu the Glorified can't attack or block unless a creature died under your control this turn.
, Sacrifice another creature: Scry 1. Each opponent loses 1 life and you gain 1 life.
, : Create a 1/1 black and red Demon creature token named Minor Demon.
, Sacrifice an artifact: Bosh, Iron Golem deals damage equal to the sacrificed artifact's converted mana cost to any target.
Whenever you cast a Spirit or Arcane spell, you may gain life equal to that spell's converted mana cost.
Whenever Brago, King Eternal deals combat damage to a player, exile any number of target nonland permanents you control, then return those cards to the battlefield under their owner's control.
At the beginning of each player's upkeep, that player sacrifices an artifact, creature, or land.
At the beginning of each player's upkeep, that player may put an artifact, creature, or land card from their hand onto the battlefield.
• Breya deals 3 damage to target player or planeswalker.
• Target creature gets -4/-4 until end of turn.
: Brigid, Hero of Kinsbaile deals 2 damage to each attacking or blocking creature target player controls.
At the beginning of combat on your turn, create a 2/1 blue Myr artifact creature token. Then you may choose a token you control. If you do, each other token you control becomes a copy of that token.
: You may put a land card from your hand onto the battlefield. If you control ten or more lands, flip Budoka Gardener.
Whenever you cast a Spirit or Arcane spell, you may put a ki counter on Budoka Pupil.
At the beginning of the end step, if there are two or more ki counters on Budoka Pupil, you may flip it.
When a creature dealt damage by Bushi Tenderfoot this turn dies, flip Bushi Tenderfoot.
: Target opponent discards two cards. Activate this ability only during your turn, before attackers are declared.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 1,482
|
\section{Introduction}
\label{section:introduction}
Galactic winds are crucial to the process of galaxy formation, ejecting gas from galaxies \citep{ham}, helping to regulate star formation, shaping the stellar mass function and the mass-metallicity relation \citep{dekel,finlator_dave,erb08,peeples_shankar}, and enriching the intergalactic medium with metals \citep{aguirre01,oppenheimer_dave06,oppenheimer_dave08}. Hot winds heated by supernovae and stellar winds or active galactic nuclei may be present in both star-forming and passive galaxies at all redshifts. Quantifying their prevalence, dynamical importance, and observational signatures is a key area in observational and theoretical studies of galactic winds.
An important outstanding question is the nature of the cool clouds of molecular, neutral atomic, and ionized gas seen in blue-shifted absorption and emission in rapidly star-forming galaxies and starbursts at all redshifts. The prevailing picture is that these clouds are driven out of the host galaxy by ram pressure acceleration from a supernova-heated hot wind (e.g., \citealt{veilleux_review}), but it is unclear if clouds can be accelerated to the required velocities before being shocked and shredded by hydrodynamical instabilities \citep{scannapieco,zhang15}. Additional mechanisms for cold cloud acceleration have also been suggested, including momentum deposition by supernovae, the radiation pressure of starlight on dust grains \citep{mqt,mqt10,mmt,hopkins12_wind,zhang_thompson,krumholz_thompson13,davis_kt,thompson_krumholz,thompson_2015}, and cosmic rays \citep{breitschwerdt94,breitschwerdt99,jubelgas,socrates_cr,hanasz}.
Here, we revisit the physics of the radiative cooling of hot, initially adiabatic flows. We argue for a picture where the cool gas in galactic winds and halos precipitates directly from the hot wind as a result of radiative cooling, based on earlier work by \cite{wang95a,wang95b}, which was subsequently developed by \cite{efstathiou2000} and \cite{silich2003,silich2004,tenorio,tenorio05,tenorio07,wunsch}, and suggested in ultra-luminous infrared galaxies by \cite{martin_lya}.
In Section \ref{section:analytic} we present an analytic discussion that shows that mass-loaded winds always cool on large scales. The key point is that adiabatic cooling lowers the gas temperature to a value where further decreasing the temperature leads to more rapid cooling (e.g., Fig.~\ref{figure:c}). In this way, the cooling timescale can become shorter than the dynamical time for expansion, the assumption of adiabaticity breaks down, and the flow rapidly cools, radiating its remaining thermal energy content. We provide simple scalings that give the cooling radius, radiative efficiency, emission measure, column density, and velocity at the cooling radius. We further identify critical values of the star formation rate surface density and mass-loading rate for the cooling instability. Section \ref{section:conduction} shows when we expect conduction to dominate energy transport in hot winds, leading to much flatter temperature profiles close to the central galaxy.
In Section \ref{section:numerical} we solve the general equations for a time-steady flow with an arbitrary cooling/heating function. The strongly cooled wind forms an extended region of cool photoionized outflowing gas. We show that the extended gravitational potential of galaxies decelerates the cooled wind on $\sim{\rm few}-100$\,kpc scales. Section \ref{section:ti} discusses the linear instability of the cooling flow. We show that it is both convectively and thermally unstable, and that the latter should enhance density fluctuations by of order $\sim100$. In Section \ref{section:discussion} we discuss our results in terms of observations of systems including NGC 253 and M82, ultra-luminous infrared galaxies with high velocity line emission, high redshift rapidly star forming galaxies, and the absorption line systems both at $z\sim0$ observed by COS-Halos \citep{werk}, and at high-redshift \citep{steidel_cgm}. We note in Section \ref{section:rebirth} that the very strong dependence of the cooling radius and efficiency on the mass-loading rate $\beta$ motivates a picture of galactic winds where the cool clouds are initially launched by ram pressure acceleration and are rapidly destroyed by hydrodynamical instabilities. This process seeds the hot flow with both a higher mass loading rate, that rapidly causes cooling on larger scales, and density fluctuations that grow rapidly as the thermal instability develops. In Section \ref{section:halo} we discuss the origin of the cool gas in halos and we calculate the cooling time of the wind after it shocks on the surrounding circumgalactic medium while blowing a wind-driven bubble. Our conclusions are presented in Section \ref{section:conclusions}.
\section{Wind Cooling}
\label{section:winds}
\subsection{Analytic Considerations}
\label{section:analytic}
One might expect that a hot spherical wind never cools radiatively since it rapidly expands to low density ($n\propto r^{-2}$). To see that it can, first note that adiabatic cooling of an ideal gas dictates that $P/\rho^\gamma={\rm constant}$, which implies that $T\propto r^{-4/3}$ ($\gamma=5/3$). As the super-heated flow initially expands from $T\sim10^8$\,K, cooling is dominated by bremsstrahlung, the cooling time is $t_{\rm cool}\propto T^{1/2}/n\propto r^{4/3}$, the advection time is $t_{\rm adv}\sim r/v\propto r$, so that the ratio $t_{\rm cool}/t_{\rm adv}\propto r^{1/3}$ grows slowly with radius. Thus, if the flow starts with $t_{\rm cool}/t_{\rm adv}>1$, it remains adiabatic as long as bremsstrahlung cooling dominates cooling.
The situation changes when the flow cools enough that metal lines dominate the gas emissivity at $T\lesssim10^7$\,K. At these lower temperatures, which the flow inevitably reaches at sufficiently large distance as a result of adiabatic cooling, the shape of the cooling function changes, and can be approximated as (e.g., \citealt{maclow,draine})
\begin{equation}
\Lambda(T)=\dot{\varepsilon}/n^2\simeq3\times10^{-23}\,T_7^{-0.7}\xi\,\,\,(10^5\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}} T\lesssim10^7\,{\rm K}),
\label{cool}
\end{equation}
where $T_7=T/10^7$\,K and where $\xi$ is meant to capture the dependence of $\Lambda$ on the metallicity of the gas: $\xi=1$ for Solar composition and $\xi\sim0.3$ and $\sim3$ for composition of $0.1$ and $3$ times Solar, respectively. In this regime $t_{\rm cool}\propto r^{-4/15}$ and the ratio $t_{\rm cool}/t_{\rm adv}\propto r^{-19/15}$ decreases rapidly with radius. Thus, even if the flow starts with $t_{\rm cool}/t_{\rm adv}>1$, on large physical scales it has a chance to become radiative. Although the temperature scaling, normalization, and density dependence of the cooling function depends on the metallicity of the medium and whether or not it is collisionally ionized, photoionized, or in ionization equilibrium, the upturn in $\Lambda(T)$ at low temperatures is unavoidable because of line cooling, even in a gas of primordial composition. An example of a representative cooling function for solar metallicity gas in photoionization equilibrium (PIE) with the meta-galactic UV background is shown in Figure \ref{figure:c}. Note that the outflows powered by supernovae in rapidly star-forming galaxies may have super-solar metallicities, depending on the mass loading and metallicity of the surrounding ISM. More discussion of metallicity in the context of radiative cooling is provided in Section \ref{section:numerical}.
\begin{figure}
\centerline{\includegraphics[width=8.5cm]{plot_c.pdf}}
\caption{Illustrative example of the cooling/heating function in photoionization equilibrium (PIE) with a metagalactic UV background \citep{oppenheimer_schaye,wiersma}. The red line shows the contribution from H and He, whereas the purple line shows the contribution from metals. Bremsstrahlung dominates cooling at high temperatures. Heating dominates at low temperatures (the ``spike" at $\sim10^4$\,K shows where $|\dot{\varepsilon}|$ goes through zero). Equation (\ref{cool}) gives a simple approximation to the cooling function between $10^5\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}} T\lesssim10^7$\,K for solar-metallicity gas in collisional ionization equilibrium (CIE) and over a range of gas densities in PIE.}
\label{figure:c}
\end{figure}
As in \cite{wang95a,wang95b} (see also \citealt{efstathiou2000,silich2003,silich2004}), we adopt the wind model of \cite{cc85} (hereafter CC85) to evaluate the importance of radiative cooling. We define dimensionless parameters $\beta$ and $\alpha$ such that
\begin{equation}
\dot{M}_{\rm hot}=\beta\,{\dot{M}_\star}
\label{mdothot}
\end{equation}
and
\begin{equation}
\dot{E}_{\rm hot}=3\times10^{41}\,{\rm ergs\,\,s^{-1}}\,\alpha\,{\dot{M}_\star},
\label{edothot}
\end{equation}
where the latter assumes that $\alpha10^{51}$\,ergs per core-collapse supernova (SN) is thermalized in the hot plasma, that there are $10^{-2}$\,${\rm SNe/yr}$ per M$_\odot/{\rm yr}$ of star formation, and that the star formation rate $\dot{M}_\star$ is measured in M$_\odot/{\rm yr}$. For a spherical galaxy of size $R$, the CC85 model assumes a constant mass loading rate and energy deposition rate per unit volume for $r<R$ and that the flow is purely adiabatic and mass-conserving for $r>R$. Given the parameters $\alpha$ and $\beta$, the solution then provides the radial run of pressure, density, and velocity as a function of radius at all $r$. The character of the solution is such that the Mach number ${\cal M}=1$ at $R$. Appendix \ref{appendix:cc85} provides a discussion of the CC85 model.
Using the CC85 wind model for a wind expanding into a fixed solid angle $\Omega$, employing the cooling function given in equation (\ref{cool}), and assuming that $r\gg R$ ($v\approx$\,constant), we find that
\begin{equation}
t_{\rm cool}\simeq3\times10^{6}\,{\rm yr}\frac{\alpha^{2.20}}{\beta^{3.20}} \left(\frac{R_{0.3}}{r_{10}}\right)^{0.27}\frac{R_{0.3}^{2}}{{\dot{M}_{\star,10}}}\,\frac{\Omega_{4\pi}}{\xi},
\label{tcool}
\end{equation}
and
\begin{equation}
t_{\rm adv}\simeq1\times10^{7}\,{\rm yr}\,\left(\frac{\beta}{\alpha}\right)^{1/2}\,r_{10},
\label{tadv}
\end{equation}
where $R_{0.3}=R/0.3$\,kpc, $r_{10}=r/10$\,kpc, $\dot{M}_{\star,10}=\dot{M}_\star/10$\,M$_\odot$ yr$^{-1}$, $\Omega_{4\pi}=\Omega/4\pi$, and we have dropped the dependence on the mean molecular weight and Hydrogen fraction in $t_{\rm cool}$ (we assume $\mu=1$ throughout this paper). In equation (\ref{tcool}) for $t_{\rm cool}$, note the importance of the star formation rate surface density $\dot{\Sigma}_\star={\dot{M}_\star}/\pi R^2$, which varies among rapidly star-forming galaxies by more than four orders of magnitude \citep{kennicutt98}. Systems with higher $\dot{\Sigma}_\star$ have shorter flow cooling timescale \citep{strickland_heckman}.
The choice of parameters for scaling the relations above (and below) is motivated in part by M82, which has $R\simeq250$\,pc and $\dot{M}_\star\simeq5-10$\,M$_\odot$ yr$^{-1}$ (e.g., \citealt{rieke,forster2001,forster2003}). In addition, from X-ray observations M82 has been inferred to have $\alpha\simeq1$ and $\beta\simeq0.2-0.5$ \citep{strickland_heckman}. These values of $\dot{M}_\star$ and $R$ are also roughly appropriate for the individual massive star-forming clumps seen in high redshift galaxies \citep{genzel_clump}. We will see below that the propensity for radiative cooling and its radiative efficiency are strong functions of the host galaxy parameters. The applicability to individual systems and collections of systems is discussed in Section \ref{section:discussion}.
Setting $t_{\rm cool}=t_{\rm adv}$, we derive the critical radius beyond which the flow becomes radiative in the $r\gg R$ limit:
\begin{equation}
r_{\rm cool}\simeq4\,{\rm kpc}\,\frac{\alpha^{2.13}}{\beta^{2.92}}\,R_{0.3}^{1.79}\,\left(\frac{\Omega_{4\pi}}{\xi\dot{M}_{\star,10}}\right)^{0.789}.
\label{rcool}
\end{equation}
Note the very strong dependence on both $\alpha$ and $\beta$. A small change in either can move $r_{\rm cool}$ significantly. However, $\beta$ cannot be arbitrarily small and still have rapid cooling. Because the cooling function has a peak and drops at temperatures below $\sim10^5$\,K (e.g., Fig.~\ref{figure:c}), and because $t_{\rm cool}/t_{\rm adv}$ decreases as $r^{-19/15}$ while the cooling function of equation (\ref{cool}) is applicable, there is a minimum value of the mass loading parameter $\beta$ such that rapid radiative cooling occurs anywhere in the wind:
\begin{equation}
\beta_{\rm min}\simeq0.64\,\alpha^{0.636}\,\left(\frac{R_{0.3}}{\dot{M}_{\star,10}}\frac{\Omega_{4\pi}}{\xi}\right)^{0.364}.
\label{betamin}
\end{equation}
For $\beta\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}}\beta_{\rm min}$, the flow remains adiabatic to arbitrarily large scales, and although $t_{\rm cool}/t_{\rm adv}$ decreases in the temperature range where equation (\ref{cool}) is applicable, it never crosses unity before the cooling function turns over at temperatures below $\sim10^5$\,K (Fig.~\ref{figure:c}). For a ULIRG with $\dot{M}_\star\sim100$\,M$_\odot$ yr$^{-1}$, $\beta_{\rm min}\simeq0.3$. For an LBG with larger $R\sim\,{\rm kpc}$ and $\dot{M}_\star\sim30$\,M$_\odot$ yr$^{-1}$, $\beta_{\rm min}\simeq0.6$. These scalings imply that radiative cooling of hot galactic winds may be prevalent both in local starbursts and in galaxies on the star forming main sequence at high redshift \citep{wuyts11}.
Equation (\ref{rcool}) can be re-written as a critical condition on $\beta$, such that $r_{\rm cool}$ obtains at a radius $r\gg R$
\begin{equation}
\beta_{\rm crit}\simeq0.73\alpha^{0.730} \left(\frac{10\,{\rm kpc}}{r_{\rm cool}}\right)^{0.342}R_{\rm 0.3}^{0.613}
\left(\frac{\Omega_{4\pi}}{\xi\dot{M}_{\star,10}}\right)^{0.270}.
\label{betacrit1}
\end{equation}
The fact that $\beta_{\rm crit}$ is similar to $\beta_{\rm min}$ implies that the maximum value of the cooling radius is $\sim10$\,kpc for our nominal scaling parameters. Substituting, we find that the maximum value of the cooling radius is
\begin{equation}
r_{\rm cool}^{\rm max}(\beta=\beta_{\rm min})\simeq15\,{\rm kpc}\,\alpha^{0.274}R_{0.3}^{0.728}\left(\frac{\xi\dot{M}_{\star,10}}{\Omega_{4\pi}}\right)^{0.273}.
\end{equation}
If the wind is mass-loaded sufficiently to cool radiatively ($\beta>\beta_{\rm min}$), it does so on a scale smaller than $r_{\rm cool}^{\rm max}$.
By setting $r_{\rm cool}=R$ we derive the critical mass loading rate such that the cooling radius collapses to the scale of the launching radius $R$. For this calculation we use the general CC85 solution valid at $r=R$ (eq.~\ref{rcool} requires $r\gg R$). We find that
\begin{equation}
\beta_{\rm crit}(r_{\rm cool}=R)\simeq1.95\alpha^{0.730}\left(\frac{\Omega_{4\pi}R_{\rm 0.3}}{{\xi \dot{M}_{\star,10}}}\right)^{0.270}.
\label{betacrit}
\end{equation}
The expressions above can be inverted to yield a critical star formation rate surface density ($\dot{\Sigma}_\star={\dot{M}_\star}/\pi R^2$) for the flow to cool at any given radius $r\gg R$
\begin{equation}
\dot{\Sigma}_{\star,\,\rm crit}\simeq 3.5\,\frac{{\rm M_\odot}}{\rm yr\,kpc^{2}}\,R^{0.267}_{0.3}\left(\frac{\rm 10\,kpc}{r_{\rm cool}}\right)^{1.27}\frac{\alpha^{2.70}}{\beta^{3.70}}\frac{\Omega_{4\pi}}{\xi},
\label{sdscrit}
\end{equation}
which is exceeded by starbursts in the local universe and by rapidly star-forming galaxies at high-redshift (e.g., \citealt{wuyts11}).
We can also estimate the critical star formation surface density required for the flow to cool at $r_{\rm cool}=R$ using equation (\ref{betacrit}), which employs the full solution to the CC85 equations at $r=R$ and not the limit $r\gg R$ (see Appendix A; see also Appendix A of \citealt{strickland_heckman}). We find that
\begin{equation}
\dot{\Sigma}_{\star,\,\rm crit}(r_{\rm cool}=R)\simeq420\,\frac{{\rm M_\odot}}{\rm yr\,kpc^{2}}\,\frac{\alpha^{2.70}}{\beta^{3.70}}\,\frac{\Omega_{4\pi}}{\xi R_{0.3}}.
\label{sds_R}
\end{equation}
is required for the cooling radius to reach $r_{\rm cool}=R$.\footnote{This value is somewhat higher than the estimate of $\dot{\Sigma}_{\star,\,\rm crit}(r_{\rm cool}=R)\simeq300\,{\rm M_\odot}/{\rm yr\,/\,kpc^{2}}$ obtained by setting $r_{\rm cool}=R$ in equation (\ref{sdscrit}), which is only strictly applicable in the $r_{\rm cool}\gg R$ limit.}
The latter is achieved and exceeded in local ULIRGs (e.g., \citealt{condon91,downes_solomon,barcos_munoz,scoville_arp220}). As a result, radiatively cooling hot winds may be a natural explanation for the fast line emission seen in these and similar systems with very high $\dot{\Sigma}_\star$ \citep{soto_fastlines,martin_lya}. More generally, much of the star formation at high redshift occurs in systems that meet or exceed the value of $\dot{\Sigma}_{\star,\,\rm crit}$ given in equation (\ref{sdscrit}), and we therefore expect the radiative cooling of their winds to be important.
The fact that the minimum value of the mass loading required for radiative cooling to happen anywhere in the supersonic flow ($\beta_{\rm min}$) and the maximum value such that $r_{\rm cool}\rightarrow R$ ($\beta_{\rm crit}$ in eq.~\ref{betacrit}) differ only by a factor of a few suggests that the velocity of the gas when it cools is constrained. This follows from the fact that the asymptotic velocity of the hot flow is $v\simeq10^3\,(\alpha/\beta)^{1/2}\,{\rm km/s}$. Substituting $\beta_{\rm min}$ into $v$ we find a maximum velocity of the cooling flow of
\begin{equation}
v_{\rm max}(r_{\rm cool})\simeq1250\,{\rm km/s}\,\left(\frac{\alpha\,\xi\,\dot{M}_{\star,10}}{\Omega_{4\pi}\,R_{0.3}}\right)^{0.180},
\label{vmax}
\end{equation}
which, for extreme ULIRG-like parameters (e.g., $R=50$\,pc and $\dot{M}_\star=100$\,M$_\odot$ yr$^{-1}$; see, e.g., \citealt{barcos_munoz}) reaches $\simeq2600$\,km/s, while for LBG-like parameters ($R=1$\,kpc and $\dot{M}_\star=30$\,M$_\odot$ yr$^{-1}$) is $\simeq1200$\,km/s. Similarly, substituting $\beta_{\rm crit}$ from equation (\ref{betacrit}) into $v_\infty$ we find a characteristic velocity for cooling on scale $R$:
\begin{equation}
v_{\rm crit}(r_{\rm cool}=R)\simeq720\,{\rm km/s}\,\left(\frac{\alpha\,\xi\,\dot{M}_{\star,10}}{\Omega_{4\pi}\,R_{0.3}}\right)^{0.135},
\label{vmin}
\end{equation}
which reaches $\simeq1250$\,km/s and $\simeq700$\,km/s for ULIRG and LBG parameters, respectively. Although subject to the uncertainty and potential diversity in both $\alpha$ and $\Omega$ in the above equations, these scalings imply that velocities inferred from absorption and emission lines in galactic winds should have a well-defined maximum. They also imply that if one selects galaxies of approximately the same size, then $v$ should correlate with $\dot{M}_\star$ weakly. We return to the issue of velocity profiles in radiatively cooling winds in Section \ref{section:numerical}, where we show that the flow can decelerate significantly on $\sim1-100$\,kpc scales because of gravity, and in Section \ref{section:discussion}, where we argue that because of potentially super-critical mass-loading (i.e., $\beta>\beta_{\rm crit}$) fountain flows with a broad range of velocities may develop. Even so, $v_{\rm max}$ (eq.~\ref{vmax}) is a well-defined upper bound to the velocity of cooled gas in rapidly cooling hot winds that can be tested with observations, and $v_{\rm crit}(\beta=\beta_{\rm crit})$ (eqs.~\ref{vmax} and \ref{betacrit}) is a characteristic velocity for hot winds (or portions of hot winds) that are sufficiently mass loaded to cool at $r_{\rm cool}=R$ and then escape to large scales.
Both the scale of $r_{\rm cool}$ and the value of $\beta_{\rm crit}$ have additional direct consequences for observables. If the flow becomes radiative, then it can give up at most its total thermal energy content, which varies strongly as a function of radius. At $r=R$, the Mach number ${\cal M}=1$ and the thermal and kinetic energy are comparable, with the total enthalpy flux exceeding the kinetic energy flux by a factor of 3. If $\beta_{\rm crit}(r_{\rm cool}=R)$ obtains, we would therefore expect the total radiated luminosity as the matter cools to be approximately
\begin{equation}
\eta_{\rm max}\sim\dot{E}_{\rm hot}/L_\star\sim10^{-2},
\label{etamax}
\end{equation}
where
\begin{equation}
L_{\star}\simeq10^{11}\,L_\odot\,{\dot{M}_{\star,10}}
\label{lstar}
\end{equation}
is the bolometric luminosity from star formation. This value of $\eta$ is consistent with observations of line emission seen in \cite{martin_lya} (e.g., their Figs.~14 \& 15). However, for $r_{\rm cool}\gg R$, the radiative efficiency of the cooling flow decreases rapidly because adiabatic cooling drives down the thermal content of the matter as the radius increases. In the limit $r_{\rm cool}\gg R$, the Mach number can be approximated (for $\gamma=5/3$) as ${\cal M}^2\simeq16^{2/3}(r/R)^{4/3}$, and the radiative efficiency of the flow can be approximated as
\beqa
\eta&=&\left.\frac{1}{{\cal M}^2}\right|_{r_{\rm cool}}=\frac{L_{X}}{\dot{E}_{\rm hot}}\simeq100 \frac{L_{X}}{L_{\star}} \nonumber \\
&\sim&5.0\times10^{-3}\,\frac{\beta^{3.89}}{\alpha^{2.84}} \,\left(\frac{{\xi\,\dot{M}_{\star,10}}}{\Omega_{4\pi}\,R_{0.3}}\right)^{1.05},
\label{eta}
\eeqa
where $L_X$ is the luminosity radiated from the wind as it cools. The subscript $X$ represents the fact that $r_{\rm cool}$ occurs at a temperature between $10^{6.5}$ and $10^5$\,K and is thus in the X-ray or far-UV regime. In Section \ref{section:coolwind} we present a more detailed discussion of the radiative efficiency in the context of the models computed in Section \ref{section:numerical}. Figure \ref{figure:dl} shows that the radiative power of the wind is not in fact dominated by the cooling region, but by the hotter, effectively adiabatic region at smaller radius (see also eq.~\ref{lum}). Even so, equation (\ref{eta}) gives a first estimate of how the radiative efficiency of the cooling material scales with the parameters of the problem. Note, though, that this estimate for the radiative efficiency is just the thermal energy of the flow at $r_{\rm cool}$. Once the temperature drops to $\sim10^3-10^4$\,K, it will be subject to phoionization heating from the UV emission from the galaxy or the metagalactic UV background. The material will then expand at approximately constant temperature, in photoionization equilibrium, and it will continue to radiate (this is the outer zone discussed in \citealt{silich2003,silich2004}). We argue in Section \ref{section:halo} that this cool outflowing material may form the cool gas seen in galactic halos at all redshifts.
The strong dependencies of $r_{\rm cool}$ on the parameters of the system translate into even stronger dependencies for the gas density, column density ($N=nr$), and emission measure (${\rm EM}=n^2r$) at the cooling radius:
\begin{equation}
n_{\rm cool}\simeq2.0\times10^{-3}\,\,{\rm cm^{-3}}\,\,
\frac{\beta^{7.34}}{\alpha^{4.76}}\frac{\xi^{1.58}}{R_{0.3}^{3.58}}\left(\frac{{\dot{M}_{\star,10}}}{\Omega_{4\pi}}\right)^{2.58},
\label{ncool}
\end{equation}
\begin{equation}
N_{\rm cool}\simeq2.5\times10^{19}\,\,{\rm cm^{-2}}\,\,
\frac{\beta^{4.42}}{\alpha^{2.63}}\xi^{0.789}\left(\frac{{\dot{M}_{\star,10}}}{\Omega_{4\pi}R_{0.3}}\right)^{1.79},
\label{Ncool}
\end{equation}
\begin{equation}
\frac{{\rm EM}_{\rm cool}}{\rm pc\,\,cm^{-6}}\simeq1.6\times10^{-2}\,
\frac{\beta^{11.8}}{\alpha^{7.39}}\,\frac{\xi^{2.37}}{R_{0.3}^{5.37}}\left(\frac{\dot{M}_{\star,10}}{\Omega_{4\pi}}\right)^{4.37}.
\label{emcool}
\end{equation}
The gas density at the cooling radius is important for assessing the applicability of CIE, the propensity of the medium to cool below its temperature in photoionization equilibrium, and in determining the final density of clouds precipitated out of the hot flow by thermal instability (see Section \ref{section:ti}). The column density is important for assessing the effective absorption optical depth of a given line for absorption-line studies of winds viewed either ``down the barrel" or at large impact parameter with background quasars or galaxies. The column is also important for estimating the scattering optical depth of Ly$\alpha$ radiation and other resonant transitions in escaping the expanding flow, and for studies of emission-line halos and their velocity profiles. Finally, the emission measure is useful for characterizing the expected surface brightness of the radiating gas in resolved systems and for emission line studies of warm/hot gas as it cools from $\sim10^{6.5}$\,K to $\sim10^3-10^4$\,K.
Note that the scalings above imply strong diversity among the observed properties of systems; even a slight change in $\beta$, $\alpha$, $\Omega$, $R$, or ${\dot{M}_\star}$ can lead to significant changes in $n_{\rm cool}$, $N_{\rm cool}$, and ${\rm EM}_{\rm cool}$. Conversely, both the range of $\beta$ and the range of velocities for a cooling flow are tightly bounded. This narrow range of $\beta$ will imprint itself on the column density and emission measures inferred from observations. For example, substituting $\beta_{\rm min}$ into $N_{\rm cool}$, one finds a minimum column density at which cooling can occur of
\begin{equation}
N_{\rm cool,\,min}\simeq3.4\times10^{18}\,\,{\rm cm^{-2}}\,\left(\frac{\alpha\,\dot{M}_{\star,10}}{\Omega_{4\pi}\,R_{0.3}}\right)^{0.18}\frac{1}{\xi^{0.82}}.
\label{ncoolmin}
\end{equation}
If the limiting total gas column density one could detect in a given survey is $10^{20}$\,cm$^{-2}$, for example, this will significantly limit the mass loading of the systems that could in principle be seen and the velocities that material would be expected to reach. If galactic winds are better described by a wide range of $\beta$ along different sightlines in a given system, as we argue may be the case in Section \ref{section:coolwind}, then the {\it fastest} cool photoionized gas (eq.~\ref{vmax}) in a given observed system with have the {\it minimum} column density, and it will be given by equation (\ref{ncoolmin}). Indeed, the strong diversity implied by our expressions for $n_{\rm cool}$, $N_{\rm cool}$, and ${\rm EM}_{\rm cool}$ may be partially mitigated if there is a self-regulating mechanism that drives $\beta\rightarrow\beta_{\rm crit}$, as one might expect if cool clouds from the host are incorporated into the hot flow, as we discuss in Section \ref{section:rebirth}.
Finally, note that although we focus here on radiative cooling on scales larger than the host galaxy size $R$, radiative cooling can also occur at $r<R$ where the mass and energy are injected (eqs.~\ref{mdothot} and \ref{edothot}; Appendix \ref{appendix:cc85}). \cite{tenorio07} have studied this problem both analytically and numerically in the context of individual super star clusters. Assuming uniform volumetric mass loading ($\dot{m}_{\rm hot}$; g/s/${\rm cm}^3$) and energy injection rates ($\dot{e}_{\rm hot}$; ergs/s/${\rm cm}^3$), and for a fixed value of $\dot{e}_{\rm hot}/\dot{m}_{\rm hot}$, they find a critical stagnation radius $R_{\rm st}$ that approaches $r=0$ and $R$ in the limit of low and high $\dot{E}_{\rm hot}$, respectively. Only material injected with $R_{\rm st}<r<R$ participates in a supersonic outflow. Material injected at $0<r<R_{\rm st}$ rapidly cools, does not enter the outflow, and perhaps gives rise to a new generation of star formation \citep{tenorio07,palous}. As $R_{\rm st}$ approaches $R$, the total amount of material ejected in the hot outflow decreases, but the specific energy of the wind remains fixed because the ratio $\dot{e}_{\rm hot}/\dot{m}_{\rm hot}$ is fixed everywhere within the injection volume.
In the models considered here, radiative cooling within the host galaxy will decrease the total amount of matter that participates in the outflow, decreasing the gas density, and thus increasing the critical mass loading rate needed for cooling on large scales. However, the magnitude of this effect depends on the energy and mas injection profile. For example, in CC85 and \cite{tenorio07} for star clusters, $\dot{e}_{\rm hot}$ and $\dot{m}_{\rm hot}$ are assumed to be constant throughout the volume, so that most of the matter and energy are injected very near $r\sim R$ ($\dot{M}_{\rm hot}\sim \dot{m}_{\rm hot} R^3$). Thus, $R_{\rm st}$ must be very near $R$ to significantly change the total mass loading of the outflow. Adopting a perhaps more realistic model for galaxy scales so that $\dot{m}_{\rm hot}\propto \dot{e}_{\rm hot} \propto r^{-2}$ ($\dot{M}_{\rm hot}(r)\propto r$; i.e., an isothermal sphere; see \citealt{zhang14}), most of the matter and energy are still injected on scales of order $\sim R$. For more realistic disk-like geometries, or for extended mass and energy injection without a well-defined ``edge" (e.g., an $R$ beyond which $\dot{e}_{\rm hot}=\dot{m}_{\rm hot}=0$), more work is required to understand both the sonic point and the mass loading.
Note that for the CC85 model and the cooling function adopted, the ratio $t_{\rm cool}/t_{\rm adv}(r<R)$ is {\it largest} at $r\simeq R$. Thus our estimate of $\beta_{\rm crit}(r=R)$ in equation (\ref{betacrit}) gives the limit such that cooling is important throughout the profile, from $r=0$ outwards. This is then the limit that $R_{\rm st}\rightarrow R$ and we expect the model to break down completely because for $\beta>\beta_{\rm crit}$ the wind is radiative at all radii.
\subsection{Conduction}
\label{section:conduction}
As an aside, we note that conduction may strongly affect the temperature profiles of low-$\beta$ outflows. In the energy equation, the net sources and sinks include both radiative losses and conduction. As we have shown above, radiative losses are important for $\beta\gtrsim1$ (eqs.~\ref{betamin}, \ref{betacrit}). The net local energy deposition rate from conduction is given by
\begin{equation}
\dot{\varepsilon}_{\rm cond}=\nabla\cdot[\kappa(T)\nabla T],
\label{cond}
\end{equation}
where $\kappa(T)\simeq5\times10^{-7}T^{5/2}$\,ergs cm$^{-1}$ K$^{-1}$ is the Spitzer conduction coefficient.
Setting aside the important uncertainty concerning the strength and orientation of the magnetic field and assuming the temperature profile follows $T\propto r^{-4/3}$ as appropriate for the CC85 model on scales $r>R$, equation (\ref{cond}) can be written as $\dot{\varepsilon}_{\rm cond}=44\kappa(T)T/9r^2$. The conduction timescale ($nk_B T/\dot{\varepsilon}_{\rm cond}$) is then
\begin{equation}
t_{\rm cond}
\simeq1.3\times10^6\,{\rm yr}\,\,\,\frac{\dot{M}_{\star,10}}{\Omega_{4\pi}}\,\frac{\beta^4}{\alpha^3}
\left(\frac{r}{R}\right)^{10/3}.
\end{equation}
The strong $\beta$, $\alpha$, and $r$ dependences of this expression stem directly from the normalization of the temperature in an energy driven model ($T\propto \alpha/\beta$), the radial dependence of the adiabatic temperature gradient, and the strong scaling of the conductive flux with temperature ($\propto T^{7/2}$).
The conduction timescale $t_{\rm cond}$ has a much steeper radial dependence than the advection timescale $t_{\rm adv}=r/v$. We thus expect conduction to dominate at small radii. Setting $t_{\rm cond}=t_{\rm adv}$ (eq.~\ref{tadv}), we derive a critical radius beyond which the flow is adiabatic, and below which the flow is dominated by conduction:
\begin{equation}
r_{\rm cond}\simeq160\,{\rm pc}\,\,R_{0.3}^{10/7}\left(\frac{\Omega_{4\pi}}{\dot{M}_{\star,10}}\right)^{3/7}\frac{\alpha^{15/14}}{\beta^{3/2}}.
\label{rcond}
\end{equation}
The fact that $r_{\rm cond}\simeq R$ in this expression implies that $\beta\sim1$ is the critical value of the mass loading parameter below which conduction begins to dominate the temperature gradient at $R$. Setting $r_{\rm cond}=R$, we find that\footnote{This estimate ignores the strong temperature gradient in the CC85 model at $R$, which is artificially steep --- and would therefore further enhance the importance of conduction at $R$ --- because of the assumed sudden and artificial cutoff in the sources of energy and mass injection at $R$ in the CC85 model.}
\begin{equation}
\beta_{\rm crit,\,cond}\simeq0.6\,\alpha^{5/7}\left(\frac{\Omega_{4\pi}\,R_{0.3}}{\dot{M}_{\star,10}}\right)^{2/7}.
\label{betacond}
\end{equation}
That is, for $\beta<\beta_{\rm crit,\,cond}$, $t_{\rm cond}<t_{\rm adv}$ on scale $R$ and we expect conduction to dominate advection out to a scale given by $r_{\rm cond}$. For $\beta>\beta_{\rm crit,\,cond}$, we expect conduction to have a minor effect on the resulting solutions. Comparing equation (\ref{betacond}) with equation (\ref{betamin}) we see that over virtually any range of $\beta$ the standard CC85 model is invalid: for $\beta\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$>$}}}\beta_{\rm min}$ the flow is radiative, while for $\beta\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}}\beta_{\rm cond}$ it is highly conductive, and $\beta_{\rm min}\sim\beta_{\rm cond}$.
A full exploration of the importance of conduction for low-$\beta$ outflows is left for a future work, but here we note that a primary effect will be to flatten the temperature profile for $r\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}} r_{\rm cond}$. In this regime we can ignore advection and radiative cooling, and thus the energy equation is integrated trivially to yield the power-law relation \citep{parker64}
\begin{equation}
T(r)\propto r^{-2/7}\,\,\,\,\,(R\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}} r\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}} r_{\rm cond}),
\label{parker}
\end{equation}
which is significantly flatter than the $r^{-4/3}$ expected for an adiabatic flow. This may have significant consequences for observations attempting to diagnose the dynamical significance of the hot flow with X-rays. For example, \cite{strickland_m82} find a flat temperature gradient potentially consistent with equation (\ref{parker}) in the M82 superwind on kpc scales. This may be due to conduction given the low value of $\beta$ implied by the hard X-ray observations of \cite{strickland_heckman} (they find $\beta\sim0.2-0.5$).
A caveat to our discussion of conduction is that the form of equation (\ref{cond}) is only valid if (1) the collisional mean free path of the electrons is smaller than the characteristic scale of the temperature gradient $\lambda_e<dr/d\ln T$, (2) the electron-proton equilibration timescale is less than the advection time, and (3) the electron thermal speed ($v_e$) is larger than the flow speed $v$. Our estimates suggest that these are all satisfied for the $\beta\gtrsim1$ models we focus on in this work, but that (1) and (2) break down rapidly for $\beta\lesssim0.2-0.5$. Thus for the minimal mass loading expected from supernova driven winds composed only of supernova ejecta ($\beta\sim0.2$) we expect the basic hydrodynamic assumption of CC85 to be invalid and a complete revaluation of the physics is in order.
\begin{figure*}
\centerline{\includegraphics[width=7.3cm]{plot_tpp.pdf}\includegraphics[width=7.3cm]{plot_tvp.pdf}}
\centerline{\includegraphics[width=7.3cm]{plot_rhopp.pdf}\includegraphics[width=7.3cm]{plot_tnp.pdf}}
\centerline{\includegraphics[width=7.3cm]{plot_vpp.pdf}\includegraphics[width=7.3cm]{plot_vnp.pdf}}
\caption{{\it Left Column:} Temperature (top), density (middle), and velocity (bottom) as a function of radius for a models with ${\dot{M}_\star}=10$\,M$_\odot$ yr$^{-1}$, $R=0.3$\,kpc, and $\beta=0.2-1.6$ in steps of 0.2, calculated without gravity (red dotted) and with an assumed isothermal gravitational potential with $\sigma=200$\,km s$^{-1}$ (blue solid). Note the rapid decrease in temperature to less than $10^4$\,K in the high $\beta$ models at the cooling radius and the decrease in velocity on large scales in the extended gravitational potential. All models assume the PIE with a meta-galactic UV background and solar metallicity \citep{oppenheimer_schaye,wiersma}. {\it Right Column:} Temperature, velocity, and column density $N=nr$ (cm$^{-2}$) versus one another for the same models. The differential luminosity as a function of temperature for these models is shown in Figure \ref{figure:dl}.}
\label{figure:t}
\end{figure*}
\subsection{A More Complete Calculation}
\label{section:numerical}
To explore the dynamics of radiative cooling in high-$\beta$ hot winds more fully, we solve the spherical steady-state wind equations with an arbitrary cooling/heating function and a simplified equation of state. Combining the Euler equations for mass, momentum, and energy conservation,
we find that (e.g., \citealt{lamers_cassinelli}; see Appendix \ref{appendix:wind_equations})
\begin{equation}
\frac{dv}{dr}=\frac{v}{2r}\left(\frac{v_e^2-4c_s^2}{c_s^2-v^2}\right)+\frac{D}{C_V T}\left(\frac{\dot{q}}{c_s^2-v^2}\right)
\label{v}
\end{equation}
\begin{equation}
\frac{d\rho}{dr}=\frac{2\rho}{r}\left(\frac{v^2-v_e^2/4}{c_s^2-v^2}\right)-\frac{\rho}{v}\frac{D}{C_V T}\left(\frac{\dot{q}}{c_s^2-v^2}\right)
\label{rho}
\end{equation}
\begin{equation}
\frac{dT}{dr}=\frac{2}{r}\frac{D}{C_V}\left(\frac{v^2-v_e^2/4}{c_s^2-v^2}\right)+\frac{\dot{q}}{C_V v}\left(\frac{c_T^2-v^2}{c_s^2-v^2}\right)
\label{t}
\end{equation}
where $D=(T/\rho)\left. \p P/\p T\right|_\rho=k_B T/m_p$, $C_V=\left.\p E/\p T\right|_\rho=(3/2)k_B/m_p$ is the specific heat at constant volume, $c_s^2=\left. \p P/\p\rho\right|_s$ and $c_T^2=\left. \p P/\p\rho\right|_T$ are the adiabatic and isothermal sound speeds, respectively, $\dot{q}=\dot{\varepsilon}/\rho$ is the net heating rate per gram (see Fig.~\ref{figure:c}), and $v_e=2GM(r)/r$ is the ``local" escape velocity (see Appendix \ref{appendix:wind_equations}). For the purposes of an exploration of parameter space, we use the cooling/heating tables assuming phoionization equilibrium from \cite{oppenheimer_schaye} (see \citealt{wiersma}), which employ the \cite{haardt_madau} meta-galactic photoionizing background. In PIE, $\dot{q}$ is a function of density and temperature. An example showing the contributions to cooling from H+He and metals separately is shown in Figure \ref{figure:c} at a characteristic density. We neglect conductive energy transport (Section \ref{section:conduction}), which should only matter for the low-$\beta$ ostensibly adiabatic models that are not the focus of the current study. Lastly, note that the wind equations above assume that the composition of the matter does not change as a function of radius, that continuity is obeyed, and that all emission is optically-thin.
We focus on solutions that start adiabatic at $r=R$ with $t_{\rm cool}>t_{\rm adv}$. In this limit, the Mach number of the flow is ${\cal M}(r=R)=1$, in accord with CC85 and the analysis presented in \cite{wang95a}. We therefore solve the CC85 model for the flow at $r=R$ and then treat equations (\ref{v})-(\ref{t}) as an initial value problem and integrate to large radii using standard methods. For the purposes of illustrating rapid radiative cooling on large scales, we assume that the wind expands into vacuum. More discussion of a surrounding ambient medium is provided in Section \ref{section:discussion}.
Figure \ref{figure:t} shows profiles of temperature, density, and velocity for fiducial models with ${\dot{M}_\star}=10$\,M$_\odot$ yr$^{-1}$, and $R=300$\,pc, close to parameters representative of starbursts like M82 and NGC 253, or the massive star-forming clumps seen in high redshift rapidly star-forming galaxies. The red dotted lines show the solution without a gravitational potential ($v_e(r)=0$) and the blue solid lines show the solution assuming a singular isothermal sphere ($v_e(r)=2\sigma$) with velocity dispersion $\sigma=200$\,km s$^{-1}$. All models assume efficient thermalization $\alpha=1$, as is implied by observations of M82 \citep{strickland_heckman}, and then a range of $\beta$ from 0.2 to 1.6 in steps of 0.2. For large $\beta$ we see the rapid decline in temperature from $\sim10^7$\,K to less than $10^4$\,K on scales of $\sim1-10$\,kpc. Here, photoionization heating balances cooling, as in the example cooling function shown in Figure \ref{figure:c}. The material then continues to flow outward at fairly high velocity. The right panels show that the rapidly cooling solutions have high column densities and a broad range of velocities, potentially explaining the broad emission line features seen in some systems (e.g., \citealt{soto_fastlines,martin_lya}; Section \ref{section:discussion}).
Note that the ``adiabatic" models shown with $\beta\lesssim0.6$ are meant for comparison with the more strongly radiative solutions. In fact, we expect these low-$\beta$ solutions to be strongly affected by conduction (eq.~\ref{betacond}) out to a radius given by equation (\ref{rcond}) and will thus have much flatter inner temperature profiles (eq.~\ref{parker}). For the $\beta=0.2$ model, $r_{\rm cond}\simeq1.8$\,kpc. A full solution with conduction and radiative cooling requires a complete revision of the CC85 model and is left for a future work.
Returning to the importance of radiative cooling, we see from the lower left panel of Figure \ref{figure:t} that the high-$\beta$ models in an extended gravitational potential slow down significantly on large scales \citep{wang95a}. Formally, the $\beta=1-1.6$ models spontaneously develop a sonic point at $\sim80$, 30, 10, and 7\,kpc. This behavior follows from equation (\ref{v}) in the limit of strong cooling. Once the temperature drops to $\sim10^4$\,K, the evolution becomes quasi-isothermal and we can approximate $\dot{q}\simeq0$ at radii $r> r_{\rm cool}$. Taking $c_s\ll\sigma$ and $c_s\ll v$ equation (\ref{v}) shows that $dv/dr<0$ and the gas decelerates because of gravity. The critical temperature at which the flow starts to decelerate is just the virial temperature $T\simeq3\times10^6\,\,{\rm K}\,\,\sigma_{200}^2$. Integrating from the cooling radius outwards, we derive the critical radius where the flow decelerates significantly after cooling:
\begin{equation}
r_{\rm slow}\simeq r_{\rm cool}\exp\left[\frac{3.7}{\sigma_{200}^2}\frac{\alpha}{\beta}\right],
\label{rsonic}
\end{equation}
where $r_{\rm cool}$ is given by equation (\ref{rcool}), and where we have used the velocity at the cooling radius in order to obtain the numerator in the exponential. We see that small changes in $\beta$, $\alpha$, or $\sigma$ change $r_{\rm slow}$ dramatically. However, note that the detailed behavior of the gas during deceleration is sensitive to our assumption of a purely isothermal dark matter halo out to large scales. In numerical experiments we find that if we put in a more realistic NFW-like dark matter profile, the sonic point moves out to very large radius, or does not occur within $\sim300$\,kpc for the parameters in Figure \ref{figure:t}. Increasing $\beta$ slightly, however, can then cause strong deceleration to reappear. As is evident from equation (\ref{rsonic}) all changes in the parameters of the wind and host galaxy will strongly affect this evolution.
The models shown in Figure \ref{figure:t} assume Solar metallicity $Z=Z_\odot$ and PIE in a \cite{haardt_madau} metagalactic background at redshift $z=0$. We experimented with changes to both. Lowering or raising the metallicity increases or decreases, respectively, the critical value of the minimum mass-loading required for strong cooling as one would expect ($\beta_{\rm min}$; eq.~\ref{betamin}). For example, for $Z=0.2\,Z_\odot$ models with $\beta=1$ and 2 are qualitatively similar to $Z=Z_\odot$ models with $\beta\simeq0.6$ and 1.4, respectively. Note that the shape of the cooling function allows for rapid radiative cooling even in primordial gas (H$+$He in Fig.~\ref{figure:c}): a model with $Z=0$ and $\beta=2$ resembles the model with $\beta=1$ shown in Figure \ref{figure:t}. Conversely, a model with $Z=4Z_\odot$ and $\beta=1$ has a cooling radius at the same location as the $\beta=1.4$ lines in Figure \ref{figure:t}. Changes to the redshift for the PIE calculation have essentially no effect on the location of the cooling radius or the critical $\beta$ for cooling, but they do change the post-cooling temperature in accordance with what one would expect for a harder metagalactic background.
Over much of parameter space the cooling radius occurs at a sufficiently high density and temperature that employing a standard CIE cooling function instead of PIE does not change the cooling radius significantly. However, the post-cooling evolution is completely different. In CIE, the gas expands adiabatically after cooling with $T\propto r^{-4/3}$, which differs qualitatively from the approximately isothermal evolution on large scales for the models shown in Figure \ref{figure:t}.
Finally, note that the flow time to $100$\,kpc is $\gtrsim100$\,Myr and we expect the assumption of steady-state conditions to be violated on this timescale, at least in starburst systems. We further expect the wind to expand into a local circumgalactic medium, sweeping up and shocking matter in a galaxy-scale wind-blown bubble (see Section \ref{section:halo}).
\subsection{Thermal \& Convective Instability}
\label{section:ti}
\cite{bs89} (hereafter BS89) studied the case of thermal instability (TI) in a dynamically evolving background, which is the case applicable to TI in galactic winds. The key radial range where TI is important is where the cooling time is less than the flow time. Otherwise there is insufficient time for the instability to grow significantly. Under these conditions, both convection and isobaric TI can in principle grow. The amplification of density fluctuations by convection is given by a factor \citep{bs89}
\begin{equation}
A_{\rm conv} \sim \exp\left[\int |N| dt\right] = \exp\left[\int |N| \, t_{\rm cool} \frac{dt}{t_{\rm cool}}\right]
\label{ampconv}
\end{equation}
where
\begin{equation}
N^2(r) =\frac{2}{5}\frac{m_p}{k_B}\left(-\frac{1}{\rho}\frac{dP}{dr}\right)\frac{ds}{dr},
\end{equation}
$N(r)$ is the Brunt-V\"ais\"ala frequency, which characterizes convective instabilities, $s$ is the entropy per unit mass, and the integral over time in equation (\ref{ampconv}) is co-moving with the flow so that $dt = dr/v$.\footnote{Anisotropic thermal conduction along magnetic fields lines modifies the convective instability criterion in dilute plasma such as the galactic winds of interest here (Quataert 2008). This does not change our conclusion that convection is relatively unimportant in galactic winds because the timescale argument applies even to conduction-mediated convection.} Evaluating this over the radial range where initially hot galactic winds cool appreciably, we find that $\int |N| dt$ is typically $\ll 1$. Thus there is insufficient time for convection to grow appreciably. This is fundamentally because the Brunt-V\"ais\"ala frequency introduced by radiative cooling in a wind is given roughly by $N t_{\rm cool} \sim \mathcal{M}^{-1}$ where $\mathcal{M}$ is the Mach number of the flow and the flow only spends a time $dt \sim t_{\rm cool}$ at radii where cooling is significant. Note that at large radii once the flow comes into PIE and $T \sim$ constant, $ds/dr > 0$ and the flow is convectively stable.
If the cooling rate is assumed to scale as $\propto n^2 \Lambda(T)$, the amplification of density fluctuations by isobaric thermal instability is given by a factor \citep{bs89}
\begin{equation}
A_{\rm TI} \sim \exp\left[\frac{3}{5} \int \left(2 - \frac{\partial \ln \Lambda}{\partial \ln T}\right) \frac{dt}{t_{\rm cool}}\right].
\label{ampTI}
\end{equation}
Evaluating this for the cases shown in Figure \ref{figure:t}, we find that $A_{\rm TI} \sim 30-300$ (for $\beta=1.0-1.6$, respectively), with all of the contribution due to the small radial range where the flow cools abruptly from $T\sim 10^{6}$ K to $\sim 10^3-10^4$ K (see upper left panel of Fig.~\ref{figure:t}). Physically, as the flow cools from high to low temperature, each logarithmic interval in temperature (which corresponds to $dt \sim t_{\rm cool}$) contributes an order unity contribution to the integral in equation (\ref{ampTI}), so that the net amplification is set by the few decades in temperature decrease during the rapid cooling of the wind.
The fact that the amplification of density fluctuations by TI is limited to a factor of $\sim 100$ implies that for the medium to develop a multi-phase structure there must be initial fluctuations of finite amplitude present in the flow. Given the high initial sound speed in the thermally-driven wind, one might expect any such fluctuations to have been erased by the time rapid cooling sets in. However, the hot outflow quickly becomes supersonic and thus expands outwards more quickly than it can establish pressure equilibrium. In addition, the mixing of cool gas into the hot flow will likely imprint large inhomogeneities on the flow, which will seed both instabilities and rapid cooling (see Section \ref{section:rebirth}). An analogous situation can be seen in Fig A3 of \cite{sharma} where an inhomogeneous cooling inflow (rather than a wind) with $t_{\rm cool} \mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}} t_{\rm adv}$ and initial density fluctuations generates significant multiphase structure.
Another requirement for equation (\ref{ampTI}) to apply is that the fluctuations must be isobaric, rather than isochoric. This implies that there must be initial density fluctuations over a spatial scale
\begin{equation}
L\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}} \frac{H_{\rm cool}}{\mathcal{M}}
\label{size}
\end{equation}
where $H_{\rm cool}$ is the width of the radial region where cooling is rapid (the temperature scale height) and $\mathcal{M}$ here is for the hot wind prior to the onset of very rapid cooling. Assuming this is satisfied, TI can in principle produce final densities of
\begin{equation}
n_{f} \sim 100 \, n_{\rm cool} \, \left(\frac{10^4 \, {\rm K}}{T_f}\right) \left(\frac{T_{\rm cool}}{10^6 \, {\rm K}}\right)
\label{nti}
\end{equation}
where $n_{\rm cool}$ and $T_{\rm cool}$ are the density and temperature of the hot flow at the cooling radius $r_{\rm cool}$ (see eq.~\ref{ncool}). However, because all of the gas ultimately cools to PIE in our fiducial models at $\sim 10^3-10^4$\,K there is nothing to maintain pressure equilibrium for gas having $n_f \gg n_{\rm cool}$. Gas at different density will have slightly different temperatures in PIE, which will allow some density fluctuations even if pressure equilibrium is satisfied. More importantly, some of the cold denser gas produced by TI may not have time to expand significantly on the flow time. One way to see this is to note that if clouds occupy a fraction of $10^{-2}$ of the area of the sphere but have densities of $\sim 100 n_{\rm cool}$, they will carry the same mass flux as the original hot wind. With that filling fraction and a post cooling temperature of $10^3-10^4$\,K the gas will not have time to expand to fill the sphere on the flow time. The proper physical picture may thus be one in which the cool clouds produced by TI in galactic winds are initially over-pressurized relative to their surroundings and then expand out at $\sim10$s of km/s, eventually re-establishing pressure balance. Alternatively, magnetic and/or cosmic-ray pressure may be important in confining the cool gas.
\section{Discussion}
\label{section:discussion}
Two interconnected puzzles persist in the physics of cool gas in galactic winds and halos that may be solved by rapidly cooling, thermally-unstable hot winds of the type presented in Figure \ref{figure:t} and Section \ref{section:winds}. The first is the presence of fast, but cool, outflows observed in emission and absorption in winds from rapidly star-forming galaxies across the universe. How is the molecular, neutral atomic, and ionized gas accelerated to the hundreds and even several thousands of km/s in some systems without being shocked, shredded, and incorporated into the hot flow? The second is the pervasive warm neutral atomic and ionized gas seen in the halos of galaxies at low and high redshift \citep{steidel_cgm,werk}. What is the origin of this gas, and how does it persist on large scales without rapidly accreting?
\subsection{Cool Gas in Winds}
\label{section:coolwind}
The prevailing picture of the cool gas in winds is that it is entrained from the host ISM in the hot wind, and then accelerated from the galaxy by ram pressure (e.g., \citealt{veilleux_review}). In accord with this view, the soft extraplanar X-ray emission from wind galaxies is often interpreted as shocked interfaces between the cool clouds and the surrounding hot medium (e.g., \citealt{strickland_ngc253a,strickland_ngc253b}). However, it is unclear how the cool gas clouds are accelerated to the velocities and large scales seen without being shredded by hydrodynamical instabilities and incorporated into the hot flow \citep{klein,cooper09,marcolini,mccourt,scannapieco,schneider}.
Figure \ref{figure:t} suggests an alternative interpretation of both the soft X-ray emission and the cool gas seen in galactic winds. In this picture, a hot flow with $\beta\gtrsim1$ expands from the energy injection region and cools radiatively on large scales \citep{wang95a}. We propose that the high velocity cool clouds seen in absorption line tracers like NaD, Mg, and other optical and UV resonance lines are precipitated directly from the hot gas in a rapidly cooling flow. Furthermore, the observed extraplanar soft X-ray emission may also originate from cooling, and may not necessarily be dominated by interfaces and shocks as in the standard picture.
Viewed ``down the barrel" towards the central star-forming host, one would expect to see fast, predominantly blue-shifted absorption lines of partially ionized gas and soft X-ray emission. Emission lines are also possible (see eq.~\ref{emcool}). Indeed, recently \cite{martin_lya} suggested that the very fast Ly$\alpha$ emission seen from local ULIRGs could originate from direct radiative cooling of the hot gas in a CC85-like outflow \citep{soto_martin_diagnostic,soto_fastlines}. We find a critical surface density of star formation in the ULIRG range --- $\dot{\Sigma}_\star\simeq500$\,M$_\odot$ yr$^{-1}$ kpc$^{-2}$ --- for rapid cooling on a scale of a few hundred parsecs, given high thermalization efficiency (eq.~\ref{sds_R}). A particularly interesting prediction of the model is a direct connection between the velocity of the material and its mass loading. Higher $\beta$ implies more rapid cooling closer to the source, but also smaller velocity since $v\propto\,(\alpha/\beta)^{1/2}$ in the CC85 model (see eqs.~\ref{vmax} and \ref{vmin}). For ULIRGs like those discussed in \cite{soto_fastlines} and \cite{martin_lya}, these scalings imply rapid cooling and bright line emission at $\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$>$}}} 10^3$\,km/s, as is observed. Although speculative, we note that for ULIRG-like parameters and with $\beta\rightarrow\beta_{\rm crit}$ (eq.~\ref{betacrit}), $n_{\rm cool}$ and $N_{\rm cool}$ (eqs.~\ref{ncool} and \ref{Ncool}) reach $\sim10^{2.5}$\,cm$^{-3}$ and $10^{22.5}$\,cm$^{-2}$, respectively, implying that after thermal instability the material may self-shield and rapidly form molecules ($n_f\sim10^2n_{\rm cool}\sim10^{4.5}$\,cm$^{-3}$; eq.~\ref{nti}), depending on its dust content. The possibility that molecules may form directly in a rapidly cooling hot flow should be assessed by following grain sputtering, grain-gas cooling/heating, and line cooling below $10^4$\,K. Much less speculative is the fast atomic line emission and absorption with maximum velocity given by equation (\ref{vmax}) that should be seen from rapidly star-forming systems if the emission measures and column densities are high enough, respectively. This picture of fast cool outflows may provide an explanation for the high velocities measured for objects like those from \cite{tremonti,diamond,sell,geach} and should be tested with absorption line observations from a large variety of systems \citep{ham}.
An observational prediction of the radiative wind picture advocated here is that the observed X-rays arise from recombination, with a well-defined temperature progression from hot to cool, possibly with non-equilibrium ionization physics, and then a spatially extended region of photo-ionized gas together with the product of the non-linear evolution of the thermal instability in a supersonic background (Section \ref{section:ti}). Such a picture might help explain the strong spatial correspondence between the H$\alpha$ and soft X-ray emission in galactic winds \citep{strickland_xa, strickland_xb}. We highlight the fact that for a medium with constant mass flux cooling from high to low temperature, there is a particularly simple relationship between the total energy radiated per logarithmic interval in temperature and the mass flux, given by
\begin{equation}
\frac{dL}{d\ln T}\simeq\frac{3}{2}\dot{M}_{\rm hot}\frac{k_B T}{m_p}.
\label{lum}
\end{equation}
This expression follows from dropping the first term on the right hand side of equation (\ref{t}) and taking the limit that the medium is highly supersonic ($v^2\gg c_s^2$, $c_T^2$).\footnote{Note that in the subsonic limit $c_T^2/c_s^2=1/\gamma=3/5$ and the numerical factor on the right hand side of equation (\ref{lum}) changes from $3/2\rightarrow5/2$ (e.g., \citealt{fabian94}).} Figure \ref{figure:dl} shows the differential luminosity $dL/d\ln T$ along the temperature profile for the models in Figure \ref{figure:t}. The expected power-law $dL/d\ln T\propto T$ in equation (\ref{lum}) is recovered for the high-$\beta$ models. We have numerically verified that equation (\ref{lum}) is precisely satisfied in the cooling region during the precipitous drop in $T$. Thus, if the cooling region is identified in observations, equation (\ref{lum}) provides a convenient expression for estimating the mass loss rate directly from X-ray observations.
As a nearby observational example of a system that might showcase the onset of strong radiative cooling in a galactic wind, we consider the $\sim0.3-0.6$\,kpc long conical limb-brightened frustum defined in \cite{strickland_ngc253a} along the minor axis to the south of NGC 253. Using the count rate in the energy range $0.3-1$\,keV (0.06 counts/s) or $0.3-2$\,keV (0.07 counts/s), temperature (0.5\,keV), and volume reported by \cite{strickland_ngc253a} for this region (assuming a distance of $D=2.6$\,Mpc), one derives a total radiated flux of $\sim6-10\times10^{-13}$\,ergs cm$^{-2}$ s$^{-1}$ and a hot gas density of order $\sim0.05-0.1$\,cm$^{-3}$ for gas with unity filling factor in CIE with Solar abundances. Comparing to the gas pressure and density derived from H$\alpha$ observations by \cite{mccarthy} in the same region, \cite{strickland_ngc253a} argue that the true hot gas density is $\sim0.1-0.3$\,cm$^{-3}$. These densities imply a cooling timescale of $\sim3-9$\,Myr, whereas the advection time is of order Myr for gas at 500\,km/s. In general, we find that our radiative wind solutions have $t_{\rm cool}/t_{\rm adv}\sim3-10$ at $r\sim 1-2R$ at temperatures of $k_{\rm B}T\simeq0.5$\,keV and that the ratio decreases rapidly at larger radius because $\Lambda\propto T^{-0.7}$, as described in Section \ref{section:analytic}. The NGC 253 observations therefore appear consistent with our models in which radiative cooling becomes important at larger scales. We thus propose that the H$\alpha$ and X-ray emission from this region and beyond may be the result of the non-linear development of the thermal instability described in Section \ref{section:ti}. The fact that the H$\alpha$-emitting clouds and the hot gas are in pressure equilibrium, and that the H$\alpha$ flux increases beyond this region (see Fig.~5 of \citealt{strickland_ngc253a}, right panel) supports the notion that $r_{\rm cool}\sim0.5-1$\,kpc. The extended multi-kpc soft X-ray and line-emitting halo would then be interpreted as the large-scale aftermath of strong radiative cooling and thermal instability on sub-kpc scales.
\begin{figure}
\centerline{\includegraphics[width=8.5cm]{plot_tdlp.pdf}}
\caption{$dL/d\ln T$ as a function of temperature for the wind models shown in Figure \ref{figure:t}. The power-law form $dL/d\ln T\propto T$ given by equation (\ref{lum}) in the limit of strong cooling is evident for the high-$\beta$ solutions. Note that these models have $\dot{M}_\star=10$\,M$_\odot$ yr$^{-1}$, which implies a bolometric luminosity of $L_{\rm bol}\sim10^{44.5}$\,ergs s$^{-1}$ and a corresponding energy injection rate of $\dot{E}_{\rm hot}\sim10^{42.5}$\,ergs s$^{-1}$, both of which are substantially off the top of the plot. Most of the injected energy goes to adiabatic losses rather than radiative cooling.}
\label{figure:dl}
\end{figure}
Note that if this feature in NGC 253 were to represent a region of strong radiative cooling, the hot gas mass loss rate implied by equation (\ref{lum}) is quite small, of order $\sim0.01-0.03$\,M$_\odot$/yr, which would imply a small value for $\beta$ if $\dot{M}_\star\sim5$\,$M_\odot$/yr and a potentially very large wind velocity if $\alpha\sim1$. However, as estimated above, $t_{\rm cool}/t_{\rm adv}\sim3-10$ for an assumed velocity of 500\,km/s, implying that equation (\ref{lum}) does not apply. The estimated $\dot{M}_{\rm hot}$ should be larger by the ratio $t_{\rm cool}/t_{\rm adv}$, which would increase the implied $\beta$ to $\sim0.1$. The additional complications made clear by the beautiful X-ray image of the frustum from \cite{strickland_xb} is first the importance of the opening angle $\Omega$ --- a naive estimate indicates $\Omega\sim0.01-0.05$ --- and second the potential importance of non-spherical areal divergence as the flow expands on $\sim0.5-{\rm few}$\,kpc scales \citep{suchkov94,suchkov96}. The latter may be especially important as the hot wind expands laterally just above the disk and encounters the halo.
Although the example of NGC 253 is illustrative, the details of what is observed will also depend on the photoionizing spectrum seen by the gas, the mass-loading along any given sightline, and whether or not the flow maintains ionization equilibrium (e.g., \citealt{breitschwerdt94,breitschwerdt99}). Clearly much more work is required to fully assess strong radiative cooling as an explanation for the H$\alpha$ and soft X-ray halos of NGC 253, M82, and other local starbursts. In fact, \cite{strickland_ngc253b} argued against the possibility of radiative cooling for the origin of the cool gas in NGC 253, but did so using the cooling times derived on multi-kpc scales, where we would argue $r\gg r_{\rm cool}$. The story of M82 may be more complicated because the wind has likely overrun halo gas from the tidal interaction with M81, or because it simply does not fit with the model of a radiatively cooling wind discussed here. Work by \cite{hoopes_OVI_M82} indicates weak cooling on large scales. In addition, \cite{strickland_heckman} found that $\alpha\sim1$ and $\beta\sim0.2-0.5$ in M82 based on the diffuse hard X-ray emission observed in its core. Although this value of $\beta$ is nominally below our analytic requirement for strong radiative cooling on kpc scales for Solar metallicity gas, the wind may undergo distributed mass-loading and entrain material on the scale of $R$, as we discuss below in Section \ref{section:rebirth}. In this way, the small value of $\beta$ measured in the core may be compatible with strong radiative cooling (higher $\beta$) on somewhat larger scales. Much more work is required to assess the possibility of a radiative flow in the many systems for which a careful comparison is possible (e.g., VV 114, \citealt{grimes_vv114}; Mrk 231, \citealt{veilleux_mrk231,rupke_mrk231}; and many more; \citealt{ham}), but the example of NGC 253 implies that very high spatial resolution observations may be required to localize the cooling region in mass-loaded starburst winds with X-ray observations.
An additional zeroth-order observational constraint on radiative hot winds comes from the total diffuse X-ray emission observed from galaxies (as in \citealt{zhang14}). As discussed in Section \ref{section:analytic}, the rapidly cooling wind can radiate at most $\dot{E}_{\rm hot}$ power, which for steady-state star formation is $\sim1$\% of the bolometric power of star formation (eq.~\ref{etamax} and \ref{eta}), $L_\star=\epsilon\dot{M}_\star c^2$, where $\epsilon\simeq7\times10^{-4}$ is an IMF-dependent constant. In practice, the radiative efficiency is much lower than this 1\%. For example, even the $\beta=1.4$ model shown in Figure \ref{figure:t} radiates $\simeq0.05\dot{E}_{\rm hot}\simeq5\times10^{-4}\,L_\star$ during the cooling phase, whereas the $\beta=1$ model radiates $\sim5$ times less. Figure \ref{figure:dl} summarizes these results. It also shows that the $dL/d\ln T$ distribution is quite flat for low-$\beta$ models and that most of the energy is radiated between $10^{6}-10^{7.5}$\,K temperatures, where the flow is effectively adiabatic, even for high-$\beta$.
These numbers for the radiative power of cooling winds can then be compared with observations of the total (integrated) soft X-ray emission from rapidly star-forming galaxies. For example, \cite{mineo} report the 0.5-2\,keV X-ray luminosity of star-forming galaxies to be $L_X\simeq5\times10^{38}\dot{M}_\star$\,ergs s$^{-1}$, implying $L_X/L_{\star}\simeq5\times10^{-5}$. In an attempt to approximate the bolometric X-ray luminosities of their galaxies from 0.5-10\,keV, they find a number $\sim20$ times higher, $L_X/L_{\star}\sim10^{-3}$. These numbers imply that rapid wind cooling could contribute significantly to the total X-ray budget of rapidly star-forming galaxies in extreme cases, but even the most mass-loaded model in Figure \ref{figure:t} does not violate an upper bound from observations. For fixed mass loading, system radius, and thermalization efficiency, we note that the radiative efficiency scales nearly linearly with the star formation rate (eq.~\ref{eta}). A particularly interesting case for a more careful observational comparison is NGC 6240 where the diffuse hard X-ray emission is larger \citep{wang_6240}.
\subsection{Distributed Mass Loading and Cloud Rebirth}
\label{section:rebirth}
The fact that cool clouds are readily destroyed by KH instabilities motivates a scenario for distributed mass loading and cool cloud production in hot winds. In this picture, supernova ejecta first thermalize on small scales and with a small hot gas mass loading factor of $\beta\sim0.2$ (as appropriate for the mass loss from supernovae alone) producing conditions for an initially metal-rich wind ($Z\sim$ few times Solar for supernova ejecta, and $\alpha$-element enriched). This picture is similar to that discussed in \cite{suchkov96}. The initially low-$\beta$ outflow is very hot and fast since the velocity and temperature of the wind scale as $v\simeq2000(\alpha0.2/\beta)^{1/2}$\,km/s and $T\simeq9\times10^7(\alpha0.2/\beta)$\,K, respectively. This wind sweeps up, shocks, shreds, and incorporates cool gas on the scale of the host galaxy $R$, with four important consequences: (1) $\beta$ increases, (2) $T$ decreases, (3) $v$ decreases, (4) $Z$ decreases. Note that the first three of these effects {\it decrease} the cooling timescale relative to the advection timescale on larger scales, increasing the chances for strong radiative cooling as in Figure \ref{figure:t}. In particular, the increase in $\beta$ drives the cooling radius $r_{\rm cool}$ rapidly inwards to small radii (eq.~\ref{rcool}). In addition, the swept up clouds seed the flow with density perturbations, which may grow via the thermal instability at $r_{\rm cool}$ (eq.~\ref{nti}). In essence, the rapid destruction of cool clouds on small scales initiates their subsequent formation on larger scales, an almost literal transmigration: self-induced cool cloud rebirth after death in an earlier form.
This scenario for producing cool fast gas from hot winds is complicated by the very strong dependencies outlined in Section \ref{section:winds}, and by the multi-dimensional and time-dependent character of real galactic winds. Indeed, rather than thinking of a galaxy as having a single $\alpha$, $\beta$, and $\Omega$, we would instead argue for a picture in which winds with a wide range of $\beta$ and $\alpha$, subtending fractions of $4\pi$, emerge from a given galaxy. Along some sightlines the amount of gas swept up is small, $\beta$ stays close to $\simeq0.2$, $\alpha$ is high, and the very fast, hot, and adiabatic wind escapes to large scales. Along other sightlines the amount of material swept up is enough to quench the wind completely: the velocity decreases sufficiently and the flow becomes sufficiently radiative ($\beta \mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$>$}}}\beta_{\rm crit}$; eq.~\ref{betacrit}), that the outflow turns into a mass-loaded fountain on kpc scales. Thus, in looking at the upper left panel of Figure \ref{figure:t} one might imagine all $\beta$s emerging from a single system, and not just one; the observations will then be a convolution of these diverse outflows and their interactions.
This picture has interesting consequences. It suggests that the most rapidly cooling material with the strongest mass-loading, and hence the highest column densities and emission measures (eqs.~\ref{ncool}, \ref{emcool}), {\it that can in fact escape to large scales} will have velocities correlated with the local escape velocity. In this way, the process of cool gas incorporation may be self-limiting: if too much gas is swept up from the host, the wind never emerges. This process may be related to the observation by \cite{martin_2005} that the wind velocity and escape velocity are correlated. Also consistent with this picture, detailed absorption line profiles from wind galaxies in general imply a broad distribution of column densities and velocities, with the fastest material having the lowest column density and vice-versa (e.g., \citealt{rupke_2005a,rupke_2005b}; but there are notable exceptions, e.g., \citealt{diamond}). A prediction of this picture is that the limiting $\beta$ would be given in terms of observables by equation (\ref{betacrit}), with a weak dependence on host size and star formation rate at fixed $\alpha$.
\begin{figure}
\centerline{\includegraphics[width=8.5cm]{plot_psp.pdf}}
\caption{Cooling timescale for the wind material as a function of radius assuming that it goes through a strong shock, as might be appropriate for the reverse shock in a galactic wind-blown bubble propagating in the circumgalactic medium for the models presented in Figure \ref{figure:t}. The sharp decrease at the end of the high-$\beta$ profiles is from the shape of the profile just before the sonic point that develops in our solutions at large radii and reflects a breakdown of the time-steady assumption used throughout this work.}
\label{figure:ps}
\end{figure}
\subsection{The Fate of Cool Gas in Radiative Winds}
\label{section:halo}
Studies of quasar absorption lines in the outskirts of foreground galaxies imply that massive halos of cool gas are prevalent at both high ($z\sim1-2$) and low ($z\sim0.2$) redshift \citep{steidel_cgm,tumlinson,rudie,werk}. Modeling the absorbing medium as $T\simeq10^4$\,K gas in photoionization equilibrium with a background metagalactic radiation field implies a total mass in this cool component that is comparable to or larger than the stellar mass of the central galaxy (see \citealt{werk} for a detailed discussion). Covering fractions of low-ionization material are high out to $\sim100-200$\,kpc around both star-forming and passive galaxies, while higher ionization OVI absorption is found more frequently around star-forming galaxies \citep{tumlinson}. The observed linewidths are hundreds (not thousands) of km/s. The densities inferred from photionization modeling, $n\sim10^{-3}$\,cm$^{-3}$, are far too low for the gas to be in pressure equilibrium with an ambient medium at the halo virial temperature, which challenges an explanation based on thermal instability within a hydrostatic hot halo (\citealt{mo_miralda,maller_bullock}; see Fig.~12 of \citealt{werk}).
We suggest that the observed absorption may originate from rapidly cooling outflows of the type presented in this paper. The calculations in Figure \ref{figure:t} show that mass-loaded flows driven from rapidly star-forming galaxies can radiatively cool on $\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$>$}}}{\rm few}-20$\,kpc scale, that these flows can reach large distances (even assuming an isothermal gravitational potential well out to $200$\,kpc), and that they slow down from their initial $500-1000$\,km/s velocities as a result of the extended gravitational potential. On these scales we expect a multi-phase medium as a result of the non-linear development of the thermal instability at $r_{\rm cool}$ (Section \ref{section:ti}) and the interaction with intervening gas (Section \ref{section:rebirth}). It is notable that for $\beta=\dot{M}_{\rm hot}/\dot{M}_\star\gtrsim1$ we expect more material to be ejected than to be formed into stars, and that it is in precisely this regime that cooling should be strong (eq.~\ref{betamin}). For the examples shown in Figure \ref{figure:t}, the column density of photoionized $\sim10^4$\,K gas is $\sim10^{17}-10^{18}$\,cm$^{-2}$ on $\sim100$\,kpc scales for the no-gravity (red dotted) calculations with $\beta\gtrsim1$. For metallicities of $\sim0.1-1Z_\odot$, corresponding metal-line column densities would be in the observable regime. The blue solid lines evolve to high column density as they ``stop" at the sonic point, but here our time-steady calculations break down and more detailed time-dependent calculations are necessary to explore the dynamics.
The low velocities and relatively high columns on large scales appear consistent with the observations, but at the cost of assuming that all of the systems probed have recently had strong mass-loaded galactic winds. This is prima facie problematic for the passive galaxy halos in \cite{werk}. However, a basic piece of wind physics might help explain why even early-type galaxies have wind-mass-loaded halos: galactic outflows do not expand into vacuum, as assumed in the calculations of Figure \ref{figure:t}. In fact, we expect a newly born, hot, fast galactic wind to expand into a pressurized circumgalactic medium (CGM), to sweep up and shock this material, forming a wind-driven bubble with a characteristic reverse and forward shock configuration. A sketch is shown in Figure \ref{figure:sketch}. In the simplest case, a contact discontinuity separates the shocked wind material from the swept up and shocked CGM. Simple arguments show that a wind-driven bubble would reach pressure equilibrium with a {\it constant pressure} CGM on $\sim40-80$\,kpc scales, assuming a CGM density of $10^{-3}$\,cm$^{-3}$ and a temperature equal to the virial temperature. In a hydrostatic gas halo with a falling CGM pressure profile, the bubble may well run away to large scales, depending on the CGM structure, the wind lifetime, and its energy and ram pressure. Importantly, wind-driven bubbles expand slowly compared to the free wind velocities that drive them. There is thus a chance that the remnant bubble formed by a cooling galactic wind would persist on multi-hundred kpc scales for Gyr timescales, thus allowing cool gas absorption around galaxies that are passive today.
Such an explanation for cool halos at low-$z$ in the sample of \cite{werk} should be fully developed. An immediate objection to such a model would be that as the cooled wind encounters the reverse shock in the wind-driven bubble it will be shocked to high temperature, potentially shutting off cooling, and thus no low-temperature photoionized gas would be observed outside of the cooling wind region. Although our calculations do not address the issue of a wind-driven bubble directly, we can make a preliminary estimate of relevance to the problem by asking what the temperature and density in the reverse shock region {\it would be} if the wind profiles from Figure \ref{figure:t} shocked at any given radius. If we assume a strong shockwave at any position, we can then compute the post-shock cooling timescale. Figure \ref{figure:ps} shows the result. We take the post-shock density and temperature to be $n_{\rm RS}(r)=4n_{\rm wind}(r)$ and $T_{\rm RS}(r)=(3/16)v_{\rm wind}^2(r)m_p/k_B$ and we then calculate the cooling time
\begin{equation}
t_{\rm cool,\,shock}=k_B T_{\rm RS}/n_{\rm RS}\Lambda(T_{\rm RS}).
\label{tshock}
\end{equation}
The models with and without gravity differ significantly. The high-$\beta$ models without gravity (red dotted) have cooling times of order Gyr only on scales below $\sim10$\,kpc. As the material goes through the reverse shock, it is heated to high temperature where the cooling time is long. The high post-shock temperature follows directly from the high wind velocities that are maintained even after cooling in the no-gravity models. In addition, at large radius these models shock at very low wind density (see left middle panel Fig.~\ref{figure:t}), which further prevents cooling.
In contrast, the models with gravity (blue solid) slow down significantly, increasing the density relative to the models with no gravity, and have lower post-shock temperatures because of the lower wind velocity. Both effects act to decrease the cooling time. We see that for $\beta > 0.8$, all of the models have post-shock cooling timescales less than the Hubble time out to scales of $\sim100$\,kpc. These conclusions would likely be strengthened by calculating the bubble evolution together with the wind evolution since the relative velocity of the reverse shock and the wind fluid will lead to lower post-shock temperatures and shorter cooling times.
Hot galactic winds with $\beta\gtrsim1$ can thus populate the halos of galaxies with cool gas in two ways. First, the freely expanding wind cools on small scales, undergoes thermal instability, and then evolves on $100$\,Myr timescales to very large scales, as in Figure \ref{figure:t}. Second, these cool decelerating winds will shock on the ambient CGM, driving a wind-driven bubble, and although the post-shock cooling times can be long, they are generically less than the Hubble time. We thus expect the post-shock region to cool radiatively. See Figure \ref{figure:sketch}.
These pictures for the origin of cool gas in the halos of galaxies do away with the notion of pressure equilibrium with a virialized hot halo. They circumvent the problem of maintaining a large mass of cool material well below the virial temperature in the CGM for many halo dynamical times.
\section{Conclusions}
\label{section:conclusions}
Following \cite{wang95a,wang95b} and a number of other works including \citealt{efstathiou2000,silich2003,silich2004,tenorio,tenorio05,tenorio07,wunsch}, \citealt{breitschwerdt94,breitschwerdt99}, we investigate how initially hot adiabatic galactic winds cool radiatively on large scales. We present approximate scaling relations for the cooling radius, density, column density, emission measure, and radiative efficiency (Section \ref{section:analytic}). Each of these quantities depends strongly on the hot wind mass loading parameter $\beta = \dot{M}_{\rm hot}/\dot{M}_\star$. However, hot winds generically undergo radiative instability --- in the sense that $t_{\rm cool}<t_{\rm adv}$ --- for $\beta\gtrsim0.5$ (eqs.~\ref{betamin} and \ref{betacrit}) with interesting and testable dependencies on the thermalization efficiency $\alpha$, the scale of the star forming region $R$, and the star formation rate $\dot{M}_\star$. In particular, as $R/\dot{M}_\star$ decreases, the $\beta$ threshold for the cooling instability decreases. Among other implications, the relatively small range of $\beta$ for strong radiative cooling implies a well defined maximum velocity given by equation (\ref{vmax}) and a minimum column density given by equation (\ref{ncoolmin}), with weak dependencies on the star formation rate and host galaxy radius.
Using the time-steady wind equations for an arbitrary heating and cooling function, we present numerical solutions showing the temperature, density, velocity, and column density as a function of radius in Figure \ref{figure:t}, both with and without an extended gravitational potential, and assuming PIE with a metagalactic UV background. Models with $\beta\gtrsim1$ (eq.~\ref{betamin}) undergo strong cooling on small scales with the temperature dropping by two orders of magnitude or more over a small fractional radial scale. The qualitative behavior of the solutions is largely independent of the metallicity of the gas and the character of the UV background.
The cooled gas forms an extended photoionized region that can reach large scales, depending on the gravitational potential. Even in the absence of a surrounding medium, the radiatively cooled wind decelerates significantly, and in our models spontaneously develops a sonic point, which signals the breakdown of our time-steady assumption. Equation (\ref{rsonic}) shows that the deceleration profile is exponentially sensitive to the parameters of the wind and the velocity dispersion of the host galaxy. For the same values of $\beta$ and the thermalization efficiency $\alpha$, this strong dependence on $\sigma$ means that cooled winds will be completely unbound from dwarf galaxies, but strongly bound to massive cluster galaxies. This physics may connect directly to the efficiency with which baryons are retained in halos of a given velocity dispersion, and thus inform discussions of the $z=0$ stellar mass function and halo mass function.
\begin{figure}
\centerline{\includegraphics[width=8.5cm]{test_wind1.pdf}}
\caption{Sketch showing the basic evolutionary stages discussed and the three distinct cool gas components. The hot fast wind emerges from the host galaxy, accelerating cool clouds to small velocities before they are incorporated into the flow. The hot wind with high mass-loading factor cools radiatively at the cooling radius and undergoes thermal instability (Fig.~\ref{figure:t}). The cooled wind decelerates on large scales, forming an extended photoionized region. The wind then shocks on the inner edge of the wind-blown bubble driven into the surrounding CGM. The cooling timescale for the shocked wind can be significantly shorter than the Hubble time (Fig.~\ref{figure:ps}). The robust existence of multiple regions of cool gas in galactic winds and halos may explain its prevalence in observations of star-forming and passive galaxies. The sketch here resembles recent results from high-resolution galaxy formation studies (e.g., \citealt{onorbe}).}
\label{figure:sketch}
\end{figure}
The primary predictions of this model for cool gas are the scaling relations derived in Section \ref{section:winds}, combined with the basic picture that the hot gas cools via recombination, with a well-defined temperature progression and differential luminosity given by equation (\ref{lum}) and shown in Figure \ref{figure:dl}. This picture of strong radiative cooling may be a natural explanation for the high velocity line emission seen in local ULIRGs \citep{soto_fastlines}, as also emphasized by \cite{martin_lya}. Equations (\ref{sdscrit}) and (\ref{sds_R}) show that the critical star formation surface density required for the cooling radius to collapse to the scale of the star forming region is in the ULIRG range --- $\dot{\Sigma}_\star\simeq10^3$\,M$_\odot$ yr$^{-1}$ --- but that strong spatially extended radiative cooling is also expected from LBG-like galaxies with $\dot{\Sigma}_\star\gtrsim10$\,M$_\odot$ yr$^{-1}$. Strong radiative cooling may also provide an explanation for the exceptionally high velocity outflows probed by Mg II absorption in blue post-starburst galaxies \citep{tremonti,diamond,sell}. For example, the system J0905$+$5759 reaches a velocity of $\simeq2500$\,km/s (Fig.~3 of \citealt{diamond}), close to $v_{\rm max}$ in equation (\ref{vmax}). Further tests of the model with ``down the barrel" absorption line studies should be undertaken. The application of the model to local starburst galaxies like NGC 253 and M82 is tentative and should be more fully explored with detailed models compared to high-resolution X-ray observations on small scales, as we highlight in Section \ref{section:coolwind}.
The rapid radiative cooling of nominally hot galactic winds may help explain observations at low and high redshift that show a significant amount of cool photoionized gas in the halos of galaxies (Section \ref{section:halo}). A particularly appealing aspect of this picture is the mass budget itself. The total mass in the cool gas halos is inferred to be comparable to or larger than the stellar mass of the galactic host \citep{werk}, which is in line with our prediction that $\beta=\dot{M}_{\rm hot}/\dot{M}_\star\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$>$}}} 0.5-1$ (eq.~\ref{betamin}) is required for strong radiative cooling.
Importantly, there are two sources of cool gas in galaxy halos. The first is the cooled wind itself, the extended photoionized regions shown in Figure \ref{figure:t}. The second is in the reverse shock that must inevitably form as the wind sweeps up the surrounding circumgalactic medium. Figure \ref{figure:ps} shows that the post-shock cooling timescales can be less than the Hubble time on 100\,kpc scales if the wind decelerates. In this way the cool gas is reborn on larger scales and at later times as the bubble moves slowly outwards.
Combining this picture of the halo with our discussion of distributed mass-loading and ram pressure acceleration of cool clouds from Section \ref{section:rebirth}, we arrive at a story of two cool cloud transmigrations. Cool clouds within the host galaxy are first swept up by the ram pressure of the hot flow. These clouds are rapidly destroyed on small scales and at low velocities $\lesssim10^2$\,km/s by hydrodynamical instabilities, as is well-documented in both idealized high-resolution numerical simulations, and in parameterized models that take into account both the gravitational potential and the radial structure of the hot flow \citep{scannapieco,zhang15}. The hot flow, now with enhanced mass loading and density perturbations, cools radiatively on larger scales, forming an extended region of atomic and ionized gas moving at $\sim10^3$\,km/s (the first transmigration) and then decelerating in the host's extended gravitational potential. The high velocity cool wind then shocks on the surrounding circumgalactic medium, destroying the cool gas again, but because of the relatively short cooling times (Figure \ref{figure:ps}) the cool gas is again reborn on Gyr timescales (the second transmigration). In this way we generically expect three spatially-separated components of cool gas in rapidly star-forming galaxies, as illustrated in Figure \ref{figure:sketch}.
Many additional issues are left to be investigated. A pressing theoretical issue with direct observational implications is the character of the non-linear development of the thermal instability in a supersonic background. The medium is thermally unstable (Section \ref{section:ti}) and the initial density perturbations may grow by a factor of more than 100 at the cooling radius (eq.~\ref{nti}), but high resolution multi-dimensional simulations are needed in order to resolve the question of whether or not the entire medium cools monolithically or whether something akin to discrete cool clouds of a characteristic size precipitate out of the surrounding hot background. Our speculations on the fate of a cool wind-driven circumgalactic bubble also need to be confronted with both parameterized semi-analytic calculations and multi-dimensional numerical simulations with radiative cooling (e.g., \citealt{onorbe,sarkar}).
A number of microphysical issues also need to be investigated. We have shown that conduction (Section \ref{section:conduction}) can dominate the energy transport in low-$\beta$ outflows (eq.~\ref{betacond}), and our estimates suggest that the assumption of collisionality and thermal equilibrium for the ionized plasma break down. A consequence of this fact is that over essentially any range of $\beta$ the standard CC85 model is invalid. Comparing equation (\ref{betacond}) with equation (\ref{betamin}) we see that for $\beta\gtrsim0.5$ the breakdown of CC85 is a result of radiative cooling, while for $\beta\lesssim0.5$ the breakdown is a result of conduction. Analytic and numerical investigations are needed to understand the temperature profile in the low-$\beta$ regime (eq.~\ref{parker}). These issues may well directly impact the interpretation of high-resolution X-ray observations of nearby starbursts like M82 \citep{strickland_m82}.
\section*{Acknowledgments}
TAT thanks the Kavli Institute for Theoretical Physics for support while preparing this work, and Ondrej Pejcha, Crystal Martin, and Tim Heckman for discussions. TAT also thanks Smita Mathur and Laura Lopez for help in interpreting X-ray observations of local starburst galaxies. We thank the Simons Foundation and the organizers of the workshop {\it Galactic Winds: Beyond Phenomenology} (J.~Kollmeier and A.~Benson) where this work germinated. We thank B.~Oppenheimer for providing the cooling/heating tables used in this work. We thank the anonymous referee for providing a timely, thorough, and helpful report. EQ was supported in part by NASA ATP grant 12-APT12-0183 and a Simons Investigator award from the Simons Foundation. TAT and DZ were supported in part by NASA Grant NNX10AD01G. TAT was also supported by NSF Grant 1516967.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,075
|
---
external help file: Microsoft.Azure.Commands.RecoveryServices.SiteRecovery.dll-Help.xml
Module Name: AzureRM.RecoveryServices.SiteRecovery
online version: https://docs.microsoft.com/en-us/powershell/module/azurerm.recoveryservices.siterecovery/get-azurermrecoveryservicesasrstorageclassification
schema: 2.0.0
---
# Get-AzureRmRecoveryServicesAsrStorageClassification
## SYNOPSIS
Gets the available(discovered) ASR storage classifications in the Recovery Services vault.
## SYNTAX
### ByFabricObject (Default)
```
Get-AzureRmRecoveryServicesAsrStorageClassification -Fabric <ASRFabric>
[-DefaultProfile <IAzureContextContainer>] [<CommonParameters>]
```
### ByObjectWithName
```
Get-AzureRmRecoveryServicesAsrStorageClassification -Name <String> -Fabric <ASRFabric>
[-DefaultProfile <IAzureContextContainer>] [<CommonParameters>]
```
### ByObjectWithFriendlyName
```
Get-AzureRmRecoveryServicesAsrStorageClassification -FriendlyName <String> -Fabric <ASRFabric>
[-DefaultProfile <IAzureContextContainer>] [<CommonParameters>]
```
## DESCRIPTION
The **Get-AzureRmRecoveryServicesAsrStorageClassification** cmdlet gets details of the discovered ASR storage classifications in the Recovery Services vault.
## EXAMPLES
### Example 1
```
PS C:\> $StorageClassifications = Get-AzureRmRecoveryServicesAsrStorageClassification -Fabric $Fabric
```
List the discovered storage classifications corresponding to the specified ASR fabric.
## PARAMETERS
### -Fabric
Specifies an ASR fabric object. The cmdlet gets the details of discovered storage classifications corresponding to the specified ASR fabric.
```yaml
Type: ASRFabric
Parameter Sets: (All)
Aliases:
Required: True
Position: Named
Default value: None
Accept pipeline input: True (ByValue)
Accept wildcard characters: False
```
### -FriendlyName
Specifies the friendly name of the storage classification object to get.
```yaml
Type: String
Parameter Sets: ByObjectWithFriendlyName
Aliases:
Required: True
Position: Named
Default value: None
Accept pipeline input: False
Accept wildcard characters: False
```
### -Name
Specifies the name of the storage classification object to get.
```yaml
Type: String
Parameter Sets: ByObjectWithName
Aliases:
Required: True
Position: Named
Default value: None
Accept pipeline input: False
Accept wildcard characters: False
```
### -DefaultProfile
The credentials, account, tenant, and subscription used for communication with azure.
```yaml
Type: IAzureContextContainer
Parameter Sets: (All)
Aliases: AzureRmContext, AzureCredential
Required: False
Position: Named
Default value: None
Accept pipeline input: False
Accept wildcard characters: False
```
### CommonParameters
This cmdlet supports the common parameters: -Debug, -ErrorAction, -ErrorVariable, -InformationAction, -InformationVariable, -OutVariable, -OutBuffer, -PipelineVariable, -Verbose, -WarningAction, and -WarningVariable. For more information, see about_CommonParameters (http://go.microsoft.com/fwlink/?LinkID=113216).
## INPUTS
### Microsoft.Azure.Commands.RecoveryServices.SiteRecovery.ASRFabric
## OUTPUTS
### System.Collections.Generic.IEnumerable`1[[Microsoft.Azure.Commands.RecoveryServices.SiteRecovery.ASRStorageClassification, Microsoft.Azure.Commands.RecoveryServices.SiteRecovery, Version=4.0.0.0, Culture=neutral, PublicKeyToken=null]]
## NOTES
## RELATED LINKS
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,011
|
(nascido em 3 de setembro, 1973) é um compositor e saxofonista japonês. Após se formar na Berklee College of Music, ele se mudou para Kansas para começar uma carreira solo como um músico de jazz. Mais tarde ele voltou ao Japão para trabalhar para a Konami, onde ele tornou-se melhor conhecido por seus trabalhos nas séries Metal Gear e Zone of the Enders.
Vídeo-game credits
Ninja Blade (2009) - música In-game, tema principal, tema final
Otomedius G (2008) - compositor, produtor
Sho Chiku Bai (2008) - compositor
Live Música by Piano and Strings: Sekaiju no MeiQ I & II Super Arrange Version (2008) - arranjador, produtor
Metal Gear Solid 4: Guns of the Patriots (2008) - música cinemática
Etrian Odyssey II Super Arrange Version (2008) - arranjador, produtor
The Outer Rim (2008) - arranjador, músico, produtor
No More Heroes Soundtracks: Dark Side (2008) - arranjador
1942: Joint Strike (2008) - música In-game
Commando 3 (2007) - música In-game
Etrian Odyssey Super Arrange Version (2007) - arranjador, produtor
Yakuza 2 (2006) - música In-game
Metal Gear Solid: Portable Ops (2006) - música In-game
Rogue Galaxy Premium Arrange (2006)- "The Ghost Ship" (Track #6)
Rumble Roses XX (2006) - música In-game
Elvandia Story (2006) - música In-game
Metal Gear Solid 3: Snake Eater (2004) - música In-game, algumas abertura de cenas, tema de abertura
Boktai 2: Solar Boy Django (2004) - música In-game
Yu-Gi-Oh! The Dawn of Destiny (2004) - abertura de filme, música In-game
Metal Gear Solid: The Twin Snakes (2004) - abertura de cenas
Boktai: The Sun Is in Your Hand (2003) - música In-game
Yu-Gi-Oh! Duel Monsters 8 (2003) - música In-game
Zone of the Enders: The 2nd Runner (2003) - abertura de cenas, música In-game
Metal Gear Solid 2: Substance (2002) – todas as músicas adicionais do jogo
The Document of Metal Gear Solid 2 (2002) - tema de abertura
Yu-Gi-Oh! Duel Monsters 7 (2002) - música In-game
Beatmania 6th Mix + Core Remix (2002)
Metal Gear Solid 2: Sons of Liberty (2001) - música In-game
Zone of the Enders (2001) - abertura de cenas, música In-game
Metal Gear: Ghost Babel (2000) - música In-game
Álbuns
Message (2008)
AKASHI (2005)
Ligações externas
Site oficial
Perfil de Norihiko Hibino na Square Enix Music Online
Norihiko Hibino na Internet Movie Database
Entrevistas
Entrevista com a GEM Impact sobre Metal Gear Solid 4
Entrevista sobre a sua companhia e trabalhos
Compositores de trilhas sonoras de jogos eletrônicos
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,420
|
Q: Android: Skip Current Fragment and loads next I am trying to implement display logic in the surveylib Library (https://github.com/AndreiD/surveylib). This is a library to implement a survey app. There will be set of questions and properties in JSON file and this library prepares a set of questions depending on the JSON file.
This library implemented the survey app by using Fragments and ViewPager. There are different fragments for different types of questionnaires (e.g. fragments folder). I want to incorporate display logic in this library which is some questions might not be displayed depending on the answers received in previous questions. I have created variables in Questions.java (in model folder).
I am struggling with skipping a fragment and load the next one. For example, I want to implement the skip routing in this FragmentCheckboxes.java file. Based on the answers of the previous question this question with checkboxes might not be displayed. I have added this part so far:
@Override
public void setUserVisibleHint(boolean isVisibleToUser) {
if (isVisibleToUser) {
// called here
Log.d("Checkbox Fragment: ", q_data.getQuestionNo() + " setUserVisibleHint");
String display_logic=q_data.getDisplayLogic();
String condition_question=display_logic.split(",")[0];
String condition_value=display_logic.split(",")[1];
String response=Answers.getInstance().get_answer(condition_question);
Log.d("Display Logic", "Question: "+condition_question+", Condition _Value: "+ condition_value+", Response: "+response);
if(response.equals(condition_value))
{
Log.d("Display Logic", "Skipping " + q_data.getQuestionNo());
Answers.getInstance().put_answer(textview_q_title.getText().toString(), "N/A");
// getView().setVisibility(View.GONE);
mContext=getActivity();
((SurveyActivity)mContext).go_to_next();
}
else
{
super.setUserVisibleHint(isVisibleToUser);
}
}
}
go_to_next() function is implemented in SurveyActivity.java. My problem is although it is calling go_to_next() function, this fragment is still visible and when I click next it skips the next question which I don't want. I wan't to skip the current question and load the next one. More specifically, Let's say we have three questions: Q1, Q2, and Q3. Q1 is radio button and can have two responses: yes and no. If the user selected yes for Q1, we will skip Q2 and display Q3. Otherwise, Q2 will be displayed after Q1.
In this particular case, how can I skip a fragment in android and load the next one?
A: 1.first u need to get the positions of the fragments in view pager adapter
2.then while calling go_to_next() pass the position of the fragment.
3.in the viewpager activity where u implemented the interface method call viewPager.setCurrentItem(pos); or pos+x
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,689
|
namespace chrome {
// Various special URLs used for testing. They have the same values as the URLs
// in chrome/common/url_constants.h, but are duplicated here so that the reverse
// dependency can be broken.
// Various URLs used in security policy testing.
extern const char kTestCacheURL[];
// The NTP is assumed in several tests to have the property that it is WebUI.
extern const char kTestNewTabURL[];
// The History page is assumed in several tests to have the property that it is
// WebUI. That's the case only with the ChromeBrowserContentClient
// implementation of WebUIFactory. With a different implementation that might
// not be the case.
extern const char kTestHistoryURL[];
// The Bookmarks page is assumed in several tests to have the property that it
// is an extension. That's the case only with the ChromeBrowserContentClient
// implementation of WebUIFactory. With a different implementation that might
// not be the case.
extern const char kTestBookmarksURL[];
} // namespace chrome
#endif // CONTENT_COMMON_TEST_URL_CONSTANTS_H_
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,843
|
tagalong
After taking stock of his supplies— a few days' worth of rations, a half-empty canteen of fresh water, a power pistol and two extra power clips, a flashlight and a spark-lighter— Max set out to explore the rest of the island. The little cub tagged along, sniffing curiously at this and that, but never straying far from Max's path. If there was anyone else on this island, he wanted to know who they were.
And whether or not they were friendly.
He took it slow since he still hadn't quite regained his land-legs.
Only a few steps, and he was startled by a flash of light, which he quickly realized was sunlight reflecting on something half-buried in the sand. As he stumbled closer, he nearly fell down in surprise upon seeing that it was his silver medallion. After a moment, he did fall down and sat there for a long moment, amazed that he even still had it. When he lost his grip and fell overboard during last night's storm, he had hung on to it for dear life, but also lost his grip on it when a particularly large wave nearly drowned him just in sight of land.
He thought he'd lost it forever. Reattaching its chain, he hauled himself to his feet and struck out again.
He hadn't traveled much farther before he came around a bend and stumbled upon more wreckage. Near the next curve in the beach were the remains of the small boat— barely seaworthy and never meant for long-range transport— that had brought him here. Just looking at the battered vessel reminded him of his wild ride; the chase and the storms had indeed been no dream.
He could tell at a glance that without a complete overhaul— including tools, parts, not to mention skills, that he did not possess— it wouldn't be going anywhere ever again. But he dug through what was left of the boat anyway. The cub poked around, too, curiosity written all over his young feline face.
Underneath the pilot's seat (or at least where the pilot's seat used to be) was a storage compartment. Max read the manufacturing label stamped on the side of the compartment: Manufactured in --------ngle Sta----y Tri-Tech, a Division of C-----------stries. He couldn't read portions, as they were blasted with carbon scarring. The name, what he could read of it, meant nothing to him.
He could only wonder how it had fallen into the hands of Cyexian pirates.
The cargo compartment itself was busted open, probably dashed against the nearby rocks on impact, and junk was strewn all over the sand near it. He immediately picked out a disrupter pistol (which was larger and heavier than his power pistol, even having an insulated handgrip along its heavy barrel, just like the one he had used the other night), another power clip, a small pan, some line and fishing hooks, an inflatable life raft (still in its tube), a spork and a survival knife, a frayed length of rope, a cracked mirror, and some clothes. He imagined all of these things might come in handy later, so he stuffed it all into his shoulder bag.
After standing there for a moment, reliving more of his wild ride than he cared to, he set out again, the mysterious cub at his side.
At first it was slow going, as he was still very sore and stiff. But the longer he was on his feet, the better he felt as his body loosened up. After about an hour or so, he was strolling along at a leisurely pace.
The whole way, he stuck to the beach. He remembered Robert explaining how, by walking around the shore of an island, one could get an idea of how big it was. And Max had paid rapt attention, wanting to know as much as he could so he could make his Dad proud when he finally got to go on that adventure he was always so certain they were going to have. As long as you're on an island, his father had said, you're never lost. No matter which way you go, you'll always find the Ocean.
Thanks, Dad… he thought quietly, I just wish you could be here to see me…
Shoving such thoughts aside, he continued along the beach, refusing to move inland until he had some idea of how big this island was. Keeping in mind that he was in unfamiliar territory, that there was no telling who or what he might encounter. He had heard of uninhabited islands in the Ocean, but even though there weren't any signs of habitation where he washed up, he had decided not to let his guard down.
There was the distinct possibility that he had ended up in Cyexian territory, or even the Triangle State. Either would be bad news; the Cyexians would almost certainly use him as a hostage to make demands on the Elders, and there was no way of knowing what the TSA might decide to do with him. Beyond those realms, it would be harder and harder to find his way back to the Islands.
But what would he say if he went back?
He started walking faster, trying to outrun the waves of grief and shame and guilt that were fast gaining on him. Part of him was still certain this was all a bad dream. That he would wake up any second now, and Mom would come and tell him everything was okay… But what would she think if she knew?
Freezing up might have been perfectly natural, under the circumstances, but it still didn't feel like much of an excuse.
Meanwhile, the cub continued to romp gleefully around, apparently happy just to have some company.
With great effort, Max pushed his feelings back, telling himself they would be of no use to him right now. He needed to concentrate, focus on what was happening right now. His very survival depended on it. If there was anyone out there, he wanted to find them, not the other way around.
To take his mind off those painful thoughts, he decided to run through everything he knew about the Cyexians and the Triangle State Authority.
Somewhere beyond the disputed Island of Kinsasha was the edge of Cyexian waters. There were eight feuding clans, but the number of islands they controlled wasn't really known for sure, rumored to be about a dozen. He knew they were a matriarchal society; his parents and Uncle Angus had told of Cyexian lands where men were slaves, and Mom had always laughed and said that sounded like a fun place to visit, as well as places where the matriarchy was more ancient tradition than current practice. Due to martial necessity, the clans here gave men certain allowances, but made it abundantly clear who was the boss. The Cyexians of these waters were pirates and scavengers, while the Layoshans, though known to dig through derelicts and ghost ships— and bring them back to the Islands, if possible— were more into salvage and trading for outside goods with the visitors they got every now and then.
The Triangle State, on the other hand, was something of an enigma. No one, not even his father (though he and Angus had once been held in the brig of one of their ships), had been there in the last forty or fifty years. Controlled— governed with an iron fist, from what he had been told— by the Triangle State Authority, what any Outlanders who had been there described as a sinister cabal of local despots who called themselves the Board of Directors, their domain was said to be three islands. And everything in between.
With so little useful intelligence— and a lot of creepy rumors— the TSA was a shadow that lurked quietly among Max's speculations.
Occasionally, he chided himself for not paying enough attention to what was going on around him. But it was hard to keep his guard up when there was nothing going on, and the cub's carefree meandering was equally disarming. The farther he went, the more he felt like the last person in the world.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,510
|
Televisie en film
Don't Look Back (Heroes)
Dont Look Back (1967), documentairefilm uit 1967 over Bob Dylan
Don't Look Back (2014), een thriller uit 2014 onder regie van William Dickerson
Muziek
Don't Look Back (The Temptations), single
Don't Look Back (Boston), single en album
Don't Look Back, van Bruce Springsteen
Don't Look Back (Fine Young Cannibals)
Don't Look Back, van Lucie Silvas
Don't Look Back, van Lloyd Cole
Don't Look Back (The Starlings)
Don't Look Back (The Remains)
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 5,669
|
Learn what is the JQuery framework is and why many web developers and designers love to this framework.
How to Use JQuery Framework?
Learn how to use JQuery framework and how place it on your website.
Writing your first JQuery function.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,157
|
Knoxville: 865-329-7269
The Tri-Cities: 423-415-1920
"We're two for you!"
A Family-Oriented
Law Firm Fighting
Have you been injured?
A Family-Oriented Law Firm Fighting For You
A Knoxville man is critically injured after being hit by 2 cars
On behalf of Fox & Farmer Attorneys at Law | Dec 14, 2018 | car accidents, Firm News
A pedestrian was struck by two cars out in front of Weigel's on E. Summit Hill Drive in Knoxville at 7:05 a.m. on Dec. 10.
The unidentified male was struck by the first motorist as he tried to cross the street in front of the convenience store after picking up some items there. That collision's impact caused him to immediately fall to the ground. That motorist pulled into the adjacent Weigel's adjacent parking lot so that he could call for paramedics to be dispatched to the crash scene.
While that first driver was making that call, another motorist fast approached and ran over the victim who was still laying in the street. That car's operator soon pulled into Weigel's parking lot as well.
The victim was still alive by the time paramedics arrived at the crash scene, although a police spokesperson admits that he was suffering from life-threatening injuries. He was transported to the University of Tennessee Medical Center where he was listed as being in critical condition.
Officers with the Knoxville Police Department haven't completed their investigation into the crash. They also haven't ruled out filing any charges against the motorists involved in the incident.
If you were hurt in a motor vehicle accident, then you may not realize just how severely injured you are until the shock wears off. By the time it does, you may find yourself having to undergo a surgical procedure that will require you to take off months from work. An attorney can help you file a claim to recover medical costs and lost income so that your bank account doesn't get depleted.
workplace injuries (20)
Almost 30% of semitruck crashes result from air brake failure
Types of ailments that may qualify you for SSD benefits
Are remote workers eligible for workers' compensation?
What is inattention blindness?
What are the rights and responsibilities of Tennessee cyclists?
Knoxville Personal Injury Attorneys Serving East Tennessee
After being seriously injured or no longer able to work, turn to Fox & Farmer Attorneys at Law for real help. To schedule a free initial consultation, call 865-329-7269 or 423-390-0000, or contact us online
We only collect attorney's fees if we secure compensation for you.
Need help from an experienced attorney?
Contact us right now for a free consultation.
8900 Executive Park Drive
Tri-Cities: 423-415-1920
115 East Unaka Ave.
Tri-Cities Law Office Map
© 2021 Fox & Farmer Attorneys at Law. All Rights Reserved.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,578
|
{"url":"https:\/\/www.emathhelp.net\/calculators\/algebra-2\/hyperbolic-cotangent-calculator\/","text":"# Hyperbolic Cotangent Calculator\n\nThe calculator will find the hyperbolic cotangent of the given value.\n\nThe hyperbolic cotangent $$y=\\coth(x)$$$is such a function that $$y=\\frac{\\cosh(x)}{\\sinh(x)}=\\frac{e^x+e^{-x}}{e^x-e^{-x}}$$$.\n\nThe domain of the hyperbolic cotangent is $$(-\\infty,0)\\cup(0,\\infty)$$$, the range is $$(-\\infty,-1)\\cup(1,\\infty)$$$.\n\nIt is an odd function.\n\nIf the calculator did not compute something or you have identified an error, or you have a suggestion\/feedback, please write it in the comments below.","date":"2021-05-13 04:06:33","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8834293484687805, \"perplexity\": 980.0423565607575}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-21\/segments\/1620243992721.31\/warc\/CC-MAIN-20210513014954-20210513044954-00178.warc.gz\"}"}
| null | null |
Janoušek je příjmení, které náleží řadě známých osobností, mimo jiné:
Antonín Janoušek (1877–1941) – český novinář a komunistický politik, otec Jaroslava
Bohumil Janoušek (* 1937) – reprezentant Československa ve veslování
František Janoušek (1890–1943) – český malíř
Gabriel Janoušek (* 1940) – československý vodní slalomář, kanoista
Jaroslav Janoušek (1904–?) – sovětský důstojník NKVD a český vyšetřovatel StB, syn Antonína
Jiří Janoušek – více osob
Josef Janoušek (1879–1935) – český právník, soudce a politik
Karel Janoušek – více osob
Miroslav Janoušek – více osob
Pavel Janoušek (* 1956) – český literární historik a teoretik
Roman Janoušek (* 1968) – český podnikatel a lobbista
Roman Janoušek (1972) (* 1972) – český fotbalista
Slávek Janoušek (* 1953) – český písničkář
Vladimír Janoušek (1922–1986) – český sochař
Vojtěch Janoušek (1897–1969) – český pedagog a historik
Janoušková
Anna Janoušková (* 1965) – československá běžkyně na lyžích
Aťka Janoušková (1930–2019) – česká herečka a zpěvačka
Markéta Janoušková – česká houslistka
Věra Janoušková (1922–2010) – česká sochařka, malířka a grafička
Podobná příjmení
Janouš
Janáček
Janeček
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Česká příjmení
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
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### Scary Monsters and Super Creeps
Also by Dom Joly
_The Dark Tourist_
First published in Great Britain in 2012 by Simon & Schuster UK Ltd
A CBS COMPANY
Copyright © 2012 by Dom Joly
This book is copyright under the Berne Convention.
No reproduction without permission.
All rights reserved.
The right of Dom Joly to be identified as the author of this work has been asserted by him in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act, 1988.
Simon & Schuster UK Ltd
1st Floor
222 Gray's Inn Road
London
WC1X 8HB
www.simonandschuster.co.uk
Simon & Schuster Australia
Sydney
Simon & Schuster India
New Delhi
All pictures provided courtesy of the author
A CIP catalogue copy for this book is available from the British Library.
ISBN: 978-0-85720-764-7
eBook ISBN: 978-0-85720-766-1
Typeset by Hewer Text UK Ltd, Edinburgh
Printed and bound in Great Britain by CPI Group (UK) Ltd,
Croydon, CR0 4YY
'Scary monsters, super creeps, keep me running, running scared'
David Bowie
For Toast (12 November 1924–2 July 2011):
'The joys of parents are secret; and so are their griefs and fears'
Francis Bacon
#### Contents
Foreword
Ogopogo
Hibagon
Mokèlé-mbèmbé
Bigfoot
Yeti
Nessie
Epilogue
#### Foreword
Before we start . . . A little admission. This book is not really about monsters at all. Don't get me wrong – I did set off on various adventures around the world to places where 'monsters' were reputed to roam, in the vain hope that I might bump into one. It's just that both you and I know that this was very unlikely to happen – and, even if it did, I was probably the worst person in the world for it to happen to.
In the UK I'm famous for being a practical joker and an accomplished liar. In my first show, _Trigger Happy TV_ , I used loads of furry costumes – including a Yeti outfit, in which I scared skiers on a Swiss ski-slope. So if I suddenly announced, in a much-hyped press conference broadcast live around the world, that I had found a 'monster', and then showed footage of said encounter, I might face some incredulity.
Done right I could probably still get away with it, though. Experience has shown that there is little our rapacious news outlets like more than a 'monster' story. Such scoops give them the excuse to endlessly replay blurry, shaky footage (does nobody own a tripod?) and get weird hairy men into the studio to talk about new discoveries of wild weird hairy men.
The former kind of hairy men often profess to be cryptozoologists. This is a posh scientific name for people who are interested in 'monsters'. A lot of cryptozoologists decide to write books on the subject. Most of these books are incredibly dull. This is because these guys are writing about something that serious scientists don't really take very seriously. So, to show how serious they are, cryptozoologists tend to write long, boring, pseudoscientific books in which they try desperately to prove to a disbelieving world that they are not nutters but actually distinguished men of science.
This is not a book like that. This is a book documenting my year spent travelling the world looking for 'monsters' and getting into all sorts of trouble with the 'super creeps'.
**Why do you keep putting the word 'monsters' in quotation marks?**
Thank you for asking. It's because I think it would be unfair to describe the creatures I've been after as 'monsters'. The dictionary definition of the word is: 'An imaginary creature that is typically large, ugly and frightening.'
This sounds more like some of my least-favourite British towns than it does anything I was going in search of.
Before I set off on my adventures I wasn't convinced that they were all imaginary. Most were definitely supposed to be quite large. 'Ugly' is a subjective term anyway and certainly not one I'm prepared to throw at an eight-foot missing link. And these creatures are frightening only because we don't know anything about them.
So far they have defeated science and managed to keep hidden(ish) from our modern world. Besides one kind of Yeti none of them are supposed to attack humans. As I set out, it seemed to me that they just want to be left alone to do whatever it is they like to do . . . If they exist at all. Confused? Welcome to my world.
**Well, we didn't start this – you called the book _Scary Monsters and Super Creeps_** _!_
Yes . . . I know. I couldn't resist. I'm a huge David Bowie fan and it was just perfect for the title. It sounded cool and I went with it – sorry. But from now on I won't put the word 'monsters' in quotation marks. It would start getting really annoying, wouldn't it?
**Why monsters?**
I love to travel – it's my obsession – but I always need a purpose. For my last book, _The Dark Tourist_ , I went on holiday to the sort of places that most people wouldn't: Chernobyl, North Korea, ski slopes in Iran, etc. For this book I needed something new.
Some people travel the world birdwatching. (Costa Rica, for example, is a place where such nerds go to holiday – it's full of twitcher couples all off to find some specific bird.) Big-game fishermen sail the Seven Seas trying to catch some special fish. I just decided to do the same: to try to find out as much as I could about the Big Six monsters of cryptozoology.
**The 'Big Six'?**
OK – I don't know why it had to be six. I guess that I was just thinking along the lines of the Big Six when you go on safari.
Some were obvious: Bigfoot and the Yeti were a given; the Loch Ness Monster was another obvious contender, though I was initially loath to choose it as this would hardly be the most exciting of foreign trips.
This left me with three others. I found some stuff about the Mokèlé-mbèmbé in the Congo. This sounded like a proper adventure and, since the big lure of my Yeti-expedition research was reading _Tintin in Tibet_ , I thought that the theme could be continued with a _Tintin in the Congo_ -type adventure (minus the hideous racism and the murdering of hundreds of animals).
Further googling – sorry, _research –_ revealed a story about a monster that lives in the hills around Hiroshima. It's called the Hibagon and one theory is that it's a man who was irradiated in the Hiroshima explosion. This at first seemed the most tenuous of stories but I've always wanted to visit Japan and it's a country with an enormous monster culture – after all, it's the home of Godzilla.
This left me with a spare. I'm married to a Canadian, love Canada and have always wanted to write something about my semi-adopted country, so I chucked Ogopogo into the mix. Thus are adventures decided.
I think my initial interest in monsters came from a book I was given for my birthday as a kid. It was called _Arthur C. Clarke's Mysterious World_ and was a companion piece to the 1980 series on UK telly. I loved this book. I read it from cover to cover, over and over again, and longed to go and find out more about the weird Yeti footprints found in the Himalayas and to find the spot where the infamous Super 8 footage of Bigfoot was filmed.
At about the same time a man came to my prep school and gave us a lecture about the Loch Ness Monster. I was transfixed. He showed us footage and photographs of 'Nessie'. Once day, I vowed, I'd go and look for her.
So that's how this book got going. I set off on all my trips with an open mind. I'd really love to think that there was a Bigfoot or a Yeti or even a Big Cat of Cirencester out there. My local newspaper in the Cotswolds loves to run a 'Big Cat Spotted' story every slow news day (there is quite a fat Siamese that sits in the window of a house on Blackjack Street but I don't think that's the one causing the commotion). Once, I . . .
**Get on with it!**
Oh great: first you ask questions and now you're actually heckling my Foreword . . .
**I hate forewords.**
Actually, so do I. I groan when I see a big, long foreword before a book. It's just padding. If a book is any good then it doesn't need a foreword. They are usually random, wistful musings from an author desperate for anything to delay him having to face that terrifying moment of actually starting to write the book. If anybody really does have any questions, then ask me on Twitter . . .
**So stop writing and get on with it.**
Alright. Here it is. I hope you enjoy the adventures.
Dom Joly, Cotswolds, 2012
### Ogopogo
'What would an ocean be without a monster lurking in the dark? It would be like sleep without dreams.'
Werner Herzog
I woke up in my bed in my house in the Cotswolds. It should have just been another normal day – but it wasn't. Today was my first day of being a monster-hunter. I woke up the kids and kissed them goodbye in a slightly formal manner, like some Victorian explorer off to deepest, darkest Africa for three years.
'Where are you going, Daddy?' one of them asked sleepily.
'I'm going monster-hunting,' I replied casually.
'What kind of monster?' they asked, slightly more awake.
'It's a lake monster in Canada . . . Sort of like a dinosaur-type thing . . .' I didn't sound very clear on the matter.
'Is it dangerous?' asked Jackson, my seven-year-old son.
'No, I don't think so,' I replied in a manner intended to convey that danger was not really something that bothered me.
'Then it's not a monster.' He seemed very sure of himself on this fact.
'Well . . . Yes, actually, it is . . . But anyway . . . Bye . . .' I wandered downstairs feeling slightly deflated and made myself a very strong coffee from the Nespresso machine that has turned my wife Stacey and me into caffeine junkies. We have terrible cold-turkey days when we run out of the little coffee capsules and hang around looking out of the window, waiting for the man. (The Nespresso man, who'll deliver our next box of coffee crack.) I sipped my perfectly frothed macchiato and started to worry a little. For someone who was off monster-hunting I felt distinctly spoilt and unprepared. Should I have gone to Millets? What did a monster-hunter need? A knife would have been good but I was flying to Canada so it wouldn't be allowed on the plane. What about a long stick? They probably wouldn't let that on board either. What did a monster-hunter wear? I quite fancied having some sort of uniform – something a touch Indiana-Jones-ish – but I'd just packed my usual Carhartt gear, the uniform of the overgrown London media wanker who refuses to grow old gracefully. Someone on Twitter once said that they thought men like me dress in the same manner as we did at our sexual peak. I'm not sure I've ever had a sexual peak but I knew what she was saying.
It was exciting: this was the first of six trips around the world in pursuit of six legendary creatures. I had decided to go to British Columbia first. It was probably the most civilized of all my destinations and I still wasn't entirely sure what the form was for monster-hunting. I was looking at this trip as a bit of a trainer mission before I hit the more difficult ones.
My destination was the Okanagan Valley in British Columbia – a popular summer-vacation destination for West Coast Canadians. Lake Okanagan, an eighty-mile-long glacial lake that runs the length of the whole valley, is supposedly the home of Ogopogo: a monster that has been chronicled and talked about for more than 200 years. Ogopogo is pretty big news in Canada – every bit as famous as Nessie.
I said goodbye to Stacey in the manner of a hunter-gatherer off to collect meat for the family: I hit her over the head with a large club and had my way with her. Actually no, I didn't. I kissed her and told her I'd be back in two weeks . . .
At Heathrow I readied myself to face the worst part of any journey: airport security. Obviously I don't want to be blown out of the sky but you can't help but look at the 'security specialists' screening you in international airports and immediately think, _Failed traffic wardens._
I'd tried to think ahead and had bought a couple of see-thru washbags so that I didn't have to put all my liquids into the see-thru bags that are always too tight and that BAA now try to sell you for £1 a go. I shouldn't have bothered. A fourteen-year-old boy with acne and an ill-fitting uniform demanded that I take all the liquids out of my see-thru washbag and put them all into one of their see-thru bags. How this might prevent terrorism was beyond both him and me.
'We are just here to help you stay safe . . .' is the repetitive mantra of airport security staff. They're not, however, there to explain anything. I'd be concerned leaving a sandwich assembly in their hands yet this is the first line of defence against global terrorism. God help us all.
Once I was through the X-ray machine a twelve-year-old Indian woman told me that I could put the liquids back in my own see-thru bag. Again, there was no explanation.
I was then free to set about my preferred, slightly OCD departure routine. First I went to buy _Vanity Fair._ As usual, I couldn't find it. As usual, I eventually checked the Women's Lifestyle section and, sure enough, there it was. _It's not a women's lifestyle magazine!_ It's a fabulously aspirational monthly fix of American snobbery, travel and good journalism. I gave up complaining about this kind of misrepresentation long ago, though. The people manning checkout counters are the younger, idiot brothers of the 'security specialists'.
I now headed for the ludicrously overpriced seafood bar to have my usual prawn cocktail and half-bottle of house champagne. This has become a ritual. If a meal could be my last on earth then I want it to be good and hang the eye-watering cost.
I was slightly tipsy as I boarded the plane to Vancouver. It wasn't very full so I ignored my assigned seat number and grabbed one in the front row with loads of legroom. This is always a tense time, waiting for the doors to close and the certainty that the row's yours. Now, at the very last minute, a bald man bustled on board and plonked himself down right next to me. I suspected that he was doing the same as me, and this wasn't his assigned seat, but I was in a fairly weak negotiating position. My new neighbour started reading Lionel Shriver's _We Need to Talk About Kevin_ while subtly trying to gain control of our mutual armrest. Unfortunately for him, he had little idea that he was taking on a hardened veteran of elbow wars. For a good twenty minutes or so we jockeyed for position while studiously 'ignoring' each other. I could feel his growing hatred of me and realized that this was my first of what I was sure would be many encounters with a super creep.
'Excuse me – has there been some mistake? Have you paid for two seats?' I lifted my elbow in mock-exasperation. This caught him totally off-guard. He wasn't prepared for direct confrontation and backed off fast. His elbow disappeared down by his side and victory was mine. I spread my arm out triumphantly over the whole of the armrest.
He attempted to disguise the defeat by pulling out his laptop (a PC, not a Mac: total confirmation that he was a super creep). I checked out his screensaver. This was a photograph of a red Rolls-Royce parked on a street in Knightsbridge. I hate Rolls-Royces – I just don't see the point of them. If this vehicle was his he was definitely a super creep. But if he'd just taken a photo of one and then used it as his screensaver he was a _sad_ super creep.
I was flying economy. Not only that but on Air Canada – and their economy is very far from the best. I loathe airline seats. We can put men on the moon but can't make a comfortable seat for international travel? Let's get our priorities right. It's physically impossible to sleep in the weird halfback position that an airline seat forces you into. I'm convinced that they use them in Guantánamo for sleep-deprivation experiments.
I'd scrounged this flight off the Canadian Tourism Commission. Canada suffers from an image problem: people tend to think it's incredibly boring. Bill Bryson once wrote that publishers went quite pale when he'd announced that he wanted to write something about the place. The Canadian Tourism Commission had high hopes that I was going to change all that . . .
I watched _Super 8_ , a Spielberg homage recommended by Mark Kermode, my all-time favourite movie critic. It was good, but not great. (Curiously given the subject of this book, I've never really been that into monster movies. _Jaws_ was good, but that's about it.) The movie over, I was bored; so I started snooping on my neighbour's emails. He was Danish and something to with paediatrics. I watched in fasciation as he spent more than an hour perfecting the dullest email I have ever read.
To all area coordinators:
A. Get idea of timelines so that inputs can be planned
B. Get an idea of how inputs should be provided
C. Get a timeline for when inputs should be provided
D. Get provision for extra input on timeline
I checked twice to see if he'd spotted that I was reading his email and was now taking the piss. Surely nobody could write an email that dull without seriously reassessing their life choices? At that moment I was _so_ glad that I was a monster-hunter. _My_ emails would never be like the dull Dane's. They would be like this:
Dear Tony,
I need that underwater sonic device as soon as you can make it. Were my designs clear enough? Gotta go – I'm picking up some infrared-camera stuff and a really big net and want to get to Millets before it closes.
Dom (Monster-Hunter)
It was just as we started our descent into Vancouver that I realized I'd left my beloved Leica camera at home. This wasn't the greatest start to my first monster-hunting trip. Now if I did actually see Ogopogo I was going to have to draw the bastard.
The stewardess offered me a glass of Dasani. It appeared nobody had yet told the North Americans that we ran this bottled water out of the country when it was revealed that the contents were simply filtered tap water.
I had to go through Canadian Immigration before catching a little turbo-prop plane to Kelowna. A smiling, friendly official looked at my passport and asked me my reasons for coming to Canada. I couldn't help myself.
'I'm here to look for Ogopogo,' I said, smiling back a little too hard and aware that I might look a bit unstable.
The official's friendly smile disappeared and was replaced with a world-weary version.
'Best of luck with that, eh? Now, what is your real reason for visiting Canada?' She had a hint of steeliness about her.
I smiled again. 'I'm here to look for Ogopogo . . .' There was quite a long pause as she sized me up, wondering whether to call security for a cavity search. Then she seemed to remember that she was Canadian and not American.
'And is that for business or pleasure?'
I thought hard for a second and then replied truthfully.
'It's purely for pleasure, ma'am.'
She stamped my passport and waved me through. I was in. I was in a foreign country hunting monsters. This was turning out to be rather Tintin-esque. Unlike Tintin, however, I have a family – not a loose, shady cabal of homoerotic acquaintances plus a white dog.
Vancouver's internal departure lounge didn't have a seafood bar so I had to change my pre-flight order to a doughnut. This made me nervous. The twin-prop was, like most twin-props, rather unsettling to the passenger. On normal planes you can't really see where the power's coming from. On a twin prop, like in a helicopter, you spend your whole time staring at the rotors imagining what would happen if they suddenly stopped turning.
An hour into the flight and I got my first view of Lake Okanagan. It's an ominous-looking thing: a long, dark stretch of water, dwarfed on both sides by steeply rising earth – a physical reminder of the immense glacial forces that once shaped it.
The first time I'd seen this lake was on _Arthur C. Clarke's Mysterious World._ Here Clarke had investigated the innumerable sightings of a 'creature' in the lake. These sightings went way back into the nineteenth century among settlers. The beast was supposed to be not unlike the Loch Ness Monster – there were reports of 'humps', long black shapes in the water and stories of a dinosaur-type creature. There were also several intriguing snippets of footage and a multitude of blurred photographs. There was definitely something in this lake that was attracting attention. Legend had it that whatever it was lived in a cave under a place called Rattlesnake Island in the middle of the lake. I decided that this would have to be my first destination.
I picked up my car from Budget. Unlike her UK counterparts, the woman behind the desk was friendly, apparently well-travelled and very helpful. I'm banned from most major car-rental outlets in Central London because I constantly get into altercations with the staff. In the UK the car-rental business seems to be designed to test just how determined you are to rent said vehicles: they'll do anything they can to prevent you leaving in a rental car. In North America it's always a joy; they even seem slightly apologetic that you actually have to pay for marching straight out into their car park and driving off in any car with the keys in the ignition.
I'd decided to stay at the south of the lake and work my way up the valley, ending up in the main town of Kelowna. My first bed for the night was in Summerland. I wondered what Summerland was going to be like in October. Not that summery, I imagined. The busy holiday season was long over and the weather was starting to get pretty cold. There was zero boat traffic on the lake, not a single vessel. As I drove south along the lakeshore I realized that I genuinely didn't have the first clue as to how to monster-hunt. I kept half an eye on the water hoping that maybe I'd get a sighting of Ogopogo immediately but I was intensely aware that this was very unlikely.
I turned on the local talk-radio station – always the best way to get under the skin of a community when in North America. The big news story of the day was about a man who had repelled an intruder to his trailer by using a screwdriver and 'bear spray'. I wondered what was in bear spray. It definitely sounded like something a monster-hunter should have in his backpack.
I kept driving and put the Kermode/Mayo movie podcast on. This has been my constant companion on so many road trips, my little slice of normality in weird surroundings. A wind had picked up and the lake was choppy and rather mean-looking. Okanagan reminded me a lot of Loch Ness. It's on roughly the same latitude and is the same sort of shape, although a lot bigger. If I'm honest, lakes have always creeped me out a bit. There's something rather ominous about their stillness and murky depths. I don't like swimming in them.
In Summerland I checked into my hotel, a rather posh beach resort that was totally deserted. Summer had indeed left Summerland, seemingly taking all the inhabitants with it.
This was the limbo season – after the summer hordes but before the ski season started in nearby resorts like Big White. I wandered the empty _Shining_ -like corridors until I got to my room. This was satisfyingly huge. As the only resident in the hotel I'd been given a suite overlooking the lake. Just below my balcony, on the lakeshore, was a hot tub. Now this was my kind of monster-hunting. I was tired from the long day travelling and slipped off my smeggy clothes and hopped into the tub. I lay back and sighed. This was the life, lounging in a hot tub while keeping half an eye on the lake for monster action . . .
I must have dozed off because I awoke with someone shaking me and shouting, 'Sir! Sir, are you OK?'
It was a security guard, probably freaked out enough to see a guest let alone a naked one passed out in the hot tub. Or maybe he just thought I was a trespasser. I tried to look authoritative but I'd been dribbling down my face and the whole thing was not a good look. I retreated to my room with as much dignity as I could muster and fell asleep. I slept like a log and woke up twelve hours later feeling a whole lot better about things. The weather had turned and it was a beautiful autumn day. I spent a cursory ten minutes staring out at the water looking for monsters until hunger took over.
I drove into Summerland proper but there wasn't anything there so I headed down to Penticton, a larger town right at the southern end of the lake. This too was like a ghost town. My spirits dropped. I rather hoped that Ogopogo hadn't also left the area for the winter.
I wandered the empty streets before opting for a place called Fibonacci on Main Street where the coffee smelt good. It tasted good as well: I had a monster latte as I surfed the Net for information about Ogopogo and Penticton. There'd been a famous sighting here back in 1941. A bunch of kids swimming off the beach saw a huge, thirty-foot-long object that looked like a snake. It was swimming just beyond the wooden-log buoys about fifty feet offshore. They all ran to get an adult but the thing was gone when they got back.
I rang the local Ogopogo expert, Arlene Gaal. She'd been on the original Arthur C. Clarke programme and had written a couple of books on Ogopogo so I hoped that she could maybe point out areas of the lake where there had been more sightings than elsewhere. I had a quick chat with her and we agreed to meet on Saturday, the day after tomorrow, at her home in Kelowna.
With no plans for the day, I decided to explore Penticton. This took about five minutes. It was like the beginning of _28 Days Later._
I wandered down to the beach where the kids had seen Ogopogo. A man in blue overalls was at the top of a ladder, putting a fresh coat of vivid-orange paint on an enormous metal peach. The peach was about twenty feet tall and looked quite cool. The Okanagan Valley has a very curious weather system. The areas to the north end of the valley get more rain and record far colder temperatures than the areas to the south, which are almost desert-like. For years the area's main agricultural business was fruit production – hence the big metal peach. Recently, however, locals have realized that the climate's perfect for wine production and this has become a boom industry, superseding the fruit business.
On the shore was a sign warning bathers of potential hazards: 'Check depth. Check weather forecast. Don't trust inflatable devices . . .' and, my personal favourite: 'Learn to swim.'
There was no mention of Ogopogo danger. Indeed, so far I hadn't come across much mention of the creature anywhere. I think I'd expected the whole area to be teeming with Ogopogo stuff and paraphernalia. I'd thought it would be the big thing around there, like Nessie is at Loch Ness. Okanagans, however, appeared to be pretty uninterested in their local monster.
I spent a futile five minutes staring at the lake hoping that Ogopogo would show up. I then wandered along the shoreline past a gargantuan old steamboat, beached like a dead whale. Before the arrival of the bridge that now spans the lake at Kelowna, vessels like these were incredibly popular with both day-trippers and locals as a means to get round the lake. In 1926, seven years before the first recorded Nessie sightings (archival records of Ogopogo sightings go back to 1872) the British Columbian government announced the commencement of a ferry service between Kelowna and Westbank. They also declared that the vessel would be equipped with 'devices designed to repel attacks from Ogopogo'.
I checked the beached steamship for signs of any such devices but could see none. I imagined some guy perched on the prow of the ship atop a terrifying harpoon scanning the dirty grey waters for monsters. I walked round a corner and finally came across my first sighting of Ogopogo – sadly this was only in the form of a badly painted depiction on a sign advertising the Ogopogo Motel.
I wandered into the reception only to be accosted by a large lady at least as scary as the legendary beast itself. She looked at me suspiciously and asked me what I wanted. Slightly caught off-guard and embarrassed, I didn't identify myself as one of the world's foremost monster-hunters. Instead, I found myself telling her that I needed a room.
'A room?' she barked in surprise, as though I'd just asked her to bare her breasts.
'How long for?'
My pointless web of deceit started to unravel. 'A week – I'm in town for a . . . convention.'
She looked at me suspiciously. 'A convention? What convention? There ain't no convention in town.'
I replied that it was the 'ZGB Inc.' convention, hoping to confuse her with the initials.
'Well, why aren't you staying at the Convention Centre, then?'
Clearly you had to be very much on the ball if you wanted a room at the Ogopogo Motel. I imagined that guests were as frequent as sightings of the creature itself.
I wanted to leave, just run out of the door screaming, but instead I continued my fantasy.
'Because I'm a . . . recovering addict and can't be near a casino or . . . I go mental.'
She looked at me funny. There was a long silence. Finally she announced that the motel was 'completely full' as all the 'orchard workers' were there. I felt like a relieved blond trucker who'd just been turned away from the Bates Motel because there was no room.
I walked quickly back to the safety of my car. The weather had turned and ominous black clouds hung low over the lake. I drove to Summerland in driving rain.
It was eleven in the morning when I got back into my lakeside suite. The rain had stopped and the lake was flat calm, like a mirror. It was seven in the evening back home in the UK so I thought I'd Skype them. By means of the kind of modern technology that Tintin would never have been able to enjoy even if he'd had a family I was soon looking at my sitting room back in the Cotswolds. Stacey started telling me a long story about something that happened to the kids at school but I wasn't listening. My attention had been drawn to some weird movement in the lake about 100 yards away from my window. Two shapes, like twin heads, were making very fast figure-of-eight motions in the water.
'Are you listening?' said Stacey, but I ignored her and jumped up, grabbed my iPhone and rushed to the window to start filming. In the background Stacey was still talking to me but couldn't see where I'd gone. With my iPhone rolling I shouted at the laptop to tell them what was going on. I backed away from the window a touch and turned the computer round so that they could see what I was seeing. The zoom on the iPhone wasn't great but I could see close enough to know that whatever was making the disturbance was not a bird. It looked like a pair of three-foot bumps sticking out of the water and whatever it was was thrashing about, as though feeding or chasing something. I couldn't quite believe it and Stacey was sure that I was joking. Below my suite, to the left of the hot tub, was a dock that stuck out very near to the disturbance. I shouted to Stacey that I was going to run down to the end to try to get some closer footage. By the time I'd got there, though, whatever it was had disappeared. But it had disappeared underwater: nothing flew up or swam away in sight.
This genuinely was a puzzling moment. As I walked back along the dock I spotted the same security man who'd found me asleep in the hot tub the day before. He'd obviously seen me screaming and running down the dock and probably assumed that I was about to do weird naked shit again. As I walked past him he nodded at me hesitantly.
'Everything OK, sir?' he asked, keeping a safe distance.
I told him about what I'd seen and showed him the iPhone footage. He looked at it for quite a while and then asked to see it again.
When it finished he looked at me seriously and said, 'Looks like you've just had your first sighting of Ogopogo . . .' He had no idea that I was the world's most famous monster-hunter or that I was looking for the very creature he was now saying I'd captured on film. I couldn't believe it. Had I really got a bona fide sighting, on film, of a monster on my very first full day of monster-hunting? Surely it couldn't be that easy? I walked back to my room and immediately posted the footage on my Facebook page. I'd just told everyone that I was off on this trip and now I was posting a sighting. Nobody was going to believe me but I swear this is exactly what happened.
Rather adrenalized by events, I drove up to Kelowna to find the main bookstore and buy Arlene Gaal's book _In Search of Ogopogo._ I wanted to do a bit of flattery research before meeting her. As I looked for it in the store I couldn't get my Ogopogo sighting out of my mind. I'd set off to find a monster and had an 'encounter' on my very first day. I was slightly buzzing but also worried that nobody would believe me. I wondered if this was what happened to other people who spotted things in the water but were worried about public ridicule.
I found the book with some difficulty and the woman behind the counter seemed surprised to be selling it. I kept the news of my sighting to myself. It was now very sunny again. I had never been anywhere where the weather changed so rapidly. I headed for the waterside park to read on a bench. There, to my delight, I spotted a statue of Ogopogo. It was green and white with humps rising in and out of the concrete ending in a slightly dopey-looking horned head with a big red tongue flapping around. It rather reminded me of Puff the Magic Dragon. I sat on a bench right beside it and started reading the book in between watching Japanese tourists drape themselves all over Puff for hour-long photo sessions. I so admire the Japanese race's dedication to having their photos taken while striking curious fictional 'gangsta-san' poses. They will think nothing of taking more than 500 photographs of a woman in an oversized sun visor giving the camera the peace sign while gurning. It must have been the sheer amount of holiday footage needing to be developed in Japan that turned their inventors towards thinking about a digital camera.
I found the Gaal book quite a difficult read. She was not the most gifted of writers and it became more of a long list of sightings. A great part of the text seemed to comprise the numbers of TV crews she'd worked with. After a while I became a little bored and nodded off on the sun-drenched bench.
On Friday I got up at the crack of dawn as I was still on UK time. I headed off into Penticton again. I wasn't quite sure how this was possible but, if anything, it was even more deserted. Back at Fibonacci I had the strongest coffee of my life so far. I'd sent an email to a man who'd promised that he could rent me a boat even though the season was over. We'd arranged to meet behind the waterfront casino at half past nine. The weather had turned again and it was unbelievably cold and very overcast. The lake looked choppy and rather foreboding.
I hadn't really brought any warm boating clothes with me. I looked around but there was only one clothes shop open, and that was in the foyer of the casino. I was surprised that the place was already open. It turned out to be the busiest place in Penticton, with about seven sad-looking individuals sitting lifelessly in front of one-arm bandits robotically feeding the voracious machine with quarters. If there is anything more depressing than a casino at nine in the morning then I haven't yet come across it.
I bought the only warm thing available: a short-sleeved fleece with the words 'Canadian Lover Man' embossed in big red letters on the back. I was sartorially mortified but had very little choice. Even the shop assistant looked at me in a weird way.
'It's getting cold . . .' I said to her almost apologetically.
_'Yeah . . . But not_ that _cold_ . . .' I sensed her thoughts.
My boat guy was all sensibly wrapped up against the cold. He didn't say anything about my outfit but you could feel a touch of slight tension once he'd spotted it. Fortunately, because of my regular summer vacationing in Ontario, I have a Canadian Pleasure-Craft Licence; this seemed to relax him a touch. We started going over what I needed to know about the boat and he asked me where I was intending to go. I told him that I was headed north, towards Peachland. I didn't mention Rattlesnake Island. I'd been told that locals were rather superstitious about the place and didn't like people going there. This guy didn't seem to be concerned about anything but payment.
I gave him my damage deposit and asked him, 'What happens if I hit Ogopogo and sink the boat?'
'Then you won't be getting this deposit back,' he said without hesitation.
'But if Ogopogo attacks me then it's not my fault . . . Do you have a special clause for that?'
He looked at me as though I was a lunatic and I didn't want to push the issue as I wanted to get on the boat. I bade him farewell and put-putted out of the marina.
The lake was rough – very rough – and my little boat and I started to be tossed about quite violently. I looked around. Mine was the only boat on the whole lake. Was this a wise thing to be doing? I looked up at the steep sides of the valley that towered high up above the dark water and I felt very, very tiny in my little vessel. I gunned the engine and the boat speeded up, bumping fast across endless advancing walls of enemy waves.
About ten minutes in and the lake got even rougher. Huge waves battered the front of the boat and I was forced to slow right down. The clouds above me darkened and the boat started to be chucked about like a piece of driftwood. I suddenly got nervous. What was I doing here? There was nobody about to help me if I got tipped over, and the water was ice-cold. It would definitely get me before Ogopogo did.
I remembered a story that I'd read in Arlene Gaal's book, about the local Indian tribe. They would always take a small animal out with them in a canoe so that, should the monster they called the N'ha-a-itk whip up a storm, they could throw the poor thing overboard as an appeasing sacrifice. I'd brought nothing with me, not even a sandwich. There'd been a McDonald's on the edge of Penticton: maybe I could have could have sacrificed some chicken nuggets. (The added problem here being that I'm not convinced those are actually made from real birds.)
To my right, the landscape looked rather nightmarish. Back in 2003 a huge fire laid waste to the forest. Now the stark grey rock is littered with the burnt skeletons of dead trees. It was crazy but I was starting to get very spooked. I put the radio on to try and calm myself down a bit but I couldn't get anything except white noise that only increased my self-imposed paranoia. The lake was crazy rough now and I tried to hug the barren, burnt shoreline to get some calmer water. The cliffs loomed over me like predatory giants and I got really freaked out by a weird noise. It sounded like howling – evil howling. Then I realized that it was just one of the ropes holding my canopy. It had snapped at the attachment and was now juddering in the wind and making an odd sound – odd, but not howling and certainly not evil. I looked down into the water: my depth finder told me that it was 600 feet deep. I tried to man up and carried on rounding Squally Point, where the lake turned right. As I did so, I spotted my goal: Rattlesnake Island was dead ahead.
I approached it gingerly. It was ridiculous but I was actually quite scared. My heart was racing and my mouth had gone completely dry. Obviously my own sighting yesterday wasn't helping. I got close and cut the engine. The waves had abated a little and I took a good look at the island. This appeared to be a barren piece of terrain, no more than a rock, really, with a lone scraggly tree clinging to it. There were no signs of either rattlesnakes or Ogopogo. I pulled out my iPhone and started filming. This was mainly so I could speak and break the silence that was starting to become a little oppressive.
Ogopogo is (or are) rumoured to live in a cave that leads into a series of tunnels starting just below the island. Sonar scans have shown a large hole down there. Ogopogo enthusiasts claim that this is why no bodies of these creatures are ever found: because they retreat inside to die. I say 'creatures' because there would have to be several. For something to exist for so long in this lake it would have to breed, start a family, get a mortgage . . .
I kept filming and started shouting, 'Hello, Ogopogo!' at the top of my voice. I had now clearly lost my mind. I gunned the motor and decided to try to navigate the very narrow channel between the island and the black, burnt mainland. This was a really stupid thing to do, as I had no charts on board and the depth finder would tell me about jutting rocks only when it was too late. I went for it anyway and an invisible current immediately caught the vessel and powered me through. I held my breath. I really didn't want to hit anything. The idea of having to get into the dark, cold water beneath me was not pleasant. I was immensely relieved when I got through and was back in safe waters. I felt like I was in some weird episode of _Scooby Doo._ I laughed out loud but it sounded a little hollow on my own. I suddenly longed for human company and started to head back to Penticton. As I rounded Squally Point the lake suddenly calmed itself. I zoomed away fast from Rattlesnake Island and kept to the western shoreline on the way back, which is populated and felt safer than the barren eastern shore. Finally I spotted the two tall buildings behind the casino on Penticton Beach and breathed a sigh of relief. I had survived my first ill-prepared expedition on Lake Okanagan. I tied the boat to the dock and called the boat guy on his mobile. He was down to meet me about fifteen minutes later. We had a little chat as he checked the boat for damage. He told me about the divers who worked on the new bridge that replaced the old floating one between West Kelowna and Kelowna itself.
'A lot of them quit. They said it was scary and that there were some seriously big things swimming around down there. Some say it's a sturgeon but nobody has ever caught one. The visibility is limited but these guys are not spooked easily. At least three I knew quit as they were so freaked out.'
I nodded and laughed as though they were weak idiots. I didn't tell him about just how freaked out I'd been about an hour ago just floating on top of the water. The idea of getting under the dark water (and I love scuba diving) was pretty unthinkable to me.
I said my goodbyes and got in the car. As I drove my phone rang. It was the aunt of an acquaintance back in London who had grown up in the Okanagan. I'd emailed her saying I couldn't find a boat to rent and she'd contacted her family. Now they were offering to take me to Rattlesnake Island on their family boat. I was too embarrassed to tell them I'd found a rental guy and had just been. So I agreed to meet them at Peachland marina at three-thirty that afternoon – I was going back out there . . .
Peachland is a pretty little community right on the lake directly opposite Rattlesnake Island. There are actually two marinas, one rather grandly calling itself a 'yacht club' and the other seemingly a little less exclusive. I parked my car and sat waiting. At almost exactly three-thirty I saw an old speedboat with two men in it zoom towards the locked marina. Since there were no other boats on the lake I presumed these had to be my guys. I peered at them through the fence and one shouted, 'You from London?' I nodded and they told me to come down. I indicated that I couldn't and so they told me to walk down the road towards the beach where they'd pick me up.
As I clambered aboard the old speedboat I noticed it was covered from front to back in weird blue carpeting. Sort of the nautical equivalent of the avocado bathroom set. The guys introduced themselves: Al and Kevin. They were both in their mid-fifties and very friendly. Al had been born in the valley and Kevin had moved there in 1985.
They had absolutely no idea why I was in the Okanagan. Al had received an email from London asking if they could help me out, so here they were. I filled them in on my mission and they both started laughing – in a good way.
Another dramatic change in the weather meant it was now a beautiful sunny day. The sun glinted off the water and the lake was almost enticing.
I asked Al and Kevin whether either of them had seen Ogopogo.
'No, but I've seen the Sasquatch. Me and my son saw one on a hunting trip only an hour and a half from here.' Kevin looked serious for a second. I asked him what the Bigfoot looked like.
'He looked like a big drunken Irishman – about eight feet tall, covered in red hair with a fast lolloping walk. He just stared right at us for a good twenty seconds and then made off into the trees . . .' This was promising: I felt these guys wouldn't think I was mad.
I asked them about Ogopogo again. Did they know anybody who had seen anything? They'd both seen big wakes in the lake but not 'the humps' – though both of them knew people who had. Kevin started talking again.
'Anyone local believes in Ogopogo. It's the fucking Albertans coming here who all poo-poo it. The thing is, nobody's scared of Ogopogo. It's not a monster; monsters kill people. It's a creature, a USO [unidentified swimming object], but not a monster.'
I nodded and looked at the fast-approaching Rattlesnake Island. It was unbelievable. In the bright sunlight it looked like a totally different place. The water lapping around it was crystal blue and sparkling. It was lovely. We motored around it, just as I had that morning.
'Do you want to see the pyramid?' asked Kevin.
'The what?' I said.
'Wait and see.'
We beached the boat in a perfect little harbour hidden behind two protruding rocks. I hadn't noticed it earlier. We hopped off the boat and clambered up the steep slope towards the island's summit. As we climbed I spotted the remains of something man-made. It almost looked like an overgrown crazy-golf course. It couldn't be, of course. This was the forbidden place, the sacred home of the monster. How could there be a crazy-golf course here? We got to the top of the island and Al was standing on a big stone pedestal. This was clearly man-made. I asked him what it was.
'It was the pyramid,' replied Al. 'Eddy Haymour's pyramid . . .' He was smiling at me.
'Sorry, I'm being thick, but I don't understand what you mean. Who's Eddy Haymour and why are there the remains of a pyramid and a crazy-golf course here?'
It turned out that back in the early 1970s a Lebanese man, Eddy Haymour, moved to the Okanagan Valley. He had emigrated to Canada and had done rather well setting up a chain of barber shops in Edmonton. He'd caused quite a stir, however, in the white-bread Okanagan community because he was, to say the least, quite a character. After a little while living in the valley Eddy noticed Rattlesnake Island and decided to buy it and turn it into a Middle Eastern theme park. Al started to tell me about the things that Eddy had either planned or actually built for the park and I just didn't believe him. He suggested that I try to find a copy of Eddy Haymour's life story, _From Nut House to Castle._
I later managed to get a copy from a second-hand bookstore in town. Here is Eddy's description of what he wanted for the island:
On the four-point-five acre island would be great landscaping with beautiful gardens. There was to be a forty-six-feet long, twenty-six-feet high concrete structure in the form of a camel with a hollow stomach where thirty-nine flavors of ice cream would be served. You could peek out of the windows of the camel's eyes, music would come from his mouth and the garbage from his tail. The washroom would be in the legs. All the cultures of the Middle East would be represented; India by a miniature Taj Mahal; Kuwait by fountains; minarets would depict Saudi Arabia; and there would be a large pyramid for Egypt. In a huge tent Middle Eastern films and other entertainment would be presented. Kids could go around the island on ponies led by Arab storytellers, or in a chariot pulled by a white horse. There would be a toddlers swimming pool with babysitters and lifeguards. I figured I needed something familiar to Canadians so I designed a unique miniature golf course, all landscaped and each hole relating to an Arab landmark. I wanted to have a small submarine to take kids down to the underwater cave, home of Ogopogo . . .
'Did it actually open?' I asked Al.
Al laughed and said that the story got even weirder. The theme park did half open – for one day. The pyramid and the camel and various other bits were built, and locals came over for the grand opening, but the local British Columbian government then stepped in and closed Eddy's island down. He had clearly rubbed some important local people up the wrong way. They tried to change the zoning for the island, refused him ferry permits to take people over there and then challenged his plans for sanitation facilities. Eddy was left with a lot of debts and no way to pay them off as it became clear that officialdom was never going to let him open the island properly. Things spiralled out of control. Eddy's wife left him and then the government inserted an agent provocateur to make friends with Eddy and get him to make threats against the government. Eddy was thrown into a nut house and the government forced him to sell them back the island at a tiny percentage of its actual value.
I couldn't believe this kind of thing went on in straight-laced, well-behaved Canada.
'It gets even worse . . .' said Al.
It was eventually agreed that Eddy Haymour would be released if he left Canada. He went back to Lebanon where, clearly, his resentment at the way he had been treated by the country he'd been so proud to make his home – and become a citizen of – boiled over. Lebanon was just starting the slide into civil war and there were armed factions all over the country. Eddy got together a couple of his cousins and forcibly took over the Canadian Embassy in Beirut, holding thirty hostages for three days. Following slightly panicky negotiations, Eddy released the hostages and, unbelievably, was allowed back into Canada, where he spent ten years fighting his case through the courts. He eventually won; Eddy was awarded damages and it was admitted that the government had behaved appallingly towards him. Eddy never got Rattlesnake Island back, though. He built a 'castle' on the shore opposite the island, which he turned into a hotel. However, he never forgot his island and eventually had a thirty-foot statue of himself built and placed outside his castle pointing at it.
I longed to know what had happened to Eddy's statue when he died but neither Al nor Kevin knew its whereabouts.
Back on the boat we cruised around Rattlesnake Island again and checked Squally Point. This was the place where the local Indians claimed Ogopogo was most often sighted. We saw nothing.
Later Kevin and Al dropped me off at the little beach. Back at my hotel I sat on my balcony with a bottle of Sumac Ridge and stared at the lake. I gave up when the bottle was empty and it started to get dark.
The following morning I checked out and drove to the big city: Kelowna. My appointment with Arlene Gaal was at two-thirty that afternoon. I wasn't quite sure what to expect but she definitely believed in the beast and was sure to have something to say.
With the morning to kill in Kelowna, I decided to wander around and get my bearings. I ambled down the main street and ended up at the lakeside park where the statue of Ogopogo was. On an adjacent patch of grass, facing a raised concrete step, was gathered a small crowd of people I don't think it would be unfair to describe as shabby. They were all listening to a black man wearing a long leather trenchcoat and dirty dreadlocks who was shouting into a microphone. Intrigued, I approached the scene. Several people were carrying placards. One of these read, 'Marx said there'd be days like this'; another said, 'Decolonize the valley'.
For some reason it really irritated me that the word 'valley' was misspelt. I wandered up to the man in a gasmask who was carrying it and pointed to the sign.
'It's spelt wrong,' I said, smiling.
'What is?' came the slightly Darth Vader-esque voice behind the mask.
'"Valley" . . . You've spelt it with three _ls_ and it should be just two.'
'Go fuck yourself,' said the gasmask.
'Only trying to help,' I replied, edging away.
About twenty yards away was another gentleman in a gasmask. His sign read, 'Capitalism – stop that sh*t'. However angry he was, the Canadian in him had forced him to asterisk the _i_ in shit.
I listened to the speeches for a while. It was only after about ten minutes that I worked out what this was all about. It was an Occupy Kelowna demonstration, taking its lead from the Occupy Wall Street demos in New York that were being reproduced all over the world. There seemed to be a general dislike of plutocracy, the rich, CEOs, bankers . . . Basically the usual suspects.
Some guy stood up to talk about some new miracle energy that could power his home. Another guy got up and started going on about the postal service and how they should all go on strike. Thankfully the stench of patchouli oil in the air was thick enough to mask the omnipresent BO.
The Rasta, who seemed to be the MC of the event, was back up speaking.
'Guys, anybody who wants to speak – anybody – please don't be shy, just get up here and say your piece. This is what it's all about.'
Nobody moved and I couldn't resist. I found myself walking past a huge earth mother who was clutching a Mega Slushy as though her firstborn. I got up on the stage and grabbed the microphone. I raised my right fist and shouted: 'Greetings, comrades . . . !' Everyone shouted greetings back. This was fun.
This isn't exactly my bag . . .' (I tried to sound like someone at Woodstock and was briefly tempted to warn the motley assembly that there was some bad acid going round.)
'I'm over here from the UK to look for Ogopogo and . . . I was wondering what you guys thought? Anybody here seen the monster? Can I see by a show of hands who thinks he exists?'
The crowd turned in a second. They started booing and one guy shouted, 'Fuck off, you stupid asshole!' It seemed that asterisks applied only to written profanities. A small Asian postal worker grabbed the mike and told me to get lost . . .
I left the stage and felt the disapproving stares of everyone around me. I even got half shoulder-barged by a man holding a sign saying, 'Eat the rich'. I decided to move on.
After walking around town for a while I had lunch in a Japanese restaurant opposite the demo so that I could people watch from safety. A woman in a Beatles cap was now on stage playing a guitar and screeching loudly. Her song seemed interminable and even the protestors seemed relieved when it was over. She, however, was nothing compared to the man in wheelchair who followed. He couldn't get up on the stage so the Rasta brought him the microphone and then told everyone to gather closer so that they could all see him. There were about 200 draft dodgers littered around the park. I reckoned if the police swooped now they could end crime in the valley in one go.
The wheelchair man was a very spiritual fellow. Rarely had I heard more bollocks spoken. He urged the surrounding crusties to close their eyes and imagine themselves to be in a happy place. Then he started chanting some mantra in cod Tibetan. It was embarrassing. To try to drown him out I asked my Japanese waitress whether she had heard of the Hibagon. Sadly, her English wasn't that great and she thought I was ordering something off the menu. It all became quite confusing.
After lunch I drove north, away from the lake and towards the mountains where, so my sat-nav promised me, Arlene Gaal lived. As I drove further and further inland it dawned on me that I'd rather expected her to be on the lakeshore with cameras trained on the waters. When I eventually got to her place I realized you couldn't even see the lake from there.
She greeted me at the door to her home, a little white puppy yapping at her feet. She was a sweet little old lady and had set up a table on which loads of photos were laid out. Some of the snaps were from strangers who'd sent them to her; others were ones she'd taken. She had become a focus for Ogopogo sightings. People who didn't know what to do with them were always directed to her. I really wanted to tell her about my sighting but thought I should be polite and wait a bit. I looked through the photos. A lot of them could easily have been freak big waves on a calm lake but there were several that were not so explainable. To me what was convincing was the sheer number of unexplained sightings, especially by people who did not want their names released for fear of ridicule. I'd assumed most people would be publicity seekers using the opportunity to get on the television. Far from it: Arlene said a lot of people were very reticent to discuss their sightings because they didn't want to be laughed at.
This was my first meeting with a cryptozoologist and she didn't seem that strange – but I noticed that whenever we talked about Ogopogo a steely determination appeared in her eyes. The 'Folden film' in 1968 had been what started it all for her. Sawmill worker Art Folden was driving along the lake-shore when he noticed something strange in the water. He pointed it out to his wife and they stopped the car and Art got out his 8mm cine camera and started filming. The object was diving in and out of the water so Folden, being aware that he didn't have much film, started to shoot every time the thing reappeared. Eventually it swam away from its initial position quite near the shore and disappeared into the deeper waters in the middle of the lake. According to Arlene this was still the best footage ever taken. She said that on the lakeshore road just near where Folden took his footage there used to be an official sign that read:
OGOPOGO'S HOME
Before the unimaginative, practical, white man came the fearsome lake monster n'aavit was well known to the primitive, superstitious Indian. His home was believed to be a cave at Squally Point and small animals were carried in their canoes to appease the serpent.
Ogopogo is still seen each year – but now by white men.
It seemed quite a patronizing sign and the government who originally put it up clearly felt the same, as it had now been removed.
I told her that I'd assumed the locals would have really used the Ogopogo story to attract tourists, but there was almost nothing visible in the valley except for the statue in the port.
She said that she loathed the statue because it was 'stupid and Disneyesque'.
I told her what hotel I was staying at and she said that there had been a great sighting from there. She advised me to sit on my balcony and watch the lake. If it hadn't been for my sighting the other morning using this 'method' I might have been more dissatisfied by her suggestion.
I'd been hoping that she might have more scientific methods for me to try. I suppose I'd always thought of monster-hunting as being a bit more exciting than just sitting staring at a lake. I wanted underwater cameras, sonars, submarine trips . . .
With my sighting in the bag, however, I was very happy just to show her the footage on my iPhone and wallow in the glory She watched it without saying a word but her eyes were sharp and focused on my little screen. When it finished she looked up at me and smiled.
'Looks like you've got yourself a sighting . . .'
I secretly wondered whether I'd make it into her next book. I sat back, hoping for something to now happen. Maybe the international news media would start swarming in? Possibly I would be asked to tell my story to packed amphitheatres? Whatever, I was sure that Arlene would know what to do. She did nothing. After some polite conversation she offered me some tea and biscuits. It was all a bit of a let-down.
I said goodbye to Arlene and drove away a little disappointed. I wasn't sure quite what I'd expected . . . But I hadn't got it. Once back at my hotel I sat on my new balcony and watched the lake.
I watched the lake for quite some time until I started to get bored of watching the lake and wandered downstairs to have a meal. I ate some duck with a bottle of local wine. It was exceptionally good wine. I asked the waitress where the winery was and she told me that it was fairly nearby, on a hill 'with an amazing view of the lake'. I figured if I was going to have to stare at the lake then I should do it from a winery rather than my hotel balcony. I thanked her and said, 'I saw Ogopogo yesterday.' She looked at me blankly and I didn't pursue the matter.
Therefore, the following morning, I found myself driving through the plush Mission quarter of Kelowna until the road started to climb out of town. Soon I was high above the lake in front of the Summerhill Pyramid Winery. To my right as I drove in was a gargantuan grey pyramid overlooking the lake. What was it with this valley and bloody pyramids?
It was an absolutely gorgeous day and I stood on the terrace overlooking a vast expanse of the lake. I could clearly see the bridge the scuba divers had resigned from working on after spotting 'large objects' down there. It was a curious design: solid, flat concrete blocks were set in the water from both banks. In the centre two arches in the shape of humps rose into the air; these allowed boat traffic through. It basically created a narrow funnel in the middle of the lake. For Ogopogo to travel from north to south, or vice versa, he'd have to swim beneath these two arches. Surely science was at a stage where two motion-sensor cameras could be placed underwater? I supposed it was money: who would pay for something like that? A monster-hunter, that's who, so technically me . . . I sat on the terrace of the winery and gorged myself on the beautiful view and a Mimosa.
I got chatting to the waitress and asked whether she'd seen Ogopogo. She said no, but she had a friend who'd seen it -although he wouldn't talk to media because everyone would laugh at him. She said she tried not to think about Ogopogo when she was swimming in the lake. A manager approached my table and asked me where I was from and whether I would like a tour of the winery. Why not?
Two minutes into the tour I remembered why not. I'd already promised myself I'd never go on another winery tour ever again. They're all identical and incredibly dull. Nobody cares how the stuff's made – just pour some into a glass and get on with it. I've always found myself trying to ask intelligent-sounding questions that I couldn't really give a shit about. It's weird: people don't go on tours of biscuit factories or tuna canneries (although both would be more interesting), so why do we go to wineries? The answer is that we all think we're going to get free wine under the auspices of 'tasting' – but so what? Are we really that cheap?
All this was going through my mind as I was shown a _kylini_ , an old Indian-style home like a huge underground yurt. We descended some steps into the cavernous room with a chimney in the centre for the smoke from the fire to escape through. There were a couple of teens with dreads cleaning up after an event the previous night.
'Was it a wine-tasting?' I asked.
'No,' replied my guide, 'a witchcraft healing session . . .'
I nodded like this should have been obvious but started to wonder whether I had stumbled into some New-Age vacation retreat by mistake. As we climbed up towards the pyramid my guide told me that it was exactly an eighth the size of the Great Pyramid of Giza. I've been to the Pyramids and, looking at this edifice, I privately doubted her claim but I kept schtum. My guide told me (though it was rapidly becoming obvious anyway) that the owners of this winery were hippies and very spiritually inclined. All the wine made on the property had to be stored for a certain time in the pyramid as 'history suggests that pyramids have magical powers over liquids'. The owner claimed to have blind-tested people on two versions of the same wine, one that had been 'pyramided' and one that hadn't. Ninety per cent supposedly preferred the pyramided wine. I didn't really know what to say. I just did my nodding thing and wondered how long it would be until I could get some free wine.
'I believe this is the only winery in the world where the wine is stored in a pyramid . . .' said my guide.
I wasn't going to argue but . . . So fucking what? I tried to steer the subject away from New-Age bollocks to something a lot more real: Ogopogo. I asked my guide if she had ever seen it. She said that she hadn't but her dad was a firm believer. He'd been out on his boat as an eighteen-year-old when Ogopogo surfaced right in front of him. He saw two huge humps and massive water displacement. It lasted for about thirty seconds and then it was gone. She said that he was loath to talk about it with strangers but all the family had heard the story many times.
We left the pyramid and got to the tasting room. I knocked back about seven glasses of sparkling wine in quick succession and started waffling on about how pyramidic it all tasted. She could tell I was taking the piss and it soon became clear that both the tour and the free wine were at an end.
All this fizzy stuff had gone to my head a touch so I decided to go for a short walk. I headed off down a quiet little road that snaked south just above the lake. There was hardly anybody about and I thought I could find a spot to sit and watch the lake for a while. After about five minutes I spotted a narrow turn-off towards a viewpoint, the edge of a tall cliff with a perfect panorama of the lake. There was nobody about except for one man, sat alone on the edge of the cliff on a collapsible camping chair. He had a flask of something by his side and was scanning the lake with a medium-sized pair of binoculars. I sensed a kindred spirit. I wandered up to the edge of the cliff and looked out as well. He noticed me and we nodded politely at each other. Emboldened, I approached him.
'Looking for Ogopogo?'
I tried to stop the moment I started saying it, realizing that, should he happen not to be familiar with the legend, this would sound very much like an offer of some rather specialist gay sex.
He wasn't familiar with the legend and suddenly looked very panicky. He replied defensively.
'No, I don't want anything like that . . . Please go away.'
Mortified, I tried to explain that I was talking about a lake monster and that I wasn't some cliff-top cruiser but this just made things worse. I eventually slipped away after forcing myself to stay and survey the lake for a couple of faux-nonchalant beats while trying to look really relaxed. He watched me walk away while shaking his head in clear disgust.
As I walked on downhill I laughed to myself at the absurdity of the situation. A bit further down the road there was a sign for a car park: 'Cedar Creek Beach'. Just beyond it was a little pebble peach and the water actually looked quite inviting – the sun sparkled off it and there was nobody around. The water was reasonably shallow but I could see where it dropped away about 100 yards offshore. The guys who'd taken me out to Rattlesnake Island had mentioned that Kelowna was actually built on a rock shelf that jutted out into the lake. I had a strong urge to swim. I was a little freaked out about swimming in these waters but something inside me wanted to try to conquer my fears. Also it was really quite hot and I knew I could do with a cool down. I could see the bottom and it was a mixture of seaweedy-type stuff and big round pebbles. I decided to go for it. I didn't have any swimming trunks or towel with me but a 'what the fuck' feeling enveloped me and I felt adventurous. I looked around: both the car park and the beach were totally deserted. I stripped off naked, took one last look around and headed off towards the water holding my left hand over my privates. I gingerly stepped on to the beach and started walking into the lake until I was about knee-deep. It was crazily cold and I realized that this was going to be a very quick dip in and out but I was determined to at least submerge myself once in Ogopogo's home. I stepped forward again and suddenly felt a searing pain in my left foot. I'd stepped on something incredibly sharp – I don't know if it was a piece of broken glass, a can, or whatever, but it had made a huge cut in my foot and I was in incredible pain. I screamed blue murder: _'Fuck . . . !_ Fuckity fuck . . . Fuck fuckity shitting fuck!' Screaming obscenities made me feel a little better.
My hand had come off my freezing privates and I was now dancing about in the shallow waters with my hands on the side of my head trying to somehow compress the pain away. The sound of my screaming echoed around the beach and bounced off the tall cliffs around me. It was only after about twenty seconds that I happened to look up. The man on the chair who'd presumed I'd offered him specialist gay sex was now standing on the edge of the cliff looking right down at my naked form hopping and screaming in the shallow waters of Lake Okanagan. Our eyes met for a second and, even at that distance, I could sense a mixture of withering pity and disgust. It was useless trying to explain.
I bolted back to the car park, grabbed my clothes and legged it. I eventually found cover in a little copse of trees, got dressed and tried to wrap my T-shirt round my bleeding foot.
This was turning into a stressful Sunday. I was bored of hotel food so decided to head into town for dinner. Sunday nights in Kelowna were _not_ rocking and there wasn't that much open. Eventually I opted for a chain place called the Keg. It was semi-buzzing and did obese portions of steak and seafood. At the back of my mind I was worried that I might bump into my cliff-top friend – and if I did, I wanted to look as manly as possible, downing pints, eating raw meat and talking to loggers . . . Actually, talking to loggers sounded a bit weird as well.
So, I was sitting at the bar nursing a pint and watching Canadian football. For some unfathomable reason this is slightly different from the American version so they can't play each other. The Canadian version has twelve players on the pitch, as opposed to the American eleven, and they have only three downs per possession whereas the American game has four.
Not that this really matters: Canadians are all about ice hockey anyway. Kelowna's ice-hockey team is the Rockets and they have Ogopogo as their mascot. The team was originally from Tacoma, just below Seattle, but they were surrounded by big cities so they moved to Kelowna for more fans. They kept the name Rockets because Tacoma was where Boeing made rockets but they adopted Ogopogo as a mascot to incorporate some local colour.
But I digress. I was at the bar in the Keg when a face suddenly came right up to mine.
'Guess who?'
It was a girl and I genuinely had absolutely no idea who she was.
'Hey, how are you?' I said, desperately trying to work out who she might be.
'How weird is this? You doing the rounds about town?'
Suddenly I clicked: this was the waitress from the winery and it looked like news of my anal cruising hadn't yet spread too far.
She moved on to join a table of friends for dinner and I returned to my pint. Then Krist Novoselic walked in. I first noticed him because he was freakishly tall – around six feet seven, maybe eight? He had put on some serious poundage since his skinny youth noodling on the bass in Nirvana but it was him all right. I was certain. He was with a girl much younger than him. She was pretty and blonde with a weird name like Skylar. I know this because the barman went a little gushy and asked Krist how he was doing. Krist nodded and introduced Skylar. The barman totally ignored me from then on and just talked to Krist. He asked Krist if he was playing at the moment. Krist nodded and mentioned a couple of gigs. Did Krist want something to eat, wondered the wide-eyed barman? No he didn't; he and Skylar were just having a drink and then going to see a friend in a band. I liked Krist. Krist seemed nice.
I moved into major eavesdropping mode. I love listening to other people's conversations, especially when those people used to be in Nirvana. It made total sense that he should be here. Seattle is only an hour or so away from Kelowna and lots of famous people have houses here on the lake: Arnold Schwarzenegger, Wayne Gretzky . . . And, clearly, Krist Novoselic from Nirvana. I pretended to play with my iPad while I listened in. Krist was telling Skylar, clearly a musician herself, that he was really impressed with her playing.
'You know the chords,' he said, smiling at her, 'but now you need anticipation.'
Skylar looked at him adoringly and Krist continued with his master class. Krist told her that when he saw a hand move over a fretboard he instinctively knew what was going to be played, allowing him to kick in bang on time, not a millisecond late.
'But that only comes with years of playing,' he said.
Skylar lapped it all up. I tried to think about what I knew of him – I thought he was Croatian in origin and he started the band with Kurt; he fought with Courtney Love over the Nirvana legacy and he was pretty political . . . Oh and he once hit me over the head with his bass guitar.
This is a true story, I swear. When I was at SOAS in London for my university years I used to help organize bands to play at the union. We once got an offer of a band I loved, Mudhoney, but we also had to take the band supporting them on their mini Sub Pop tour. We weren't happy but said yes, as we really wanted Mudhoney. That support band was, of course, Nirvana, who'd just finished their first album, _Bleach_ , and were relatively unknown. Come the day of the gig, I told them that they had five songs and then they were off so Mudhoney could come on. They ignored me and played on into a sixth song so I pulled the plug on their PA. The band went mental and Krist swung his bass guitar at me, and it glanced off my head. Never mind, though – my job was done: we wanted Mudhoney on, not these grungy losers . . . Whatever happened to Mudhoney?
I looked over at Krist again wondering if I should just bring this up. We'd all laugh about it and he'd invite me along to the gig tonight? But Krist was now busy tongue-sandwiching Skylar so conversation was difficult. Whatever, I needed proof of this encounter. I waited until they came up for air and then brought out my iPhone. It was in a casing designed to look like an old cassette tape so it was quite subtle. I turned the camera on and manoeuvred the thing so that it was pointing to my left – right at Krist. I paused then pressed the button. To my horror the flash went off. I'd turned it on to get a photo of Arlene Gaal in her dark house. Krist and Skylar both looked up with a start and stared at me. I went into panic mode and started fiddling with the phone as though it was faulty. I made the flash go off a couple more times in my face to make it look like I was just an idiot trying to get the camera to work. There was no way I could talk to him now but at least I had proof. I surreptitiously checked the shot and the photo was clear. I wolfed down a New York striploin and then left about the same time as Krist, who towered over Skylar as they walked off arm in arm. I briefly considered following them to their gig but realized this might be a tad creepy. I'd once done this when I spotted Mick Jones from the Clash in my local Tesco in Portobello Road. I followed him all the way to Holland Park and watched him browse through the paperback section in a charity store. I thought I was being subtle but, years later, when _Trigger Happy TV_ was at its height, I ended up at his house with a group of people that, weirdly, also included Kate Moss and Sadie Frost (I know – clang, clang – who dropped those names? But it was just a weird night). Anyway, I'd got talking to Mick Jones and he was bit pissed and he ended up saying, 'You followed me once all the way home from Portobello – I thought you were doing some hidden-camera stunt on me.' I was mortified and slipped out soon after.
I drove back to my hotel and immediately googled a recent photo of Krist. It definitely wasn't my guy. I couldn't believe it. I was angry. What was this fraudulent bastard doing swanning around pretending to be Krist Novoselic? Then I remembered that he'd never claimed to be him. It had been me who'd made that supposition and, not for the first time in my life, I felt a bit of an arse.
The next morning I awoke early and snuck up to the curtains, whipping them open with some force as though I was going to somehow surprise Ogopogo and catch him mid-feed with a red face staring up at me. But there was only a lone duck who proceeded to 'duck' down leaving only his feathered ass wiggling insultingly in my direction.
It was my final day in the Okanagan and I'd agreed to go on a little road trip with Al, one of the two guys who'd taken me out on the boat. He was going to collect me in his enormo-pickup truck and we were going up to Myra Canyon to check out the old railway line. There was little chance of lake monsters in the mountains but Al said it was something that I had to see – and, besides, there was a chance to see cougars, bears and wolverines. I wasn't quite sure what a wolverine was. I thought they were fictional creatures? (In the eighties Brat Pack flick _Red Dawn_ , the Russians invade the USA and the kids from the high school run to the hills and become resistance fighters – calling themselves 'Wolverines'.) Al didn't know how to describe a wolverine but he settled on ROUS (Rodent of Unusual Size). He said it was like a huge chipmunk, the size of a goat with big teeth and long sharp claws. 'They are mean sons of bitches,' said Al.
At the entrance to the trail was a large handwritten sign warning that a bear had been spotted with cubs. 'Under no circumstance should you run away from a bear unless you have somewhere to go . . .' was the very curious advice here.
I'd seen other signs in the valley that suggested you take a bell with you on hikes and ring it frantically should a bear approach. I decided that, should we be faced with this predicament, I'd stand directly behind Al and cower.
The old railway line spanned the entire canyon and used to be used to transport gold from mines in the hills. It was a spectacular feat of engineering that had burnt down in the huge forest fire of 2003. The bridges had been restored and now comprised part of a cycling and hiking trail. We walked and walked and walked. I had no idea how far Al intended to go but I didn't want to look like a wimp. This was a big day for my left foot. In 2011 I broke three metatarsals on a TV show in Argentina and this would be the first big test of my recovery. At the sixth mile I couldn't walk much more and had to sit down and take my boot off. I think Al was secretly quite chuffed that he'd 'broken' me and went a bit easier on me as I hobbled back towards his pickup.
Sadly there were no signs of bears, cougars or wolverines. The Okanagan Valley is not one to easily give up her fierce creatures – real or fictional.
We drove back down to Kelowna, where Al left me to a final spot of monster-hunting. Just along the shoreline from my hotel I'd bumped into a guy who had a boat he could rent me for the afternoon. It had a depth finder _and_ a fish finder. This kind of sonar device could possibly help me spot an unusually large creature in the water beneath me.
Having left a hefty deposit, I roared off over the lake. It was like a mirror: 'perfect Ogopogo conditions', the renter had said just before I set off. Once again mine was the only boat on the entire eighty-mile expanse. I wanted to head north this time and this meant going under the new bridge. I put-putted under the left-hand arch and looked down into the black water. The fish finder was not seeing anything. I headed out into the very centre of the lake. The depth finder told me that it was 356 feet deep. Not bad, but there are areas near Squally Point that supposedly go down to 800 feet.
I turned the engine off and all was silent. I floated quietly on Lake Okanagan, all alone save for a solitary loon staring at this loony cockily. I peered earnestly at the fish finder but it could find no fish let alone a monster. I could see how an obsession with something in this lake could drive a man insane after a while. The more you told people about your obsession the more determined you'd become to prove it so they'd stop referring to you as 'that monster guy', the loony who believed in Ogopogo. Now _I_ was a loony who thought he'd seen Ogopogo, endlessly propping up bars showing people the footage as they attempted to shuffle a couple of stools away. After an hour or so, I gave up. I switched on the engine and turned for home. As I docked the boat I wondered what had happened to Eddy Haymour's statue after his death. Here was an inhabitant of the Okanagan who'd been driven crazy by a different obsession. There was no record of Eddy ever having seen Ogopogo; he'd simply used the story to help him with his brilliantly crazy project. It was funny, two people – Eddy and myself – both born in Lebanon, both fixated by something in this curious stretch of water so very far away from the distant cedars of our homeland. I turned towards the lake for the last time and gave a little nod to both Ogopogo and Eddy Haymour. Things and people like these are what make life so interesting.
### Hibagon
'Godzilla is the son of the atomic bomb. He is a nightmare created out of the darkness of the human soul. He is the sacred beast of the apocalypse.'
Tomoyuki Tanaka
I remember asking my dad whether he'd ever killed a man. Obviously I was asking about the war, and not hoping that he'd suddenly buckle under my interrogation and admit to a string of grisly murders. He flew in the Fleet Air Arm against the Japanese in the Pacific in the last two years of the Second World War. Like many of his generation, he wouldn't talk about that kind of stuff very much; but one night, after a couple of drinks, he let on that he'd shot down a Zero. He hadn't seen a parachute.
At Heathrow Airport, about to fly off to Japan, I felt slightly odd knowing this. My father had passed away just four months previously and to me this trip was something of a connector mission with him.
The departure lounge was, unsurprisingly, stuffed with Japanese, all uber-trendy, some reading cartoon books and most wearing those curious surgical masks. Until recently I'd always thought this demonstrated some sort of national OCD, a Wacko Jacko fear of germs. But then I'd read that this is all a cultural misunderstanding. They're actually worn in politeness: so as not to spread germs. I had a feeling that I would come across a lot of misunderstandings like this over the next ten days. I'd never been to Japan before and already had that heady feeling I get when I'm about to visit a new country.
Having wangled my way into Virgin Upper Class, I slept most of the way – and there is little that makes a traveller much happier than this. I did, however, make the mistake of watching the Steven Soderbergh film _Contagion_ , about a pandemic spreading round the world. The moment the film was over I felt desperately keen to join the Japanese and purchase a job-lot of surgical facemasks. I listened to the latest Kermode/Mayo movie podcast. There was a great interview with the film director John Landis, of _Thriller_ and _An American Werewolf in London_ fame. He had just written a book about 100 years of movie monsters. He talked about how Godzilla was such an obvious metaphor for the atomic attacks on Japan when it was first written, in the early 1950s. The father of the guy who wrote it had actually survived one of the explosions. Monsters feature heavily in Japanese life. Their mythology is chock-a-block with them and a quick trawl through the Internet revealed a veritable cornucopia. Here are some of my favourites:
Aka Manto: a malicious spirit who haunts bathrooms and asks the cubicle occupants if they want red or blue paper
Akaname: a spirit that licks untidy bathrooms
Mujina: a shape-shifting badger
Hikiko: the ghost of a girl who was treated badly by her parents and bullied by her classmates
Ittan-momem: a possessed roll of cotton that attempts to smother people by wrapping itself around their faces
Kasa-obake: a paper-umbrella monster
Sazae-oni: a turban snail that turns into a woman
Zorigami: a possessed clock
I could carry on with a list of more than 200 but you get my drift.
There are a couple of internationally famous monsters that initially seemed perfect for my quest. The best-known of these is Issie, a lake monster who apparently resides in Lake Ikeda at the far southern end of Kyushu Island. However, I found a much more interesting prey. I'd read about the Hibagon, a creature that is supposed to roam the mountains around Hiroshima. A lot of people refer to it as the 'Japanese Bigfoot'. Others, though, say it's some sort of mutant man who survived the atomic bomb dropped there in 1945 by the Americans. This really intrigued me. The Hibagon seemed to be a very Japanese type of monster and the bonus was that I'd also have an excuse to visit Hiroshima itself. My father had flown over the city a day after the bomb was dropped and the experience had affected him deeply. So it was decided like that. I was off to Japan to find the Hibagon.
As the plane started its descent into Tokyo I spotted in the inflight magazine the worrying suggestion that an International Driving Permit is required when renting a car in Japan. I hoped this was bollocks because I'd stupidly left mine at home.
My landing card asked me to give my reason for visiting Japan. 'Monster-hunting', I wrote proudly.
The very first thing I noticed after we'd landed was that the machines that registered your fingerprints and took your photo at the passport desk were framed in an almost childish, Hello Kitty-type cartoon design. This made a nice change from the usual stern-looking welcome.
Once through I caught a train going to Shibuya, the area of Tokyo famed for its crazily populated intersection and where my hotel was. On the train I noticed built-in combination locks so that you could attach your suitcase to the luggage rack. This is a genius idea. Why don't _we_ have that, so I don't have to spend the whole time on the Worst Great Western's Kemble-to-Paddington ride hoping some chav hasn't pinched my bag?
Also, despite the Japanese famously being half our size, there was a lot of legroom, maybe double what you get on the Worst Great Western. All in all, this was a supremely civilized experience. The only slight problem was quite a pungent smell of BO, but this couldn't be blamed on my hosts. Two nearby Spaniards had clearly just spent fifteen hours in economy and were not looking well on the experience.
The train flashed past a mish-mash of bamboo and electricity pylons as we entered the outskirts of Tokyo. I arrived at Shibuya Station feeling very smug. Everyone had told me that getting round Tokyo was a nightmare. Obviously not one of them was an experienced traveller like me. I walked outside and hopped into a cab. I showed the driver my hotel name and he nodded enthusiastically. Strangely, however, my door suddenly opened automatically in a very _Total Recall_ manner. The cabbie indicated that I should get out. I did as he asked and he pointed vaguely in a direction in which I started walking. I had zero idea of where I was going and there were no English street signs. I got to the Shibuya Crossing. I guess if Tokyo has a 'sight' then this is it: Piccadilly Circus times ten with supersized neon. I wandered lost as a lost person in Lostland. I tried to ask a couple of people but each one sent me in a different direction. This was not a good start.
I walked down what looked like Carnaby Street on acid. Alien noise was coming from everywhere. A giant video screen showed a band called the Funky Monkey Babys, while the warblings of what sounded a little like George Michael emanated from behind a window displaying rubber clothing.
I felt big, dwarfing the pedestrians around me, as I trudged through the streets, already Lost in Translation. I spent a good hour and half wandering aimlessly about, hopelessly lost. Every time I returned to Shibuya I could see where I was on my map but I'd then head off in the wrong direction again. Finally, in complete despair, I tried another cab. I hopped in and attempted to pronounce the street address in my best Japanese. Amazingly the electric door stayed shut and we set off. He seemed to know where I was going. Ten minutes later we got back to Shibuya Station, where the first cabbie had refused me entry. We continued on in the exact opposite direction to the one the first guy had pointed me in. The hotel was about three minutes' walk from the station up a narrow lane. I stumbled into reception, deliriously happy. To my great relief the receptionist spoke a little English.
'Mr Jory Your woom is weady It is larger than one you weserve – no extwa charge . . .'
I thanked him profusely and got the key.
I got into a lift that was pitch-black but I managed to find the button for the third floor. The corridors were also barely lit and it took me some time to find my room. When I opened the door I assumed there had been some mix-up in the reservations for a hamster and me, as the room was so small I could barely get in it. I rather longed to see the smaller one from which I'd been upgraded. There was about half a foot between the tiny bed and the wall and I edged myself down to the window and dumped my bag. I then edged back and squeezed into the bathroom, which resembled something I'd seen in a doll's house. I needed the loo but this was not simple. I had to keep the glass door open so that I could sit on it and push my legs into the room. There was a frankly terrifying control panel to my side that appeared to show the various methods in which this contraption could give you an enema. I left the panel alone, did my business and then pressed a button from the safety of the bedroom. A jet-like spray fired up from inside the bowl and I thanked the Lord I hadn't been sitting on it at the time: it looked like this process could remove skin. I closed the glass door and decided to deal with all this later.
Downstairs in the restaurant the menu comprised a simple list of sentences in Japanese. It was giving very little away. I took pot luck and pointed at one for the waitress. Three minutes later a basket of goodies was placed before me. Among the various little bowls of stuff I immediately recognized tofu and possibly some cabbage, but the rest – besides an extremely fishy fish in the middle – amounted to total wild cards. I tucked in anyway. It was all quite edible and I felt pleased with myself for being so 'local'. I sat back and looked around the restaurant. The Japanese family next to me were chowing down on bowls of spaghetti Bolognese onto which they were shaking Kraft 100% Parmesan. Opposite me a spotty fat woman was also devouring some spaghetti Bolognese. It looked rather good.
I headed out of the hotel and back towards Shibuya Station. Now I had my bearings I was very happy: I have a pretty good internal compass, which was now set to magnetic north. Outside the station hundreds of people were crammed in to the outdoor smoking area, puffing away like it was going out of fashion. Here also stands a statue of Hachiko: the dog who sat and waited every day for his master to return, not understanding that he had died. I think Richard Gere made a film about it . . . (Or was that a gerbil? I forget.).
I crossed the Shibuya Crossing again and spotted a Starbucks with a perfect view over the pedestrian maelstrom. Upstairs the window seats were all jam-packed apart from one section that was curiously empty save for one woman. I sat down and started snapping away at the crowd below. It really was a ringside seat for people-watching. Then I realized why the section was empty. The lone woman next to me was scribbling furiously in a book in red ink (always a giveaway). She looked up, spotted me looking at her, and let loose a stream of furious invective. Froth and spittle appeared around her lips. I had no idea what she was saying but it was clearly not 'Welcome to Japan, stranger!'
She started to get quite violent and mock-punched me a couple of times while screaming at a new level. I held my ground and pretended to ignore her, which was exceedingly tricky as the whole place was now ignoring the people river on Shibuya Crossing and watching the loon and me.
I wondered whether I'd maybe broken some social code, committed an awful cultural faux pas. I took a quick look at the loon. She was quite attractive but now dribbling all down her face, which somewhat offset the look. In the end I gave in and moved to a vacated spot in a more peaceful area. I was then free to watch a young American approach the crazy woman and ask her if the seat next to her was free. I sat back and enjoyed the fireworks . . .
Below me a camera crew was filming four pretty young girls, all clad in leopard-skin and, on cue, sauntering across the crossing trying to look carefree yet sassy. They had the look of a band. Maybe they were the female equivalent of the magnificently named Sexy Zone, a boy band whose huge poster looked down on the crossing? Curiously, it appeared that – unlike anywhere else in the world – the Japanese give photography a miss at home: the only people snapping away here were tourists.
I walked out through a music shop where I discovered some more great names for groups. I was particularly pleased to see that Bump of Chicken had a new album out. So did Heartful Voice with Tackey and Tsubasa. I had a listen; _Hurtful_ Voice would have been more apt. Sadly I was unable to find any of Bump of Chicken's work but I googled them later and found out that they were on their seventh album, so no flash in the pan.
Once back in my matchbox, I googled the Hibagon to see where I could start. There was very little information available. The epicentre of Hibagon sightings seemed to be based around Mount Hiba, 125 miles from Hiroshima in the north of the prefecture. I needed to get to Hiroshima and then get a car. I also realized that I would definitely need an interpreter. I turned to Twitter for help. I asked my followers, a random bunch, for help finding a good, interesting interpreter in the Hiroshima area for five days. Within about five minutes someone had Tweeted me that his friend ran 'Get Hiroshima' and could definitely help. I Tweeted them and they gave me the name of a lady called Koizumi who would be available. I emailed Koizumi with my requirements and hoped she wouldn't think I was a nutter.
I lay scrunched up on my bed and watched _Lost in Translation_ , one of my favourite films anyway and absolutely perfect to watch on your first day in Tokyo.
I left the hotel at dusk and went to look at the Shibuya Crossing lit by neon. Tokyo is a city best seen at night, when it gets seriously _Blade Runner._ I entered a lively looking counter-service place. The food was spectacular: pork and noodle ramen with a gratifying excess of chili. Through the window I watched the neon city fizz with energy and loud, bad pop music.
Back at the hotel I checked my emails. There was one from Koizumi, who was very up for being my interpreter and extremely excited about hunting the Hibagon. She wrote that she had contacts in the Hiroshima prefecture with whom she was going to set up interviews. This was very good news. I was asleep before my head hit the midget pillow. The room was incredibly hot and I kept waking up and thinking I was in some weird washing machine. Come the morning I packed fast while standing on the bed before going downstairs for breakfast.
A waitress doing what seemed to be a pitch-perfect impression of Minnie Mouse served me with a bowl of soup and some cold meats. Ryuichi Sakamoto's soundtrack for _Merry Christmas, Mr. Lawrence_ floated hauntingly in the ether.
I had to get to Shinagawa Station to catch a Shinkansen (Bullet Train) to Hiroshima. I looked at the Tokyo subway map. At first it looked as though I might need an Enigma machine to crack it but I toughened up and worked out that I needed the green line to Shinagawa. I was pretty sure that the numbers by each station corresponded to the amount I needed to pay. I pressed 160 yen and put in the money. A ticket was spat out and I was off. On the platform a mass of humanity awaited the train. This was a whole different ball game from the peaceful airport shuttle. For a moment I worried that I might be on some private penal line. The cars were full and I mean _really_ full. Little faces sporting white facemasks were squashed right up against the door glass and it looked very tricky to breathe. As the door opened a tsunami of people swept on to the platform, all elbows and shoves, the usual Japanese politeness totally abandoned. I valiantly fought my way in and blockaded myself into a corner using my suitcase as a perimeter wall. An official in white gloves was physically pushing people into the carriage until nobody could move either arms or legs.
I was fortunate in being a good foot taller than anybody else so I could see over the chaos. On the walls of the train ads were playing on a multitude of screens. The first one was of a smiling woman sitting on the loo. She had what appeared to be a mini-basketball net positioned in the sink and she was bouncing a ball and trying to get it into the net. When she succeeded, she went totally mental and seemed to forget that she was on the loo. Next up: what appeared to be an ad for an erotic barbershop. Western men were portrayed having their faces massaged by very beautiful women wearing barbers' smocks. No haircutting seemed to take place and the camera lingered over the girls' bodies as they floated just above the man's submissive face. At the end, the man walked out looking very happy and a beautiful blonde in the street stopped and stared at him lustfully.
It reminded me of being in Moscow filming a scene in which a naked woman gave me a haircut. The entire crew was so British and awkward about the whole situation. It soon became clear that we were in a brothel and girls lined up outside waiting for the filming to finish. When it did, we all skulked out giggling like schoolboys. I don't think we did the sexual reputation of the British any good.
The final ad was brilliant. It was for a new Nintendo game called _Monster Hunter._ The Japanese voiceover kept shouting 'Mooonster Hunttter' over and over. I looked around but it was clear that nobody knew that they were in the presence of the world's greatest monster-hunter.
At Shinagawa I fought my way on to the platform and joined yet another tide of humanity cascading into the main station. I found the Shinkansen counter: I needed a Nozomi Shinkansen as this was the super-express. Amazingly I seemed to be able to purchase a return, with reserved window seat, mainly through the art of drawing. My artistic skills being on a par with those of my cat, Colonel Mustard, I hoped that I was OK. Twenty minutes later and I was standing confidently on platform 21. Every car had its own gate and I was at the far end, car number 16. I filmed a Bullet Train coming in. They really are beautiful: sleek, impressive pieces of engineering, probably one of the most iconic sights in Japan. I had to admit to being quite chuffed with how easy it had all been. Again, I'd been warned that it was infernally complicated – but not for me. Maybe I could become a sort of unofficial Shinkansen expert? Shinkansen-San, they would call me. My train arrived and I got on, taking my reserved place by the window next to two Japanese businessmen in Wacko Jacko masks.
We were back in the world of massive legroom and comfort. I was very happy. I was on my way to monster-hunt in Hiroshima, somewhere I'd wanted to visit since I was a boy. It was all too perfect. The train stopped and everyone got out. This was even better. I had the whole carriage to myself. I sat and waited. Three women in pink uniforms got on and started cleaning. One of them shouted at me and indicated the platform in the very same fashion as the Tokyo cabbie had done.
'Hiroshima?' I asked.
_'No!'_ she screamed.
I got my bags and got off – mortified. I'd got on the wrong train and was now at Tokyo Central, the world's most confusing train station. What a bloody idiot.
Fortunately the Shinkansen system is infinitely more efficient than both Worst Great Western and me and I was quickly able to get on one heading the right way at the speed of a bullet.
Twenty minutes out of Tokyo and I spotted Mount Fuji. It was a crisp, sunny day and the summit was absolutely clear, which is apparently rare. I saw this as a good omen. I was going to get a second monster spot: first Ogopogo and now the Hibagon. I was feeling confident.
The ticket inspector entered the carriage. As she did so she bowed to everyone inside. The woman with the food and drinks carriage did the same. Could you even imagine this on Worst Great Western? In between every carriage was a little tiny glass capsule in which you could go and smoke. It was like a scene I'd done for _World Shut Your Mouth_ when I'd had people in glass cases full of smoke in the middle of a park.
Outside, Japan flashed past my window. For some reason there seemed to be an extraordinary amount of graveyards by the track. Fortunately the Shinkansen had a 0 per cent accident rate, which is pretty impressive. After three hours of silent, ground-level flight we pulled into Hiroshima. As I stepped on to the platform birdsong was being piped through hidden speakers, which was rather eerie. It was almost as though the city wanted you to know that there was still life there. The Lonely Planet, in its infinite wisdom and clearly used to dealing with idiots, wisely warned visitors that the place was not 'rubble' but 'a thriving city'. No shit . . .
As I walked through the station building I saw a big 'Welcome to Hiroshima' sign, behind which was a garishly vulgar McDonald's sign. It looked like the American occupation was still ongoing. I caught a cab and, fabulously, the driver knew where my hotel was. I checked in and got a room on the fourteenth floor in which I could actually swing a cat. I liked Hiroshima already.
The hotel rented out bikes so I grabbed one and used it to pedal about town. Despite Hiroshima being almost flat the bike was motor assisted, which made cycling a dream. I was careful in the traffic, though – the irony of being killed in a bicycle accident in Hiroshima was not lost on me. I felt pretty certain that the Japanese must have a 'possessed-bicycle' monster – they definitely have one for everything else. Like a two-wheeled version of Christine it would have a mind of its own and propel the hapless rider over the nearest cliff at the first opportunity.
I cycled down Peace Boulevard until I got to a bridge spanning a river and saw the Atomic Dome on my right. This is the remains of a factory built by a Czech architect in 1915. The bomb exploded 600 metres above it and the walls and dome partially survived -unlike 70 per cent of the rest of the city, much of which was made of wood. Eighty thousand people were killed instantly.
I pedalled around the building and then cycled on through the Peace Park and up and down lively alleyways. As in Tokyo, when the sun set and the lights came on, the city really came alive. Everywhere I looked was blazing neon and gaudy lights. Old Japan built architecture of exquisite, intricate beauty. New Japan builds with neon and chutzpah. What looks drab and unexciting in the day becomes an exhilarating assault on the senses by night.
I headed for Okonomi Mura, a three-storey building filled with dozens of little places solely cooking the local speciality: _okonomiyaki._ Although sounding like a name for yet another type of monster, this is a kind of Japanese pancake filled with cabbage, meat, egg and noodles. I chose a spot run by a wizened old lady who looked like she knew exactly what she was doing. She cooked it on a hot steel counter right in front of me. It was phenomenal. I waddled away sated and content in the direction of my hotel. I was still jet-lagged and buzzing from the excitement of possibly 'bagging' another monster. The following morning I would meet my guide, Koizumi, and my hunt could begin.
I awoke at four in the morning and couldn't get back to sleep. I watched a very good Scorsese documentary on George Harrison before going downstairs to find some breakfast.
I really wanted to attempt the numerous Japanese options but most were squidgy, oily-looking things in cubes and my heart wasn't in it. I opted for scrambled eggs but had a green tea instead of coffee to 'keep it real'.
I sat down and looked at my watch. It was 8.15 a.m., the exact time that this city was wiped off the map on 6 August 1945. So far there was not a hint of this traumatic event in the air. The psyche – the 'feel' of this city – was entirely positive. I was curious to find out how a city dealt with being defined by its destruction. The only other place I'd come across like this was Halifax, Nova Scotia: a city that was totally flattened when a French munitions ship exploded in the harbour in 1917. Up until then it had been the largest explosion in history. I'd had the misfortune to be snowed in at Halifax once and, believe me, Hiroshima has way more going for it.
After breakfast I met my guide, Koizumi. She was a lady of about my age and bubbling with energy and smiles. She was very excited about the Hibagon hunt. She said that the locals of Mount Hiba had worried that everyone had forgotten their story so were honoured to have one of the world's foremost monster-hunters visiting them. She had organized a trip there for the following day. Today however, she wanted to show me round her home city. I was a little disappointed: I was perfectly capable of exploring the town myself and was itching to go monster-hunting, for which I needed her help. There was little I could do, though, and I didn't want to cause offence. So off we went.
Minutes later we were on Peace Boulevard again, where I'd cycled the previous afternoon. Koizumi stopped by a clump of trees. They all had little yellow signs attached to them, indicating that they'd survived the atomic bomb. I would never have noticed these natural 'survivors' had Koizumi not pointed them out to me. Maybe having a guide wasn't too bad?
We crossed the bridge and went into the Peace Park, a large area with various statues and memorials as well as a museum all dedicated to the atomic explosion. The museum was utterly fascinating. What particularly grabbed me was the information about how and which cities were targeted. Four large cities had been chosen. The Americans had wanted the drop-site to be cloud-free so that they could analyse what happened. The selected cities were Hiroshima, Kokura, Niigata and Kyoto. However, the then Secretary of War, Henry Stimson, had honeymooned at Kyoto many years before, and had been captivated by its beauty, so when he saw that it was one of the targets he had it removed from the list. Nagasaki was chosen instead. Such are lives changed.
Hiroshima was an important port city and therefore the target for the first raid. Unluckily for the city and its inhabitants, the skies were clear on the chosen morning: the bomb was dropped and exploded 1,900 feet above the city. In addition to the 80,000 people killed instantly, 70,000 more suffered appalling injuries. Whatever my views on what the Japanese did in the Second World War – and the fact that the dropping of the bombs may have shortened the war, and possibly saved my father's life – this is a shocking event to consider. We looked at burnt items of clothing, molten metal and an extraordinary exhibit of the stone steps of a bank with a stone wall behind it. Someone had clearly been sitting on the steps when the bomb went off, because his shadow was burnt into the wall behind him.
I looked at a map that showed the area where the atomic 'black rain' fell. This stretches way out into Hiroshima province. As I've mentioned, one theory about the Hibagon suggests that it's a human who was transformed by the radiation following the explosion. The radiation produced by the bomb here was not long-lasting. Unlike the stuff unleashed by Chernobyl, which has a half-life of about thirty years, almost all radiation in Hiroshima dissipated within six days. Was it possible that in those six days something had somehow been so heavily irradiated that it had mutated into the Hibagon? This kind of story is very common in Japan (Godzilla was 'created' in much the same way) but it seemed pretty unlikely. Maybe it was less spectacular than some glowing green radiation monster? Maybe it was some form of deformed beast, a bastard product of some radiated animal? I had no idea but it was intriguing and I could find out a lot more the following day.
We left the museum and walked around the park. In its centre was a huge bell that you could ring using a pole suspended on ropes. This was a memorial bell for all the victims of the bomb. I pulled back the pole and gave the bell a hefty thwack. The peal resonated deep and long around the park. I stood there for a second absorbing the shock waves and thinking about my father. I wished he'd been a proper dad. Moving on, we reached the Atomic Dome that I'd cycled round the day before. It was right by the T-shaped bridge that had been the precise mark for the bomb, though wind had pushed it slightly off target and therefore the hypocentre was about 200 yards away directly over a hospital. Our atomic tourism over, we hopped on a boat to Miyajima, an island in Hiroshima Bay and home to one of the most famous shrines in Japan. As we put-putted through the city Koizumi told me that I should call her Naoko, as Koizumi was her surname.
It was a gorgeous day so we stood on deck to have a look at the oyster-shucking machines on the bank. In hindsight this was probably a mistake as we went under four or five very low bridges and Naoko had a mini-heart attack every time, thinking we were going to hit them. As we approached each bridge she'd scream, 'Dom-San, get down!' and then pull me to the floor in a slightly hysterical manner. If bridges freaked her out, I wasn't sure she was of the right temperament to go monster-hunting with me.
As we left the river and entered the bay proper the boat roared across the delta to the island. Once docked, we meandered through the streets of the tiny town in the direction of the shrine. The builders had sunk the posts of a huge red gate deep into the seabed so it appeared to be floating. It was rather soothing to see something old and cultural after the shock of the new of my first couple of days in Japan. Naoko wanted to show me the temple. You had to give a financial donation to do so and the first thing we saw when we entered was an entire corridor of colourfully decorated sake barrels. Life is good if you're a monk. We climbed up through exquisite ornamental gardens on the way to the cable car. I wanted to get right to the top of the mountain for a view over Hiroshima. Naoko kept trying to put me off and eventually admitted that she prayed for bad weather when bringing clients here as the climb after the cable car was quite arduous and she was lazy.
Once off the cable car we started to walk. As we did so, Naoko filled me in on what she'd done so far to organize a Hibagon hunt. She told me that she'd contacted the Mount Hiba district council and that they were very excited about our visit as there hadn't been much press interest lately.
The whole Hibagon office is at our disposal,' said Naoko laughing. 'We will meet with the man responsible for the Hibagon and he will take us to the places where there have been sightings. He is happy to answer all questions and he is very knowledgeable. I am very happy we are going to hunt for a monster, Dom-San. This is a big adventure.' She paused for a second with glorious comic timing. 'This, however . . . We don't have to do this . . .' she continued, puffing and looking up the path. I teased her and suggested that she should advertise herself as a 'lazy' guide. Eventually we got to the top. I felt good. It was from this very spot that an incredible photo was taken of the mushroom cloud six minutes after the bomb went off. I'd seen it in the museum. What must the photographer have thought as he raised his camera? It must have seemed like the end of the world. Had the Hibagon, whatever it might be, been created at that very moment? Had it maybe gazed too long at the fiery skies over the city, transfixed, as so many were, by the sheer enormity of what was happening around it?
We were very happy on our summit and stayed there for quite a while. Eventually Naoko took me a down different way because there were several holes in the rock that supposedly had certain powers. One hole, for example, cured 'itchy body' – unless you were 'bad', in which case it actually _gave_ you 'itchy body'.
Back at the bottom, we stopped in a Shinto shrine while Naoko prayed for a while. I stood on an outside platform, smelling the incense and gazing down at my feet – ever the awkward atheist. I have to admit to feeling very calm there. These sorts of places seem to work better the older you get.
We caught the boat back to Hiroshima. I'd done enough 'normal' tourism. It was time to crack on with some monster-hunting.
The next morning I was up early and ready to bag me a Hibagon. But first, breakfast: monster-hunting is always more productive on a sated belly. The dining-room walls were wallpapered with black-and-white photographs of New York. Some of them showed New Yorkers flying the Stars and Stripes and cheering in a ticker-tape parade. I rather hoped it wasn't a war-related celebration and marvelled at the insouciance shown by Hiroshimites towards their destructors.
I met Naoko and we set off to rent a car. Having been told that I needed an International Driving Permit I was rather worried but I needn't have been. Renting a car in Japan was akin to being greeted on a royal visit. Every production of a form or a credit card was met with much bowing and smiling and we were soon sitting in a tiny little Toyota Box (not the real name but it would be apt) while the entire staff of the rental agency lined up to bow as we drove out of the garage. We were headed for Saijo, seventy-five miles north-east of Hiroshima. Naoko had arranged for us to meet a Mr Maeda, who was head of the town's tourist association and also ran the 'Hibagon office'. As with the low-bridge experience, Naoko was a very nervous passenger and I worried for her fingers as she gripped the door handle for dear life. Not that we could do much speed in the Box. I managed to get it up to about sixty miles an hour but that was it.
Saijo is an unremarkable little town at the foot of a small range of mountains, the highest being Mount Hiba. We parked up outside a building in the centre of town and were met by Mr Maeda, an unassuming man who seemed incredibly pleased to see us. He invited us into his messy office where he handed me his business card. It had a cartoon drawing of a rather cuddly-looking creature that Mr Maeda confirmed was the Hibagon. It turned out that he had designed this endearing image to make the Hibagon more tourist-friendly, as the original UMA (unidentified mysterious animal) was not quite so sweet.
We sat around a table and I asked Mr Maeda to tell me the history of the Hibagon. He produced a map that marked every sighting of the beast as well as a photograph of a footprint, taken about seven miles from where we were sitting. As I looked at these he told me the story, with Naoko translating:
On 20 July 1970, very near Mount Hiba, a farmer reported seeing a big ape wandering through his field. The ape measured about two metres seventy centimetres, had dark reddish-brown hair, a big head like a cone and was walking on two feet. They had small monkeys in the area but nothing remotely like this. After the farmer's original sighting there followed about ten more sightings within a month – all fitting the same description. A local newspaper wrote the story and coined the name 'Hibagon' after the nearby mountain. Things then got a bit hysterical, with schoolchildren having to be accompanied to school by the police and a lot of locals becoming incredibly nervous about going out. The story exploded all over Japan and all sorts of weirdoes came from all over the country to hunt the creature. Universities set up Hibagon student-exploration clubs and groups would come and roam the mountains for weekends. The initial panic lasted for about a month but then things started to calm down a little. Journalists came from everywhere and the locals started getting a bit annoyed because Japanese magazines and newspapers started making fun of them.
The local government set up a special Anthropoid Section' and they got a budget to deal with press and inquiries.
Then, in August 1974, a local, a Mr Mitani, took a photo of the Hibagon. He had stopped his car on a road in the mountains when he spotted the beast in foliage at the side of the road. This was the only photograph that had ever been taken of the Hibagon and it kicked off the story again. More sightings were reported and then started to dry up.
The Anthropoid Section was eventually closed in 1975 and the town declared the 'end of the Hibagon', as they'd had enough of all the attention. Despite these protestations, though, soon products as varied as noodles, washing powder and sweets appeared bearing the Hibagon name. Mr Maeda was very much still a believer in the beast but recognized that it was a scary thing that might not really attract too many tourists. This was when he came up with the cartoon version of the animal. This logo was now used on everything related to the town, from the Forest Commission to tourist literature and hiking guides.
I was just going to ask more about a movie called _Dear Hinagon_ that had been shot in the town when the door opened and a very fat man waddled in. He was a journalist from the local paper in Chugoku and he'd been tipped off that one of the world's most eminent monster-hunters was in town. He asked if I'd mind doing an interview and then have him follow my investigations. I agreed and we all decided that we should set off in Mr Maeda's car to check out some of the places where the Hibagon had been spotted.
We drove to the place where a rice farmer spotted the Hibagon. He'd been driving up a remote road when he saw the beast crossing the road in front of him. When it heard the car, it ran off up into the woods above the road. The farmer's description of the beast broadly matched that of the first sighting, but he also said the creature had a vaguely human face.
I got out of the car and looked around, not exactly sure what I was supposed to do about something that had happened forty years previously. It was very unlikely that the Hibagon was still hiding in the bushes above. I was mindful that the local journalist was watching me intently and taking loads of pictures, however, so I felt that I needed to act out the part. I knelt down beside the road and ran my hand through the earth in a questioning manner. I then went and smelt the bark on a nearby tree for quite a long time before nodding and writing stuff on my iPhone. This seemed to satisfy the journalist, who took more photos.
We got back in the car and drove on up a river valley until we got to the farm where the original sighting had taken place. We got out and trudged over a ploughed field until we reached a particular spot in between the farmhouse and some woods.
Mr Maeda told us that the farmer had been working in this field and had been just about to stop for the day as it was getting dark when he saw a figure approaching him. He said that the first thing he'd noticed was that there was a terrible smell. At first the farmer thought it was his elderly neighbour and shouted out a greeting. (I thought this didn't say much for either the looks or the personal hygiene of his elderly neighbour, but I kept quiet.)
When the farmer shouted out the figure stopped moving and the farmer walked towards it. He said the smell became even worse and he saw that the creature was not his elderly neighbour but a tall, hairy man 'like a caveman ape'. As he approached, the creature bolted back into the woods at the same time as the farmer ran to his neighbour's house.
Presumably when he knocked on the neighbour's door in panic he left out some of the details, and didn't just blurt out, 'I just saw a really ugly, hairy beast in the field; it stank to high heaven and I presumed it was you but it wasn't . . .' Whatever, the farmer was absolutely terrified, in quite a state of shock, and refused to go back to his house.
I wanted to speak to the farmer but he had since died. Mr Maeda had spoken to all these witnesses at length, though, and saw absolutely no reason to disbelieve them. Mr Maeda said that Mount Hiba was a 'holy' mountain and had been sort of off limits to people before 1970. At that time, there had been talks of developing the area for tourism and it was then that the initial sightings happened. Some said the Hibagon was angry about this invasion of his territory and this was why he was coming closer to humans.
Hungry from all this monster-hunting, we stopped at a mountain lodge to have lunch. The local reporter started to interview me. It turned out that he was something of a monster aficionado himself and had been to the home of another of Japan's big monsters, Issie, at Lake Ikeda. I told him that I intended to try to go there and check it out for myself. He was quite encouraging and said that there were boats that took people out to search for the beast and that he had spoken to several local fishermen who had all seen peculiar things.
We all headed back off down the mountain to the city, where Mr Maeda showed us the roadside signs that he'd had erected. These all featured the cuddly Hibagon welcoming people to the town. He then took us to a local bakery where they made Hibagon sweets. It was all a little desperate and I was quite glad when it was time to finally say our goodbyes and head off back to Hiroshima. There was something a bit depressing about the whole Hibagon affair. It felt like the creature had left town a long time ago. I'd sort of hoped for more of the irradiated-man angle but Mr Maeda didn't even seem to factor this in as a possibility. The journalist told me that foreign journalists had heard about the Hibagon, noticed it was in Hiroshima prefecture and put two and two together to make five. I tried to look bewildered at how people could be so stupid while subtly emphasizing that I was not one of those idiots. I'm not sure that I was entirely successful.
Mr Maeda waved at us sadly as we disappeared down the road. I drove back towards Hiroshima while Naoko chatted away about her travels to Europe. She told me that she'd been terrified while on the sleeper train from Venice to Nice because someone had told her that people gassed sleepers and stole their stuff. She and her husband had barricaded their compartment and refused to let the ticket inspector in because they were sure he was a baddie.
As we re-entered Hiroshima I glanced at our little rented car's incomprehensibly complicated sat-nav system. I noticed a plethora of swastikas dotted all around the city. For a moment I worried that these might denote secret Nazi bases in town but Naoko laughed and told me that they represented Buddhist temples – the swastika being an old Buddhist symbol before the Nazis swiped it for their own nefarious purposes.
Back at the hotel, I said goodbye to Naoko and gave her a copy of my last book. She promised to find the article by the local journalist when it appeared in the newspaper and to send me a translation.
'Soon you will be famous in Hiroshima, Dom-San,' she joked. I loved her calling me Dom-San – I don't know why -and I wondered whether I could persuade Stacey to start doing so too. It being a mark of respect, I somehow doubted it. Naoko and I bowed to each other and she wandered off smiling to the end.
Despite the appealing prospect of my forthcoming celebrity status in Hiroshima, I had decided to leave the city the following day. The Hibagon had not been as exciting a prey as I'd hoped. I'd had an actual sighting of Ogopogo in Lake Okanagan whereas here so far I hadn't even spotted a shape-shifting badger. I decided that I was going to take a Shinkansen and get down to the very bottom of Kyushu Island to see if I could learn anything about Issie the lake monster.
That evening I headed out to a Yakitori joint that Naoko had recommended. When we'd been wandering round town she'd popped in and introduced me to the owner. Had she not done this, I think I'd have been too nervous to walk in on my own – it was a low-ceilinged room packed with Japanese patrons. My entrance through the sliding door caused some raised eyebrows until the owner waved at me and beckoned me to sit down at the lone remaining seat. I was immediately given a plate piled high with raw cabbage covered in soy sauce and tons of pureed garlic. An enormous glass of chilled Sapporo was plonked in front of me. I didn't order anything. Every time the owner cooked a round of skewers he'd walk up and down the bar placing one on each plate. When you finished a skewer you put it in a cup opposite you and this was how you were charged at the end of the meal. Life was good. I ordered some cold sake and the owner suggested I have some from a beautiful-looking bottle. Some rice wine can be slightly vomit-inducing but this was perfect – and incredibly strong. Within half an hour I was showing the whole bar photographs of my wife, kids, dogs, cats, mother. The reserve was down on both sides.
On the way back to the hotel I stopped at a portrait painter's and had him paint me. When he'd finished, he handed me a portrait of an elderly drunken Irishman. I presume that it was supposed to be me but nobody who has seen it has ever yet guessed this fact.
The following morning I had a sore head and found the lift ride down particularly tricky. There was some sort of convention happening in the hotel and it was even fuller than the Tokyo metro. However calm and 'Zen-like' the Japanese character is supposed to be, there's always one man hammering away at the 'close' button in every lift. Should he get out at a certain floor, then another man just steps up to take his place. At breakfast I checked my emails. There was one from Naoko. The fat journo's story was in the paper and she had translated it for me.
Dom Joly, 44, an English comedian, visited Saijo Town, Shobara, to write in his book about Hibagon, which was seen there in around 1970. He is writing a travel book on six UMAs in the world, including the Nessie and Yeti. Hibagon has been included in the six.
He energetically reported the area where Hibagon was first seen and the vegetable field where the local resident met Hibagon at a very close distance. He said that the nature and the atmosphere in Saijo is similar to that of the forest in California where Bigfoot roams.
Mr Tadanori Maeda, 44, Secretary General of the Saijo Town Tourist Association, said, 'It is our honour that Hibagon has been included in the top six UMAs in the world. I'm glad to know that, even today, Hibagon has been paid attention to.'
Mr Joly will soon go to look for a dinosaur in a deep forest in Africa in January, and Yeti in the Himalayas.
There was a photograph of me, Naoko and Mr Maeda. I was in the middle and pretending to be inspecting something. Naoko wrote, 'You are very famous now in Hiroshima, Dom-San; all my friends have seen this.'
Monster-hunters like myself, however, do not do this for fame or women or free food: we do it for science.
I caught the Shinkansen and, an hour out of Hiroshima, we pulled into Kokura. I gazed out of the window at the city This was the 'B' target on the day of the Hiroshima bombing. It survived purely because of clement weather over Hiroshima. The train flew on, like a silent projectile (a bullet, if you will), through Japan. An overly helpful man insisted that I get off at Hakata.
'You need to change here for sure.' He smiled, almost tugging on my sleeve.
Are you sure? I thought the Shinkansen went all the way to the bottom of Kyushu?'
'No . . . You change here one hundred per cent. I go where you go.'
'To Lake Ikeda? I want to see Issie.' I tried to make what I considered to be the international sign for 'lake monster' but was not convinced it worked.
'Yes, yes, we must hully, please . . .' he smiled and I followed him off the train meekly. I had to admit that my Japanese train-getting had not been brilliant so far. The man took me through some barriers and we entered what looked like a distinctly less-salubrious train platform.
'Shinkansen?' I asked, looking concerned.
'No Shinkansen . . . Local train.' The man seemed absolutely convinced and the signs were now in Japanese so I relinquished all control and got on a smaller train. I sat down and the man sat opposite me. I sort of wanted to be alone but he seemed to have taken me under his wing. I hoped that I wouldn't end up in some gimp basement and surreptitiously got my penknife out from my bag, just in case. The train rolled along the coast and the scenery became rather beautiful. Occasionally my new friend would point out of the window and say something like 'Sea . . .' while pointing at the sea.
I would nod and say 'Yes, sea . . .' back to him and then stare intently at the sea as though I had just noticed it.
'Tree . . .' he said, pointing at a rather nondescript tree.
'Yes, tree . . .' I said, now longing for an escape route.
The train rolled on for a good two hours. We stopped at several places but I didn't have a map so I had no idea how far we had gone or how long it would take. The train started to slow down and the tannoy lady was very vocal for a while.
'We are here,' said my friend. 'Velly good.' He smiled at me and I smiled back. He had been very helpful and didn't seem to have any hidden weapons or chloroform at the ready. As the train pulled into the station, I thanked him.
'Thank you,' I said.
'Welcome to Nagasaki,' he said.
I'm still not sure how my 'friend' ever got it into his head that I was going to Nagasaki. Certainly I never mentioned it. All I could think was that he assumed every foreigner on a train was bound for his hometown. Maybe they had a very proactive visitor programme? Whatever, there was very little I could do. I checked the map and I was way off course. Nagasaki is on a peninsula on the westernmost part of Kyushu and it would take an age to get back to a line that would take me down to near Lake Ikeda. I had limited time left in Japan so I decided to go with the flow and visit Nagasaki. To my knowledge it has no monsters – but it _was_ the site of the dropping of the second atomic bomb in the Second World War and so, in a Dark Touristic way, it sort of made sense that I be here.
I was joking about my friend on the train thinking that all foreigners must be going to Nagasaki but this actually used to be the case. The city was 'opened' to the world by the Portuguese in 1571. It was a flourishing trading port and the centre for all Christian missionaries in Japan. In 1641, however, Christianity was banned and the Portuguese were chucked out of the country. Japan's only connection to the outside world was the Chinese settlement in Nagasaki and some Dutch merchants who were quarantined on Dejima Island in Nagasaki Bay. This alienation from the outside world lasted for 200 years and made Nagasaki far more liberal and cosmopolitan than the rest of medieval Japan. I was expecting great things from the place and hoped that I might stumble on a monster story or two.
I found a hotel called the Monterey. A whole section of the city was built in a European style by the foreigners who lived there after being released from their island exile. I dropped my bag off and immediately set off for some more atomic tourism. Surprisingly the museum and hypocentre aren't in the centre of the city but are in Urakami, a little suburb to the north. Cabs in Japan are eye-wateringly expensive and my experiences in them so far hadn't been wonderful, so I opted for the tram. We headed through Chinatown and started trundling up towards Urakami, where the bomb exploded at 11.02 a.m. on 9 August 1945 – three days after Hiroshima.
I remembered the terrible story of a man who had been on business in Hiroshima when the bomb was dropped. He was wounded but survived and managed to make his escape from the devastated city. He headed for home – you guessed it: Nagasaki. He arrived there just in time to be hit by the second bomb, which he also survived. He'd only recently died.
So, I was off to my second atomic memorial site in a day. I hopped off the tram, crossed a big road and entered a park -and within seconds was standing at the spot above which the bomb went off.
The A' target city had been Kokura. Lucky old Kokura, however, had been too cloudy so, after making a couple of circles, the plane headed for the secondary target, Nagasaki. The drop site was supposed to be the centre of town, near my hotel, but there was more cloud here and the pilots were by now very low on fuel. Then they spotted the Mitsubishi factory that was situated in this industrial suburb and dropped the bomb there instead. Because Urakami lies in a valley, the centre of the city was spared the very worst effects of the bomb. That's why there's much more left of 'original' Nagasaki than there is of Hiroshima.
The hypocentre memorial features a big black column shooting up into the sky surrounded by a set of concentric rings spreading out like the ripples of a blast wave. Right next to it are the remains of the Catholic cathedral that was completed in 1925 and had, until it was vaporized, been the largest in the Orient. I looked up into the sky above and tried to imagine that moment. There would have been no sirens wailing: this was a lone plane and wouldn't have been seen as a threat. The city would have gone from total normality to an inferno of hell in a millisecond.
I climbed the nearby steps to the Nagasaki Atomic Bomb Museum. As I entered the foyer a woman approached me and asked me if I needed a guide. I politely said no and tried to move on but she seemed not to understand and started ushering me towards a door on the right.
'Thank you, but I am happy to be alone,' I said, smiling and half-bowing.
'I learn English and am honour to be my guide for your tour.' She bowed back and almost pulled me towards the door.
'I really do not need a guide, but thank you for your most kind offer.' I bowed again, slightly lower, smiled again and tried to move in the other direction.
'Gratuities will of course be at your discretion; we am volunteer guide but we am also housewife . . .' She blocked my move. There was no bow this time – just an iron will that would not truck with dissent.
I gave in and followed her towards the door. What was it with Nagasakians and their insistent dealings with visitors?
We set off on the 'tour'. Her initial pitch had been understandable but she now lapsed into virtual gibberish with an almost comical Japanese over-accent. She reminded me of a rather stern woman who had shown me round the Museum Dedicated to the Evil Work of the Imperialist Pig-Nation, America, in North Korea. (I don't think that was the actual name but it was the gist of the place.) This tour was mind-blowingly bad. My housewife guide just approached every exhibit and read out the English blurb on the wall, but in a language that I didn't recognize.
'Vis fologlaf dispray effect of ladiation on wesidnt of rbble tin . . .'
If I tried to move too fast or look somewhere else, she would scold me and I'd be pulled back hard on the leash. It was hell.
I remember very little of the tour, as I spent most of the time trying to plot my escape. The one fact I did take in was that the bombs' special antenna, which allowed them to register their altitude and to explode at precisely 500 metres, were invented by a Dr Yagi Hidetsugu, a Japanese scientist from Osaka.
My guide seemed to think that I would be fascinated by anything to do with Christianity. There was a little section of the museum – by far the least interesting, in my opinion – about a Japanese Christian who lived in the city at the time of the attack. There were moments when I felt that was where I was to spend the rest of my life, listening to a tiny, unintelligible woman waffle on about 'Jesus Clist'.
She also had a tendency to go on and on about 'enemy pranes'. While I accepted that they were the enemy to her, she was actually talking about my father and his friends and it started to really get on my nerves. I wandered off and read a notice informing me that, on the fiftieth anniversary of the bomb being dropped, the city of Nagasaki had bought the original colour footage of the bombing from the Hooper Institute in the US. Whatever the Japanese did in the war, I sort of thought that the Hooper Institute, whatever that might be, could have just given them the footage of their city being wiped off the map rather than selling it to them.
Eventually I couldn't take it any more and told my guide that I was feeling ill and needed to leave immediately. Unfortunately she got very concerned – too concerned.
'What is wong wiv you? Where is pain? I call doctor?'
'No, no . . . I just need to go back to my hotel and rest a while. Thank you so much for your tour, though; it was . . . good.'
'You sit down. I call doctor now.' She got on her mobile and started jabbering away while motioning me to a nearby chair. I started to panic. I was never going to lose this infernal woman. She wouldn't ever let me leave.
'I speak to doctor – what is sympton?' She held her hand over the speaker and kept the doctor hanging as she discussed my fate.
'Please, I just want to go back to my hotel.' I was almost shouting at this tiny woman, and several Japanese visitors to the museum wandered past tut-tutting.
'OK, I tell doctor to come now . . .' She started jabbering into the phone again. I looked around and saw some stairs just round the corner from where we were standing. The little lady was now deep in animated conversation with the doctor. I took my moment and bolted. I was up the stairs faster than Charlie Sheen out of rehab and I didn't look back. I came to a fire escape and pushed it open. Alarms started to ring and a curious wailing siren, remarkably like an air-raid siren, sounded out.
Once you make the decision to run you must commit. The last thing you want is to be caught or bump into the other party concerned. Otherwise you have to start pretending you've lost your mind or are on strong hallucinogenics and it all spirals ever further out of control.
'Oh, these tangled webs we weave . . .'
I could hear the siren wailing and could imagine my hyperactive little guide describing what was happening to a by now very confused doctor.
'He bleak thlough door; silens they wail rike clazy coyote . . . Now he lunnning away like clazy man! We leed sedative gun fast . . .'
At the bottom of the hill I spotted a taxi and jumped in and we roared away as fast as the little Japanese Box could roar.
Back at the Monterey hotel I lay on the tiny bed in my tiny room for a while. The tiny pillow had something very weird inside it. It felt like beans. Not beans as in a beanbag; beanbags are quite comfortable. These felt like actual beans: uncomfortable, hard beans that needed to soak for forty-eight hours. Why would anyone put beans in a pillow?
There's so much about Japan that I didn't get. It's a truly unfathomable place to the casual visitor. It's a country obsessed with modernity and cutting-edge gadgetry and yet still so steeped in tradition and mythology. I wondered whether their relentless surge towards the future has made the Japanese cling to beliefs about things like monsters more than people in most countries. The shock of the new, the 'Year Zero' effect of the Second World War and the subsequent rapid modernization might have left them with a need to hang on to old superstitions. When your country is set on fire, nuked and invaded by a civilization that you then aspire to you must need something to blame for stuff . . . And maybe to make uncomfortable pillows for visitors by way of subtle revenge.
I headed out into town to try to find a YO! Sushi type of place. I wanted somewhere I could sit and watch food go past me on a conveyor belt, choosing whatever took my fancy. I got the name of the best one off the Internet and tried to give this to a cab driver. He wasn't interested in my desired destination and dropped me off somewhere wholly wrong and quite insalubrious. I had learnt not to bother complaining to Japanese cabbies. I waited until he drove out of sight and then hailed another one and tried again. The new cab, driven by an elderly woman who seemed to have no idea how to change gears, took a look at the name of the place I wanted and took off. We drove for about twenty minutes and I soon knew that we were not going anywhere I wanted as we appeared to be leaving the centre of town. I decided to give up and simply go with the flow. Wherever this lady wished me to have supper would be where I did so. She eventually came to a stop down a tiny, smoky alley straight out of a kung-fu film. The electric door opened and she indicated that this was where I was going. It certainly wasn't but I was now committed. I got out and she screeched off, leaving me alone in the alley. I half-expected a large gang of martial-arts clichés to suddenly appear out of the smoke and say, 'So, Mr Jory, now we shall decide who is the master . . .'
The gang didn't appear and I looked around to get my bearings. Most of the buildings were unpromising, with the occasional fire escape and a lot of dustbins. One door looked to be vaguely inviting, however. It had a light over it as though it expected people to stand there. I knocked and a little hatch opened at eye level. A grumpy pair of eyes stared suspiciously back at me. The owner of the eyes said something in Japanese. I looked puzzled and pointed at my mouth. There was a long pause and then the door swung open to reveal a man in stained chef-type clothes. This was a start. At least I hadn't knocked on the door of some Yakuza heroin gang. Having said this, Chef, as I shall call him, looked very shifty, as though he had just been in the middle of doing something really terrible and had quickly hidden the evidence. The room was minuscule, like a cupboard (or a medium-sized Japanese hotel room). After a moment he indicated that I should enter. I hesitated but I spotted some beer under a table and I was thirsty and tired from my taxi adventures. I squeezed through the door and got past him. Between us there was not much room left for oxygen.
He stared at me for a very long time without doing anything. I wondered whether this might not be a terrible mistake. He pointed at a stool and I sat.
He produced a menu in Japanese. I looked at it helplessly.
'Biru,' I said in my fluent Japanese. He poured me a glass of beer and stared at me as I sipped it. It was very good beer. I raised my glass to him in a salute but he just stared at me as though sizing me up.
Amewica?' he suddenly said in a very threatening manner.
Just what I needed: to be stuck in a cupboard down a back alley with a mad Japanese man who was going to blame me personally for the atomic destruction of his city.
'No . . . not Amewica . . . Boo to the USA . . . No, I am from . . . Brazil . . .' I've no idea why I plumped for Brazil. I suppose nobody dislikes the Brazilians. It did the trick, though.
'BRAZIILLL!!!' Chef was ecstatic. 'Brazil, Rio, goal, Ronaldo . . .' He had exhausted his whole Brazilian repertoire but he seemed happier than his homicidal appearance had started to suggest moments earlier.
'Goal, Pelé, Amazon, São Paulo, Ronnie Biggs!' I shouted, exhausting my own Brazilian knowledge and knowing as I said it that Ronnie Biggs wasn't going to cut the mustard.
I was right: he looked confused.
I stopped being Brazilian and studiously pretended to look at the menu. Chef jabbered at me in Japanese in a manner that made me fairly certain that he was asking me what I fancied. What I actually fancied was getting out of here but I was stuck now and determined to go through with whatever was to come. I pretended to look indecisive for a moment before pointing at two things on the menu decisively. He looked at my two chosen things and then at me quizzically. He asked me something in Japanese that sounded a little like, 'Are you sure, you flucking idyot?'
I nodded and indicated that, yes, this was definitely what I wanted – whatever it was. I then looked down and concentrated on my beer. I felt him continuing to look at me for quite a while before finally crouching down and starting to fiddle with stuff in a cupboard on his side.
After about five minutes he produced a bowl of what looked like raw cat sick and placed it in front of me. He then returned to his larder and fiddled a little bit more before producing a tiny bowl of slimy pickles. He then stood staring at me expectantly. Dinner was clearly served.
I looked at the meal and then up at Chef. Chef's eyes moved fast from the cat sick to me and back again. I stalled for a while but very soon I had to face the inevitable. I looked down. The cat sick had a putrid fishy smell and it was becoming quite overpowering in the cupboard. I gingerly lifted my chopsticks and tried a pickle. They were not terrible – revolting, but edible. I smiled at Chef but he didn't smile back. Chef looked at the cat sick again and then at me. I started to get hot and panicky. What if this actually _was_ cat sick? Maybe that was what he was doing when I'd interrupted him? Forcing his fat fingers down some poor cat's throat.
Eventually I could stall no longer; I had to dig in. I picked up a small amount of cat sick and reticently put it in my mouth. Like uranium, you clearly needed only a tiny amount for an explosive reaction. This was a taste so awful, so utterly heinous that I genuinely have no words to describe it. I suffered an instant gag reflex and found it almost impossible not to projectile vomit. Thankfully some inner survival mechanism made me keep it in. Chef did not appear to be a man who appreciated being vomited on. This was the secret ingredient they needed on the Bush Tucker Trial on _I'm a Celeb . . ._ It would be TV gold.
Eventually I swallowed this fiendish mouthful but weird things lingered in my teeth. I looked up at Chef, who had a big smile on his face. Was he laughing at me or just pleased that I was enjoying his cat sick? I hate to say this but he was inscrutable.
I downed an entire glass of beer. There was nothing on God's earth that would persuade me take another mouthful of that crap. I looked at him and did the _X Factor_ sign with my arms to indicate that I was finished and wanted my bill. Chef looked at me in astonishment. He shook his head and pointed at the cat sick. I looked around subtly for any knives. There was a chopper lying on the counter about a foot away from him. I just wanted to get out of there and run away again.
Chef shouted something at me in a low guttural growl. I hated this situation. I didn't want to be here so why was I here? I was a grown man. I didn't have to do anything that I didn't want to – but how to get out? For the second time in the same day I feigned terrible illness. Having just consumed the cat sick, this was not tricky. I grabbed my stomach and started to make terrible noises. Chef looked startled. I pulled out a 1,000-yen note, dropped it on the counter and then stood up, pretending to stagger. I hit the door hard before sliding it open. I managed to get half into the alley but Chef grabbed me and was shouting stuff. This was becoming a terrible day. I pointed up to the sky and he let go for a second and I took my moment. I bolted, running as fast as I could, and I didn't stop running for five minutes. When I eventually did, I vomited all over the pavement. I was certain that this was the moment where my guide from the museum would walk by but fortunately I was alone. A hollow husk, I hailed a taxi that I had to direct myself to the Monterey hotel for a night battling the evil pillow monster.
My time in Japan was running out. I had to get back to Tokyo the following day to catch my flight home without a sniff of a monster.
The next morning, I caught a cab to the station and, to my surprise, the driver asked me in polite, stilted English whether I minded if he chatted to me. He'd learnt his English from the Internet and was keen to practise it on somebody.
'Sir, why are you visit Japan?'
'I am here to hunt the Hibagon.'
'Slow please . . . I no understand.'
'I . . . am here . . . to find the Hibagon . . .'
'Why are you in Japan? For tourist purpose?'
'No . . . To hunt the Hibagon . . . The Hibagon – the monster . . .'
'Slow please . . .'
'I'm a monster-hunter . . .'
_'Ghostbusters?'_
I looked out of the window and prayed for the cab ride to end soon. I'd really had it with Japanese taxis.
I didn't have long in Tokyo before flying out and I was seriously annoyed that I'd been unable to find a monster in the land of Godzilla. I wondered whether Japanese monsters were more subtle than others. The Hibagon had garnered international attention because it seemed to be more of a traditional 'Western' type of monster of the Bigfoot variety that we could understand.
Most Japanese monsters are more understated and can probably only really be understood or 'discovered' once you understand the Japanese psyche – something I was still a long way away from doing. I did a whistle-stop tour of the capital. I went to visit the weird manga kids in Harajuku who were all dressed up as zombies and scary nurses and freakazoid Goth characters. I found one who spoke English and asked her what she knew about the Hibagon.
'Hibagon he big hairy gorilla monsta in Hiroshima . . . He cool.'
I asked her whether there were any monsters in Tokyo.
'Monsta everywhere in Japan. Tree monsta, road monsta, shop monsta . . . Lot of monstas.' She gesticulated all around us and I nodded, trying to look like I understood.
I left her and her friends posing like the world's best tourist attraction while screaming at every tourist who tried to take their photo.
I headed for the Imperial Palace and spent ages wandering around the grounds trying to find the palace. It took me about an hour to realize that there isn't one. Nobody had bothered to mention this in any of the guides.
I ended up in 'Brand Street', a huge boulevard bursting with every big-brand store you can think of. The place was packed with rampant shoppers seemingly having a great day out. It was my idea of total hell and I was about to head back to my skyscraping hotel room when I spotted the Ginza Lion. On a whim I wandered in and discovered a complete gem. Slap in the middle of Tokyo, this is an extraordinary Munich-style bierkeller. It was warm and atmospheric and buzzing with people: a Gothic retreat from the Metropolis. It was teatime and the place was full of gossipy tables of Japanese ladies. Instead of tea, however, everyone was brandishing pints of beer. The Japanese customers were all impeccably turned out, the men in suits or tweed and the women looking smart and hip. It wasn't traditional dress but, as with much else, they had copied Western style but done it properly.
In the centre of the hall were a couple of tables of Westerners bedecked in sweat tops, hoodies and scruffy T-shirts. Had we moved on as a culture or had we lost something? I thought back to the cabbies in uniform, the smiling train conductors bowing to the carriage. I thought we'd lost something.
I sank my two-and-a-half-pint Sapporo with ease and ordered another one. I felt a lovely warm feeling come over me. Near the middle of the room I spotted a white-faced lady of about seventy. She was beautifully done up and in full traditional kimono garb. She had a healthy-looking glass of beer in front of her and was trying to eat beef and potato croquettes with chopsticks, a tricky task that she was accomplishing with much grace.
The following morning, and I was on the plane sitting on the runway at Narita airport nursing a rather splendid hangover. The stewardess was one of the same ones who'd been on the flight out.
'Did you have a nice time in Japan?' she asked.
'Yes, I did,' I replied.
'Business or pleasure?'
'A bit of both . . . I'm a monster-hunter.'
She gave me that look that people give to mental people who stagger up to them in the street, then moved on a little too fast to the next passenger. I looked out of the window and wondered whether she was right. Maybe I was a nutter and this was all a supremely pointless exercise? Then I remembered my Ogopogo spotting, the Super 8 footage of Bigfoot, the black-and-white photos of Nessie, the Yeti footprints in the Himalayan snow. Who knows what is right and what is wrong? I was on a great adventure and I was as likely to find something as anybody else. I took Robert Frost's words for encouragement:
Two roads diverged in a wood, and I—
I took the one less traveled by,
And that has made all the difference.
I was off to the Congo next and that was definitely a road less travelled. As we taxied off the ground crew were all lined up and waving goodbye to our plane. In Heathrow they'd be watching porn on the laptop they'd nicked from your luggage.
### Mokèlé-mbèmbé
'"But what if the monsters come?'
"Fancy." Kit looked away from the drama to stare at her sister, surprised. "We are the monsters."'
Dia Reeves, _Slice of Cherry_
I sat on the rotating stool at the seafood bar in Terminal Four and tried to ignore the Russian couple sucking face right next to me.
I was heading to Brazzaville, capital of the Republic of Congo, by way of Nairobi. Stacey had been freaked out by the Congo. I'd tried to explain that I was going to the 'good' Congo as opposed to the 'bad' Congo. 'Good' Congo being the old French colony, the Republic of Congo. Yes, it had just had a civil war, but it was OK at the moment. 'Bad' Congo is the Democratic Republic of Congo, the old Belgian Congo. Experience has taught me that any country calling itself 'democratic' is always anything but (see East Germany and North Korea). The good and bad Congos face each other over the mighty river of the same name.
To make things worse there had been a Code Red terrorist alert in Nairobi on the eve of my departure. I had absolutely no idea what a Code Red was or what I should do about it. I'd tried to mouth some platitudes about this making the airport even safer but she was already in a tizz about the whole trip.
'You're going off on your own into the middle of bloody nowhere to look for a dinosaur? For fuck's sake, Dom, you've got responsibilities . . .'
She was right, of course: it was unnecessary. But then again, if you play by those rules then everything is unnecessary – and she had to remember that I was one of the world's foremost monster-hunters and knew no fear . . .
The more I looked into the Mokèlé-mbèmbé, the 'Blocker of Rivers', the more intrigued I became. In the vast swampy borderlands between Cameroon and the Congo is said to exist an aquatic creature that looks very much like a dinosaur. There have been reports of sightings going back to the very first Western explorers and local tribes have stories that go a lot further back in time. If there's something undiscovered in the world, this is the sort of place where it might reside. The area is almost inaccessible and very few Westerners have ever been there – the combination of war and remoteness has kept most people away. This was going to be a proper, middle-of-nowhere adventure: just a guide, porters and me heading off into the heart of darkness.
My two destinations were the village of Boha and then Lake Tele, where the creature is supposed to live. Information was scarce but it seemed to be a two-day walk after an EU-blacklisted flight, a seven-hour car ride and an indeterminate boat journey. This was it. I was an adventurer, an intrepid explorer . . .
'Excuse me . . .' A sixteen-year-old blonde girl interrupted my reveries. 'Were you on _I'm a Celebrity . . . Get Me Out of Here!?'_ She looked at me enquiringly having clearly been egged on by her pre-shaving companions in the queue for the plane to Nairobi . . .
'No . . . It wasn't me,' I muttered.
'Yeah, it was. Weren't you the one who pooed on the camp toilet seat and didn't clean it up?'
'That wasn't me. That was the stupid Playboy Bunny throwing false accusations about.'
'So it _is_ you! What's your name? You were a comedian, weren't you?'
I started to almost run for the plane.
It turned out my neighbour on board was also on an adventure. He was an ex-army officer who now worked on pirate patrol. He spent eight weeks on, eight weeks off cargo vessels defending them against Somali pirate attacks.
We landed in Nairobi and as I got off the plane I felt the clammy wall of African heat. I made my way to my gate. The flight was going to Brazzaville and Kinshasa. It seemed that good and bad Congo shared a flight. There was a vast amount of people in the departure gate and they were a tough-looking bunch. There were three Westerners, a couple of Arabs, a handful of Chinese and the rest were Africans. It looked like the sort of plane that James Bond might hop on to and look around suspiciously.
Next to me sat an enormous Congolese man with a very annoying mobile-phone problem. Every two or three minutes his infernal ringtone would go off. It was a rather creepy voice whispering very loudly, 'Boss, you have received a text message.' To my right, a man was reading a local Nairobi newspaper -'Moustachioed woman robs neighbourhood!!' screamed the rather wonderful headline.
The plane was an hour late and the general consensus seemed to be that this was rather better than usual. I went for a wander to get a cup of coffee. At the bar sat a Sloaney English girl yacking away on her iPhone.
'Yah, no, I'm in Kenya, on my way to Rwanda . . . I know . . . Totally weird . . . Absolutely . . . Yah . . . I'm meeting Milo . . .'
After an hour the packed room was explained: there were three delayed flights waiting – one for Zanzibar, one for Kigali and mine. I was secretly pleased that my lot looked decidedly rougher than the other two. The three Westerners were on other flights so I was the only non-African on mine apart from a lone Chinese man who looked decidedly shifty .
On the plane a surly African youth wearing the international costume of hip-hop twat was slouched in my seat. I explained the situation but Chocolate Ice just stared at me through dull, sullen eyes and didn't move. Fortunately he was only about fifteen so I was able to be slightly more assertive than I might have been with an adult. He moved to the middle seat and I sat down only to find myself in another elbow war. Fortunately, there was quite lot of turbulence and mini-gangsta almost shat his overly baggy pants. To show how unconcerned I was I fell asleep on the window. I woke up just as we were coming in to land in Brazzaville. Through my dribble I could see rows and rows of houses with rusty corrugated-iron roofs. An eerie, misty cloud hung over the city.
Once on the ground I walked through surprisingly clean corridors, with every Congolese I met welcoming me to Brazzaville. It all seemed very friendly and organized. Maybe I'd got the wrong end of the stick?
Then I got to passport control.
I handed over my passport and vaccination forms to an official staring at me in exactly the same way as the sullen youth had on the plane. He spoke to me in heavily accented French. This was to be the same throughout my journey in the Congo. Fortunately, having grown up in the Lebanon, I'm bilingual. (I have therefore translated all conversations into English for the purposes of this book.)
I gave the official my passport and he flicked through it rather contemptuously. He got to the page that had my visa.
'Where is your letter of invitation?' He looked at me accusingly as though I was hiding it in my pants.
Anyone applying for a visa to the Congo needs a letter of invitation from someone based in the country to support their application. I told Grumpy that mine was with the Congolese Embassy in London, as they'd needed me to send it to them so that they could issue the visa.
'So why don't you have it?' asked Grumpy.
I repeated that I had sent it to the Congolese Embassy in London and that that was how I'd got the visa that he could see in my passport. No letter, no visa.
'So can I see the letter?' Grumpy looked me straight in the eyes.
This was all starting to get a bit _Catch 22._ I repeated, slowly, that the letter was with the Congolese Embassy in London and added that I must have had one or I wouldn't have been issued the visa . . .
'But you do not have a letter of invitation. You need a letter of invitation.'
We stared at each other in silence for a while, neither of us backing down. Eventually I told him that someone outside was meeting me. Could I leave my passport with him, go and find the man who was meeting me and hopefully he could sort everything out? To my surprise Grumpy accepted this idea. I wandered past the luggage hall and several armed soldiers and into the morass of humanity waiting for passengers to come through. I looked around. I was supposed to meet a Cameroonian called Jean-Pierre. He'd come highly recommended and had travelled all over the Congo, although he'd never been to the area that we were off to. I scanned the crowd but there was nobody showing any interest. I tried to go back in to talk to Grumpy but an armed soldier placed his AK47 in front of me and shook his head.
'Nobody goes in: this is the exit.'
I explained that I had just come out and needed to go back in to get my passport but he was not interested.
'Nobody goes in through here.'
I was buggered. I was just wondering how I could deal with a trip to the Congo with no passport when a smiley face burst through the crowd.
'Monsieur Dom?' It was Jean-Pierre. I was saved. Jean-Pierre had a word with someone he knew and we both wandered back towards Grumpy un-hassled.
I introduced Grumpy to Jean-Pierre and told him that this was the man who had written my original letter of invitation. Grumpy looked at Jean-Pierre.
'Show me the letter . . .'
I groaned. We were back at square one.
Jean-Pierre started the same complicated set of explanations that I had attempted. He handed his passport over to Grumpy and explained our mission here. Grumpy looked at Jean-Pierre's Cameroonian passport.
'You are from Cameroon. You are not from the Congo.'
Jean-Pierre nodded in agreement at this statement.
'If you are from Cameroon you cannot invite someone to the Congo.'
This conversation went on for roughly half an hour. I believe money exchanged hands, though I never actually saw it happen – but, whatever, we were finally through and Jean-Pierre took me to the luggage hall.
'What does your luggage look like?' he asked.
'I don't have any more; it's on my back.' Jean-Pierre peered at my small, grey rucksack and then back at me with the look of a man who could not decide whether I was an idiot or an exceptionally talented packer.
We hopped into a cab and headed downtown to Mikhael's Hotel. I was dog-tired. During the Second World War, with his country under German occupation, General de Gaulle briefly made Brazzaville the capital of Free France. As we drove through it now the streets were awash with activity. Every time the car stopped people would shove useless things through the window, trying to make us purchase them. Fake-silver photo frames seemed to be a big favourite.
Jean-Pierre was very excited about our adventure. He told me that he had always wanted an excuse to travel to the part of the Congo we were going to and had been thrilled when my request came through. He warned me that we might not see the monster – 'but there are plenty of pythons and crocodiles'.
We drove on past a couple of markets, our little green taxi doing well negotiating the chaotic traffic. Suddenly we rounded a corner and there it was, the River Congo. It's enormous, far bigger than I'd expected. The water was dark and grey and I could see Kinshasa, capital of bad Congo, on the opposite bank, what looked like about half a mile away. Apparently this is the only place in the world where two capital cities are within sight of each other across a river (Buenos Aires and Montevideo on the Plate estuary are much, much further apart).
I gazed at this extraordinary expanse of muddy water that has both fascinated and thwarted so many explorers for so long. The Congo is a bit like Everest, one of those things that featured in so many stories of derring-do I remember reading as a schoolboy.
Kinshasa is huge, with a population of twelve million, whereas little Brazzaville is home to barely a million.
'It is crazy expensive here,' said Jean-Pierre as we arrived at the hotel that turned out to be run, like most of Brazzaville, by a Lebanese merchant class.
I tried to get some sleep but couldn't so I had a shower and then went to find the restaurant. I ordered the plat du jour – a uniquely Congolese dish (not) – Couscous Royale. It was delicious and I sat back contentedly and tried to eavesdrop on a tableful of five women whom I guessed to be Brazilian. The room was packed with Brazzaville's 'ladies who lunch'. They not only lunched but also smoked like it was going out of fashion and constantly showed each other videos on their laptops with the music turned up to the max.
There were still Christmas decorations hung up around the place despite it being late January. Christmas in the Congo: who'd have thunk it?
A man in an orange suit brighter than the sun, subtly offset with a fluorescent-blue shirt, wandered around the room permanently on his mobile. He was Belgian and was talking to someone on his phone about the fact that he was off to Moscow the following day for three days.
I ordered an Um Bongo to try to fit in. The waiter looked desperate to please but eventually returned to ask me to repeat my order.
'I'd like an Um Bongo please.'
He disappeared again but returned quickly, shaking his head in a disconsolate manner. He admitted that they had no such drink.
I wasn't going to let this go.
'I understand that you don't have any Um Bongo on the premises. Perhaps you have run out due to the high consumption rate? When will you be restocked?'
The waiter looked mortified to have to admit that he had never heard of Um Bongo.
'Are you a Congolese?' I asked him. He confirmed that he was born and bred in Brazzaville. And you have never heard of the soft drink Um Bongo?'
The waiter shook his head and slunk away. I was dumbfounded. All those years when I'd been taken in by the Um Bongo TV adverts with the catchy song: 'Um Bongo, Um Bongo, they drink it in the Congo . . .' It was all lies. The company responsible for the drink – the sinister-sounding Gerber's Juice Company Ltd (known as Libby's, to make them sound friendly) had been lying to us all. Nobody drunk Um Bongo in the Congo. Nobody had even heard of Um Bongo in the bloody Congo. I smelt a lawsuit and ordered a beer instead.
I got an Ngok beer that had a very familiar crocodile on the label. Later, I asked Jean-Pierre what type of crocodile it was and he told me it was called a Lacoste. You couldn't make it up.
I sat on the terrace by the pool. For about the first time ever I was by a pool where swimmers weren't treated like retards. There were no signs anywhere telling you there was no lifeguard around. Nobody was telling you that you couldn't heavy pet should you fancy it. There was not even an indication of depth or the almost compulsory 'No Diving' sign that seems to be on every public pool in the world. Not here. The Congo is a place that relies on you to make your own decisions. It's somewhere that allows you to be a grown-up. It was hot – very hot – and the water looked inviting. I stripped off to my swimming trunks and dived into the cooling water. I hit my hands on the bottom of the pool so hard that it partially dislocated my right wrist. The pool was only three feet deep. It appeared that I had stupidly dived into the shallow end. I swam to the other end only to find, to my astonishment, that it was about a foot and a half deep. Had I dived in there (and it had been 50:50) I would have broken my neck for sure.
Three young and coquettishly beautiful African girls appeared and started swimming hesitantly. They had clearly been there before as they eased themselves in feet first. The only other people by the pool, an elderly Belgian couple, looked on rather disapprovingly. They were clearly locals and I couldn't help thinking that things must have certainly changed since the days when this area was the 'European district' and Africans swimming in their pools would have been punishable with the dreaded _chicotte_ , a nasty leather whip.
Jean-Pierre came round and suggested a little tour of the capital. We grabbed a taxi and rented him for two hours. We drove slowly through Poto-Poto, the old 'native area' and now a bustling market full of life. Then Jean-Pierre showed me the Basilique – an extraordinarily modern church built by the French in 1943. The green malachite roof can be seen from most of the city and is a very useful landmark. We popped inside. I'm not a churchy person but this is a remarkable building: one vast, vaulted space with not a single column for support. Two Congolese choristers were practising and their haunting voices echoed beautifully around the space. It was an unexpected moment of serene calm in this most un-calm of countries.
Next we visited the artisans' market. I was after Tintin stuff for a souvenir but was unsure whether there would be any. _Tintin in the Congo_ is now widely acknowledged as a very racist tome full of negative stereotyping, where every Congolese is represented in an overly caricatured manner as either evil or very dim and almost childlike, needing the wise assistance of their Belgian colonial overlords (not to mention that, in the spirit of the era, Tintin blasted away at about 200 animals, including a rhino he drilled a hole into then blew up with dynamite).
Hergé was simply of his time but I wondered what the actual Congolese made of it all. I presumed that Tintin had visited the Belgian Congo and not the French one but I was still curious to see what they might have. The moment I entered the little market I was faced with walls of scary tribal masks and figurines. Nestled in between these, however, were what I was after. In the same style as the masks and figurines were depictions of Tintin (almost always tied up) as well as Snowy and Professor Calculus. I spotted a particularly rubbish attempt at the moon rocket that I fell instantly in love with. The best, though, was yet to come. The ultimate Tintin/racist/tourist trophy was a gloriously bad version of the _Tintin in the Congo_ book cover. The name Tintin had been left blank and the guy offered me the opportunity to own this artwork replete with my name on it instead of Tintin. I was hooked and haggled him half-heartedly down to about six quid. The deal done, he shuffled off to get the artist to do my name. I left thrilled with my booty.
Back in the car we went past the old president's house, the scene of much fighting. Cameras had to be put away as stern-looking soldiers with mirror shades and mean faces tracked us with their machine guns. They were used to trouble here. Here is my attempt at a potted history of the place:
The Republic of Congo used to be the French region of Middle Congo. King Leopold II of Belgium had been desperate for a colony for his little country and, by fooling the great explorer Henry Morton Stanley into helping him, he created a huge private fiefdom in 1877. Instead of this being the philanthropic exercise he had promised the world, however, Leopold turned the whole area into a horrific slaughterhouse. Estimates go up to ten million Congolese killed as they were forced to produce first ivory and then rubber for the coffers of the big-nosed Belgian king.
The reason that France got this part of the Congo was that the Italian-born French explorer Brazza claimed the northern bank of the Congo for his country from right under the nose of Stanley.
The Republic of Congo became the present-day country upon independence from France in 1960. There was a coup in 1968 and the country turned into a fully fledged Marxist experiment closely allied to the Soviet Union. The current leader, Denis Sassou Nguesso, took power in another coup in 1979. The country was oil rich, with the largest oilfields in Africa lying off her coast, and predatory foreign com panies were quick to exploit this.
Under heavy pressure Nguesso finally introduced multiparty politics in 1990 and was subsequently defeated in the 1992 elections by Pascal Lissouba.
In 1997 things really came to a head when Lissouba's men (the Ninjas) engaged Nguesso's private militia (the Cobras). Lissouba accused Nguesso of trying to stage a coup. A devastating four-month civil war ensued, which tore Brazzaville apart. Finally with the help of socialist Angolan troops, Lissouba was unseated and Nguesso reinstated. He has been in power ever since. Confused? Welcome to the 'good' Congo.
The local theories are that everything was about oil. Lissouba had done a deal with the American company Occidental Petroleum and the French oil companies that Nguesso was in bed with weren't happy about this.
Back on our tour of the city and we'd reached the banks of the Congo, where thin dugout canoes (pirogues) supported precariously balanced fishermen. Little unofficial ferries constantly crossed the river to and from Kinshasa. This was the city where, in 1974, Muhammad Ali fought George Foreman in the Rumble in the Jungle.
_'Ali, boumbaye! Ali, boumbaye!'_ the little kids had chanted over and over while they ran next to Ali as he jogged along the banks of the Congo. In English: Ali, kill him!' They probably meant it literally.
Jean-Pierre got the car to stop at the edge of the city and we wandered down to the riverbank and on to a plastic-bag-strewn beach.
_'Plastique – c'est le nouveau SIDA_ [AIDS] _d'Afrique,'_ said Jean-Pierre sadly looking around us. We'd come there to see the rapids that turn this mighty river – beaten in tonnage of water only by the Amazon – into raging, angry foam. Most of the river is on an inland plateau but upon reaching Brazzaville it drops 1,000 feet to sea level in the space of about 200 miles. The water is forced through narrow canyons and more than thirty-two different 'cataracts' until it finally reaches the ocean, where its sheer force has carved out an enormous trench in the sea floor.
It was this final stretch of water that prevented early explorers from sailing up the river. The mouth of the Congo was discovered in 1482 but it was only in the middle of the nineteenth century that Stanley managed to navigate the whole length of the river, crossing the continent from East to West.
There are some islands just below the first rapids. One of them is called Devil's Island. My cab driver told me that couples used to take pirogues out there to make love. Unfortunately peeping toms started to do the same, to pry on the passionate couples. To counter this, pirogue pilots now only take couples out to the islands. These days if you want to be a peeping tom in Brazzaville you need to get organized and pair up with someone who shares your interest.
Beyond Devil's Island, across the river, is the Democratic Republic of Congo. We could just make out some figures on the other side.
'You would be stripped naked in two minutes over there,' laughed the cabbie.
'Two minutes? That's in the good areas . . .' laughed Jean-Pierre, a little too hard for my liking.
We drove back into town and changed some money, as this was probably the last place where we could do so. We were also taking food, drink . . . everything with us as we had no idea what was available where we were going. I started thinking about the fact that Jean-Pierre had never actually been to where we were going. This did seem to me to be a basic flaw in his role as a guide. He seemed pretty relaxed about the whole thing so I rolled with it. After all, as the guy who'd recommended him to me back in the UK had said: 'Listen, he has a satellite phone and with that you can get help wherever you are if it all goes tits-up.'
I asked Jean-Pierre how long the battery on his sat phone lasted and how he powered it up in the middle of nowhere.
He smiled ruefully. 'I have no sat phone any more – the humidity steams up the screen and I break three in three years – so now I just pray to God that all goes well.'
This was not the most comforting news to an awkward atheist but I tried to remember how I 'rolled' and attempted to laugh in what I took to be an overly manly fashion. In reality I felt a bit sick.
I'd been rather hoping that a shopping list for a monster-hunting trip into the African heartland might include:
A gun
A bigger gun
A big net of some sort
Machetes
A helicopter
A really stupidly big gun just to be sure
Sadly none of these seemed to be on JP's list. It was more like water, corned beef and rice: slightly less glamorous.
We popped into a supermarket, the biggest in Brazzaville, to do our shopping. It was a Casino, like the ones in France. Well, sort of like the ones in France – if the ones in France had gone back to 1820. Just to be a 100 per cent certain, I scoured the shelves for any sign of Um Bongo but there was none. Jean-Pierre asked me what I was looking for but I was too embarrassed to explain.
What I definitely needed was sunblock. Casino didn't have any.
'There is not much call for it here,' laughed Jean-Pierre, pointing at his jet-black skin. 'Is too late . . .'
We walked through town trying to avoid being run over by the relentless stream of green and white taxis. I noticed a couple of signs on the walls: 'Il est interdit d'uriner ici.' I wondered whether there were pee police to enforce this rule.
Finally we got to a tiny chemist. I asked the chemist for sunblock and he appeared to be totally bemused. He looked around slightly randomly before pulling something off a shelf. It was a cream used to prevent brown blotches appearing on the skin. I looked around myself and finally found a cream that was to protect babies from the sun. I bought it and the chemist looked at me as though serving a paedophile.
I was now as ready as I could ever be. My only other problem was power. I'd bought a little folding solar panel with a USB outlet but it turned out not to be supported by either my iPhone or iPad. I'd also bought two USB-powered batteries that could recharge my iPhone. I plugged them into my laptop and charged them up as much as possible.
I headed for the hotel restaurant, where I joined a group of tables full of rather depressed-looking white men drinking beer way too early in the day. We all sat drinking Ngok beer and smoking cigarettes, each one of us quietly wondering what strange twist of fate had brought us here.
We had our last supper in Brazzaville on a terrace overlooking Kinshasa at Mami Water, a French-themed restaurant in a kind of marina.
'It's for the Brazzaville jet set,' said JP, pointing to various speedboats and jet skis lying around. We ate pizza served by a very grumpy waitress who brought a whole new meaning to the phrase 'couldn't give a shit'. Out on the river fishermen floated by in their pirogues as the lights of Kinshasa twinkled gently over the water.
JP told me about sitting where we were five years ago and watching tracer bullets arc over the river from fighting in Kinshasa. It made me think of Beirut.
We started talking about the trip and for the first time I realized that I was quite scared. The Congo is a creepy enough place, even in the capital. I had no sense of intuition in how to judge whether something was safe or not. I couldn't read people's faces as I could in more familiar surroundings. At first glance everybody looked rather intimidating and unfriendly. Also where we were going there are enough scientifically validated 'monsters' without worrying about a Mokèlé-mbèmbé: leopards, crocodiles, pythons, chimps, hippos, elephants, wild dogs, green mambas, black mambas, scorpions . . .
JP started talking about wading through waist-high swamps and all the things that could slip into various orifices but he ended on a positive note: there were no lions in the area.
I asked him what antidote he used for snakebites.
'Pray God,' he said in English.
Jean-Pierre was a most relaxed individual – a little too relaxed for my liking, but I didn't want to judge until we saw what happened up north. He went through our plan. We would fly to Impfondo and meet members of the WCS (the Wildlife Conservation Society), who had a base somewhere near the river we needed to go down. We would then find a boat and head off towards the village of Boha, whose inhabitants 'own' Lake Tele – the lake where the Mokèlé-mbèmbé is supposed to live. In the village we'd have to negotiate access, then get porters, a guide and then hike for two days to the lake. That was the plan. I was on a self-imposed tight timetable. I had exactly a week up north and had to be back to Impfondo in time to catch the weekly flight back to Brazzaville. This I needed to do because I had to be back in London for the most crucial meeting of my TV life. ITV would decide whether to go ahead with my new TV series or not. If they did, it was Saturday-night prime-time for me. If they didn't . . . Well, maybe I could apply for a job reading the news on Congolese TV? I was pretty sure that this was not the normal type of problem that international monster-hunters faced. They were probably more worried about having contracted some hideous disease or smuggling unusual skulls across borders.
JP and I shook hands outside the hotel and agreed to meet the following morning at seven. As I walked into the building I spotted a pack of wild dogs taking it in turns to pee into the hotel's main air-con vent . . . Which was nice.
The following morning, on the way to Maya-Maya, the airport from which our EU-blacklisted plane was to depart, the cab took us down a long wooded avenue bordered by desolate concrete buildings.
'This used to be the zoo but they shot all the animals and ate them during the civil war,' said JP ruefully.
The Chinese were building an extension at the airport but for the moment it looked like total chaos, despite JP assuring me that it was 'the best airport in Central Africa'.
Nevertheless, if you are of a nervous disposition then the domestic-departures area of Maya-Maya Airport is most definitely not for you. It was like a huge mosh-pit. People queue-barged from so many sides that the queue itself became non-existent. A lone Lebanese man who seemed to be nominally in charge hurled abuse at every passenger, flatly refusing their demands to have everything from huge fridges to flat-screen televisions, all wrapped in brown cardboard, allowed on board. One man ignored the Lebanese man and simply tried to hurl his cardboard box through the flap at the end of the conveyor belt. The Lebanese man did not hesitate: he punched the offender hard in the face and the guy went down like a sack of potatoes. The Lebanese man looked around triumphantly, as though daring anyone else to try something. The tide was stemmed for a minute or so but the battle was soon back on as the unconscious man was dragged away by a relative. To my great surprise we appeared to be flying 'Canadian Air'. I was pretty certain that Canada had very little to do with this outfit but what could I do?
The Lebanese man seemed almost shocked at how little luggage we were taking with us and he looked around suspiciously as though smelling a rat. He gave me my boarding pass with some hesitancy and snarled at JP, who gave him one of his beaming smiles.
While waiting at the departure gate I started re-reading the notes I had on the Mokèlé-mbèmbé.
The earliest reference to the creature seemed to be in a book by the nearly appropriately named Abbé Bonaventure in 1776. Bonaventure was an early French missionary in the Congo and wrote about seeing 'huge footprints, about three feet in circumference'.
In 1909 the famous big-game hunter Carl Hagenbeck wrote in his autobiography, _Beasts and Men_ , about hearing from several independent sources of a creature living in the Congo described as 'half elephant, half dragon'. Meanwhile the naturalist Joseph Menges told him about an animal that was 'some kind of dinosaur, akin to the brontosaurus'.
In 1913 German Captain Freiherr von Stein was asked to do a report on German colonies and wrote about what was now Cameroon, just on the other side of the border from where we were headed. He too described reports of a mysterious creature:
The animal is said to be of a brownish-gray color with a smooth skin, its size is approximately that of an elephant; at least that of a hippopotamus. It is said to have a long and very flexible neck and only one tooth but a very long one; some say it is a horn. A few spoke about a long, muscular tail like that of an alligator. Canoes coming near it are said to be doomed; the animal is said to attack the vessels at once and to kill the crews but without eating the bodies. The creature is said to live in the caves that have been washed out by the river in the clay of its shores at sharp bends. It is said to climb the shores even at daytime in search of food; its diet is said to be entirely vegetable. This feature disagrees with a possible explanation as a myth. The preferred plant was shown to me, it is a kind of liana with large white blossoms, with a milky sap and applelike fruits. At the Ssombo River I was shown a path said to have been made by this animal in order to get at its food. The path was fresh and there were plants of the described type nearby. But since there were too many tracks of elephants, hippos, and other large mammals it was impossible to make out a particular spoor with any amount of certainty.
In 1976 herpetologist James Powell went on an expedition during which he showed villagers illustrations of various animals both alive and extinct – the natives suggested that the diplodocus was the nearest match.
In 1979 Reverend Eugene Thomas claimed that the Bangombe tribe near Lake Tele had constructed a large spiked fence in the Tele tributary to keep Mokèlé-mbèmbé away from fishing. One broke through and was killed and the natives ate it and died from food poisoning. This was supposed to have happened in 1959.
In 1988 a Japanese TV crew flew above Lake Tele and filmed a large wake in the water . . .
I read these little nuggets of information over and over again. The truth is that nobody really knows much about the area we were going to and that was exciting enough in itself. It's a rare thing nowadays to find somewhere in the world that's still properly off the beaten track. If Canadian Air delivered then we would soon be heading off into just such a place. I had a very dry mouth. This normally happens when I'm nervous; it's a weird mix of excitement and nerves. I was excited about monster-hunting. I was nervous about the state of the plane, the flight, the landing, the insects, the animals, the heat, the cold, the unknown . . . It's the unknown that always scares us the most.
JP had been on his phone and announced that he had managed to get through to the WCS office in Impfondo. They knew we were coming and had confirmed that a car of some sort would take us to Epema, where we could get hold of a boat. They also confirmed that it would be possible to borrow two tents. This sounded fairly promising but JP just gave a fatalistic shrug. He'd travelled long enough in Africa to know that nothing was real until it happened.
We boarded the plane through a very narrow tunnel and the organization was clearly provided by the same people who'd dealt with the last helicopter off the roof of the US Embassy in Saigon. It was actual, physical fighting to get on board. At first I was a bit reticent. I was a visitor here and didn't want to behave badly. This was just taken as a sign of weakness by the other passengers and I was soon being shoved and elbowed to the back. It was sink or swim, so I thrashed and punched my way to the plane door in an almost hysterical manner. A woman was in a big argument with a soldier who was not letting her on board. He had drawn his handgun and was pointing it at the woman's chest but she seemed a lot less phased by this than I would have been. I was boiling hot and covered in sweat and starting to have a little panic attack. I wanted to tell JP that I didn't want to go to Lake Tele. I wanted to go back to my hotel in Brazzaville where there was CNN and the comfort of the Internet. Locals said that to go into the 'forest' was like going to war: you had to be prepared for anything to happen. I've never properly been to war – I've been in one but never actively gone towards one. I was really panicking badly and couldn't breathe. JP was looking at me and smiling and I tried to smile back, to mask my weakness. As ever, the mask worked.
The plane flew over dense, impenetrable forest for what seemed like hours. There was not a house, a hut, no sign of human life beneath us except for just occasionally a wisp of smoke escaping through the trees.
The plane landed in Impfondo, which seemed to suddenly appear out of nowhere beneath us. The landing was heavy and very fast. Both JP and I were sure that we were going to overshoot the runway and braced ourselves dramatically for impact. We survived and got off the plane into what seemed like complete wilderness. We immediately had a cigarette on the tarmac and watched as several passengers attempted to retrieve their luggage out of the cargo hold. One actually climbed into the plane's belly and was unceremoniously hurled out by a soldier. Another was grabbed and hit hard in the back of the head with a rifle butt. We decided to wait for due process. As we smoked, hundreds of bees swarmed around us forming a thick yellow cloud. I sprayed some Deet on to my arms and legs and wished I was back home.
We tried to enter the luggage hall and had our passports taken away by a man who disappeared into the crowd. There was very little we could do about this and we both hoped that he was some kind of official. We stood by the lone, broken carousel and waited for JP's luggage. I looked around. The room was packed with both the arrivals from our plane and the departing passengers waiting to get on it. I was the only white man in the building and felt that I was really sticking out. I could feel everyone staring at me and I buried my head in a book.
Finally JP's bags came through and we chucked everything on a trolley and tried to head out while looking for the man with our passports. The soldier at the gate took one look at me and directed us to a police room in the far corner of the terminal, where four men lounged about in virtual darkness. There's no electricity in Impfondo in the day and rarely any at night unless you have a generator.
The eldest of the four men stood up and shook my hand. He indicated that I should sit down in a chair opposite him. This I did while he perused my passport, which had suddenly appeared in front of him. He flicked through the pages for a while before looking at my visa.
'Where have you come from?' he asked.
'Brazzaville,' I replied. This being the only flight each week, the question seemed a touch unnecessary.
'How long have you been there?' he asked.
'Two days,' I replied.
'And what are you doing here?'
'I'm going to Lake Tele – I'm a tourist.'
His eyes suddenly lit up. 'A tourist? Your visa is a _visa ordinaire_ not a _visa touristique.'_
I shrugged and told him that I had let the Congolese Embassy in London know what I was doing and this was the visa they had issued me with. The man smiled unpleasantly, as though talking to a thick worm.
'Monsieur, you have a _visa ordinaire_ but you are here as a tourist – therefore you are here illegally.' I felt myself about to lose it. I was hot and tired and stressed and I hate bureaucracy more than anything else in the whole world. I started to argue in French and I could feel my voice rising. JP stepped in and started to explain in overly flowery French. He was charm incarnate. He gave me a glance to indicate that I should step away and I did what I was told. It was clear that these officials smelt money and were not going to let go. I sat on a chair just outside the room and watched two men scream at each other nonstop at a counter on the other side of the hall. One was trying to get a piece of luggage away from the other. The whole place was utterly chaotic. Normally I'd enjoy this sort of thing but I was really on edge. This didn't bode well.
Back in the little room where my passport was, the four officials were now arguing with each other while JP stepped out for a cigarette with me. It was all mind-blowingly pointless and not the best welcome to the Likouala Province. Eventually, after much negotiation, another man arrived, who was – judging by his puffed-up manner and arrogant swagger – a boss of some kind. He turned out to be the regional-tourism official for the local government and he took little time to inform us that we were in big trouble.
My visa allowed me into the Congo but, for me to do any thing touristic, I should have got permission from the Tourism Ministry in Brazzaville. This we hadn't done, and we were now in the region illegally and could be arrested.
We were marched out of the airport by this new guy whom I shall call 'King' as he had that air of self-importance about him.
We were bundled into the back of a pickup truck and driven into town. We stopped at a wooden shack that revealed itself to be the Centre Pour Le Departament De Tourism Du Likouala. I'm guessing that this is probably one of the least busy buildings in Africa.
King marched us into his office, which was like a sauna. An imposing photograph of Nguesso started down at us from the wall. King shouted at his secretary, who was sitting in an anteroom full of books, saying his office was a disgrace and asking why was it so untidy.
He was clearly trying to lay down the law and show that he was an important man. We nodded and looked suitably impressed. He picked up a mobile and rang his boss, the head of the prefecture, because (as he kept repeating to us like some demented mantra): _'On a un hiérarchie ici, et il faut le respecter.'_ We nodded in agreement. With his boss on the line he informed him in puffed-up terms that he had two strangers here with no papers and that they were proposing to go to Lake Tele. He told his boss that the WCS had once again broken their agreement about being purely a scientific organization. They were now organizing tourist trips. He got off the phone and told us that we were to be taken to the prefecture. We nodded and smiled like this was the best news ever but JP looked worried. For about fifteen minutes we were marched through town, down dusty tracks and back alleys. The sun was burning hot and my rucksack was starting to cut into my shoulders.
Eventually we arrived at an unpainted concrete building with a terrace running around a little garden. Off this terrace were dozens of little offices full of official-looking people. We were ushered into the secretariat, a boiling-hot room in which sat three secretaries listening to music on a mobile phone. They were singing along and totally blanking JP and me.
We sat there for about twenty minutes with nobody saying anything to us. Eventually it got too hot and we escaped to the relative cool of the terrace. Half an hour later and King finally came out of an office looking a bit flustered and being a tad more friendly. He had clearly been given the brush-off.
'My boss is too busy to see you but he says we should go back to my office and we will do the necessary requirements . . .'
This sounded a bit more promising. We walked back down the sweltering dirt streets towards the shack with King's assistant, Noel. King had got a lift back in a car but we didn't mind as Noel was much friendlier. We started to talk about beer. Noel, it turned out, was a huge fan of Guinness. I pretended that I also was the world's biggest Guinness fan and we both made vague sounds of Guinness appreciation. When we got back to the shack King was still not there and Noel took us to the bar next door, where we had locally brewed Guinness and decided that no country with oil could ever be happy.
After a couple of resuscitative pints of the black stuff we returned to the shack to find King looking very miffed at Noel for slacking on authority. We sat back down in his sauna/office and watched as he spent ages filling out two official-looking forms replete with lots of rubber stamping and copies for various in and out trays that nobody would ever read. He'd occasionally look up and ask us a question, like how long we intended to be at Lake Tele. We took educated guesses but we really had no idea whatsoever and he knew it. Eventually he brought up the subject of money. Technically, he said, we should be paying a fine of 100,000 Congolese francs each but – and here he raised himself to his full pomposity – he did not operate that way and so we would only have to pay 50,000 Congolese francs (about €80) simply for the permit that we needed to be tourists. The whole charade was total nonsense but there had been hints of overnighting in a jail and we were both immensely relieved. JP paid the money, we got a receipt and we were allowed to go on our way.
There had been no sign of the WCS people and it was now too late to attempt to take the road to Epema, as it was getting dark. We asked around and found a guesthouse called Le Rosier where we could stay the night. It was fairly clean but had no electricity or water. We threw our stuff on our beds and I lay down for a rest.
JP's bank card had not worked in Brazzaville but he had got some money wired from Cameroon to the post office, so he headed straight off into town to try to get it. He was back pretty quickly because it turned out that the post office had closed at two in the afternoon. We were on a fairly tight schedule and this delay in Impfondo had already set us back, but JP had a plan. We could get to the WCS in Epema the following morning and set off straight away downriver. We hopped on to the back of two motorbikes ridden by local kids and found a restaurant called Tropicana right on the banks of the Oubangui River, a wide offshoot of the Congo.
As we were finishing up a man turned up at our table. He was Hermes, the driver from WCS, and he'd been looking for us all day. Hermes was with a friend who had actually been to Lake Tele. The friend told us that it was two days' walk from the village of Boha and that the water en route was not too deep at the moment. He said that it came up to your knees at the worst parts. This still sounded totally horrendous but it was better than what I'd read that we could expect. JP told them that we wanted to leave early the following morning after he had got his money from the post office. Hermes nodded in agreement. It looked like we were back on schedule.
We sat on the grass outside the Tropicana where a makeshift screen had been set up and a very dull French soft-porn film was showing. Our surly waitress was not happy with the choice of film and started shouting at the men watching. Someone changed the channel reluctantly. Suddenly we were watching Southampton vs Tottenham. God how I loathe football . . . But JP loved it. It turned out that he had been a very promising player in his native Cameroon but his father had disapproved and forbade him to play. JP had sneaked away and kept on playing. He eventually played in two international league matches but his father then heard his name on the radio and that was that. I told him that I loved cricket and he asked me who I support: India or South Africa?
We walked slowly back to Le Rosier through the town. It was pitch-black but we could hear sounds of life all around us. Occasionally a motorbike would appear out of the darkness and roar past us. JP and I talked about the Mokèlé-mbèmbé. His personal view was that it was more of a bogeyman-type thing that was used to keep order – i.e., 'behave or the Mokèlé-mbèmbé will get you'. I looked disappointed at this and he smiled at me.
'There is a thin line between reality and mystery in Africa, Dom.'
From somewhere nearby came the sound of a group of girls singing together. It was either a church or maybe a party. It was powerfully beautiful.
Back at Le Rosier I slept fitfully as it was hot and the bed was almost deliberately uncomfortable, but I was well aware that tomorrow this would feel like a Mandarin Oriental. At least there were no mosquitos. The rainy season had finished about two months before but I was still on hyper-alert, having been given a quick run-down of all the things I could catch from those buzzing plague-ridden bastards.
JP woke me up very early the next morning and we headed off through town for breakfast. I marvelled at how rapidly one adapted to places. Only yesterday we were under arrest and being marched through these streets by King and I'd wanted nothing more than to go home. This morning, however, I didn't want to go anywhere except off on our adventure.
JP and I shared a generous bowl of _ndongo_ (chillies) with our breakfast omelette. He told me that I was the first Westerner he had ever travelled with who could pronounce words like _ndongo_ and Impfondo correctly. I was incredibly chuffed.
After breakfast we sauntered down along the river towards what passed for a commercial district. I tried to walk slowly as it was already very hot. After ten minutes we got to the post office, which also served as the town bank and MoneyGram office. JP had been told that it opened at seven-thirty in the morning and he admitted that before I was awake he had already been and found it closed.
It was now nine-thirty and the place was finally open. We entered to find a lone woman sitting in a darkened room at a dirty wooden desk. JP told her that his assistant in Cameroon had paid in money at her end and he was here to pick it up. We needed the money to pay for boats and porters. No money, no trip. The woman gave JP a form to fill out. He completed it carefully and handed it back to her. She looked at it long and hard. JP was asking for 750,000 Congolese francs (about €1,100). After a long silence she looked up.
'The moneyman is not here. You must come back later.'
JP was annoyed by this and it showed. 'You are a bank. How can the moneyman not be here? This is your business!'
'You must come back later.' The lady was not for turning.
We asked her at what time the moneyman would be there.
'Two, maybe three hours,' she replied in a frustratingly noncommittal fashion.
JP turned on the charm and told her that we were on a very tight schedule as we had to get to Epema in time to catch a boat to Boha before sunset. She looked spectacularly uninterested.
We went and got a coffee and kicked our heels for an hour and a half. Then, unable to wait any longer, we went back. To our delight there was now a man in a passably smart shirt and trousers sitting next to the woman. JP asked him if he was the moneyman. He nodded gravely and confirmed that he was indeed.
JP produced the form that he'd filled in earlier and gave it to Moneyman. Moneyman looked long and hard at the form – far longer and harder than the woman had done before. Again there was total silence. Finally, after what seemed like about five minutes, Moneyman looked up at us.
'We have no money,' said Moneyman matter-of-factly.
JP looked shattered. 'No money? But . . . You are a bank . . . How can you have no money?'
Moneyman shrugged his shoulders. 'Yes, we are a bank with no money. We had to pay the Americans yesterday. They have very big salary so we have no money.'
We asked him what Americans he was talking about. He told us that there was a UNHCR (UN Refugee Agency) camp just outside of town. We hadn't seen any foreigners and were completely unaware of their existence. JP reiterated to Moneyman just how important it was that we got some money. We had only limited time to get to the lake because . . . Moneyman shrugged and JP stopped his explanation. It was no use. As we started to leave there was a hint of guilt from Moneyman. He told JP that he could ask around the market and see if he could borrow the money. I looked out of the door at the motley collection of stalls selling little plastic bottles of petrol and assorted bicycle parts and thought this was most unlikely.
Nevertheless, Moneyman said he would ring us if he was successful or if any money came in. We returned to the guesthouse to wait but already it looked like we wouldn't make Epema that night, let alone Boha. There had been no sign of Hermes and his friend from the WCS. They had been supposed to pick us up at nine but there was no answer from their mobile. Time was clearly a very relative concept in the Congo . . .
As we sat waiting outside our rooms at Le Rosier, I realized that I was secretly quite pleased with the delay. It meant one day less in the swamps, one day less of hardship. This is quite a common sensation for me. Whenever I was driving around looking to do a hidden camera stunt I was always relieved when something went wrong with the set-up and we had to delay things. Once I was actually in the thick of it I loved it, but the pre-tension was unbearable.
Finally JP's phone rang at around one p.m. It was Moneyman and he had managed to scrape together 350,000 francs. Someone had come in and paid some money over to someone in Brazzaville. It was not enough but JP hoped he could pay WCS by wire transfer and use the cash for porters and negotiations with the village chief. However, the problem was that we still couldn't get hold of the WCS driver to take us to Epema. It was incredible enough that there was mobile phone reception in Impfondo, but this was all part of the process of 'municipalization' that the government had implemented in the last six years in an attempt to ensure that all towns in Congo have at least one tarmac road and better public buildings. At the time of my visit the road from here to Epema had been built only four years previously, with Brazilian money, and was the only tarmacked road in the entire province.
JP went off to get the money from Moneyman before Moneyman lost his money. I started reading _King Leopold's Ghost_ by Adam Hochschild. Finally at around five p.m., the WCS people arrived at our guesthouse. They seemed entirely unconcerned with being almost a day late. Hermes had brought his boss, a Rwandan and a man called Sylvestre who had been to Lake Tele four times. He told us it would be a two-night camp from Boha, with us arriving at the lake on the third morning. Today was Friday and we had to be back in Impfondo by Thursday morning to get the plane to Brazzaville. This left us with very little time to play with.
We went through the finances of the trip. Sylvestre wanted 5,000 Congolese francs per porter per day. He was going to come with us and his fee was 10,000 Congolese francs a day. He estimated we needed to pay around 100,000 Congolese francs each to the chief and that we would need 80,000 for food and supplies.
We went to the market for the supplies. We needed two bottles of water per person per day, tinned tuna, hamburger buns, pasta, rice, tomato sauce and some Babybel cheese. On our way back from the market we spotted that a large boat had docked in town and we went down to have a look. On deck was a white man: a rotund Frenchman who seemed rather surprised to see us.
The boat turned out to be a bi-national cooperation between the Central African Republic and the Congo. He had been working on it for seven years and knew the rivers very well. They would trawl up and down the river from Bangui to Brazzaville putting in buoys and arrows to show boats where to navigate. The Frenchman told us that it was a never-ending task, as the power of the river shifted the sand constantly.
I asked him if he had ever seen the Mokèlé-mbèmbé. He looked puzzled. He didn't know what it was. A Congolese crew member sitting on an oil drum and having his head shaved with a rusty-looking razor looked up.
'Mokèlé-mbèmbé? _Le dinosaur?'_
I nodded and ask him if he'd seen it.
'I haven't, but I know plenty of people who have.'
He got very animated and I feared for his scalp. He told us about a place on the river charts where everyone said there was a Mokèlé-mbèmbé. He said that all Congolese avoided this bend because of the beast. Another deckhand spoke up. He claimed that planes didn't fly over the Lake Tele area because the beast had a magnetic power that dragged them into some sort of aerial whirlpool before crashing them. While I was happy that they knew about my quarry, I hate stories like theirs. These were so ludicrous they made the reality of a Mokèlé-mbèmbé seem unlikely.
The Frenchman laughed and promised us that he'd keep an eye out. It was another four days downstream to Brazzaville from Impfondo.
Bangui to Brazzaville on a boat down the Congo – now _that_ was an adventure. But sadly it would have to be for another time. We said goodbye to the Frenchman and scrambled back up the bank past a mother washing four little naked kids on top of an upturned dugout canoe.
Once back at the guesthouse we were finally ready to go. The Rwandan was driving a big white Toyota Land Cruiser, the vehicle of choice for charities and relief agencies worldwide. We squeezed into the back and set off out of Impfondo along the road that the Brazilians built. Now, I know little of the Brazilians' road-building capability, but let's just say that I hoped this wasn't the jewel in their crown. The road was essentially a series of joined-up potholes with the forest attempting to reclaim the route from all sides. We had to go at a snail's pace to avoid breaking the Land Cruiser's suspension.
Sylvestre was sitting opposite me in the back and I asked him if he believed in the Mokèlé-mbèmbé. He said that he hadn't seen one but he knew of many people who had. This seemed to be a stock answer round here and I hoped that I might get some first-hand experiences from someone in Boha.
It took about two hours to get to Epema and most of that time was spent listening to the government _conservateur_ of the park having a screaming argument with the others about the Bible.
His French was very heavily accented and I couldn't follow everything but the main thrust seemed to be about vegetarianism and how Daniel, because he was a vegetarian, was not eaten in the den of lions as he didn't smell of meat. At one juncture they asked me what I thought and I admitted that I had no idea but I had heard that the Mokèlé-mbèmbé was vegetarian. They all nodded and said that, yes, it was a herbivore.
'I hope so,' I said, mock-nervously, and everyone roared with laughter. Outside the vehicle was total darkness. Occasionally I could spot the glimmer of a fire outside a hut through the thick trees. Every so often the headlights would catch the surprised face of someone walking in the pitch-black along the road. Where were they going? Come to think of it, where were _we_ going?
We rolled into the WCS compound in Epema, where JP and I were given a very basic room with two beds. The sounds of the forest were all round us.
'I'm afraid that I snore,' I warned JP.
'No worry – I make gas,' he replied.
Sylvestre promised that he would wake us at five in the morning and that we'd immediately set off in a boat for Boha. We woke up at seven. Nobody had woken us up and we stumbled out and tried to get things going. Predictably everyone was still asleep and it took another good couple of hours before we were ready to get on the boat. I glanced at my iPhone and was astonished to find that I had reception.
I once did a show for Radio 4 about how extraordinary the mobile-phone boom has been for Africa. It's allowed fishermen and traders to check where they can get the best price for their goods as well as keeping migrant families in touch with each other. In the days before mobiles this was a major problem and would have made our already tricky trip almost impossible to coordinate. Now, though, men in day-glo orange vests patrolled the streets of every town we'd been in trying to sell people 'credit'. All the government had needed to do was erect some mobile-phone masts and the money started to roll in. Every African 'rich' kid wanted to get in on this boom business.
I sent Stacey a final text:
I'm getting on the boat and heading off into the unknown – laters xxx
Finally, Sylvestre, JP, the boat driver and I (plus all our supplies) set off down the misty river of Likouala aux Herbes in search of dinosaurs.
The river was bordered on both sides by a swampy savannah and you could see the high-water marks from the rainy season. Kingfishers swooped and dived all around us. On the banks perched large herons and the occasional vulture. Beneath us were crocodiles, hippos and possibly a Mokèlé-mbèmbé or two. I felt very vulnerable in our tiny boat.
Every so often we would pass fishermen standing tall on their long, thin dugout canoes, the rim just inches above the water level. As seems to be the international boat convention, everyone waved at each other frantically. Admittedly this was a friendly thing to do but why does this just happen in boats? Why don't we all wave at each other every time we see someone in another car? I made a mental note to start doing this when I got back home.
The sun rose higher in the sky as we ploughed on down the river at a steady but unexciting pace. The river was still and trees were reflected in the water as though in fantastical mirrors. After a time my eyes started playing tricks on me and tree stumps became men and logs metamorphosed into giant crocodiles. The herons all sat ramrod-straight on dead branches, seemingly contemplating life as they knew it. They appeared Zen-like in comparison to the hyperactive kingfishers, swooping up and down looking for the slightest hint of a fish. I felt incredibly peaceful: to travel is better than to arrive – and I was slightly dreading the arrival. It was the calm before the storm.
Our boatman waved at a man sitting on a log by the riverside. The man beckoned us over and we turned sharply towards the shore and beached the boat. The man lived in a tiny hut with his wife and two little kids. He would catch fish and then smoke them on a wooden trestle table that hung over a slow-burning fire. Every month he would make the trip to Epema in his pirogue to sell the smoked blackened fish. We bought 2,000 Congolese francs' worth, which JP chucked into the front of the boat where their blackened, dead eyes stared balefully at me as we proceeded on down the river.
An hour or so later we rounded another corner and came across two men in their pirogues. One, who bore an extraordinary resemblance to Snoop Dogg, was holding an antique rifle and had a large dead python wrapped round it. The other, whom I shall call Bulldog, was a more physically intimidating-looking man and held a long, nasty-looking spear.
'We are in luck,' said Sylvestre quietly. 'These are the very men we need to talk to . . .'
It turned out that Snoop Dogg, who looked very young for the role, was the village chief, whereas Bulldog was a tribal elder (although not that elderly).
Snoop Dogg seemed to be in a good mood on account of his python kill, whereas Bulldog was friendly enough but a little distant. Aware of the constraints of time, we offered to tow them back to the village. They agreed and each sat in their dugouts holding on to the sides of our boat as we became a kind of DIY catamaran.
After ten minutes we turned right off the main river and down a side tributary. To our left we began to see the village half-hidden in the trees. Our arrival sparked much interest and most of the village rushed down to the river to watch us disembark. We climbed a steep mudbank and entered the village right by the chief's hut. He went inside and produced a couple of low, home-made chairs that he beckoned us to sit in. The chief, it quickly became apparent, was also the government's man in the village. He told us that he was responsible for any white man in his area and he wanted to be sure we had no ulterior motive for visiting the lake. He seemed quite smart, rather charismatic and young – maybe thirty? We told him why we were there and that we were on a very tight timeframe and needed to leave as soon as possible on that day to have a chance of reaching the lake and returning on time. He nodded and said it was possible to organize porters and do the trip, but that he needed his 'chief's fee' for this to happen. We discussed the fee and it was within our budget so we handed the money over and we shook hands. This had really been too easy. It looked like the travel gods were on our side. The chief stood up and said that he'd start thinking about who would go with us. While he did this, he said, we should go and greet the village elders and get their blessing for the trip. This was all going swimmingly and we set off through some quite thick forest to where the elders were assembling.
On the way I asked Sylvestre about Boha and how it was that the people here looked after Lake Tele since they were so far away from it. He said that the tribe used to live around the lake but they, like everyone else, were forced to move to the riverside so that the colonial authorities could keep an eye on them. So the tribe had moved to Boha but they were still custodians of Lake Tele and anybody going to the lake had to go through them.
Sylvestre pushed his way through a thick bush and we found ourselves in another little village. It was all part of Boha but this area had definitely been built with a view to keeping a distance from the rest. This was where the elders lived and we could see a couple of their wives pounding manioc, their staple food, in the doorways of their huts. We rounded a corner and walked into a central area that was clearly used for meetings as there were two long low benches on each side. It was indicated to us that we should sit on one of these benches, and this we did. I looked around us. Directly opposite us were the elders. There were about five of them, including Bulldog, and they were all holding the rather nasty-looking spears we'd seen earlier. A couple also had machetes hanging by their waists. In the middle of them was a man who seemed to be a lot older than the elders. He looked fairly ravaged by life. His eyes were bloodshot and slightly crazed-looking. He wore a tattered old combat jacket that was open to the waist. He stared at us with a look that didn't immediately scream 'Welcome to the jungle.'
To our left sat about fifty men from the village all settling down as though about to watch a good match . . . Which they probably were.
JP, though normally pretty cool about things, was visibly quite unsettled by the amount of weapons on show.
'Will this take long?' I whispered to him. We needed to get cracking as soon as possible if we wanted to get any distance towards the lake before we had to camp.
'I have no idea what is going on,' replied JP.
A young, very tall man stood up. He was holding a wicker brush and another long spear. Sylvestre explained that he was the porte-parole. He would stand in between the elders and us and relay any messages. We were not to speak to the elders directly. Everything had to be directed through Porte-Parole. I couldn't believe this system at first but actually it wasn't that dissimilar to MPs directing their remarks to the Speaker in the House of Commons. Supposedly it helped to avoid full-on arguments.
Proceedings started with Sylvestre, speaking through Porte-Parole, greeting the elders and telling them that we had come a long way to go and see the lake and wanted their blessing for the trip.
Porte-Parole relayed this to the elders, who all nodded and grunted in what looked like a fairly amenable manner.
Then the crazed-looking man in the combat jacket, whom I shall call Crazy (because, frankly, he was), started to speak. I say speak; it was more a series of shouts and gesticulations. Porte-Parole listened and, after a little pause, informed us that Crazy was happy that we'd come to see them and that they would give us their blessing if we paid them the sum of 250,000 Congolese francs. This was totally out of the question. Firstly, we had just paid Snoop Dogg for the privilege of making the trip to the lake and we were not about to pay twice. More importantly, we just didn't have anything near that sort of money.
Sylvestre stood up and thanked the elders for their kind offer but hinted that this was a little more than we had been expecting to pay. Both JP and I hissed at him that it was a _lot_ more than we had expected to pay, since we had already paid. Porte-Parole took this all on board and passed it on to the elders.
The elders went . . . apeshit. Crazy started waving a machete at us in a distinctly unfriendly manner and Bulldog was shouting at the other elders and pointing at us.
I asked Sylvestre what was going on. Why were we negotiating to pay more money that we didn't have when we'd already paid the chief? Sylvestre explained that Snoop Dogg was chief but he was the government's man, whereas traditional tribal authority rested with Crazy and the Elders (good band name). They had no interest in what we'd negotiated with Snoop. As far as they were concerned he was an irrelevance. They were the top dogs and we needed to pay them for access. I asked Sylvestre why he had not mentioned this before and he shrugged in that infuriating African way. I could see JP subtly looking at how much money we had left and making a quick calculation of what we could afford. I already knew that it was not much and the clock was still ticking.
JP took the floor and did his flowery-French thing. He dropped the fact that he was a prince back home and said he knew how these things worked and didn't want to offend. He then went on to explain that we'd had a lot of unexpected problems on the trip and this had left us short of both time and money. He started off on quite a long allegory about a hunter going out into the forest and chopping down trees that were too big and took too long to chop down, and he didn't have enough provisions so he went home without the trees because it was dangerous in the forest at night. I was just about following this and hoped that Porte-Parole could convey it in full. Porte-Parole did his best but, when the gist became clear that we were not going to pay anything near the amount they wanted, Bulldog exploded. He ignored Porte-Parole and started screaming at JP and me while waving his spear about.
'If you don't like the price then go back to where you came from. We have no need for you here. Anyway, the _gros bébé_ will never make it to the lake – he will die on the journey . . .'
It took me a moment to work out that the _gros bébé_ in question was me. It took me a little longer to confirm that he was not threatening to kill me but was sure that the forest would . . .
I looked at JP and he looked at me with a sense of foreboding. We agreed to leave the clearing and discuss what we should do next. Curiously Snoop Dogg had now turned up and insisted on joining our discussions. I was already annoyed with him for taking a payment that gave us nothing, but joining our secret negotiations was a bit much. JP and I talked, and worked out that 50,000 Congolese francs was the best we could do and they would have to take it or leave it. Snoop Dogg nodded at this offer and said they would listen to it. I asked him whether he had at least organized the porters so that, should this sort itself out, we could leave immediately. He looked at me vacantly and I guessed the answer.
We returned to the clearing and I told Porte-Parole what we could offer. Bulldog would now not even look at us and Crazy had gone nuts again. He shouted and screamed for about five minutes until even the other elders looked a little disturbed by whatever it was he was shouting. Porte-Parole looked a little embarrassed and gave us what was clearly an edited version, saying, in effect, that we should get back on our boat and leave, pronto.
JP stood up again and laid on some more flowery prose. This was becoming like a weird poetry slam. He laid on the flattery very thick and basically said that we were now in a situation where they either got something or nothing. A couple of the elders seemed to accept this logic but Crazy and Bulldog were now competing with each other to shout at us. Finally JP played his last card. He offered 80,000 Congolese francs and said that this was the final offer: they should either take it or leave it – but, whatever, if we didn't leave immediately there would be no point in any of this. Porte-Parole spoke and I sort of felt that he pleaded our case a little as Crazy and Bulldog calmed down a little. Porte-Parole told me they were going elsewhere to discuss things. All the elders got up and disappeared into a nearby hut. Porte-Parole stood outside a little awkwardly as a guard.
They spent fifteen minutes in there and then emerged to return to their seats. Crazy stood up and spoke quite pompously for about eight minutes. Porte-Parole listened intently and then turned to us and said, 'They accept.' He smiled at me.
It had taken about two hours but we finally had a decision and we were good to go. We looked over at Snoop Dogg and asked him how long it would be before we could get going. Snoop looked at his watch, except he didn't have one, and then said, 'Soon, but first we celebrate.'
Porte-Parole had slipped into a hut and was now coming out with about five bottles of something very dirty and visibly home-made.
'Oh no . . . Jungle gin,' said JP 'We are all supposed to drink to the agreement.' He grimaced.
Now, having done a whole TV series for which I went round the world drinking revolting local alcohol, I do have some form in this area and I doubted anything would be more lethal than the 90-per-cent-proof Samagon that I drank outside St Petersburg. I was wrong.
Snoop poured me some jungle gin and I sipped it. It was quite horrific and I felt dizzy. JP pretended to drink but didn't swallow anything. The rest of the elders weren't so reticent and started knocking it back like it was Happy Hour at Hooters. More worryingly, so did the whole village. We tried to leave the circle but this was considered bad form so we had to stay and watch as everyone got blind drunk and started stumbling over spears and fighting and generally behaving like they were at a lock-in at the village pub. After a desperate hour of this, during which we pretended to drink along, we managed to slip away back to the main village to get ready.
We were done in ten minutes and we then stood around outside Snoop's hut looking hopeful. Our boat driver wanted to know if he could leave, as he wanted to get back to Epema before nightfall. I realized that once he'd gone we would have no way of contacting him until he came back to pick us up in five days' time. There had been no mobile reception since Epema and I was very wary of letting him go until we knew what was happening. I asked him to wait. He wasn't happy about this but I had a feeling that things were going to go weird.
I sat down with JP and Sylvestre and tried to make a plan. The big problem seemed to be that nobody was prepared to be specific about how long it took to get to the lake. Some people were saying two days while others were saying three. If it was three days each way, then we wouldn't have enough time to get back to catch the plane. I needed someone to give us a definite answer so we could make a decision.
Snoop Dogg turned up with three men whom he informed us were our porters. They were all totally drunk and stumbling about. It was becoming very clear that whatever happened nobody was going anywhere today. I needed to know if it was possible to do the trip if we left the following morning but Snoop wouldn't give me an answer. Finally I'd had enough and sat Sylvestre down in front of me. He'd been to the lake three times and was our best bet.
'Sylvestre, can we make the lake in two days' walk? Yes or no?'
Up until now he had been hinting strongly that it was possible in two days but we had never really pinned him down about it. He looked shifty and started to talk about how we'd need some manioc for the porters . . .
I ignored him and asked him again: 'Sylvestre, can we make the lake in two days' walk? Yes or no?'
Sylvestre's eyes flickered around in panic looking for a distraction but there was none. He looked at me, and his face dropped.
'No, it is impossible in two days; it is at least two days and a half.'
We were fucked. JP and I both knew that this news, plus the fact that we couldn't leave today because the whole village was blind drunk, meant we were not going to be able to get to Lake Tele.
JP and I went for a walk around the village to discuss our next move. I told him that I was there to get information about the Mokèlé-mbèmbé. We had to accept that we couldn't go to the lake but, since there was nobody living around the lake, I could still get some answers from the villagers . . . If they sobered up.
JP looked at me with worried eyes. He knew I was right – that we couldn't make the lake – but he'd promised to get me there and I could see he felt bad. I told him not to worry: shit happens. The important thing was to see what I could get out of the villagers about the Mokèlé-mbèmbé. JP agreed and we decided that, whatever happened, we would spend the night in the village and take it from there.
We returned to Snoop Dogg's hut, where a quite drunken crowd had now gathered and were hanging about. JP announced to them that the trip we had spent three hours negotiating was now off. There was an immediate air of great tension and too many machetes and spears suddenly appeared for my liking. Snoop went mental and asked us if we were mad – which was a fair question. JP tried to calm people down by explaining how our series of delays – first with King, then Moneyman, then the WCS, and now the drunken initiation – had left us with no time left to do the journey. He then attempted to explain the concept of Western time to the villagers, who looked absolutely bemused by the whole idea.
'You wanted to go today and you can go today?' said the chief suspiciously.
'We wanted to go today, early this morning, not today, this evening. We have no time left to go to the lake and back,' said JP.
'Why not? You can be there in three days?' said another villager.
'Because we have only five days in total for the journey now . . .' said JP desperately.
Everyone nodded wisely and then asked what time we now wanted to leave.
JP calmly explained again that we would not be going to the lake but that we wanted to talk to them about the Mokèlé-mbèmbé. Snoop Dogg now realized that we were serious and seemed to sober up quite quickly. JP and I sat down and opened a couple of beers while the Snoop discussed this peculiar new state of affairs with the drunk porters. We watched them all try to make sense out of this group who had suddenly arrived out of the blue one morning, organized porters, spent a spirited three hours negotiating forest access with the tribal elders, paid the money, and then cancelled the whole trip. It was clearly something that they would talk about for years to come but, for now, most of them were still a little ripped to the tits on jungle gin to really think about it.
An hour passed during which we had a bit of lunch from our plentiful supplies. Then a stressed-looking Porte-Parole appeared to inform us that the elders wanted to see us again. We groaned inwardly but trooped over through the forest to their part of the village. As before, the elders were sat around waiting for us. Bulldog, however, was not there – but Crazy was and he was waving an empty bottle of jungle gin. Crazy launched straight into Porte-Parole for about four minutes. Porte-Parole then told us that Crazy was confused by this extraordinary situation and that he was worried that they had taken money and not done the job. This was technically illegal and they feared that the police would come.
This was a lot more rational than I'd anticipated and, through Porte-Parole, we assured Crazy that this would not be the case. We told him we would not complain and that circumstances had just conspired against us to make us run out of time. He could, of course, simply return the money should he so wish . . .
JP launched into some flowery French again. He told Crazy that sometimes travel plans go wrong. Maybe God himself did not want us to make it to the lake and we would have been struck down by giant pythons?
Crazy nodded and confirmed that January and February were the very worst time for pythons.
JP was now on a roll: 'Maybe one day the elders have gone into the forest to cut down a tree. They cut a little one down and carry on. Then they find a medium-sized tree, cut it down and carry on. Finally they come across a huge tree and start to try to cut it down but it takes ages and they there all night and their food and water run out and they are are in trouble . . .'
Crazy nodded at this story which appeared to be the same one that JP had told earlier. Crazy said that he understood the point – although he personally did not cut down trees, as he was a tribal elder.
The elders then asked for a final promise that there was no problem between us and we assured them that all was fine. They could even keep the money: all we asked was that we could talk to them about the Mokèlé-mbèmbé.
Crazy agreed immediately and told us that the village was at our disposition. He went further: he said that he would personally get a group of elders together at five p.m. and they would tell us everything they knew about the Mokèlé-mbèmbé. He warned me that I would run out of pages in my notebook as he personally had so much to say on the subject. Porte-Parole told him that I had a magic notebook that used electronics and had no limits. Everyone nodded. Porte-Parole slammed his spear into the earth: the meeting was over.
We returned towards the main part of the village where tents were to be set up near Snoop's hut for us to sleep in that night.
On the walk back JP apologized for talking too much.
'I have to talk in pictures here – like with the tree story – it is the way an African speaks and it can be very long-winded.'
We laughed and followed Porte-Parole along the open scrub area that linked the parts of the village. We walked past a cool, shady spot under a tree where they were making palm roofs. Five kids ran out of a house and started shouting 'Hello' at us.
'My kids,' smiled Porte Parole. _'Petits_ Porte-Paroles.' We all laughed.
Once back at Snoop's hut we sat down at a table that Snoop had put under the cool shade of the tamarind trees. Snoop was looking concerned again and asked if he could speak to us. We nodded wearily.
He consulted a little child's maths exercise book in which he had written down our names.
'Mr Dominic Joly _et_ Mr Jean-Pierre Samon . . . I will not talk for long . . .' he said, before talking for a very long time. The gist of it was that the village of Boha did not receive many tourists (sixteen in the last thirty years) and he wanted us to help him start a tourist industry there. JP and I nodded enthusiastically while looking at each other and thinking that this was probably not the best time to have this discussion. After a good twenty minutes of non-stop fast French patois it was as though we had been hit with a verbal machine gun.
'I am not the sort of person to talk for an hour . . .' he said, looking like he was winding himself up for another round. I told him that I had to talk to villagers about the Mokèlé-mbèmbé and slipped away, leaving poor JP to round two.
I talked to a man in a bright-yellow shirt emblazoned with photographs of the president. I had not seen this man before and he appeared to be the most sober man in the village. His name was Mandzamoyi Marcelin and he said he was the secretary of the village, whatever that might be.
He told me that Lake Tele was originally a little pool that their tribe, the Bakolou, would all hunt and fish around. Then the lake started to grow in size because the Mokèlé-mbèmbé would dig channels to allow themselves in and out. These channels obviously allowed more water into the lake and it got bigger and bigger. The Bakolou were unhappy with the Mokèlé-mbèmbé coming into the lake because they ate all the fish. So, in the time of Marcelin's great-grandfather, the tribe built nine wooden dams in an attempt to stop the Mokèlé-mbèmbé.
They soon spotted a Mokèlé-mbèmbé trying to come in and it managed to break through eight dams before the tribe managed to spear it to death at the ninth. People dived in and cut bits of the flesh off the huge body. The whole village celebrated the kill by cooking a great feast with the Mokèlé-mbèmbé meat. Unfortunately, everyone who ate the meat died. Marcelin said that it killed more than a hundred people. He said the channels were still visible around the lake.
JP came over as the story was finishing and I filled him in on the rest. I asked him why they wouldn't have dragged the creature out of the water on to land. He said that this was a very Western attitude. If an elephant was killed in the forest people would come with knives and cut off the flesh to take back home to cook.
Sylvestre suddenly piped up. He had been very quiet since the whole cancellation debacle. He said that on his first trip to Lake Tele he had brought a man from Congolese TV. They had got to the lake and had just made camp when they spotted a huge shape in the middle of the lake. He said the Congolese man had filmed it and the footage was often shown on local TV. I thought this sudden admission to be a little odd, since I'd already asked him if he had seen the Mokèlé-mbèmbé and he'd said no. I wondered whether he was now trying to get back into our good books. (I have since searched for this footage online but have found nothing.)
At around five p.m. Snoop Dogg asked us to sit down at his table. Our boat guy had zoomed off to get some beer. Forty minutes later, he returned with a carton of Congolese red wine made in Pointe-Noire called, rather confusingly, Baron of Madrid. It was the single most revolting thing I have ever drunk – and I've had Irn-Bru.
Crazy limped his way over and sat down. He looked angry, but then he always looked angry. He spotted the wine and announced that it was a woman's drink. He then poured himself a large glass and downed it. Snoop launched into a long introduction as to why Crazy was _the_ man to tell us about the Mokèlé-mbèmbé. Crazy had another large glass of wine and looked out towards the river disinterestedly.
When Snoop finished, Crazy started speaking. He spoke fast with loads of gesticulations and I got quite excited, feeling sure he was telling of some epic encounter with the Mokèlé-mbèmbé. When he eventually drew breath I looked expectantly at Sylvestre for a translation.
It turned out that Crazy was very annoyed that he had invited us to the elder's area and was then told that he must come to Snoop's hut.
_'Lui, c'est l'état!'_ he screamed in French in case I hadn't understood his beef.
Crazy announced that he would not say a single word unless we came over to his area. Snoop laughed, obviously quite enjoying Crazy's discomfort. He said that it would soon be dark and that he did not go over to the elders' area in the dark because it was via a forest path and he worried about his security at night. Crazy shouted back at Snoop that us being at Snoop's table was an insult to the elders. Snoop told him to leave if he didn't like it, but we were his guests. Crazy crossed his arms in an overly dramatic gesture and sat sulking.
Nobody was going to give in. Then a man in another vivid-yellow Nguesso shirt stepped in and, in perfect French, explained that this had become a fight between the state and 'tradition'.
He looked at me and said: 'Get in your boat and leave. Your mission has failed and you are causing a fight between the chief and the elder.'
Crazy stood up and hurled abuse at Snoop before storming off with Porte-Parole. We all got up and followed him, including a very reluctant Snoop. Rather than walk the five minutes it took to get to Crazy's area, we all trooped down to the river-bank – where our boat guy was ordered to take everyone about 300 yards upriver, where we disembarked and walked into Crazy's clearing. The whole thing was getting ridiculous but we were now at least in the correct place and were going to get some great Mokèlé-mbèmbé stories.
Crazy sat down in his favourite place under a tree and grabbed a spear for effect. He started by admitting that he had never actually made it to Tele himself because of his bad foot (and attitude). He said that we had gone about the search for the Mokèlé-mbèmbé in the wrong way. We had come in a hurry and wanted to leave immediately. The correct way was to come and spend three four nights (the very thought!) in the village and then set off on our trip with their blessing.
At this juncture Snoop interrupted and shouted at Crazy, saying that he was not setting the story up properly. Crazy ignored Snoop and carried on talking.
'In the old days the Badzama, who lived by the lake, would put their manioc in the water but every day it disappeared and they couldn't work out why. They accused another tribe . . .'
At this point an elder sitting next to Crazy tapped him on the shoulder and pointed to the sky. It was nearly dark and he reminded Crazy that tradition dictated that they should stop telling stories at this time. This was a full ten minutes after we had finally begun. I started to wonder whether I was the victim of Congolese _Candid Camera._ You couldn't make this farce up. The elders disappeared into a hut for some more jungle gin and we were left to troop back through the forest with a nervous Snoop.
I now just wanted to go to bed. I'd had enough of Boha politics. We were told that we could return for more stories the following day but both JP and I were reaching the end of our tether. The boat guy had set up our tents while we had been away. One was badly broken but the other one was OK save for a large hole in the roof. We ate a sullen meal with Snoop. His wife provided some smoked river fish with rice with peanuts. There was also a bowl of the local staple food, cassava. It's very easy to grow but has little nutritional value. The leaves are edible and known as _saka-saka_ but it's the tubers on the roots that feed the Congo. First they're soaked, then cut up and dried until they're white and brittle and then they're pounded into flour by the women and made into _foo-foo_ , the bland dumplings on the table in front of us. It is extraordinary how little the Congolese grow for themselves. As we came down the Likouala aux Herbes JP had been marvelling at how perfect the swampy flatlands are for growing rice.
'If the Chinese came they would be in rice heaven,' he'd said.
Snoop started ranting again about Crazy but we'd had enough and called it a night. We got into our tents. It had got seriously cold and I had no warm clothes and no cover. I lay on a mat and stared at the stars through the hole in the roof. I could hear something slithering around just on the other side of the canvas. I became convinced that it was a python and stayed rigid in the middle of my mat for about an hour as whatever it was slithered all around the tent. Unable to sleep, I used my head torch to read the part in Redmond O'Hanlon's book about when he visited Boha. This was a big mistake. There's a particular passage where O'Hanlon asks his friend where they should pitch their tents. His friend replies that only a crazy man would camp in Boha. He insists that they sleep in a hut with someone guarding the door.
'I'm not going to be axed through the canvas in the night . . .' says the friend.
I started to imagine a spear suddenly slamming through the thin canvas into my sides. I wondered what it would feel like. Would it kill me immediately or would I go slowly, groaning, my lifeblood draining away into the sand?
As it so happened, I was neither speared to death nor bitten by snakes in the night but I was woken by the sound of terrible, terrible singing very close to me. It was worse than my mother-in-law on a road trip. (She is a wonderful woman who loves to sing but nature has blessed her with the voice of a tone-deaf hobo. It's a cruel fate – like adoring animals but finding out that you're allergic to them.) And, like my mother-in-law, the singer here was not going to stop.
I got up and clambered out of the tent. It was about five in the morning and the sun had just risen. I strolled down to the riverbank where villagers were already setting off for dawn fishing trips on their pirogues. A thin mist hung low over the water. Birds sang lustily and, for a moment, Boha was almost a pleasant place. I climbed up from the riverbank and walked down the dusty main drag. On a whim I turned left near Porte-Parole's hut and followed a little track. To my astonishment and delight, the first hut I came to had the words 'Boha – Pilote – Dinosaur' daubed on the wall in fading white paint. I bumped into Porte-Parole on the way back and asked him about it. He said that the owner of the hut had done it about thirty years ago, when the first interest in the Mokèlé-mbèmbé had surfaced. He'd hoped that there would be a flood of visitors he could guide to the lake.
And what happened?' I asked.
'Nobody came,' replied Porte-Parole ruefully.
I returned to the tents to find Crazy and Snoop standing outside Snoop's hut and having a furious shouting match. It was seemingly never-ending, like being stuck in a nightmare loony council meeting. JP was up and we went for a little walk and decided that things were getting a little out of hand and we should probably beat a retreat back to Epema. Bulldog had turned up again and was looking like thunder at us.
We returned and interrupted Crazy and Snoop to let them know that we were leaving. They immediately stopped fighting and Snoop produced a bottle of jungle gin and announced that, before we left, we must drink to celebrate. If I was honest, it was perhaps not the ideal breakfast drink and it burned my throat quite badly.
As we sat, Bulldog started accusing us of all sorts of things and got quite nasty. JP whispered that we needed to get on the boat and fast. Suddenly there was a commotion beyond Snoop's hut. A man appeared brandishing a machete. He was bare-chested and had cut himself all over his chest and arms and was approaching us fast and didn't look friendly. Blood was pouring from his wounds and it looked like a scene from hell. I recognized the man as one of the porters assigned to us the previous day. Had we left yesterday we would be in the middle of the forest with this man going crazy. It didn't bear thinking about.
He had an insane look in his eyes and they were focused right on me. Fortunately for me, two villagers grabbed him and there was quite a tussle with the man flailing away with his machete. He was eventually subdued and tied to a tree with rope.
We didn't wait any longer. JP and I headed for the boat with the entire village following us, shouting and screaming at each other and at us. We didn't hang about. Our boat guy was already in the boat with Sylvestre and JP shouted to him to turn on the engine, which he did. We hopped on and shouted 'Go!' to him. He needed no further urging: he'd been looking very uncomfortable throughout our stay. We pushed off and were soon free of the reeds and in deeper water. Back on the shore the crowd had got into a huge argument and were screaming at each other again. The whole thing was more Asterix than Tintin. As we left the reedy channel and joined the main river we all breathed a huge sigh of relief.
On the way upriver towards Epema we stopped at the village of Mohounda to get some more petrol for the boat. Despite the fact that a young thirty-five-year-old man had died of a heart attack in the night, they were very welcoming and petrol was provided and we were soon on our way. As we made our way upriver there were a lot of pirogues on their way down. Many were full and precarious with seven, eight, nine people in them.
They are all going for the funeral in Mohounda,' said Sylvestre. I asked him how long the ceremony took.
Three days and three nights of dancing and then they bury the body and everyone goes home.'
I was astonished at how quickly the 'grapevine' had informed everyone, all the way up to Epema, about the death.
We finally got back to the WCS camp and went to see the Rwandan boss. He said that he'd known that there was a problem when the boat hadn't come back. He didn't seem at all surprised. Carefully ignoring his part in our delay, he started slagging off the villagers – saying that if they behaved like this they would lose their rights to control access to the lake and that he would find another way in. He then told us that, six years previously, four Americans went to the village but refused to pay the huge sum they were asking and came back. It was nice of him to tell us all this now.
JP and I didn't want to be stuck in Epema. It was a total ghost town. Fortunately the director was embarrassed enough to offer to drive us to Impfondo himself. We were very grateful. After a minimal wait (in Congolese terms) of about two hours, while nothing seemed to happen, we were off.
The first part of the road out of Epema was in quite good nick and the director drove like a mentalist.
'This is the only road in the whole province of Likouala –' he said, half looking at the road and half at me – 'twenty-three thousand square miles and only a hundred miles of road . . .'
Up here, of course, the rivers were the real roads and this was what first excited the French and the Belgians: a ready-made artery of infrastructure for them to transport first ivory and then rubber to the coast for transport to the West.
We eventually arrived in Impfondo after a long talk about Pygmy discrimination. They were seen by locals as ' _sous-humain'_ and appallingly discriminated against.
We got dropped off at Tropicana and discovered that they had rooms as well as a restaurant. After Boha, it seemed like a five-star resort, and we had a fabulous meal of lamb, potatoes, rice and a lot of _ndongo._ We passed on the sautéed antelope.
Everyone was much more relaxed and Sylvestre was like a new man. The director asked us our plans. We were uncertain but worried that we might bump into King, who would surely make us pay more money as we did not have a _permis_ to _not_ go to Lake Tele. Everyone laughed. If you can't laugh, then Congo travel is very much not for you.
Sylvestre told me about a Swiss man who had come to Epema and wanted to see gorillas. They'd stayed in the forest for six days and didn't see a single one. The Swiss man went totally mental and was blaming Sylvestre and threatening him.
'I must make remote-controlled gorilla so I can control them,' laughed Sylvestre.
Sylvestre and the director left to head back to Epema. We headed for our new rooms to chill out. They were much better than those at Le Rosier, which had felt like they were modelled on a prison exercise yard. Our new rooms had a TV, air con and a shower. I was excited. I turned on the air con as it was boiling hot – but there was no electricity. JP went and made enquiries and found out that it would come on between six and nine in the evening. It was four-thirty, so I lay on my bed in a pool of sweat and read a book. At six-forty-eight p.m., the electricity came on. I know this because there was a sudden violent flash and some smoke from two bare wires hanging out of the wall above the bed. I tried the lights and a light bulb turned on but I received a moderate electric shock as penance. I turned on the air con. There were no life signs: it didn't work. I plugged in the TV very gingerly but that didn't work either. I went into the bathroom and turned on the shower: a tiny trickle appeared and I stood under it, desperately trying to get wet. After two minutes I had enough water on me to start lathering up and I attempted to get the forest off me. Just as I'd covered myself in soap, the water stopped completely and the electricity went off. I was left stumbling around with foam all over me – this was not good. I managed to find a towel and wipe as much off as possible. I then felt my way to the bed and lay down on it, still soapy and wet. The electricity came back on and another small firework display burst from the wall wires. I was wet and that doesn't mix well with electricity. I carefully returned to the shower but there was no more water. I gave up and lay back down on the bed. I was foamy and naked but I didn't care any more.
I continued reading my book. For the next hour or so I could hear quite a commotion on the terrace that ran past my room. There was much laughing and shouts and it felt like there was quite a crowd out there. I ignored it and carried on reading. The man I called Crazy turned out to have been the chief of Boha when Redmond O'Hanlon visited in 1995. There was even a photo of him, bare-chested, clutching his favourite spear and looking quite the young warrior. He looked about forty and fit as a fiddle. There was simply no seeing him as the haggard, rotting old man we had been dealing with. Among its many other properties, it was clear that jungle gin was not good for eternal youth.
At eight-twenty there was a knock on the glass door of my room. Both the door and window of our rooms were mirrored, which was quite disconcerting. I caught a glimpse of myself in the door. I looked like I'd just been to a San Franciscan foam party. I opened the door an inch to find JP looking a bit embarrassed.
'I think it's best you close your curtains – you are becoming something of a town exhibit . . .'
I didn't have the foggiest what he was talking about so I put on some shorts and stepped out on to the balcony. I looked at the window to my room which, in the daytime, had been mirrored and impossible to see in through. Now, however, with the lights on inside, the mirror had turned into a sheet of clear glass – as had my door. Unbeknown to me, for the last hour and a half I had become the equivalent of the women exposing themselves in windows in the Reeperbahn. The show, although catering to quite specialist tastes, had apparently been very popular. I was mortified and when I got dressed and entered the restaurant the whole place was awash with smirks, glances and laughter. I felt violated in a most curious manner. I ate an omelette and some _ndongo_ as quickly as I could before heading off to bed. This time I closed the curtains – not that this mattered now, as the electricity had cut out again for good.
About an hour later I started to feel unwell – very unwell -and I spent the rest of the night sitting on the loo with a head torch while I shed half my bodyweight into the Impfondo sewage system (which I'm guessing meant the river just below the hotel).
The following morning at breakfast I was still feeling delicate and very, very drained. Fortunately I had thought to bring some Dioralytes and these helped a lot. JP said that if we didn't make a journey it was because God didn't want us to. I'm not a religious man but I thanked the Lord for not allowing me into the swamp forest with a man so crazy he had to be tied up while I was shitting my insides out over holy ground.
JP was trying to work out how we would get back to Brazzaville. We'd checked the two boats in town the night before, including the one owned by the Frenchman we had met previously. However, both boats were going upstream to Bangui, so this option was a no-no. JP needed to try to get some money from Moneyman as there were rumours of a rogue plane headed for Brazzaville the following day.
We said hello to Moneyman who explained that, sadly, he once again had no money – but that he was fairly confident he'd have some at one p.m. The most common phrase you hear in the Congo is this: _'Vous savez, avec ça le grand probléme c'est . . .'_ and then a reeling out of all the possible problems ahead of you in whatever you're wishing to do. Actually scratch the word 'possible' – the problems will definitely happen . . .
Back in Tropicana I lay on my bed in the infernal midday heat. I would not have liked to be here in the wet season. Then, according to the Rwandan boss, the whole town – a grid of dirt tracks and shacks – becomes a mud bath and almost impenetrable.
At about two-thirty we got good news. JP had got money from Moneyman and thought he'd managed to swap our previous airline tickets for two on a flight to Brazzaville the following day. The man in question said that he would bring the air tickets to us at six that evening. If I was a betting man, I wouldn't have put a single Congolese franc on this happening.
Come six p.m. we were seated in the Tropicana garden having a beer. Nobody showed up. There was, however, a bit of drama unfolding to keep us entertained. A little man in a shiny suit came in, surrounded by three bodyguards. The man was a caricature of a self-important African oligarch. He was on his mobile and very upset. He was talking so loudly that he made my _Trigger Happy TV_ character look like an amateur. He was complaining to the local police inspector that he had just been beaten up by a crowd in the street. I knew nothing of the affair but my sympathies were almost immediately with the crowd. I tried to get some footage of this buffoon but one of the bodyguards spotted me and charged over to demand why I was filming. I professed total ignorance and claimed I was cleaning my camera. There was a brief stand-off but the guy backed off. Next, a large group of excitable youths entered the garden and surrounded the buffoon. I rather hoped that they were the crowd in question, here to finish him off, but he stood up and gave them a rousing speech at which they all roared their approval. His speech finished, he proceeded to lead the youths out of the garden as though to battle.
Suddenly a new character – a huge, fearsome-looking woman – entered the garden in a voluminous, multi-coloured, all-encompassing dress with a repeat pattern made of the colours of the Congolese flag and the face of the president. The woman squared up to the buffoon, right in his face, and started screaming that, if he wanted to cause problems, he should do so in parliament in Brazzaville – not here, where she had to live.
The waitress whispered to us that the buffoon was a senior member of the ruling party and this woman was his sister. He backed away as she continued her verbal assault. The buffoon was having a bad day. A policeman turned up but appeared to be nervous of both parties. He tried to separate them and they both turned on him. Eventually the whole circus poured out into the street and marched off shouting at each other.
As if on cue a tall, thin Congolese man wearing a Pete Doherty cast-off hat pressed 'play' on a PA system that he had been setting up in the garden. A gritty rap song blared out. The hook phrase, endlessly repeated, was, 'I know you niggazzzz wanna fuckkkkkk meeee . . .' On the grass three five-year-old kids danced innocently a yard away from the speakers, apparently transfixed by the hypnotic groove of this appalling tune.
JP and I downed another Ngok and prayed that the plane would take us away from all this tomorrow.
The rap stopped and now Phil Collins's Against All Odds' polluted the African night. The girls looked disappointed and stopped dancing: some things have no frontiers. We called it a night at ten. There was still no sign of the airline-ticket man, whom I was now convinced was a close relative of Moneyman. Whatever, this was the Congo and tomorrow would be what it would be . . .
My stomach was feeling a little better, possibly because it could find little else other than my vital organs to get rid of. I had weird dreams about being on the shores of Lake Tele watching Motörhead playing 'Ace of Spades' on a floating platform before they were attacked by a Mokèlé-mbèmbé and drowned. The subconscious is a curious creature.
I awoke at eight a.m. and watched the fishermen out on the river. Lonely silhouettes on calm waters. JP was in the restaurant and he had still not heard from Ticketman but he was confident that we could get on the flight. The only thing I could be sure of was that you can't be sure of anything in the Congo. We had a final _ndongo_ -peppered omelette. The only other person in the restaurant was a rather shattered-looking man from Benin. He'd been on our flight up from Brazzaville and was supposed to stay for a week but, like us, was going back today (although, also like us, he had no ticket).
'It is impossible to do business here – it is like the Stone Age. I have not managed one meeting, one discussion; nobody turns up for anything. It is beyond belief. This country is dead.' He wandered off looking distraught and quite frazzled.
We said goodbye to the nice waitress at Tropicana, who had never understood why we were there or what we were doing but had delivered the closest to decent service that we'd encountered in this country.
At midday we hopped on to the back of a pickup and roared off to the airport. There it was predictable chaos. JP found Ticketman, who wanted more money for the tickets but gave in when JP went apeshit. We then fought our way through the laughable screening process but were pulled out of the line by the same officials who'd hassled us on our arrival. They were now demanding money for an 'exit tax'. JP was close to nuclear explosion. The officials said that we could not get an exit stamp without paying. JP pointed out that, as were not leaving the country, we didn't want an exit stamp.
Everything kicked off big style. JP and I eventually marched out of their room and joined the line again. The officials followed us and were by now really hassling us, but there appeared to be a VIP in front of us and he became interested in what was happening. The officials backed off and we got through the screening.
A soldier checked my passport and thought my Iranian visa was my ID page. He spent ages checking it was in order. In the end he nodded and we were on the tarmac waiting for the plane to land so that we could get the fuck out of Dodge.
The mystery plane eventually landed six hours late and only after it missed the actual airport twice and overshot the runway once. There was another rugby scrum to get on to the plane.
Onboard every other passenger seemed to have a plastic bag full of smoked fish and the smell was absolutely astonishing -like an abattoir in a fish tank. A woman sat down next to us with an open bag and I very nearly passed out and thought I was going to vomit. Fortunately we were by the emergency exit and the stewardess told her to move: women are apparently not considered responsible enough to operate emergency exits. Normally this kind of sexism would appal me but, on this one occasion, I let it slide. It would be no exaggeration to admit that, for the duration of this flight, I became a little more religious. It lasted only an hour but both JP and I were absolutely convinced something else was going to go wrong.
In between thoughts of imminent death I reflected on the problems inherent so far in monster-hunting. I certainly wasn't dispirited. There was no rulebook for this. I was flying by the seat of my pants and trying, in a series of short trips, to do what some people spent years doing. The main problem seemed to be trying to distinguish between native beliefs and cold, hard facts. The stories of the monsters in the Okanagan and the Congo had come from the indigenous peoples and had been subject to osmosis by settlers or travellers who seemed to have some problem in interpreting what was real and what was spiritual. At least the Hibagon was a relatively new (though short-lived) creature: very rare in Japan, the land of a million ancient monsters. Whatever, I had at least had a sighting -something that many firm believers in these creatures would never have. It was all turning out to be as varied and weird and downright exciting as I'd hoped it would be. It was good to be alive and I couldn't wait to set off on my next expedition, one of the daddies of the monster world: Bigfoot.
That evening, safely back in Brazzaville, JP and I headed out to the best restaurant in town – a posh riverside place called Terminalia. We both ordered pizza and sat gazing over the mighty Congo at the lights of Kinshasa dancing on the moonlit waters. We were happy and in a slightly euphoric post-adventure mood. We went over some of the 'highlights' of the trip and the alcohol flowed freely. The pizza took a while but we were busy chatting and it was only when a terrible tiredness came over us both after an hour or so that we realized that nothing had arrived. The waitress wandered past and I asked her if there was any sign of our pizza.
_'Oui, oui – ça arrive tout de suite,'_ she said, walking on.
Another twenty minutes elapsed and finally Jean-Pierre went inside to remonstrate. He came back out half-laughing, half-stressed.
'The oven is not working,' he said.
I asked him why they hadn't just told us that from the start.
'It's the Congo, my friend: nothing is normal here . . .'
We got up and wandered out through the gates and into the streets of Brazzaville.
'Do you feel like visiting Kinshasa tomorrow?' asked JP playfully. I looked at him and he was laughing.
'Maybe next time, Jean-Pierre, maybe next time . . .'
### Bigfoot
'Darwinian man, though well behaved,
At best is only a monkey shaved.'
W.S. Gilbert
I settled into my seat on the flight to San Francisco. As I did so the one in front of me was tilted back violently by a very fidgety Indian gentleman. We were still on the tarmac and I knew my rights: no tilting until after take-off. I gave his seat a shove back and battle commenced.
'I have the right to put my chair back – this is why it is designed in this manner!' shouted Fidget.
'Not until we take off. Seats must be upright until we take off; that is airline law,' I countered.
'You are British? This is why you lost the Empire: for this sort of arrogant behaviour. It is why you lose the cricket as well.' This was a very left-field argument but it stumped me momentarily. I considered mentioning that it was a shame we had not taught our former colonial subjects the basics of travel law, but realized that I'd be hoiked off the flight by the steward.
'As a matter of fact, England are currently the best cricket team in the world.' It was checkmate.
'Go fuck yourself.' He had run out of steam and he knew it.
His chair stayed up until we'd taken off, when he immediately titled it back to the full extent. Sometimes I hate travelling.
Arriving in San Francisco I braced myself for the usual torment of entering the United States. In my experience, if you happened to have been born in an 'enemy' country, as I was (Lebanon), then you could expect a five-hour wait, a dumb interrogation and, if you're lucky, an internal examination. To my utter joy, however, it turned out that the whole NSEER system has finally been abolished, having caught precisely zero terrorists and turned everyone who went through it into an anti-American since its hasty introduction by panicky Neo-Cons post-9/11. I sailed through and was soon riding the futuristic monorail to the car-rental agencies. I wondered how long it would be before monorails stopped looking futuristic.
There can be few things more liberating than driving over the Golden Gate Bridge with good music blaring, off on a road trip into the badlands of Northern California. When the purpose of said trip is to find Bigfoot, it's off the scale.
I'd done most of this journey before. The very first travel piece that I ever wrote was for the _Independent._ In it, I set off to find the Giant Redwoods I'd heard so much about. Like most tourists, I stopped at Muir Woods, about twenty minutes north of the bridge, and walked around craning my neck upwards and marvelling at these natural monsters, unaware that they were mere minnows compared with what awaited me further north.
California is an extraordinary state. To me it's a microcosm of the United States, with the deserts and lush hills of the south, the seaboard cities and then the wild of the north that even includes a 'Lost Coast'. My destination this time was Willow Creek: Bigfoot Central and home to a museum dedicated to the Sasquatch, as the Native Americans call him. Willow Creek is inland from Eureka, a large(ish) town on the coast about 250 miles north of San Francisco. As you drive north from San Francisco you go through several areas. First you pass the outskirts of Sonoma and Napa, wine country. Then it's into beer country, dotted with hundreds of microbreweries. Finally, you enter marijuana country and all bets are off. Willow Creek is in Humboldt County, famously the home of hundreds of reclusive weed farmers who hide their crops in the woods and are renowned for not liking strangers wandering about. Maybe unsurprisingly, Bigfoot territory lies slap in the middle.
I'd left San Francisco at about three-thirty in the afternoon so I wasn't going to make Eureka that evening. I drove up Highway 1 for a while. This is one of my favourite drives in the world: it's like rolling through an enormous Hitchcock set. I got to Bodega Bay, where he filmed _The Birds_ , before turning inland and stopping in Ukiah, a little beer town, for the night.
I'd stayed here before on my Redwoods trip and had been billeted in a bed and breakfast with a woman who lived with a giant turtle. I hate bed and breakfasts. This time I checked into the Economy Inn Motel. The reception smelt of curry and a tiny woman gave me the key in exchange for forty bucks. The room was like every motel room you've ever seen in American movies: slightly grotty, but sort of strangely exotic at the same time.
The last time I was in town I found an oasis called the Ukiah Brewing Company & Restaurant – a haven in a town full of beauty parlours and combat gear. It was only three blocks from my motel so I popped in to get a drink and some supper. There were several groups of draft dodgers in various corners and I installed myself at a high table in the window. I ordered a beer and steak and watched the Patterson/Gimlin film of Bigfoot on my laptop. This is probably the most famous piece of monster footage ever. The film was shot on a sixteen-millimetre camera on 20 October 1967 by Roger Patterson and Robert Gimlin on the Klamath River, near Orleans in Northern California. The film shows a large female (she has big bosoms) Bigfoot walking away fast along the bottom of a creek. The creature has come to be known as 'Patty' by cryptozoologists. It's either a hoax or the most important piece of wildlife footage in history. The Wikipedia entry about the encounter goes like this:
In the early afternoon of October 20, Patterson and Gimlin were at Bluff Creek. Both were on horseback when they 'came to an overturned tree with a large root system at a turn in the creek, almost as high as a room.' When they rounded it they spotted the figure behind it nearly simultaneously, while it was 'crouching beside the creek to their left.' Gimlin later described himself as in a mild state of shock after first seeing the figure. Patterson estimated he was about 25 feet (8 m) away from the creature at his closest. Patterson said that his horse reared upon seeing (or perhaps smelling) the figure, and he spent about twenty seconds extricating himself from the saddle and getting his camera from a saddlebag before he could run toward the figure while operating his camera.
He yelled 'Cover me' to Gimlin, who thereupon crossed the creek on horseback, rode forward a while, and, rifle in hand, dismounted (presumably because his horse might have panicked if the creature charged, spoiling his shot). The figure had walked away from them to a distance of about 120 feet (37 m) before Patterson began to run after it. The resulting film (about 53 seconds long) is initially quite shaky until Patterson gets about 80 feet (24 m) from the figure. At that point the figure glanced over its right shoulder at the men and Patterson fell to his knees; Patterson would later characterize the creature's expression as one of 'contempt and disgust.'
At this point the steady middle portion of the film begins. Patterson said 'it turned a total of I think three times,' the first time therefore being before the filming began. Shortly after glancing over its shoulder, the creature walks behind a grove of trees, reappears for a while after Patterson moved ten feet to a better vantage point, then fades into the trees again and is lost to view as the reel of film ran out. Gimlin remounted and followed it on horseback, keeping his distance, until it disappeared around a bend in the road three hundred yards away. Patterson called him back at that point, feeling vulnerable on foot without a rifle, because he feared the creature's mate might approach.
Next, Gimlin rounded up Patterson's horses, which had run off before the filming began, and 'the men then tracked it for three miles, but lost it in the heavy undergrowth.' They returned to the initial site, measured the creature's stride, made two plaster casts (of the best-quality right and left prints), and covered the other prints to protect them. The entire encounter had lasted less than two minutes.
Those two minutes had a profound effect on me as a kid. I first saw only still images taken from the film and reproduced in the Arthur C. Clarke book. This was obviously in the days before the Internet and there was no way for a schoolboy in Oxford to access the footage. I finally saw the shaky, mysterious film on a repeat of the Arthur C. Clarke show on television. I watched it once and was transfixed. Now you can go on the Internet and see it endlessly looped, slowed down, analysed . . . Back then, however, it was an elusive experience and I longed to know more about it. Was it a hoax, as many had claimed? One of the problems was that Patterson had rented a very expensive sixteen-millimetre camera and couldn't remember what film speed he'd filmed it on. This made reconstructions quite difficult. I wanted to know more about the film, to try to get near to where it was shot, speak to people who knew the pair. A couple from New Jersey seated next to me – presumably because I was the least dangerous-looking person in the room – were eager to chat. I told them that I was hunting Bigfoot and they told me that there was one in New Jersey. I tried to look sceptical, as though I was something of an expert in the field. They appeared impressed – but then again, they also appeared impressed by metal cutlery.
I told them about my wine/beer/marijuana Northern California division and informed them that they were right on the beer/marijuana frontier. They looked very excited and asked me whether I could sell them any dope. I patiently explained that I was 'not from round here' and they nodded, unconvincingly They ended up asking the waitress, who told them about medical marijuana. It was readily available: all you needed was a note from a doctor. They asked her if she was a doctor. She replied that, no, she was a waitress. I started to weep quietly.
Just before I left, I went to the loo and found the old New Jersey guy talking to the chef, who appeared to be a doctor because some deal was clearly going down.
My motel room would have made a fabulous murder scene in a TV show. I turned on the telly but Piers Morgan popped up so I turned it off feeling dirty and violated.
Surprisingly I got a great night's sleep, despite the man next door to me snoring so loudly through the paper-thin wall that I thought this might be my first earthquake. I awoke to the sound of the snoring man weeping loudly. I'd had enough exotica and headed out for breakfast before hitting the road north again.
The scenery just got more spectacular the further north I went. Mist clung to the hills and the moss-covered coastal oaks had a haunting air about them. I was now most definitely in weed country. I drove into Garberville, a town that boasts a 'Cannabis College'. The town looked like a movie set from 1972, with long-haired hippies shuffling about and sitting on the sidewalk strumming guitars tunelessly. Everywhere I looked were smoke shops and shady-looking men in pickup trucks buying disproportionate amounts of lighting equipment. It was all a bit _Twin Peaks._
Just to the north of Garberville is the magnificent Avenue of Giants – a truly majestic stretch of road about twenty miles long that is bordered and overlooked by some monster redwood trees. Unless you have actually seen a redwood or a giant sequoia, it is difficult to describe just how mind-blowing they are. It is no coincidence that a group of them is known as a cathedral. There is something immensely spiritual about them.
Would you like some redwood facts? Here you go then.
They are the world's largest trees in terms of total volume. They grow to an average height of about 280 feet and 26 feet in diameter. Record trees have been measured to reach 330 feet in height and more than _55_ feet in diameter. The oldest known giant sequoia based on ring count is 3,500 years old. I thank you.
Back in my enormo-vehicle, I was really enjoying my leisurely cruise down the Avenue of Giants. At times the trees were so tall and so thick that it was almost dark. Then a lone shard of sunlight would find a way through the foliage and light up a section of the road like a movie spotlight. I kept taking photographs; I just couldn't stop as the trees were so impressive. The problem was that it was almost impossible to take a successful photograph because I could never get the whole tree in my viewfinder unless I lay on the road and looked straight up – and even then it was a bit weird.
Suddenly, to my left, I spotted a fallen giant. The trunk was about thirty feet in circumference, and its decaying root system looked like a small city's infrastructure. The Avenue was totally devoid of traffic so I stopped right in the middle of the road and took some more photographs out of the window.
Then, from out of nowhere, a beaten-up old truck came speeding down the Avenue honking violently at me. There was no problem in getting past me and the driver would have spotted me from a good 200 yards away, so they were just trying to be jerks. I ignored them and carried on taking photos. I think that my fearless attitude might possibly have been slightly emboldened by my new status as a monster-hunter. Canoeing down the Congo changes you somehow: honking pickup trucks do not phase you . . .
The truck drove up alongside mine. I looked into it. To my surprise all three occupants were women. They were in their mid-twenties and looked like they spent their weekends beating up lumberjacks.
Asshole!' they all shouted in unison before giving me the rigid digit and zooming off.
Assuming that these 'ladies' were not simply people who didn't appreciate my TV work, I considered this to be totally uncalled for. In fact, I'm ashamed to admit, a red mist descended upon me in this, the calmest and most reflective of nature's places. I gunned my engine and set off in hot pursuit of the rogue females. After two or three minutes I'd caught them up and observed with pleasure the driver's startled look in the mirror. I started honking and flashing my lights. I was behaving like an idiot, but I was really angry. I had no idea what I'd do if they stopped; I just wanted to register my displeasure. I imagined Stacey in the seat next to me screaming, 'For fuck's sake, Dom, stop . . . What are you doing, you bonehead?'
But Stacey was not here and I was a monster-hunter and my pride had been dented. We careered on down the Avenue of Giants, nose to tail. Suddenly they screeched to a stop, forcing me to steer madly to the left before halting just behind them. I pulled myself together and thought about remonstrating with them but feared that they would laugh at my English accent.
I can do quite a good 'Southern' accent – a hybrid Mississippi/ Louisiana thing I use when I'm online playing _Call of Duty._ I like to assume the role of a moronic redneck and tease Brit gamers by feeding them every American stereotype in the book.
The girl in the passenger seat got out of the car and I wound down my window and, in my finest hick, shouted, 'What the heyyl do yooo asshooles think you're all doing getting all uppity and sheeeet . . . ?'
The girl ignored me and went straight to the boot of the car. She leant in and pulled out a crowbar. I didn't hang about. This cowboy got out of Dodge as fast as his steed would take him. I hit the gas and shot off past them. In the mirror I saw her jump back in and they started chasing me.
I really panicked now. I was being chased by a truckload of homicidal killer ladies. We roared down the Avenue of Giants aping a scene from _Need for Speed._ I thanked the Lord that California has tougher gun laws than say, Nevada, where the woman would have probably pulled a bazooka out of the back. I looked in the mirror again. They were now right behind me and I could see their faces, twisted in anger. I was genuinely scared. This was like a cross between _Deliverance_ and _Duel_ (with women). It would actually have been quite a strong movie synopsis had I not actually been living it out for real. We burst out of the redwoods and into a town called Myers Flat. I saw a sign advertising a tree that you could drive through and, right in front of the sign, a huge wooden statue of Bigfoot waving at me. It was my very first sighting and he was taking the piss. I shot through town with the murder gang still on my tail. I sped up and rounded a corner, almost on two wheels.
I was slightly ahead now and, for a moment, they couldn't see me. I spotted a little track to my right and I acted instinctively. I pulled hard right on the wheel and skidded off the tarmac. I prayed that they hadn't seen me turn. I kept driving for about five minutes but there was nobody in the rear-view mirror. I turned off the track into a little grove and parked the car behind a big tree so that it couldn't be seen. I sat in silence for five minutes. I was sweating and my heart was beating fast. Nothing happened and I started to relax a little. I opened the car door and listened for the sound of murderous banshees. There was nothing, just an eerie silence. I got out and closed the door gently. I locked it with my key fob. It beeped to let me know that it, unlike me, was now safe. The beep reverberated like an explosion in the silence of the redwoods. I tensed up and waited but everything stayed silent. I didn't want to go back on the road for a while so I decided to go for a bit of a walk into the forest. Maybe I'd get lucky and find Bigfoot first go? I wandered off, away from my car, along what seemed to be a vague path. The trees surrounding me were even bigger than the ones on the Avenue. I felt tiny and very alone. I was utterly dwarfed by nature.
I kept walking, the only sound being the muffled _clump, clump_ of my Reeboks on the spongy forest floor. As before, occasionally the sun burst through and lit up a little clearing, but predominantly it was dark and primeval.
My mind started to play tricks on me and I could easily see how people might believe that monsters lived here.
I started to hum to myself as the silence had become quite oppressive. I hummed A Forest' by the Cure: 'Suddenly I stop/ But I know it's too late/ I'm lost in a forest/ All alone . . .'
I kept on walking deeper and deeper into the magical forest. It was like being on the set of _The Hobbit._ Every tree seemed to be bigger, thicker, taller than the last . . .
Suddenly I stopped . . . I spotted the unmistakable outline of a bear about fifty yards ahead. Fortunately for me it was looking the other way, down into a leafy crevice. I froze to the spot and then started to lower myself to the floor in tiny little movements. This was totally crazy: I'd been chased by killer lady rednecks and now I was about to be eaten in a forest by a bear. The bear did not move and nor did I. I lay perfectly still on the soggy mushy ground for what seemed like an hour but was probably no more than a couple of minutes. The bear was totally immobile; it seemed to be focused on some unseen prey. My broken foot started to ache and I crawled forwards a little until I was behind a bush. I waited there for another couple of minutes or so thinking about my options. I could get up and run away. Then I remembered the sign in the Okanagan: 'Under no circumstance should you run away from a bear unless you have somewhere to go . . .'
I had nowhere to go. Other signs I'd read in bear country suggested having a bell with you and ringing it if you saw a bear. I'm not sure if this is to scare the bear or to alert the search party to where your remains could be found. Whatever, I didn't have a bell on me now so I was stumped. The bear was still not moving. I crawled a little bit nearer . . . It was a weirdly shaped log. Actually it was a scary bear-shaped log. Your mind plays extraordinary tricks on you out there.
The balance between human and nature is completely inverted among these natural giants. The place gives you a constant slightly freaked-out feeling, an age-old instinct telling you to be on the _qui vive._ It had been the same in the misty mountains of Japan, the steamy forests of the Congo and the dark, impenetrable waters of the Okanagan. When nature decides to turn on the creepy mood music it's incredibly effective.
I was by now exhausted from this adrenaline roller coaster. I retraced my footsteps and found the car. I got in and locked the door. I needed some normality. My phone had no network so I plugged it into the car and played a Kermode/Mayo podcast. Never had the Good Doctor's ranting (this one about some arsey new film by Gus Van Sant) been so reassuring. I drove back up the track and on to the Avenue of Giants. I turned left and headed back the way I'd come. Very soon I was back in Myra Flats and spotted the statue of Bigfoot. I wanted to park and get a good photo of it but as I was about to pull up outside a bar I spotted the three women. They were sitting outside drinking beer with three men who looked like they enjoyed a spot of butchery at the weekends. Somehow, in a country in a county full of hippies, it looked like I'd angered the Manson family. I drove by trying to look as inconspicuous as possible. Fortunately my Chevy blended in and couldn't have been that memorable as they didn't seem to recognize me. One of the women looked up and stared at me as I drove past but I just looked straight ahead and headed out of town. The moment I was through I hit the gas and didn't stop until I got to Garberville, where I ducked into a bar and had several glasses of chilled Sauvignon Blanc to soothe my frayed nerves.
In the space of about two hours I had gone through every American-backwoods cliché that I'd ever seen in a movie except bumping into the Ku Klux Klan and some incredibly racist fat police chief. I looked out of the window to check for burning crosses or a roaming cop car but there was nothing but a dread-locked hippy strumming a guitar under a tree. I needed some medicinal marijuana to calm me down – the only problem being that it always makes me paranoid and I didn't need that right now. I got up and left. Outside I had to step over two parody hippies complete with headbands, guitars and John Lennon glasses. They were seated on the pavement just staring vacantly at the traffic. I had a huge desire to tell them that Vietnam was over and that whatever they were running away from could all be sorted out with a haircut and a good bath. I realized that I'd become 'the Man': I was a 'suit' and a 'total square'. I walked on without saying anything but slightly depressed. I passed two hemp shops, got into my car and drove a couple of miles to where I was staying, the Benbow Inn, a mock-Tudor building that felt rather incongruous slap-bang in the middle of Hippy Country. In its heyday this place had served as a bolt-hole for Hollywood luminaries looking for some privacy and a spot of fishing. Guests had included Spencer Tracy, Clark Gable and, more recently, the King of Jordan and Cher (though sadly not at the same time).
My room was a joy, with a four-poster bed and a stone balcony overlooking an old bridge over the Eel River. Sometimes even monster-hunters need a bit of downtime.
I posted some video footage of me lost in the woods on Facebook. People asked whether this was for a TV programme. I told them that it was for a book called _Scary Monsters and Super Creeps._ Inevitably, a cyber-twat started accusing me of 'ripping off' David Bowie. He seemed to think that he was the only person who had spotted that the title of this book was lifted from a Bowie album. I told him that this was not exactly a secret and, as a huge Bowie fan myself, it was a nod to one of my heroes.
He wasn't having it: 'You're a relentless asshole riding on his coat-tails.'
I was genuinely unsure as to what his beef was. If it was with me personally, then why on earth was he following me on Facebook? If it was with the appropriation of the name, did he think I was trying to con people into buying my book by making them think it was actually a David Bowie album?
I hit the bar and ordered a Grey Goose up with a twist. All was good with the world. After a couple more beige birds I was ushered into a rather fussy dining room where I was constantly asked what I thought of the food by a waitress called Bambi.
The food was very good but, like in the UK in 1972, Bambi started putting chairs on tables at nine p.m. Out of spite I hung around playing with my overly elaborate pudding until nine-thirty. (I'm a bit rock 'n' roll like that: no sleep till bedtime.) I eventually left the dining room as the lights were about to be turned off. As I walked through the hotel I started to look at it properly through drunken eyes. There were teddy bears everywhere. Back at the bar, a solitary gentleman propping up the corner and nursing a short turned out, on closer inspection, to be a seven-foot teddy bear. It's a universal truth that a building with more than two teddy bears on display is telling you that its owner was probably sexually abused and is now a predatory serial killer. I hurried to my room and locked it securely.
I left early the next morning. It didn't take me too long to make it to Eureka. This is a weird city, quite industrial in parts and with an area containing some quite extraordinary Victorian Gothic mansions. The main one, the Carson Mansion, is an unbelievable piece of fantasy architecture reputed to have provided the blueprint for some of Walt Disney's ideas for Disneyland. I was staying at the Carter House Inns, a hotel owned by Mark Carter, a local bon viveur and wine aficionado. One of the hotel buildings is another example of crazed Victoriana. Painted bright yellow and orange, it's an exact reproduction of a building that stood in San Francisco until it was destroyed in the Great Quake of 1906. I checked in and was shown to a rather magnificent cottage that turned out to be entirely at my disposal. I had an enormous living room and kitchen, a master bedroom, two bathrooms (because one is never enough) and a terrace. This was going to be hard to leave.
On the table was an envelope with my name on it. I opened it.
Mr Joly
Welcome to Carter House Inns. We hope you enjoy your stay here. Your companions, Corey and Kirsten, left a message saying that they would be arriving at midday.
Reception
I reread it wondering what on earth they were going on about. I had no 'companions' and wasn't expecting any. I wondered if this was hotel code for letting me know that a couple of complimentary hookers were being provided. Corey however, sounded like a man. This was California so maybe they were just hedging their bets? I checked my watch; it was ten-forty a.m. Whoever my companions were, they were arriving in twenty minutes.
I opened my laptop and got online. I searched for the names Corey and Kirsten on my Facebook pages. The name Corey came up and I clicked. Now I remembered. Back in the early days of Facebook, when _Trigger Happy TV_ was out in America, this fifteen-year-old kid called Corey had contacted me and we'd chatted occasionally about stuff like music and comedy. He'd been going through a tough time (his parents were divorcing) and I'd felt bad for him. He'd also given me a lot of cool music recommendations. I now also remembered that his pages had always been littered with photos of a girl called Kirsten. When I'd been in San Francisco a couple of years before Corey had contacted me and tried to meet up; I'd been filming, so we'd been unable to. He'd said he was at Humboldt State University, in a town called Arcata, just ten minutes from Eureka.
Before I left for California this time I'd put out a 'can anybody help me' message on Twitter and Facebook and it looked like Corey had answered the call. I checked his Facebook page. He was now married to Kirsten and living in Sacramento, the state capital – about seven hours' drive from Eureka. I hoped to God that they hadn't driven up from there. If they had, then they would be expecting to accompany me on some serious monster-hunting – and I normally prefer to do this kind of thing on my own. I'm an inherently selfish traveller and like to do things when and where I want to, without worrying about others.
Fifteen minutes later there was knock on my door and I opened it to find Corey and Kirsten staring at me. He was a tall, thin and slightly Goth-looking guy of about twenty-three and Kirsten was slightly shorter and pretty .
'Hey,' I said.
'Hey,' they said.
'You came,' I said.
'We came,' they said.
'Not from Sacramento?' I asked.
'Yeah, it took us seven hours,' they answered.
'Shit,' I said to myself quietly.
I let them in and we all sat down slightly awkwardly. Actually, _I_ was awkward – they seemed to think that this was the most normal thing in the world. They _oohed_ and _aahed_ about my hotel 'complex' and I nodded as though this was my life all the time. I longed to show them the photo I'd taken of the Economy Inn Motel in Ukiah but I thought I'd keep up the pretence of being 'Dom Joly' for a while. They told me that they were in town for two days. I nodded and they asked me what the plan was. Obviously the plan was to find Bigfoot, but I'd heard that nearby Arcata was the US's hippest town and who knew what this could turn up? So I asked them to show me around.
Corey had spent three years at the university and was the ideal guide. The campus was dotted all over the little town. The main square comprised weird little shops around a grassy park in which very stoned students were all playing Frisbee. I counted nine sets of dreadlocks and five didgeridoos in plain sight. Corey told me that he'd shared a room for a year with a guy who did nothing but smoke dope and play the didgeridoo. Personally, I would have thought these were grounds for justifiable homicide.
We popped into a 'head' shop that sold innumerable variations of bongs and . . . Frisbees. According to Corey everyone just got stoned and then wandered off to play Frisbee golf in the woods. I didn't believe him so we headed off into the woods where, sure enough, we found a Frisbee-golf course and a large amount of stoned students aimlessly holding Frisbees.
This seemed a bit weird to me. Surely Frisbee golf isn't the obvious first thing that comes to mind when you're stoned? I remember my own light dabblings involving watching a lot of really rubbish TV and eating crap. Apparently whatever they smoked up here gave you Frisbee cravings. Corey told me that, apart from Frisbee golf, sport's not really that big a thing at Humboldt State. The university football team is called the Lumberjacks and Corey told me about a headline in the local paper that had simply read, 'THEY WON!'.
Our tour of Arcata complete, Corey and Kirsten looked at me expectantly. The plan was for me to drive (alone) to Willow Creek the following day and start my hunt there. If you look at a map of Bigfoot sightings in the Northern California area then Willow Creek is the geographical epicentre and also the home of the Bigfoot Museum. This was the idea for the following day but right now I needed something to do with Kirsten and Corey.
Then I remembered reading about a place called Tall Trees Grove about an hour north of Eureka. It's supposed to have some of the largest trees in the world and there have been two Bigfoot sightings nearby. I figured we could start there. I asked them if they knew where it was; Corey rang a friend who told us how to get there. The decision was taken and the hunt was on. We were off to Tall Trees Grove for my first Bigfoot hunt.
We drove through Orick, a 'town' that seemed to consist solely of four roadside stores selling extremely weird wooden statues. A couple of miles later we turned off the scenic coast road and started to drive up into the mountains. We drove for about twenty minutes and then the tarmac road became a dusty track. We kept climbing higher and higher.
'I hope you're not axe murderers?' I said semi-jokingly This was exactly what you were not supposed to do on an Internet first date: drive miles and miles from anywhere into deep woods. To divert myself I read a warning leaflet that we'd got from a park ranger we'd met at the bottom of the mountain. It was called 'What to do if you meet a mountain lion'.
1. Do not run!
2. Do not crouch or bend over
3. Remain calm
4. Yell loudly wave arms and throw objects
5. If the animal attacks – fight aggressively
You had to wonder who wrote these things. There was nothing about what to do should we meet a Bigfoot.
We found the trailhead and parked the car. The trail descended very steeply into what looked like the Lost Valley. We walked down and down and down. It took us about half an hour to reach the bottom. When we did, it was like stepping back in time. Towering ferns bordered gargantuan trunks of monster trees that soared high into the sky like vast wooden spires. All the trees had massive burls, growths that resembled gargoyles, their twisted shapes metamorphosing into hideous creations. It was further proof, if needed, of how spooky surroundings could really feed a hungry imagination. We all instinctively started to talk in hushed voices as though in church. At the very centre of the ring of tallest trees, we stood in complete silence for a moment and listened to the earth's heartbeat . . .
I know this all sounds like I'd either had a huge spliff or become a hippy overnight, but I have never, ever been so profoundly affected by the sheer presence of nature. These are the tallest trees you can visit anywhere in the world. There are individual taller ones, but most are on weed farms and protected by armed and paranoid hippies. One interesting fact (despite my not really liking _Star Wars)_ is that the moon scenes of Endor, when we meet the Ewoks in _Return of the Jedi_ , were filmed here. George Lucas lived in Northern California and it's very obvious that the idea for Chewbacca was influenced by tales and descriptions of Bigfoot.
We spent about an hour just chilling in this magical place before realizing that it was getting dark. We didn't want to be stuck down here at night: it would definitely get very spooky. There was no sign of Bigfoot on the long, hard walk back to the car. Fortunately, we didn't meet a mountain lion either.
Back in Eureka, Corey and Kirsten headed off to find their hotel while I chilled in my cottage at Carter House Inns. I had an hour before I had to meet a guy called Richard, whom I'd contacted on the Internet, at the Lost Coast Brewery. Richard was the media manager for the Humboldt County Visitors Bureau and had promised that he could help me with accommodation and introductions to some Bigfoot enthusiasts who were known as 'squatchers'. It was through his influence that I'd got the beautiful cottage to stay in so I liked him already.
The Lost Coast Brewery was a big shed-like building about four blocks away from my hotel. I walked in and a good-looking, urbane guy of about my age stood up and greeted me. This was Richard and he'd clearly done some research on me, as he knew what I looked like from the Net. He looked a bit like a news reporter: square jaw and round glasses with a soft Southern lilt that betrayed his roots in Georgia. He was a very easy-going guy and we chowed down and got talking about Bigfoot. Richard said that a lot of Bigfoot 'scat' was found in the woods. A local Indian had confirmed that this was no human scat as it would have 'split a human apart'. Some stools had supposedly been found that measured up to two feet long. Local wags claimed that this accounted for a lot of the screams heard at night in the forests.
He also told me about Bigfoot 'nests'. These are very similar to those gorillas build in Africa: thick branches bent over and made into a type of bed. This really got my interest and Richard showed me some photographs. They did indeed show quite intricate constructions with thick branches having been weaved together into a rather cosy retreat. Were Bigfoots really sitting and constructing such things? It was another thing to look out for.
We poured over a very detailed map of the Bluff Creek area where the Patterson film had been shot. The place looked very remote – two hours' drive from Willow Creek on tarmac road, an hour of off-road driving and then a two-hour hike. Richard seemed keen to come with me and, as he knew the area and could get me permissions, I was all for that.
Corey and Kirsten turned up and I introduced them to Richard, who couldn't get his head round the fact that they'd come all the way from Sacramento despite never having met me before. I think it somewhat increased my kudos as a world-famous monster-hunter. We had a couple of drinks before I called it a night and headed back to the hotel.
The next morning, Richard insisted that Corey, Kirsten and I went to the Samoa Cookhouse for breakfast. This was a place that used to feed the hungry loggers three times a day and then, as now, they didn't really go in for calorie counting. The food was homicidally filling, all doughy biscuits and beans and bacon and thick black coffee. We met the owner, Jeff Brustman, who told me that the local Indian tribes used to call the Sasquatch 'Omah'. They were convinced that the creatures were a lost Indian tribe. They were also convinced, laughed Jeff, that hippies were living proof that Bigfoot had mated with humans. I told him that hippies were too skinny to be descended from any missing link – though they did share a pungent odour with the Sasquatch.
He told me about the very first Bigfoot sighting, before he was even known as Bigfoot. It was reported in a letter to a columnist on the local paper. I found it later online.
_Humboldt Times_ , 21 September 1958
I am writing regarding a queer situation my husband has encountered while at work. I have read your column for a long time and have noticed that you often dig into things of various natures. This happened when my husband recently took a land-clearing job up on Bluff Creek, near Weitchpec.
The rumor started among the men at once of the existence of a 'wild man'. We regarded it as a joke and even added fuel to the story by passing on bits of information. It was only yesterday that my husband became convinced that the existence of such a person (?) is a fact.
On their way to the job, the men found tracks going down to the road. The tracks measured 14 to 16 inches in length. The toes were very, very short, but there were five to each foot. The ground was soft and the prints were very clear. In soft places the prints were deep, suggesting a great weight. The tracks were quite wide as well as long and things such as fruit have been missed by the men camping on the job. There are at least 15 men that will swear this is true, among them, my husband. Have you ever heard of this wild man?
The newspaper columnist replied:
Well, honestly, no! I wonder if anyone else knows about this. Please help. Maybe we have a relative of the abominable snowman of the Himalayas, our own Wandering Willie of Weitchpec.
Weitchpec was on our way to Bluff Creek and it was cool to hear about such an early sighting. I wasn't quite sure why they'd called it 'Willie' but assumed that it was just good alliteration. Whatever, the report was a classic Bigfoot encounter. I couldn't wait to get over the hills and into Bigfoot territory proper.
But first I went for a wander round the 'old town' with Corey and Kirsten. They had made such an effort to join my curious adventure and I wanted to spend a little more time with them.
As we walked around the four blocks of nice old buildings and reasonable shops that constitute 'Old' Eureka, I hoped that they weren't too disappointed; but they were a quiet couple and didn't say too much.
We popped into a great bookstore and started browsing. I checked out some books from the seventies on Bigfoot. They were all pretty much the same. Like the ones on Ogopogo, they tended to be self-published, rather dry accounts of sightings. It was all part of this desperate need to be taken seriously, but it took all the fun out of it. In the far corner I spotted Corey and Kirsten looking at a book excitedly. I went over. They were the most animated I'd seen them yet.
'What's the book?' I asked.
'Only the greatest book ever written,' replied Corey.
I looked at the cover and a sudden, nauseous feeling swept across me. The book was called _The Fountainhead._
Wasn't this that book so beloved by extreme right-wing Aryan Brotherhood types? I seemed to remember Louis Theroux doing something about it. I looked at Corey and Kirsten again. She was of Scandinavian origins while Corey had piercing blue eyes and slightly Preacher-esque sideburns. Oh . . . My . . . God . . . Were these two Nazis? Had I unwittingly been hanging out with a pair of white supremacists? On the way to Tall Trees Grove I'd found a Christian religious book in the back of their car but I hadn't mentioned it. Thinking about it, they didn't swear and seemed to eat only fish – but surely if they were white supremacists they would eat only raw meat? Corey had been to university in Humboldt State, hardly the epicentre of Nazi power. Maybe the hippy with the didgeridoo had sent him crazy? I didn't want to ask them straight out but how could I broach this kind of thing?
'Yo, what about them Negroes?'
We left the bookstore and continued our walk but I found myself actually distancing myself from them. A Chinese man walked past us and I subtly watched to see if they showed any sign of disapproval. They didn't and we walked on.
'There's a great bagel place here?' said Corey.
'The best in the USA,' said Kirsten enthusiastically.
I spotted my chance.
'That's weird – why here? After all, bagels are _a Jewish_ thing . . .'
I really over-emphasized the _'Jewish_ thing' but they didn't seem to flicker. We went into the bagel place. It was called Los Bagels. In the window there was a sign in Spanish saying that the staff spoke English. The staff were all Latinos. I tried again.
'Jesus Christ, this place is totally run by Mexicans. Don't any Americans work?'
I looked across at them for a reaction but they were looking at me a bit weirdly. Maybe they thought I was one of them now? We walked out, munching on our Jewish-Mexican bagels, and headed back to my hotel. I was totally freaked out now. I wanted to dump the Nazis. I told them that I had to pack and head off for Willow Creek. This was something I needed to do alone, I said. They took a couple of photos of us all together and then said goodbye. They were going to drive all the way back to Sacramento . . . Probably for a Klan meeting.
I waved from my porch as they drove off. I very nearly gave a Hitler salute to see if they responded but decided against it. The moment their car disappeared round the corner, I rushed inside and googled the book.
_The Fountainhead_ is a totally legitimate 'classic' about an architect who fights against the system. There's nothing right wing or dodgy about it at all. The book I was thinking about was _The Turner Diaries._ I've absolutely no idea why I confused the two. Corey and Kirsten were not Nazis and I was a complete idiot who appeared to be in the early stages of mental illness.
I packed up and checked out of the hotel. I got into my car and drove the hour inland to my final goal: Willow Creek, considered to be Ground Zero for squatchers. There was the Bigfoot Museum to go to and Richard had kindly set up a meeting with Al Hodgson, a local Bigfoot expert who had made plaster casts of Bigfoot feet. He was the guy who told Patterson that Bigfoot had been spotted in the area and prompted him to set off on his expedition.
The town's pretty much a one-street affair, so as I drove in I quickly spotted the museum: there's an enormous wooden statue of Bigfoot outside that must be thirty feet tall. I parked up and was let in by a woman called Peggy McWilliams, who let me look round. She told me that the museum was normally closed from October to May but that she opened up for visitors if they contacted her. She was very chatty and told me that there had just been a Bigfoot sighting over the border in Oregon. A woman saw a Bigfoot cross the road right in front of her car. I was thrilled and pumped her for more information – had anyone taken a photo? She didn't know much more about it.
I asked her whether there had been a dip in sightings.
She made the point that the only dip had been in footprints, because more and more roads were now tarmac and early sightings were all on fresh-cut dirt roads that had been built into new areas for logging.
She also told me that Al Hodgson was a bit doddery and had just lost his wife so he might be a bit late. I told her not to worry and started looking around the museum. The exhibits were mostly casts of big feet, which was probably to be expected. There was a tape on loop playing an old black-and-white TV show about Bigfoot. It was quite interesting and showed some new footage that I hadn't seen of a hairy figure running fast over a prairie-type landscape.
Having finished watching the tape I read some of the newspaper clippings that were pasted on the walls, one of which was the article that coined the term 'Bigfoot'.
Al Hodgson eventually turned up after Peggy phoned him. He was a nice guy but a bit deaf and, yes, a touch doddery. I asked some pretty stupid questions and he seemed to be on autopilot. He told me that at the time of the Patterson film there had just been a huge flood in which Bluff Creek had been stripped bare. He said that, should I try to go there now, I'd find it was very different and totally overgrown. It would be very difficult to find the actual spot where the Patterson film was taken. He understood why I wanted to go to the actual site but, if I wanted to see a Bigfoot, I had as much chance anywhere in the surrounding area. He told me about a recent sighting only four miles from where we stood. A local lady driving a produce vehicle had spotted a Bigfoot on the road at four in the morning. He knew this woman and said she had no reason to lie about it. I asked him if this was the same incident that Peggy had just told me about. He replied that no, this was a different one and happened very near to where we stood. I was really excited. I was definitely in with a chance of at least spotting Bigfoot. This was the stuff that little boys' dreams are made of.
Al started asking me about where I was from. I got the distinct feeling that he was a bit bored of talking Bigfoot. He told me that his family was originally from Leeds. I nodded as though Leeds was my favourite place in the world. In the end the conversation went a bit dry and I thanked Al for meeting me and he pottered off to talk to Peggy.
I had another look round the exhibits before wandering over to where Peggy and Al were shooting the shit. Al showed me a plaster cast of a Bigfoot print that he had taken himself at Notice Creek in 1955. I fell in love with it instantly. This had to be the best travel souvenir I'd ever seen and had to have one. I asked him if it was for sale and he said that he had one I could buy. For twenty dollars, this was the best thing I'd ever bought on my travels. Al's lift turned up and we said goodbye but I stayed on chatting to Peggy who was turning out to be a lot more interesting than Al. She said that Hoopa, the nearby Indian reservation, had the most Bigfoot sightings and that I should go and talk to people there – although, she warned, they were not that keen on 'snooping strangers'. A local man came in and said hello to Peggy. On a whim I asked him if he had ever seen a Bigfoot.
He looked at me and said, 'Nope – but then again, I've never seen a mountain lion and they exist. There's something out there; I know too many people who've seen stuff.'
I was getting hungry and left the museum and had a look round town for somewhere to eat. I walked past the Bigfoot Motel, which looked suitably awful. It had a large cage in the car park supposedly there to capture a Bigfoot should one wander into town. After lunch in a little Mexican place I went into a coffee shop to get an espresso. On the television hung on the wall behind the counter was a show about the 'Skunk Ape', another creature that's supposed to roam Florida. I presumed the show was on a loop and that this was part of the local monster industry. It wasn't, though: it just happened to be on the telly, which was a weird coincidence. I'd read all about the Skunk Ape when I was thinking about where to go on my trips. Florida certainly has a lot of 'wildlands'. I wondered whether an ageing Bigfoot, bored of living among the weed farmers of California, had decided to retire to Florida?
I had very little left to do for the rest of the day. Willow Creek is a town that can be 'done' in well under an hour. Richard had organized for me to stay in one of a series of rather posh cabins that were normally for people coming up to fish in the nearby river. He was joining me the following day so that we could head off on our trip to Bluff Creek.
I checked in and sat outside my cabin reading until it got dark and then went to bed to watch yet another Republican Presidential Hopeful debate. When it had finished I despaired for America.
The next morning, Richard turned up in a decidedly dodgy-looking car. It was quickly agreed that we should take my rental on our expedition as his had seen better days.
We first went to my Skunk Ape café, where we had some muffins and coffee. We looked at maps again and Richard started talking quite loudly about Bigfoot, causing several local heads to swivel round to look at us suspiciously. Richard certainly seemed to know his stuff. It turned out that his father-in-law had a cabin near the mouth of Bluff Creek. I had lucked out and seemed to be with something of an expert. Before we set off, I popped into the local mini-mart and bought some beef jerky and water. There was a huge sign outside saying that they sold buffalo meat. I took a picture of the sign. When I got to the checkout the woman working there was looking at me distrustfully.
'They say you took a photo of the store . . . Why?' I think she suspected me of some Federal Government snooping.
I over-Brit-accented and told her about how weird it was to see buffalo for sale. She immediately warmed up and started to chat.
'Do you guys not eat buffalo, then?'
I told her that, no, we didn't and that this was clearly our loss. I asked her my default question: 'Have you ever seen Bigfoot?'
'No, sir, I've never seen him, but I know a lot of good folk who have round here.'
This appeared to be the default answer to the default question.
I joined Richard in the car and we set off out of town on the so-called 'Bigfoot Highway'. We hadn't been driving for long when Richard asked me to pull over at a forestry station. He wanted to find out which roads were open and which were closed, as there was still snow in the mountains. Forest Ranger Jim was behind the desk and was busy issuing permits to a group of Mong (a displaced Burmese border tribe, not a bad Ricky Gervais joke) who were off mushroom picking. The Mong were taking ages and quibbling about the fee.
Forest Ranger Jim eventually finished with the Mong and we had a chat. He told us that we were better off asking at the forestry station in Orleans as it was much closer to our destination.
I asked him about Bigfoot and he went very weird with us.
He announced that he didn't speak to the media since the National Geographic Channel had come to talk to him.
'They reduced my hour-long interview to just twenty seconds, then they got my name wrong, and then the bastards ignored all my referrals to serious people who have really seen Bigfoot. Instead they talked to some woman who claimed she fathered a Bigfoot in Arizona. So I'm sorry, but I don't talk to media no more . . .'
He then proceeded to talk non-stop for about half an hour about various sightings. He told me that when the original logging tracks were first built into the wilderness, a logger friend of his complained that something was messing about with his machines. They were being vandalized, with something hurling big heavy rocks at them. He also found several sets of footprints.
Forest Ranger Jim was on a roll.
He told us about how many creatures previously unknown to science there were discovered every year under the sea. He mentioned the recent discovery of a type of deer that scientists had thought extinct.
It was safe to say that Forest Ranger Jim was a believer but we couldn't stop him talking. It felt like the whole world was being sucked into some time-space vortex from which we would eventually have to extract ourselves if we were to go anywhere that day. Despite this I was quite stoked to find a forest ranger who believed in Bigfoot. These guys were officials who had spent all their lives in this environment. It seemed to me that he was a pretty credible witness.
We drove on through Hoopa, the Indian reservation. Richard's father-in-law had been the dentist here in 1969, after he'd left the Marine Corps, and so he knew quite a lot of the people. Richard had rung the tribal museum to ask whether we could come and talk about Bigfoot, but they'd hung up on him. They didn't like people cheapening the Omah story and without their cooperation it was very difficult to gain access to any valid information on the reservation. We drove on through. There was trash everywhere and shanty-type trailers with old cars and junk lying about. To be honest, it was a bit of a shit-hole.
Once out of the reservation we soon came to the town of Weitchpec. This was the site of the very first reported sighting of Wandering Willie of Weitchpec. We stopped at the local store, which had more than a touch of _Deliverance_ about it. We bought turkey jerky, which was a first for me. Richard chatted to the old guy behind the counter, who looked like Uncle Jesse from _The Dukes of Hazzard._ He was the descendant of a Norwegian who'd married a squaw and he very much gave out the impression that he was in charge. I asked him whether he knew anything about Wandering Willie. He nodded slowly as though he knew everything about it. I waited for him to expand but nothing came and it was fairly clear that nothing would.
We drove on again, the road running above the raging river and through some staggeringly beautiful scenery. As I looked up into the thick woods bordering the roads it was easy to see how something could stay undiscovered there if it so wished. We got to a bridge that spanned Bluff Creek where the creek ran into the main river. We parked the car and clambered up the sides of the hill to get a better view. As we climbed, I noticed a weird, unpleasant smell. Was it Bigfoot or was I being paranoid? We climbed higher and I had to push my way through some thick bushes. The smell was really bad now and the hairs on the back of my neck were standing up. I bustled through another thick bush and the ground flattened out into a little plateau on which stood a loo. It was weird – we were about fifteen minutes from the road, in the middle of nowhere, and suddenly we'd stumbled on a loo. It stank to high heaven and I opened the door while holding my shirt over my nose. It was a mess. It looked like somebody had exploded in there. What on earth was this thing doing here in the middle of nowhere? Did bears shit in the woods? I was definitely on the lookout for Bigfoot scat but hadn't expected this. Richard was totally confused. He had no idea what this was doing where it was. It was oddly freaky and we climbed back down to the car quite speedily. I looked up the creek for a while and imagined spotting a creature down on the shoreline fishing for salmon. It would look up startled and then move off in that curious gait, back into the thick undergrowth. This was just what a fisherman had seen on this river very near Willow Creek. Bigfoot was not showing himself for me, however. I was rather pleased. Nobody would believe me if I spotted one anyway, but in Bluff Creek of all places it would be ludicrous.
We got to Orleans, once a serious conurbation with a population of 6,000 inhabitants but now a bit of a ghost town with a population of less than 600.
We found the forestry office and popped in. An elderly woman was sitting in the back and seemed very surprised to see someone. We asked her a question but she couldn't hear too well, so she got up and started walking towards us so slowly that I estimated it took her about seven minutes to traverse the twelve feet or so between us. She was also wearing glasses so thick that you couldn't see the eyes behind them. When she eventually reached us we asked her about roads and whether they were open. We wanted to try the Go Road. Originally built to join Orleans with the next valley, this was the closest proper road to where we needed to go. Local Indians had complained that it crossed over sacred ground and building work had stopped. Now the Go Road just went to a dead end miles up in the hills. It was therefore known by local wags as the No-Go Road. The Forest Ranger Lady knew nothing about the road but offered us some posters and Smokey Bear goody bags. Out of politeness we said yes and to our horror she announced that she would have to go find them. She started to move off on what looked like the beginning of an extraordinarily long trip. Richard and I glanced at each other in despair.
As we waited, an unkempt man called Bud turned up and seemed to know all about which roads were closed and which were not. He said that the Go Road had snow and was not really passable. I asked him the default question.
'Have you ever seen a Bigfoot?'
'I have not personally, but I know a lot of people who have . . .' This was getting to be very repetitive.
He told us about having to rescue 'some idiots' from very near the Patterson Gimlin site. They had hiked down there having heard a story of a Bigfoot massacre. Forest rangers had supposedly killed ten of the creatures and were covering the whole incident up. The hikers had got completely lost and had no food with them. Bud told us that they were lucky to have been found.
'It's a big place out there . . .'
I asked him, just for clarity to assure me that forest rangers had _not_ carried out a hushed-up massacre of Bigfoots and he confirmed that, no, this had not happened. I asked him what had started the story.
'Who knows? Some wacko writes something on the Web and then it's fact. Those sorts of things get in the way of real sightings.'
I later looked up this theory and could find only a confused story claiming that Patterson and Gimlin, on an earlier trip, had stumbled on a clan of Bigfoots and had shot dead several of them. They then panicked at their similarity to humans and buried the bodies. Supposedly, when they returned and shot their famous film, 'Patty', the Bigfoot featured, had been in the process of digging up the bodies.
The world of cryptozoologists is indeed a weird one.
We thanked Bud and got back into our car. I was a bit depressed about not being able to go up the Go Road. This meant that we could not get near enough to the Patterson Gimlin site to hike in. Richard looked at me like I was crazy.
'Sure we can go up there. He said what he had to say, I said what I had to say, but the body language was all about understanding that we were going up there anyhow.'
I hadn't picked up on any of this at all; I just heard that the roads were closed and that was that. But Richard clearly knew the area and I trusted him. We filled up with gas at a petrol station run by Indians. There was a sign on the door with a drawing of a nasty-looking gun:
If you are found here tonight, you will be found here dead tomorrow
The Indians seemed pleasant enough but you certainly wouldn't mess with them. With a full tank we headed uphill on the Go Road. It was a rather magnificent thing and we cruised up it in bright sunshine in high spirits. We spotted mile markers, which were useful because we needed to turn off at the seventeenth mile; there we would start heading down to the end of Cedar Creek Road and walk from there. I couldn't help feeling that we were a little underprepared but Richard didn't seem worried and he knew the area so I rolled with it.
I adore driving, especially in snow or sand – the fun stuff. We saw nobody on our way up, although there were supposed to be the paranoid weed farmers hidden away all over the place and they don't like strangers. On we drove until, at the ten-mile marker, we started hitting some patches of snow. It was mainly lying on the side of the road and not too bad so we cracked on. Any part of the road on the south side was exposed to the sun and had no snow but we slowly came across more and more stretches of road that were on the north side and in shade with a lot more snow on them.
There were a couple of tracks through the snow, and my rental was an all-wheel drive, so we carried on. I motored through a particularly deep bit, just keeping the car under control and on the road and preventing us from falling off the drop on the right. I got a massive adrenaline rush and Richard seemed impressed with my driving. The short snowless stretches gave us confidence to face the next lot. I started to drive and film at the same time and this made Richard a little more nervous. He hinted that, should I need a cameraman, he could be of assistance. On we ploughed as the road got steeper and darker and snowier. As we passed the fifteen-mile mark we came round a corner and the road got very steep and the snow extremely deep. I gunned the motor and flipped into low gear. We made steady progress but it was tough going. About half a mile further up the slope I veered out of the tracks and into deeper snow. The SUV stalled and we were stuck.
We were not overly concerned. We both got out to film the situation and made jokes about what idiots we were. It was noticeably colder than down in Orleans, even though it was only around one in the afternoon. Richard was all for calling Bud and getting him to come up and help us out but I was rather embarrassed that we had not heeded his advice. My manly pride kicked in and I thought that we should try to do this ourselves. I had a look at the car. We were not that stuck and I thought that, if we dug the snow out from under the car, we should be able to move back into the tracks and reverse back down to a better spot. Richard didn't seem so convinced. I was slowly coming to realize that he was about as practical as I was: i.e., not at all. We were two idiot city boys stuck in the middle of nowhere.
We set about digging the car out and soon had all four wheels fairly clear of snow so I got back in and started the engine. Richard was at the front pushing and, after a while, a wheel gripped the snow and we moved back on to the tracks. I was elated and whooped with excitement. Richard shouted at me not to stop and to keep reversing downhill until I got to an easier spot. Over-adrenalized, I shot off way too fast. For a while I managed to keep the SUV on the road but suddenly I skidded very badly and the car screeched into a ditch on the side of the road and smashed against the bank. We were now well and truly stuck. Richard came down looking concerned and I announced ruefully that we were not getting out of this one. It was clear from his face that he already knew this. I said that maybe it was time we called Bud. He picked up his mobile and peered at it for a while before putting it back down.
'You not going to call him?' I asked.
'We haven't got any reception,' he answered. The first shiver of uncertainty shot through my body. I'd seen this before in Ray Mears's TV shows – the gentle progression from adventure to big trouble. Nobody knew that we were up here, unless Richard's 'body language' theory was right (which I was starting to doubt). We had no way of contacting anybody and we had only four hours of daylight left. We had to get the car out. I looked around and spotted a fallen tree. I told Richard that we had to get some of the thick bark on the trunk and then put it under the tyres to create traction. He looked rather impressed by this. I waded into the deep snow and ripped some bark off in long strips. We put them under the car and I tried to rock back and forth using the engine but the car was stuck fast.
We struggled for a while but it was useless.
We needed to make a decision as to what to do. We were stuck in the middle of nowhere, very high up a mountain, and it would get extremely cold at night. Richard thought that we should start walking down the road and he would soon get cell reception. I couldn't help remembering Ray Mears's golden rule of survival: always stay by the vehicle . . .
That, however, was surely when you were lost in the middle of nowhere? At least we knew that there was a town seventeen miles away down the road we'd come up. Sure, there were bears, mountain lions, paranoid and armed weed farmers and possibly a Bigfoot, but there was a town at the end. We got our coats out of the SUV. We put our three bottles of water and a pack of turkey jerky in a bag and headed off. Confidence at this stage was fairly high. We would simply walk downhill into the sun where we would get cell-phone coverage.
For the first mile or so we kept stopping to hunt for the elusive single bar, but there was nothing. I told Richard that we needed to be near Orleans before we got any signal. He admitted he hadn't had any there either. I started to realize that we were not going to get cell coverage. We needed to find someone to help us or it was going to be a very long trek.
I thought humans averaged about four miles an hour walking speed so I estimated that it would take us about four hours until we hit Orleans. It would be dark in three. It was doable but we had to be careful. Richard said that weed farmers tended to 'SSS': shoot, shovel and shut up.
Many of the local Indians were also a law unto themselves and not really who you wanted to bump into. I'd heard a story about Jeeps Colgrove, a physically imposing Indian woman who used to wander about with a razor stuck in her hair. She'd committed several murders and used to pick people up off the side of the road and then ask them, 'Do you know who I am?' She was from the Hoopa reservation, which was not a huge distance away.
We really didn't fancy any of this. I'd stupidly left my passport in the car and was worrying about that as well. It was also getting darker quicker than we anticipated. I had a little panic. Had we really fucked up here? Were we going to become the poster boys for idiots from the city? I was wearing a T-shirt and a Prada anorak – what a twat. Then I remembered another Ray Mears tip: survival was all about keeping a positive frame of mind.
We started telling stories. Richard and I were actually quite similar: he used to work in Congress and for CNN whereas I used to work in Parliament and for ITN. We were both passionately interested in American politics and we discussed the upcoming presidential campaign. We both thought that Obama would scrape back in, but only because of the lack of serious opposition.
I was quite pleased with my pace, especially considering the old foot injury I have.
We reached the twelve-mile mark only to find a long uphill slope that we didn't remember driving over. The slope sapped our strength and we both realized how far we still had to go.
Had this been a script, this would have been the perfect time for Bigfoot to find us after we'd fallen unconscious by the roadside. I would awake in a Bigfoot nest to find the beast gently breastfeeding me.
We got to the ten-mile marker at four-thirty. We had one more hour of light, if we were lucky. It was getting very cold and I longed to see a car: even if we got _Deliverance-d_ maybe they'd let us go afterwards?
Richard kept saying, 'You've been in worse scrapes, right?'
And I kept thinking, _Yes, but with people who knew what to do._
I tried to imagine the headlines: 'Idiot comedian dies in Californian hills looking for Bigfoot. But is it a hoax?'
We reached the eight-mile marker and my feet were really starting to hurt. It was almost dark and getting rather creepy. We spotted some tracks going straight up a steep bank of red clay. We couldn't work out what had made them and, for a moment, we forgot our plight and got quite excited. Richard tried to clamber up the hill beside them and they were a big size. There seemed to be smaller tracks of something like a deer followed by far bigger tracks, bigger than a human foot would make. What puzzled us, however, was the steepness of the slope. It was almost vertical. Richard was having difficulty getting any traction but whatever had made these prints had gone straight up and over. Richard re-joined me on the road and we started trying to figure out what might have made them. It was not a bear, as there were no claw marks; these tracks had toes and did look pretty humanlike. As we stared I heard a voice – and then, a miracle. I spotted someone coming out of the woods further down the road. He was wearing a blue T-shirt and was carrying a basket. Was he a crazed weed grower? He didn't appear to be and I couldn't see a gun. To my shame we forgot all about the tracks and approached our potential saviour like desperate castaways to a friendly boat. He gave us a big smile as though he'd been expecting us. His name was Peter and he was picking mushrooms to eat. He had a car hidden out of the way down a side track because, 'There are dangerous folk around here.'
I asked him what he meant and he listed a series of people whom we really would not have been thrilled to encounter: local rednecks, homicidal Indians, bears, mountain lions, armed weed farmers . . . It appeared that we had been picked up by the only safe man in Humboldt County.
He seemed more than happy to give us a lift and we walked towards the car. I had never been so pleased to see anyone ever in my life. Peter seemed like a really cool guy. His parents ran a 'dude ranch', a place that harked back to the tradition of 'soft' East Coasters coming west for a vacation where they could rediscover their manliness by taking part in cattle drives, shooting guns, riding horses, etc. He was polite enough to leave the matter unsaid, but it seemed clear that he felt that Richard and I were prime candidates for this experience. Surprisingly, Peter had not seen Bigfoot himself but he knew lots of people who had . . . I told him about the tracks we'd seen and he said that they were probably deer or bears. By this stage I was too tired to care.
Peter clearly realized that we probably shouldn't be left alone again and he insisted on driving us the twenty miles to Richard's father-in-law's cabin. I'd expected this to be some small two-room hut. How wrong I was. Seemingly in the middle of nowhere, a set of gates swung open and led us up a drive and past two fish ponds to a giant log cabin complete with swimming pool. It was idyllic.
Richard's father-in-law was a very charismatic Irish ex-Marine. He was most definitely a man's man and I felt that, on the inside, he had to be howling with laughter at his effete son in law and his Brit girlfriend.
He was polite enough to keep his thoughts to himself and ushered us inside where a roaring fire was burning and several bottles of rather good red wine awaited us. Never had I been so cosy or so happy. We started to tell him about our experience and make light of it but I realized that we had made a really stupid mistake; it could have all gone very wrong.
We headed out for dinner with his neighbours. I sat in the back of his pickup truck on a veritable arsenal of guns and loose bullets and hoped that nothing would go off.
The neighbours were extraordinary people – uber-smart Berkeley types who had got away from it all. We were fed ginormous steaks and given more wine. Out of adversity come the best things.
At about ten I looked over the table and saw that Richard was nearly asleep. Our adrenaline rush was fading and we were cream-crackered. My foot was hurting really badly. We headed back to the cabin and, after a final tumbler of single malt, I hit the sack and slept the undeserved sleep of kings.
The next morning we woke early and headed back up the Go Road to retrieve our car. We had two big pickup trucks equipped with powerful winches and strong, tough men. As we reached the halfway mark a pickup truck roared down past us going towards Orleans at a rate of knots. Everyone joked that the Indians had found the car and pillaged it.
We passed the point where St Peter picked us up and I asked the guys to stop so we could look at the tracks – but it had rained hard overnight and the slope was washed clean. We kept on climbing. It was a very long way up to the car and even Richard's father-in-law was quite impressed by the distance we'd covered. Eventually we got to the car. It was safe; nobody had broken into it. Everyone had a good laugh at my bark-track attempts and Richard and I were gently moved aside as the real men winched the vehicle up and pulled it out. I then drove it gingerly backwards for a mile down the hill until I found a turning spot. We were free.
It felt really good to be under our own steam again and not reliant on others. Also, I had learnt a valuable lesson: I am a moron.
We headed back to the cabin for some beers in the sunshine. Then we went on a tour of the property. Richard's father-in-law was a very devout Catholic and he'd built a beautiful chapel on the edge of a cliff. Richard had met his wife at a retreat here. I realized how different Americans are to us. Back in the UK, if someone's Christian it kind of defines him, whereas here in the USA it's a sort of given.
'Have you seen Bigfoot?' I asked Richard's father-in-law, anticipating the stock answer.
'No. I don't believe in that stuff. The Indians will tell you any old story you want to hear.'
At last, a different perspective.
We said our goodbyes and headed back to Willow Creek. I had been going to drive to Garberville to talk to the head of the Cannabis College but she hadn't emailed me back so I figured that I'd stay on in Willow Creek for another night. When we got there, however, my room was empty. All my stuff had gone and I panicked, thinking it had been stolen. It turned out that the owners had rented the room to some fishermen and moved my stuff out. They'd booked me into the Bigfoot Motel. Suspecting that even Bigfoot wouldn't stay there, I passed and decided to drive through the night to San Francisco.
Richard and I went to the Mexican place for a last meal. I asked Gonzales, the owner, if he had seen Bigfoot.
'No, I have not personally, senor, but I know many, many people who have . . .'
On the way out of town Richard suggested that I pop into Bigfoot Books, a small bookstore that was the centre of the new generation of Bigfoot hunters. I spoke to the owner, Steve Streufert, who was initially a little suspicious of me but eventually warmed up. He showed me some photographs of a trip he'd made to the Patterson Gimlin site in Bluff Creek. It was unrecognizable, but he was convinced that they'd managed to pin it down. There were various arrows pointing to trees and other identifiable things from the original film. I tried to take a photograph of the photograph on his computer but he got a bit shirty about it all. It reminded me of people hiding their JFK assassination 'evidence' from others. Surely we all just want an answer?
I asked him the question.
'No, but I know a lot of folk who have . . .'
I picked up a copy of the _Bigfoot Times_ and said goodbye.
Outside in the car park I bade farewell to Richard. We'd shared quite the adventure and there's nothing that bonds people together quicker.
On the drive down to San Francisco I had time to think about matters Bigfoot. Before I'd actually gone to the area, my main problem had been that it was simply impossible for a largish tribe of Bigfoot to be roaming an area of the United States without them having been scientifically discovered. Now, having been right into the heart of this vast region of almost impenetrable and empty forest, I was more convinced. The sheer volume of sightings by people who lived in the area and were not likely to mistake a tree stump for a bear, along with the disinterest in publicity a lot of them shared, made me think that there was something out there. As for the Patterson/Gimlin film, I remained uncertain. I really wanted to believe it, and that was maybe the problem: it's almost too good. A man goes off in search of Bigfoot with a sixteen-mil camera and hits the jackpot? It would be like a comedian sitting in his hotel room and spotting a monster in the lake through the window . . .
Everything is possible.
Six hours later I spotted the flickering lights of the Golden Gate Bridge through my windscreen. I had a little bit of time to kill before heading off to the airport so I went for some breakfast at the farmers' market in the ferry-terminal building. A lot of tourists had confused it with the nearby Occupy San Francisco shantytown. I watched as one woman peered into some grebo's tent and was spat at as she tried to take photos. Meanwhile a weathered-looking Indian woman was screaming at a policeman who had taken a piece of paper off her. She had been brandishing it in his face while refusing to vacate a seat reserved for 'Seniors for Peace'.
'This is my land! I was here thousands of years before the blue eyes came . . . !'
Everyone gathered round to watch the fight. Meanwhile some Maoists were having a huge argument with some anarchists. It turned out they were both supposed to have a stall in the same place and nobody was giving an inch. It was a truly depressing place. The only unifying thing about the whole encampment was the noxious odour of the great unwashed that floated off them all.
In a way they kind of reminded me of monster-hunters. Everyone obsessed with their particular theory or exclusive photo and not wanting to share it or listen to anyone else. It was Crazyville.
I headed off for a pulled-pork sandwich. Rather aptly it was called the 'Pigfoot'.
### Yeti
'There is precious little in civilization to appeal to the Yeti'
Edmund Hillary
The call for my flight came over the Virgin Lounge Tannoy.
'Would all passengers bound for New Delhi please proceed to Gate 22.'
I was off: first to New Delhi and then on to Kathmandu, where I'd catch another little plane high up into the Himalayas and then trek up into thin air looking for the Yeti, the Abominable Snowman. The Himalayas are so vast that I'd been slightly at a loss as to where to go for this particular quest. There's a monastery in Khumjung, within sight of Everest, that claims to have the actual skull of a Yeti. It would be a hard slog to get there and I eventually figured that this should be my destination; I would learn what else I could along the way.
My favourite Tintin book has always been _Tintin in Tibet._ Quite why it's called _Tintin in Tibet_ has never been clear, though, as Tintin clearly landed in Kathmandu. I presume _Tintin in Nepal_ just didn't sound as good. Maybe Hergé couldn't resist the alliteration? Whatever, Tintin had headed off to Nepal to try to find his Chinese friend Chang, who had been in a plane crash. The Yeti made a guest appearance in the book. I was rather hoping that he might make a similar appearance in mine.
Along with the Loch Ness Monster and Bigfoot, the Yeti is probably the most famous monster in the world. Pre-Buddhism a lot of Himalayan peoples reportedly worshipped a 'wild man', an apelike figure said to carry a stone for hunting. From the moment Westerners began to attempt to climb the peaks of the Himalayas, starting in the early twentieth century, reports came both of sightings of a bipedal, apelike figure and discoveries of footprints. Probably the most famous of these were the photographs Eric Shipton took of huge footprints. Shipton was a respected mountaineer who was attempting to climb Everest. He took them at an altitude of about 19,500 feet and the footprints looked human except for the size and the fact that they were made by someone or something wearing no shoes. I came across these photos as a kid and was blown away by them. I think it was the fact that so many reports and sightings came from well-known mountaineers like Edmund Hillary (he saw footprints in 1953) who were not in the business of self-promotion. A lot of them were serious, scientific types and this gave the sightings great credibility. Of all the monsters I'd been after so far, to me, the Yeti was the most credible given the extraordinary remoteness and inaccessibility of the Himalayas and the constant drip-drip of sightings from visitors to the region.
The flight to India was uneventful except that Virgin seemed to have kindly upped my minor-celebrity status as at least three people came up to my seat, shook my hand, and hoped everything was fine on board. Then, when we landed in New Delhi, a lady was assigned to meet me at the plane and take me through immigration to get my luggage. There was no queue jumping and no special doors; just the normal procedure but with someone in uniform standing next to me all the time. It just made me feel like a rather simple child being shown how an airport worked.
Flights to and from Kathmandu are notoriously late so I hadn't booked any connecting flights on the same day. I'd therefore decided to spend the night in the Indian capital, which was no real hardship.
I awoke the following morning at four and packed before heading downstairs to the lobby. As I got out of the lift a man standing in the middle of the lobby wearing a trendy leather jacket said, 'Mr Joly?'
I nodded at him and settled my bill before following him outside. I got into a tiny old Ambassador, the signature car of 'old' India. I slumped down in the back seat as another man drove us out of my hotel, amusingly named the Claridges. An armed guard at the gates bowed as we exited on to the relatively empty streets of pre-dawn Delhi. The airport was only twenty minutes away and we drove in silence for about ten minutes before the man in the leather jacket turned to talk to me from the passenger seat.
'We will stop somewhere for breakfast on the way,' he said.
I thought this very kind but a touch unnecessary.
'Thank you,' I said. 'I'm OK, I'm not hungry.'
'OK, we shall see later but it is up to you.' Leather-Jacket Man wobbled his head about in that peculiar Indian fashion.
The car drove on and I started to think about Nepal. It was late February and, technically, still too early to trek: things normally don't get going until March as the weather is too cold before then. I was going to climb to about 13,000 feet (that's two and a half bloody miles straight up into the sky, if you're reading this on a beach). I worried about whether I was fit enough to do this.
'First time in India?' Leather-Jacket Man was talking to me again.
'Uumm, no . . . I've been twice before: Delhi, Mumbai, Goa, Hampi and Agra.'
The man looked at me in surprise.
'You have been to Agra before?'
I nodded, not sure why this was odd.
'The Taj is truly magnificent but normally people do not revisit; there is so much to see in India.'
I half-nodded, not really taking in what he was saying.
'Is there any specific part you would like to visit that maybe you were unable to do so on your last visit?'
His sing-song voice suddenly caused me concern.
'Sorry what do you mean?'
'I am saying that if you would like, we can organize a different itinerary than possibly the one you did on your last visit to the Taj . . .'
'I'm going to the airport . . . To get a flight to Kathmandu . . . You know that, right?'
'You are Mr Crawley?'
'Joly . . . I'm Mr Joly.'
'Oh blimey . . .' said Leather-Jacket Man. 'You are the incorrect fellow.'
By this time we'd been in the car about thirty minutes and were well on our way out of Delhi towards Agra to start Mr Crawley's tour of the Taj Mahal. Leather-Jacket Man was very kind. He ordered the driver to take me to the airport while he telephoned the Claridges to tell a no-doubt concerned Mr Crawley that he was on his way.
I was flying on an airline called JET to Kathmandu, which was a short hour-and-forty-minute flight. The flight was two and a half hours late, though – which, according to the guy sitting next to me, wasn't too bad. Apparently you need 8,000-feet visibility to land in Kathmandu, as there's no facility for an instrument landing. Visibility was currently 160 feet. This was a predictable problem when you wanted to land in one of the world's highest capitals. I was disappointed to learn, however, that Kathmandu, at an altitude of some 4,400 feet above sea level, only just slips into the top-ten list. What's the world's highest capital city, then? The answer is quite complicated. It should be Lhasa, the capital of Tibet, at 11,975 feet, but since the country has been annexed by China Lhasa is now technically not a capital. The honour, therefore, should really go to La Paz, in Bolivia, at 11,811 feet. Unfortunately, although La Paz is the default capital of Bolivia, with most of the government institutions there, the official capital is actually Sucre. So the world's current highest capital is Quito, in Ecuador, at 9,350 feet. There's one for the pub quiz.
The first thing to hit me as I got off the plane in Kathmandu was the sweet smell of incense and a sense of restrained chaos. While filling out my form for a visa I managed to leave my iPhone on the desk. I discovered this only in the luggage hall. In most countries that would be it: there would be no way you'd be allowed back into the immigration department. Nepal, however, is a relaxed place and I wandered back through the airport without anybody even questioning me. I gratefully retrieved my phone and exited the terminal building. To my delight the weather was glorious: very sunny with clear blue skies. There was no sign of the mysterious thick cloud that had delayed our arrival.
I hopped into a car that had been sent to meet me and headed towards my hotel, the appropriately named Yak and Yeti.
The hotel was very plush and filled with Buddhist monks all bustling around a sumptuous buffet in the ground-floor restaurant. I chucked my bags into my room and went out to have a look round the city. I needed to get some hiking equipment for my trip and I'd been told I could rent it. I wanted a warm down jacket, a walking stick, a decent hat and a couple of maps. I turned on to the Old Kings Way and walked down to the Royal Palace before turning left and heading into Thamel, a bustling area of little streets containing hundreds of shops groaning with trekking stuff, hippy gear and Nepalese/Tibetan art shops. The streets were packed with tuk-tuks, motorbikes and little cars all hooting and barging their way past pedestrians. I loved it. Every hundred yards or so, however, men walked past very close and whispered, 'Grass?' or, 'Wanna smoke something?' or, 'Marijuana sir?' This was definitely not my bag so I walked straight on, trying to look as though I knew my way around. This is always the secret in these types of places. If you look hesitant for even a second then you'll be swooped upon. Fortunately there were far greener horns than myself wandering about and I was mainly left alone. There was one ratty-looking little guy who spotted me and made a beeline towards me. I tried to swerve and move but he was fast and right next to me in seconds.
'You want tiger balm?' enquired my new friend in hushed tones.
I looked surprised. 'Tiger Balm?' I asked him.
'Shhh, police! You want tiger balm? I have best tiger balm in Kathmandu.'
I was confused: wasn't tiger balm some muscle relaxant easily available at any major pharmacy? Why was this guy 'dealing' tiger balm when it wasn't even illegal?
'No, thank you. I'm OK for tiger balm right now.' I tried to dismiss my new friend with a lofty wave.
'My friend, this is best tiger balm in Kathmandu – premium gold standard, sir . . .'
I really didn't know what to say and just kept walking until he finally gave up and picked on someone else. I walked on wondering whether I'd possibly misheard him. Maybe he'd been selling tiger bum? Maybe this was a new dastardly area of Chinese medicine now that they'd finished with tiger penis and dolphin nose? Were they experimenting elsewhere? Sadly I shall never know.
I rented a thick down jacket, a water flask and a walking pole for my Himalayan monster-hunt then walked back to the Yak and Yeti. There I had to meet Robin – an Englishman who had been in the Gurkhas and had then driven out to Nepal in 1978, literally moments before Afghanistan and Iran made that particular trip inaccessible. Robin had lived in Nepal ever since, and was helping me with my expedition. He'd arranged a Sherpa guide for me. His name was Mingmar and he'd come along with Robin to meet me. The plan was to fly into Lukla, one of the world's most spectacular airstrips, perched on the edge of a cliff. From there we were going to trek up the Khumbu Valley towards Everest and the town of Namche Bazaar (two days' walk). From Namche it was half a day's walk to the monastery at Khumjung where I hoped the monks would show me the scalp of a yeti. Mingmar had been born in Khunde, a village right next to Khumjung, and knew everybody up there. He had a very wide smile and seemed to be happy that we were trekking at this time as there would be very few people about.
Robin warned me about altitude sickness. It is a problem above 6,500 feet and could affect anybody, especially those who don't acclimatize and climb too quickly. You could be an incredibly fit marathon runner and it still could affect you whereas 'someone like you' – Robin looked at me slightly disparagingly – 'might totally get away with it; you just don't know.' He recommended that I take a pill called Diamox twice a day. Although not actually designed to help with altitude sickness (it's for glaucoma and epilepsy) this thins your blood and climbers have been using it for ages. Robin said it would help but warned me once again that you never knew how altitude was going to affect you. If he was trying to freak me out then it was working.
I said my goodbyes to Robin and, as he walked away, wondered whether I should have mentioned my intense loathing of walking uphill. It was too late, however: he'd disappeared into the Kathmandu night.
I certainly wasn't going to find a Yeti sitting around the hotel. They did have a rather pathetic footprint in a rock in the garden, which a sign claimed 'had been found when the hotel was being built . . .' Yeah, whatever . . .
I sat down in the lobby to read a bit of the only book I could find on the Himalayas: the one written by Michael Palin to accompany his TV series. The book was, like Palin, very charming and enthusiastic. I'd met Palin once at a show we were both doing in memory of Peter Cook. He was utterly charming, like a rather lovely uncle whom you could be fairly certain wouldn't abuse you.
I was pleased to see that he'd also stayed in the Yak and Yeti. I started to read about his director being kidnapped by Maoist rebels while they were trekking but couldn't really concentrate as there was a pianist playing 'Baa-Baa Black Sheep' in the centre of the lobby. He then moved on to murdering a ropey version of 'Let It Be'. Why are there always pianists in bloody bars? Nobody wants them there. It's like music in lifts. Who decided that lift music could be either pleasant or necessary? Were 'they' scared that, left alone with our thoughts between the first and third floors, we might find it all too much and blow our heads off with concealed handguns?
Maybe lift music accounted for the biggest news story to hit Nepal since Everest was conquered? (George Mallory, the English climber who disappeared on Everest in the 1920s, hated the term 'conquered'.) In 2001 the crown prince shot dead the king and queen and seven other members of the Nepalese royal family before killing himself. Perhaps he had been in a lift with no music? More likely he'd had to sit in the lobby bar of the Yak and Yeti listening to this God-awful pianist. Whoever was to blame for that tumultuous event, it had led to the deposing of the monarchy and the establishment of a democratic system that was currently still finding its feet. The poor Maoists who kidnapped Michael Palin's director had lost their mass appeal as they found themselves to be just another potentially corrupt political party. Being a guerrilla was so much more fun.
I got to a section in Palin's book in which he describes asking his Sherpa about the Yeti. The man told him that Yetis liked to drink so locals attempted to catch them by leaving out dead dogs full of alcohol. Unbelievable: Bigfoot loves menstruating women and the Yeti is an alcoholic.
The pianist was now playing 'Delilah'. It was definitely time for supper and bed. My adventure started the following day and I wanted to be in good spirits. I had an early start and reading about Palin's altitude sickness was making me nervous again.
I joined the seemingly endless hordes of Buddhist monks pigging out at the hotel buffet. It appeared that, like their Benedictine counterparts, Buddhist monks don't do the ascetic thing. They like the good life. I opted for à la carte as I can't trust myself with buffets. Humans, in my experience, when faced with unlimited food, will just keep eating until they can't walk. It's the inner hunter-gatherer instinct within us all. This is a basic truth as relevant to Buddhist monks as much as to porky comedians from the Cotswolds.
I slept well and left the hotel at five in the morning, having left my main suitcase with the concierge for the duration of my trek. I had managed to get what I needed into a smallish rucksack. A very tiny man, flirting with the frontiers of dwarfdom, picked me up and drove me through the deserted streets of Kathmandu. Occasionally the headlights would pick out groups of Nepalese police deployed in strategic corners and covered in protective riot gear. My tiny driver was perched on two big cushions that just about enabled him to peer over the dash at the pockmarked road ahead.
The domestic terminal was a place full of exotic-sounding airlines like Buddha Air and Yeti Airlines. Yeti Airlines were sadly in the process of changing their name to the less interesting Tara Air, which was a shame. I was pretty sure that this had something to do with the terrible crash they'd had at Lukla's Tenzing-Hillary Airport, the very airstrip we were now headed for _._ It happened on 8 October 2008: there was very heavy fog but, for some reason, the pilot still tried to land on the tiny (1,500 feet long and 65 feet wide) runway. He missed and smashed into the cliffs below, killing all eighteen passengers. The pilot was the only survivor. I was flying on Tara Air and hoped that they'd done more than just change their name. As my Sherpa guide, Mingmar, joined the scrum for tickets I realized that I was feeling rather out of control. I worried about how I would fare with the altitude – Robin had said that 99 per cent of people suffered from it and around 20 per cent were totally incapacitated. I opened my bag and necked my first Diamox. I felt a bit wimpy. Even at our maximum height, at Khumjung Monastery, we would be at roughly 13,800 feet -only half the height of Everest. I have a neighbour in my village back home with the unlikely name of Kenton Cool. He has apparently climbed Everest loads of times. I should have popped in to see him before I went, except I would have probably felt even more of a wimp.
The plane was a Twin Otter – a name that I presumed represented the total power of the two tiny engines. There were twenty of us crammed into the thing: five trekkers (obviously one of them being a Kiwi: international law forbids any interesting journey taking place without a Kiwi) and the rest red-cheeked locals. The plane took off very steeply and, as we flew higher, we bounced around as though we were in a fairground ride. The cockpit door was wide open and I could see no sky through the windscreen, just monstrous snowy peaks that seemed to loom above us even though we were miles up in the air. I thought of Chang, Tintin's friend whose plane crashed high in these mountains. Despite Hergé having never travelled, his depiction of Kathmandu was remarkably accurate. I hoped that this would not extend to this plane trip.
We landed on the sloping runway at Tenzing-Hillary It was eight in the morning and cold. We were at 9,350 feet and I almost immediately felt a little dizzy and disorientated. I wasn't sure if it was altitude, hypochondria or nerves. It was not unlike the feeling I'd had when I wandered into the _I'm a Celebrity . . ._ camp. Then, as now, I'd found myself in a desperate private battle to retain control. There was another bunfight for the luggage as the departing passengers were bundled on to our plane and it rocketed off down the runway to be flung into the abyss by the ramp at the end. They didn't muck around up here. The total turnaround time from landing to take-off was about seven minutes. I presumed they didn't want to risk a build-up of ice on the wings.
We walked out of the one-room airport and climbed a path that ran above the runway and then down into the village of Lukla itself. Mingmar signalled that I should follow him into a guesthouse, where he was warmly welcomed. I had some cheesy scrambled eggs and hot coffee and read a rather alarming leaflet on 'Acute Mountain Sickness'. I was still definitely feeling a little light-headed but I hoped that it wouldn't get worse as the symptoms listed were 'extreme nausea, vomiting, unconsciousness and death'.
Cheery stuff. We had a four-hour walk ahead of us to where we would be spending our first night in a village called Monjo.
The secret to trekking at this altitude, said Mingmar, was to go very slowly. This was absolutely fine by me and I assured him that he would get no speed out of me whatsoever. The beginning was rather nice as it was downhill to the valley floor below us. The only problem was that I was acutely aware that everything I descended I had to climb back up again.
The first half-hour was pretty easy-going and I just took in the scenery. Locals had to carve steps out of the steep landscape to enable them to grow anything. These descended like a curious set of giant stairs to the raging river far below.
The sun came out and it got rather hot so I immediately started shedding the layers and layers of protective clothing I had donned that morning in Kathmandu. I asked Mingmar if Yetis were ever seen this low. He replied that there were occasional sightings but that most occurred above 13,000 feet. He told me about a girl from Khumjung who had been attacked by a Yeti and survived. She had said that it smelt really bad. It had ripped out a huge clump of her long hair and thrown her into a nearby river. She had played dead and the Yeti left her alone and killed three of her yaks as she watched out of her half-closed eyes. Mingmar said that she would probably talk to me but that she would want money. I supposed that it was a way to earn a living telling tales of being attacked by the missing link. One of the unifying themes in the creatures that I'd gone after was that they smelt bad. Was being hygienically challenged a must for any self-respecting monster? It crossed my mind that the smell might not actually come from the monster but actually emanate from the witness, their body switching into automatic fear-response pong mode.
Mingmar told me that there were two different types of Yeti: a yak attacker and a man attacker. I asked him which type this one had been as it seemed to have done both. He said that it was a yak attacker – otherwise she would not have survived.
We walked on and on down the path, slowly descending towards the azure-blue river roaring beneath us. Every so often we'd come to a stupa that tradition dictated we had to pass on the left. Most had prayer wheels that you spun as you passed: 'for clean soul', said one sign. There were also enormous rocks decorated with multifarious symbols; most were money rocks and were supposed to bring good luck.
Keeping a steady pace behind us was another trekking couple – a crusty-looking Australian and his German companion, who seemed to be constantly weeping. We stopped for a breather and they did the same, with the German girl ripping off her Converse shoes and attempting to puncture several nasty-looking blisters. They were headed for Everest base camp, a full seven days' trek away. I was highly doubtful that they were going to make it.
In one village we passed a group of four British-Asian trekkers. They did a double take as I went past and then, when I was a hundred yards away, one of them bellowed at the top of his voice, 'Hello! No, I'm in Kathmandu and it's _rubbish_ . . . !'
Technically this was stupid as we were nowhere near Kathmandu, but that's British geography skills for you. I ignored them and walked on as they all creased up in peals of laughter. Mingmar was very confused and I tried to explain that it was from a TV show I'd done but I don't think it got us anywhere.
Going downhill was OK except you really had to watch your footing. It was the occasional uphill parts that really took it out of me, and this was the easy day. I'd been getting fairly fit in the previous months, having bought a running machine, but the foot I broke in Argentina was still giving me big problems. The thin air left me breathless and my foot was starting to ache; this was not a good sign but I cracked on and we reached Phadking in two hours, which was fairly good going. We stopped for lunch and I decided not to have a beer but went for a Coke (sugar energy) and the rather wonderful option of yak and chips. It was pretty good – a little stringy if I was being picky, but tasty.
We sat around for an hour chatting. I asked Mingmar what he thought my chances were of seeing a Yeti. He smiled and said that everything was possible. I was already starting to smell like one and hoped that this might attract one to me. We set off again and soon crossed a long and rather wobbly suspension bridge to the other side of the 'Seven Rivers Join' river (it possibly lost a bit in translation). I noticed that every Sherpa who passed by had a long red line on the top of their cheeks as though wearing rouge. I worked out that, because they had such angular faces, the strong Himalayan sun hit their cheekbones hard. This wasn't going to be a problem for me.
We started walking along the left bank of the river and I really began to understand the sadistic logic of the topography. If the path went down for a bit, it inevitably started going up soon after. After four hours of walking my legs were starting to feel like dead weights and my broken left foot was screaming in agony. I had to stop and indicated so to Mingmar, who pointed to some flat stones overlooking the river. We sat and I quizzed Mingmar more about the Yeti.
I asked him whether he had ever seen one. He laughed and said no. I asked him why he was laughing. Did he not believe in the Yeti?
'No, no, Eti he exist – just I no see him.' He said he'd heard the Yeti howling and made an echo-ey, throaty sound that reverberated across the valley to demonstrate. 'When we hear this we burn juniper branches – Eti no like.'
I'd seen villagers burning pine all along our path through the valley but this juniper Yeti-preventer tip was good.
'Also Yeti kill my father's yak.'
This was excellent stuff. I enquired further. They'd had three yaks in the family and one went missing. He and his father went into the mountains to look for it and they found it dead.
'It was rip apart, in two pieces. Eti kill yak.'
I nodded and asked him if he by chance had any photos of this.
'No, it was before mobile phone with camera. Now everyone have camera. My cousin he take photo with mobile phone of Yeti footprint in snow at 13,000 feet.' He drew a huge footprint in the sand.
I asked him if it could have been a bear or a yak footprint.
He said, 'No: this very big.' This was promising. He said I could talk to his cousin in Khumjung. I was stupidly excited.
We moved on and every step was torture now. I kept having to take more and more frequent breaks. Mingmar was really sweet and pretended that he was exhausted, but he was a rubbish actor. Finally, after five and a half hours' walking, he pointed to a village high up on the hill above us.
'Is Monjo – we sleep there.'
I could hardly breathe, both with the excitement of the news and because I was near death. We came round a bend and the path snaked steeply down back to the river where a little bridge crossed it. On the other side I could see the path wind suicidally up an enormous hill towards Monjo. It was the final push but I had hit the wall and it took me a good twenty minutes and several breaks before we got up the hill and finally walked into town. Mingmar pushed open a little gate to a sweet guesthouse. We'd made it.
I felt unbelievably relieved. The crusty Aussie and his weeping girlfriend (who was no longer weeping) were already there as they hadn't stopped for lunch and had pushed on through in a supreme effort that made me reassess their abilities to reach Everest base camp in Converse trainers.
We greeted each other like warriors back from a battle and agreed to meet for a beer a little later. I found my room, a sweet little prison cell with MDF walls and a rickety bed. I flopped down and immediately started to worry about the next day. The path up to Namche Bazaar was straight up the mountain for a good three hours. I reckoned I'd make it in about eight if I didn't start getting ill from altitude sickness. My left foot had swollen up and I could barely get my boot off. This was not looking good. I had to get up to a serious altitude to find this Yeti scalp. I wasn't going to have another disappointment like the one in the Congo. I limped over to the main building and ordered a beer. It was a real _Ice Cold in Alex_ moment as I downed the cool, lovely liquid in one go. The Aussie came in and, despite sounding like a New South Wales sheep farmer, turned out to be British and had only been working in Australia for two years. This was a man clearly desperate to change countries, as there was zero sign of his English upbringing. They were travelling on a shoestring and I felt very spoilt. His German girlfriend, Carina, soon joined us. She seemed in much better spirits and appeared to have forgotten her earlier unhappiness. I had some Eccles cake (my secret drug of choice for the trek – I'd got it from an outdoors shop in Cirencester. It's basically pure sugar and gives you a real kick) and I gave them one for the trip. I also gave them a couple of Diamox as I had way more than I needed. They clearly had very little money and they reminded me of what Stacey would have been like when she trekked here in the early nineties. They'd already been for a wander round the village and said that I shouldn't bother. There was nothing to do or see.
'Just a place renting a horse to anyone who doesn't want to walk. Can you bloody imagine?' The Aussie/Brit looked disgusted.
He asked me something else but I wasn't listening any more. A horse? That was the answer. I'd rent a horse. I wasn't here to bloody trek; I was a monster-hunter and I needed to get to where I was going as fast as possible. Also I knew that the less exertion you went through the less chance you had of getting altitude sickness. Also my foot was really hurting and . . . I knew I was just making excuses but all I wanted to do was to go and rent the bloody horse. I was too embarrassed to broach the subject with the Aussie/Brit and German so we had supper together in the communal room. A wood stove was giving out great heat and a couple of cute little cats wandered about stealing our food.
I had a second culinary first in the same day: water-buffalo curry. It was good, but the yak and chips just edged it. After supper I slipped out of the room and wandered down to where they'd said the horse lived. I found the lady owner and we did a bit of bartering. I managed to get a horse to take me up to Namche Bazaar for forty pounds. It was the best money I would ever spend. I returned to my (by now freezing) room and got into the first of my two sleeping bags and fell asleep almost immediately.
Mingmar woke me up early the next morning and we had a breakfast of eggs and black coffee. I broached the subject of the horse slowly with him. I told him that, as he knew, I was a leading scientist here to do some serious investigative work and I couldn't have my injured foot prevent me from reaching my goal. I told him that I'd come across the lady with the horse and had rented it for the day. He seemed totally astounded by this but nodded politely and said this was fine, although I could almost see the words 'You total wuss' appear on his forehead. Once breakfast and my embarrassing admission was over, we packed up and I prepared to meet my horse.
The horse was walked through the village to the guesthouse, as though for an execution, by the lady owner. Her name was Tiza (the horse, not the lady) and she had one look at this large foreigner and took an instant dislike to me. Undaunted, I hopped on and grabbed the reins. I'm a pretty good horse rider. I did quite a bit as a kid in Beirut and I've ridden all over the Atlas Mountains so I was certainly going to show Tiza who was boss. I urged her on but she refused to move. I gave her a couple of prods with the stirrups but she ignored me and wouldn't budge. I did the weird 'click click' sound that horse riders around the world have variations of. Nothing, Tiza was going nowhere.
The horse's owner meanwhile, a thin scary lady, was sizing me up and already regretting her decision to rent me the horse. There are temples in China on top of steep hills where lazy pilgrims can be carried up the innumerable stairs in hammocks suspended on a bamboo pole between two porters. The only catch is that the porters charge per weight of the pilgrim. While very sensible on the porters' behalf, this is rather humiliating for the larger pilgrim when being put on the scale and having their fee shouted out for all to hear.
Scary Lady had had enough of my equestrian demonstration. She grabbed the reins off me and set off ahead leading a still, reluctant Tiza.
Great, I was going to be led all the way up the Himalayas like a fat child at pony club. I'd envisioned more of a macho riding role for myself. This really made me look seriously Kenton Uncool.
Off we trod through the street of Monjo. Trekkers were preparing for the day ahead, sorting out their poles, stuffing their backpacks. All to a man just stopped what they were doing and stared at me as we trudged past. I could hear snickering from some and words in many languages that didn't sound complimentary. I looked straight ahead as though thinking about some great mission ahead of me but it was no use. It was a little like being paraded through the streets with the word 'paedophile' slung around your neck. The sense of general disdain was palpable. Thankfully we were soon out of the village and going along the path.
We came to a gate where there was an army checkpoint that wanted to see our papers. As the soldier took my passport he said something to Mingmar, looked at me and laughed. Mingmar laughed as well. As the soldier started to carefully peruse my passport I got off the horse and wandered towards a sign I'd spotted on the gate. It was in English and welcomed visitors to this 'special area'. It then went on to urge visitors to:
1. Refrain from taking life
2. Refrain from anger
3. Refrain from jealousy
4. Refrain from offending others
5. Refrain from taking excessive intoxicants
Bugger: this place was going to be no fun if I couldn't take lives and offend people. I decided to rely on a sensible amount of intoxicants.
On we plodded until we came to another wobbly metal suspension bridge that crossed the raging torrent below. UK horses would have been literally shitting themselves looking down through the thin metal lattice. Tiza, however, was made of sterner stuff and crossed over without a hint of concern. Once over the bridge the path climbed steeply up the mountain through thick pine forests. Mingmar told me that this was a new path: just three weeks ago massive winds had knocked down hundreds of trees in the valley and left the old riverside path impassable. I was amazed at how quickly it had been built. Since this was the only way up the valley and eventually to Everest, the livelihood of the whole valley depended upon it -and this was a powerful incentive. We climbed and climbed and I could feel Tiza breathing very heavily so I made us all stop and take a breather. I offered Tiza some Eccles cake but she wasn't interested. I was convinced that she was plotting about how best to chuck me off the vertiginous slopes to our left.
We rounded a corner and came across the Aussie/Brit and the German. They had left far earlier than us and were therefore unaware of my equine conversion. We passed them as they struggled up the steep hill. I waved an embarrassed hello. They were sweet enough to wave back but you could see that they thought I was a total arse.
We came to another bridge. This one was built by Edmund Hillary, who had clearly done a lot for the people of this valley. Mingmar had been educated at the Hillary School in Khumjung, which was how he learnt to speak such good English.
At the bridge I got off, as the exit from it was a ludicrously steep drop straight down which then immediately started to climb back up again. This was the beginning of the arduous three-hour climb up to Namche. The first bit had just been a warm-up.
When I say arduous, I know that I was on a horse and that any keen trekker reading this is poo-pooing it as a Sunday stroll. Well, my Sunday stroll is up the hill to the pub – a distance of about 300 yards with an altitude differential of about 13 feet. This was a three-hour steep climb going up more than 2,000 feet while already nearly two miles in the sky. We were already approaching the height of some of the highest peaks in Europe. Rant over.
I got back on to Tiza. I was totally out of puff from my five minutes on foot. This was pathetic but it also made it clear that if I wanted to see this Yeti scalp I was not going to do it without Tiza.
Up and up we (Tiza) climbed. The path took a sharp zigzag pattern with very little let-up. Tiza was clearly finding it quite tough as she had started farting profusely. Every time she did so Scary Lady looked back at me with an accusing glare. I smiled back, assuming she would recognize her own horse's farts. By the third such instance, however, it was obvious she was convinced it was me.
'Not me: horse . . .' I said, pointing at Tiza's arse. Scary Lady just shook her head in disgust and trudged on.
We passed a descending pair of German trekkers. They looked at Tiza and then they too looked at me in disgust. I tried to indicate that I had a broken foot but they continued on down, confident in their moral superiority. After an hour and a half's steady climb we reached a tiny plateau where a lone Sherpa woman sat with a bowl of tangerines for sale. I dismounted and bought one. Considering the effort she had made to get them there, it was the least I could do.
'You want to see Everest?' enquired Mingmar, as though he was showing me an interesting bird.
I walked past the tangerine lady and there, through a gap in the pine trees, was the tallest mountain on earth, the roof of the world. We were unbelievably lucky: it was another clear day and the peak was clearly visible, with a thick plume of cloud being blown off the summit towards Lhotze, the fourth-highest mountain in the world. I was dumbstruck. Everest is so much part of schoolboy folklore and here I was standing next to my flatulent horse looking right at it with my own eyes. I just stood and stared for about five minutes. Suddenly there was a new arrival on our little plateau. It was an Australian who had been on the same flight as me up from Kathmandu. He was crazily fit and had been full of talk about all this being quite easy compared to 'two weeks in the Bush'. He'd set off from Lukla at the same time as me with a Turkish guy he'd met. They'd gone at breakneck speed and told us they were aiming to get to Namche Bazaar in the same day. Mingmar looked very doubtful and warned them about altitude sickness and how long the trek was. They hadn't listened and set off confidently. The Aussie had even strapped a heart monitor on to his chest and he had a watch that constantly beeped at him to relay various medical information.
Now, halfway through the next day and here he was: alone and clearly not in Namche Bazaar. He looked extremely surprised to see me ahead of him. I didn't tell him about the horse for a while and asked him where the Turk was. He told me that he'd got terrible altitude sickness just out of Monjo, the village where we'd spent the night. He'd felt dizzy and was vomiting and they'd been forced to return to Monjo and overnight there. The Aussie had left the sick Turk and headed on alone that morning. I showed him Everest and he completely freaked out. He started filming himself and narrating at the same time. I left him to his video diary and clambered on to my horse. As I said goodbye, the Aussie looked up from his camera and noticed the horse. His face told me that our brief bonding period was over.
We climbed for another hour until finally we rounded a corner and I got my first glimpse of the curious village of Namche Bazaar. Set in a half-bowl on the mountainside, its multi-coloured buildings cling to the steep slopes in symmetrical rows. I rode into town praying that no Westerners would see me. A German couple did, but they looked the types to have several prisoners incarcerated in their basement back home. We locked each other in mutual stares of contempt. They changed tack and tried to give me a condescending look but I'd figured that the attitude to take now was that you only walked if you couldn't afford a horse. I was a king riding into town saluting his poor pedestrian subjects.
I was eager to try this new approach on others but it was Saturday and everybody was at the bazaar, of Namche Bazaar fame. People come here from as far as Tibet to barter and trade their goods. I got off Tiza gingerly and walked down the main street giving the distinct impression that I had a cucumber stuck up my arse. I found a place that would give me cash off my credit card and I paid Tiza's owner. It was money I would never regret spending.
We checked into the Yak Hotel and I had lunch, some _momos_ (Tibetan dumplings) and a bottle of sugary Orange Fanta. As I sat alone in the wooden dining room, I spotted a photo on the wall of the Dalai Lama. He was being led through some snowy mountain pass while seated on a yak. Not only that, but he was carrying an umbrella to keep the sun off him. This all made me feel a little better about my horse problem. If it was good enough for the Dalai Lama then it was certainly good enough for me.
I still felt absolutely fine, although I was very aware of the thin air and how it makes you behave a little like an old man. I shuffled around Namche to have a look at the place. It was the biggest village in the Khumbu but there was still not much to do. As in everywhere on earth and no doubt, when we eventually get there, on Mars, there was an Irish bar. I have no idea how the concept of global Irish bars started. Was there somewhere in the world an enormously rich Irishman who kicked all this off? Who was behind this worldwide conspiracy?
The other staple of world travel is, of course, the Kiwi. I wouldn't be surprised if there was an Irish bar run by a Kiwi on the summit of Everest. I hadn't been in Namche for more than two hours when I spotted my first one. He was wearing his All Black rugby shirt (it is illegal, as a New Zealand citizen, to wear anything else abroad) and wandering vacantly down a little alley. He spotted me.
'Excuse me, mate – do you happen to know a spot selling toilet paper? I'm bloody desperate and the guesthouse doesn't provide any. I think – pardon my French – if I don't find some soon I'm going to drop the kids off right here in the street.'
I pointed to a little shop down some stairs that I had just climbed up. It seemed to sell everything. He thanked me and ran towards the place in a cautiously desperate manner. I continued on toward the far end of town, where very little was going on. The day of the bazaar, always a Saturday, is also a holiday in town so lots of places were shut. I started back towards the Yak Hotel. Every step was a bit of an effort and I felt a little like an asthmatic pensioner. I shuffled into the street of the Yak Hotel and bumped into the Aussie/Brit and the German. They were both looking remarkably chipper. It had taken them only four hours to do the trek. I had clearly completely underestimated their stamina. There was no mention made of my horse; it was the elephant in the room.
I suddenly felt very tired. Just the walk around town had wiped me out. This is one of the symptoms of altitude sickness and the reason that I needed to acclimatize there. I spent the afternoon in bed, sleeping and reading. At six Mingmar knocked on my door and came in. He asked me if I was OK. I said yes but he didn't believe me and told me that it was important to tell him if I wasn't. I insisted that I was fine, just sleepy, and we went down to the wooden dining room, very like a European-style ski chalet. Two climbers were watching _Touching the Void._ Although a brilliant film, it really isn't the one I'd watch before going climbing. In five days' time the Everest climbing season (March and April) would start and there would be several expeditions going through Namche.
I still felt fine but incredibly lethargic – everything was a bit of an effort and I went upstairs and got into my sleeping bags and read some more Michael Palin. He was now in Tibet and his soundman had been hospitalized with acute altitude sickness. I felt very lucky and drifted off to sleep but I had terrible dreams. A Yeti smashed my window and dragged me outside. He put me over his shoulder like a rag doll and bounded up the mountainside. I didn't seem overly concerned about the Yeti's intentions but I kept shouting at him: 'I must acclimatize! I simply _must_ acclimatize.'
The Yeti didn't seem interested and we eventually ended up in a cave covered in blue ice, where he threw me down in a corner and started to watch _Downton Abbey_ on a television. When _Downton_ finished the Yeti was weeping loudly and he came over to my corner and started shaking me . . .
I awoke to find Mingmar shaking me and looking concerned.
'You OK?' he asked. I nodded, blinking in the bright morning sunshine. 'You were screaming.'
'Bad dream, but I'm good . . .'
I really was feeling OK and a lot less lethargic than the day before. I had another full day acclimatizing in Namche ahead of me before we set off for Khumjung. Mingmar wanted to take me to the top of the mountain behind Namche to visit the Sherpa Museum. We walked up a set of steep steps that seemed to go on forever. Every step in this thin air was torture. Eventually we got to the top and I was rewarded with an epic view of Everest. Once again the sky was swimming-pool blue and a thick plume of wind roared off the peak like a mini-tornado. Mingmar told me that I was very lucky to get this sort of weather in February.
We visited the Sherpa Museum, a lovely place commemorating all things Sherpa and especially their climbing achievements. Mingmar introduced me to Lhakpa Sonam, his cousin. He ran the museum and was a veritable fount of knowledge. He was, however, very deaf – something very common among the Sherpa people and put down to iodine deficiency. He asked me to write down any questions I had about the Yeti.
I started asking him any questions I could think of. He told me that the name Yeti was a Sherpa word, _'ye te'_ , meaning 'mountain monkey'. He was convinced of its existence as so many people had stories of encounters. The Yeti, he said, was supposed to have huge breasts – so if you came across one you should run downhill, as these breasts tended to knock it off balance. If you ran uphill the Yeti would sling the breasts over its shoulder and could climb very fast. It was supposed to have brown hair and be very similar to a large monkey. Sightings by locals all claimed it was bigger than a gorilla and he said that it existed on both meat and berries. He also repeated what Mingmar had told me about there being two types of Yeti: one that attacked yaks and another that attacked humans.
The vast majority of footprints and sightings were found between 16,500 and 19,500 feet. He said that when Hillary found footprints he became fascinated in the whole story. I asked him about the Khumjung scalp. He said Hillary had negotiated its loan from the monastery and it was taken to London to be examined. There it was ascertained that it was definitely not a bear. They said it had to be a very large creature but they did not know what it was.
I had to see this scalp.
I walked back down into Namche Bazaar and spent the rest of the day sitting in the sun outside my hotel, watching people go by. I saw the Aussie/Brit and the German again and we agreed to meet later for a drink.
Come five p.m. we headed for the Irish bar, but it was closed. Instead we went to a nearby bar from which ear-splitting house music was coming. Inside, the walls were festooned with T-shirts signed by visitors from around the world. Nearly all were friendly and funny – except the British ones that were invariably of the depressing 'Lads on tour'/ 'Smash it up!'/ 'Foreign bastards!' variety.
There were two locals playing snooker and a hectic international field-hockey match on the blurry TV. We sat at the bar and I must plead guilty to taking excessive intoxicants. The German told me that they'd climbed up the steps to try to take photos of Everest but when they'd got to the viewpoint they hadn't been too sure which peak it was. This was hardly surprising as it turned out they'd based their identification on a comparison with the mountain on the Toblerone packaging -they were convinced that was Everest.
I had to tell them that it was actually the Matterhorn.
We said our goodbyes and I wished them well. They were heading off towards Everest tomorrow and I was off to Khumjung. I'd spotted another horse for rent and hadn't been able to resist. I didn't tell my new friends.
I headed back to the Yak. That night was the coldest yet and I slept with my hat on with only my nose peeking out from beneath my two sleeping bags and three blankets.
I was awoken to the sound of very loud Buddhist chanting from one room next door and the heavy smell of dope from the other. Some trekker was clearly having a rest day. I felt on top form, particularly knowing that I had a horse on hold. My new steed was a lot better-looking than Tiza and went by the unusual name of Hermann. Apparently Hermann used to be owned by a German baker who plied his trade in Namche to hungry trekkers. Whatever, I was very pleased to mount Hermann as the route out of Namche was a veritable Kamikaze climb. I had given up all embarrassment about riding a horse: anything that got me to where I wanted to go without killing me was fine by me. A helicopter back to Kathmandu was the ideal scenario the moment my investigations were over. Sadly, this was not an option unless I fell off a mountain. It looked like I was going to have retrace my steps all the way back to Lukla.
Hermann and I set off uphill and, after fifteen minutes, passed a group of three Germans who looked to be very near death. They were at a stage way beyond contempt and I could now see a vicious, desperate sort of jealousy in their vacant eyes. They looked capable of ripping me off Hermann and claiming him as theirs. I kept his adopted nationality quiet and remembered the Dalai Lama. I passed by them quickly with a regal wave.
We climbed and climbed and Hermann made amazing progress. After an hour and twenty minutes we reached a plateau where Hillary had built the highest airstrip in the Himalayas: Syangboche.
This is a dust track ending in an aircraft-carrier type ramp from which to fling the planes into the void. He'd had it built to help evacuate stricken climbers. What I hadn't known was that, just three months after its completion, his wife and daughter died in a plane crash.
Hermann, Mingmar and I plodded over the deserted runway and entered a forest of short, stubby juniper trees. The ground was now an endless lawn of coarse grass and it reminded me very much of plateaus in the High Metn in Lebanon where I used to picnic as a child. On we plodded until we came to a corner with yet another magnificent view of Everest. We continued through patches of rhododendron, pine and juniper; it was by far the most incredible scenery of the trip so far. Here I was, riding a majestic German/Nepalese steed through the unexplored Himalayas. They would surely write books about my bravery in years to come. Only the very boldest made it here . . .
Suddenly up ahead there appeared an elegant, low-slung modern building. I asked Mingmar what it was.
'Everest View Hotel . . . Many Japanese, they come for one night to see Everest . . .' He looked slightly appalled by the concept.
I was dumbstruck.
'But, how do they get here?' I asked plaintively.
'They fly little plane from Lukla to Syangboche and then taken to hotel for one night. Is for very lazy tourist.'
'More lazy than me?' I asked, sitting on Hermann.
'Much badder than you . . .' Mingmar grinned from ear to ear.
'But you can't just fly into this altitude and not get bad altitude sickness?' I asked.
'No, they have oxygen in rooms but many get very sick; is big problem.'
'Well, I suppose, since it's here, we might as well go in for a cup of tea,' I said.
'OK, but very expensive,' warned Mingmar.
I got off Hermann and looked around vainly for valet parking. I tied him to a rhododendron bush and we marched up the imposing steps into reception. It was all minimalist swank inside and we were soon ushered on to the Everest-View Terrace for the pièce de résistance. The view was un-bloody-believable, possibly the best I've ever seen from a hotel, and once again I found myself staring slack-jawed at Everest.
We sat down and had a cup of tea. We were not alone on the terrace. To our left was a group of about twenty Japanese residents. The waiter told us that they had arrived only an hour ago. Looking at them, most seemed to be quite near death – but they were still bravely trying to rustle up enough energy to strike some gangsta camera poses. It was clear, though, that their hearts weren't in it. As we watched, one man dropped his camera and ran to the edge of the balcony and vomited profusely over the railings. A couple of the group started taking photos of the vomiting man while several started to suffer from the inevitable gag reflex. It was time to move on.
We walked down from the hotel through a small rhododendron forest until we reached the bottom of the valley and entered the village of Khumjung. I was really chuffed. Hermann dawdled through the dusty streets. He'd clearly had enough and knew that the end was in sight. We came into the main square in which sat a guesthouse run by Mingmar's brother. On the other side of the square was a stupa and the Hillary School where Mingmar had been a student.
We had lunch, the ubiquitous vegetable curry and rice. While we ate Mingmar told me that the woman who was attacked by a Yeti and thrown into a river would talk to me . . . For 6,000 dollars.
I nearly spat out my curry. I politely declined and suggested that I could possibly go to thirty dollars. Mingmar apologized but said that some Japanese TV crew had paid her this sort of money for an interview and she now refused to talk to anyone who wouldn't stump up the same sum. Bloody Japanese, they were really ruining this area . . .
I told him that for 6,000 dollars I wanted an exclusive interview with the Yeti himself. Mingmar laughed but he was obviously a bit embarrassed about the whole affair. He said that she had gone a bit doolally since the attack anyway. The Japanese crew had brought a Yeti costume with them, as they wanted to film a reconstruction of the attack. It turned out that they hadn't bothered to mention this to the woman in question. When the fake Yeti appeared she went totally mental.
Realizing that this interview was never going to happen I asked Mingmar if we could go to the monastery. He nodded, pleased to get off the loony-Yeti-attack-woman subject.
We left the guesthouse and Hermann, who was tied up outside, visibly flinched. I patted him on the head and assured him that his work was over before walking through a maze of waist-high stone walls towards the monastery, which I could see at the top of the village. We passed by locals sitting in the warm sun doing their washing or chopping wood. There was a house whose roof was ripped off in the terrible winds two weeks ago. About ten people were hard at work repairing it and it looked like a sort of Sherpa barn-raising ceremony.
After five minutes or so we arrived at the monastery, the Khumjung Gomba. A large money stone was positioned right outside the entrance. This was rather appropriate as it turned out that nobody could see the Yeti skull without paying a 'donation' to the monastery. This Yeti business was . . . A business. A warty Buddhist monk stood outside the main door and greeted us with a beatific smile as we entered a courtyard filled on three sides with wooden benches and surrounded by cloisters. Mingmar whispered that once a year the whole valley came here for a five-day festival.
Then an elderly looking man, not in monk clothes, made his way slowly down to greet us. He pointed to a hidden door covered by a golden drape. He lifted the drape and unlocked two stiff padlocks. He opened the door and it creaked open in a rather satisfyingly _Scooby Doo_ manner. We stepped inside behind him. It was a shrine and a rather beautiful one at that. Large multi-coloured Buddhas lined the rear wall and formed the backdrop to the central shrine. Hundreds of sticks of incense burnt everywhere in the room, creating a pungent, mystical fog. On both sides of the room were hundreds of little cubicles in the wall with the edges of prayer cloths hanging down from them. Just to my right an enormous gong hung from the ceiling. Both Mingmar and the old man set about bowing and lighting candles while my eyes darted around the room looking for the elusive Yeti skull. They eventually alighted on a locked lime-green metal cabinet that seemed at odds with everything else in the room. I waited until Mingmar had finished his ritual and then pointed at the cabinet.
'Is that it?'
'Yes.'
'What now?'
'Please make a donation.' He pointed to a slot in the side of the cabinet that I hadn't spotted before. I pushed a tightly folded 500-rupee note through the slot and the Keeper of the Skull (this appeared to be his title) unlocked the cabinet and dramatically swung open the doors. Inside was a glass box with the words 'Yeti Skull' daubed in white paint on the wooden frame. The box had a white silk shawl draped over it. I bent down and lifted the shawl off the box. Inside was a cone-shaped object, about twelve inches high. It looked like someone had lopped the top off the head of a cone-headed animal. The hair was a reddish-brown colour and, on first impression, it looked pretty convincing.
I asked Mingmar to ask the Keeper whether he could bring the box out of the cabinet. The Keeper said no, we couldn't touch it; it was forbidden. Mingmar spoke to him for a while and then told me that the Keeper had kindly agreed to get the box out of the cabinet and put it on the top for a mere 200 dollars. I was starting to get a bit hacked off with the financial nature of anything Yeti. It was an expensive business, this monster business. I told the Keeper of the Skull that for 200 dollars I wanted the case unlocked and for me to be able to put the skull on my head while levitating. Mingmar communicated this to the Keeper, who didn't seem very amused. We eventually agreed that if I gave 200 rupees to the Keeper personally, as opposed to making a donation, Mingmar would be allowed rotate the skull in the cabinet so that I could photograph and film all sides of it. The Keeper told me that this was a special honour and that I was not to tell anyone, so . . . You didn't hear it from me.
It was a bit annoying because I was pretty sure that I remembered footage in the original Arthur C. Clarke TV show in which a much younger Keeper of the Skull danced around the courtyard with it on his head.
I handed over the money and Mingmar gently rotated the box a full 360 degrees as I snapped away. One side of the skull had a vertical split all the way down it. Mingmar showed me a piece of paper on which was written, in dodgy English, the history of the skull.
Before the Khumjung monastery was established the peoples of the valley celebrated the festival of Dumji in the village of Thane. A dispute arose over who should organize the festival and the people of Namche, Khunde and Khumjung went it alone with Khumjung chosen to be the new host. As the new hosts it was expected of the people of Thane to give a worthy present in tribute. They gave them the Yetis Skull. They were so offended by this gift that they kicked it all the way home (hence the split) the skull was kept in the monastery and it was only in the twentieth century that its significance was realized.
I asked the Keeper if he had seen a Yeti. The Keeper said he had never seen a Yeti but he had heard them many times. He said that they sounded like a crying baby and that he often heard them at night. He did the cry for my camera and I have to admit that it was a touch spooky.
He then told me a story about a local villager who was at 19,600 feet with his Yaks. It started snowing really hard and the man wanted to move down lower into the valley. On the way down he saw a figure ahead of him in the snow. He thought that it was someone from his village and he shouted and the figure stopped. As he approached it the Yaks went crazy with fear and he smelt the creature (you guessed it: it was the Yeti) and it was not a good smell. The creature disappeared into the blizzard. When the man got back to the village he was crazy with fear and got very ill.
The Keeper now looked at us expectantly, like someone telling a ghost story to a bonfire of Scouts. There was silence for a moment. I wanted to hold the skull in my hands and it wasn't going to happen. The Keeper locked up the cabinet and ushered us out into the courtyard. We thanked him and he began the long, slow shuffle back up to his cloistered quarters.
We exited the monastery and walked back down to the square. I had a look round the Hillary School. It was very impressive. Mingmar went up to Khunde to visit his parents and I sat in the sun writing up my notes and soaking up the silence.
That evening Mingmar returned with his brother, the owner of the guesthouse. It was another freezing night and we sat around the communal stove drinking beer and talking. They told me about their other brother, who had climbed Everest. He had taken photos of a Yeti footprint in the Makalu region -again at the seemingly preferred Yeti altitude of 19,600 feet.
Then, out of the blue, the brother in front of me started telling me about a trip he'd made up the Holy Mountain the previous October. The 'Holy Mountain' is the name for the mountain that stands right behind the village; Western climbers are not allowed on to it.
The brother told me that they had built a big drinking-water construction project on a ridge on the other side of the Holy Mountain. One night, they were camping up at the site when the temporary water supply that they'd set up stopped working.
The brother climbed uphill to where they had set up a big water tank, only to find that it had been knocked over. Nearby were a set of huge footprints in the snow, just like the ones his brother had photographed. He said that he took two photos of the tracks on his mobile and then scarpered, as he was very afraid.
I asked him where the photographs he'd taken were. He said that they were on his computer. I asked him whether I could see them. He nodded and beckoned me through into the family bedroom. In the corner he had an old computer set up on a table. He fired it up and, when the home screen appeared, clicked on a file in the bottom right of the screen. There, on his computer screen, were two photographs of a set of large footprints. They were not the best quality, and he hadn't put anything next to them for scale, but they were clearly large footprints and he said that they were not of any animal he knew of. I saw no reason for him to lie.
Back on my night out in Namche, I'd noticed a painted Yak skull in the bar and I had told Mingmar how much I liked it. His brother had one that he'd bought in Tibet and he and Mingmar presented it to me as a present. I was so chuffed: it was a really beautiful thing, painted in yellow with Tibetan script on it.
I thought about it all night and in the morning I had to tell them that I couldn't take the painted skull: I'm always bringing stuff back from my travels and I worried that this might well end up being the straw that broke Stacey's back. Also, I was unsure as to whether I could get a skull through UK customs. Hillary apparently got the Yeti skull back to England with the help of the actor Jimmy Stewart's private jet. (Stewart happened to be holidaying in India at the time and helped Hillary out.) I had no such high falutin' assistance (as Jimmy might have said).
That morning it was crazily cold and there was a heavy frost on the ground. The plan was to walk all the way down to Monjo without a horse. We waited until the sun snuck over the nearest peaks and then set off. I'd rather hoped that Hermann would be waiting outside for me but his owner had retrieved him like a horse thief in the night.
The first twenty minutes were awful with a steady slog up a set of very steep steps. I huffed and puffed like a big bad wolf but, once we reached the top and passed a little stupa, it was downhill all the way. And I mean _downhill._ We crossed back over the dirt airstrip and kept going down. We threaded our way through the dwarf-juniper forest dotted with the occasional bulbous rock. Soon – ridiculously soon, in fact – we got to a point overlooking Namche. The view was wonderful but the descent into town was perilously steep and my knees were really starting to hurt as they took all the downhill strain.
A short history of my left leg
I suppose I'd better take you through the history of my unlucky left leg as I keep grumbling about it. Back in 1987, when such things might have seemed a touch cooler and I was still sporting guyliner, I was the proud owner of a pink Honda Camino Scooter (49cc). I was on my way down the Gloucester Road to my girlfriend's house when I overtook a bus and was hit by a Sloaney woman in a Peugeot 205. Her bumper went straight into my left knee and I went flying off the bike and into the doorway of a pub. I ruptured all the ligaments in my knee and had to have quite an operation that left me on crutches for three months. It also left me with quite a cool scar shaped like a question mark that I tell my kids was the result of a great-white shark attack.
Then, in around 1996, I was on a Greek island called Evia, visiting my lovely sister who lives there. I went with my then girlfriend, who had a PhD, allowing her to call herself 'Dr' Burr. (Weirdly, I have dated two PhD 'doctors'. One, the aforementioned Dr Burr. And the other? Dr Gurr. I kid you not.)
Anyway I wanted to rent a scooter but the Greek guy at the rental place persuaded me to take a motorbike instead. I had absolutely no idea how to ride a motorbike. We were in the town of Styra and Dr Burr was on the back when we came to a stop at some traffic lights. Dr Burr, who was also unused to motorbikes, started wobbling and the whole bike fell over, crushing my left knee.
Dr Burr was uninjured but I my kneecap was smashed into four bits. I was forced to fly back to the UK where (proper) doctors wired it up and tried to fuse the thing back together. While recovering in hospital I became rather attached to a button that would give me a hit of morphine every time I squeezed it. This helped a lot when Dr Burr came to see me in said hospital and dumped me unceremoniously.
Then, in 2011, I agreed to do _Celebrity Total Wipeout_ in Argentina. For those of you not 'up' with shit TV, this is an insane assault-course-type competition where you are thrown into water and bounce off huge red balls for the sadistic pleasure of the viewing public. I agreed to do it because they flew me out to Buenos Aires club class and, as I was going to Antarctica afterwards via Patagonia, it was all going to work out nicely. Sadly, in the qualifying round, I found myself in second place and took an ambitiously competitive leap into the void and landed very, very awkwardly on my left foot, snapping three metatarsals. It's a wonder, frankly, that I can still walk. And, yes, I am both left-handed and left-footed.
Anyway, back in Namche, the descent had killed my knee so we stopped in a store that was doing a roaring trade in knee supports and painkillers. I slipped two on and two in and I felt much better. We continued on down towards the river. Ten minutes out of Namche we walked past some panting trekkers.
'Don't worry, only twenty minutes to go,' I said to them and they smiled and I smiled back. We were all trekkers together in one big happy trekking world and nobody needed to know anything about my horse problem.
Then the painkillers kicked in and I felt a bit woozy and evil. We spotted another pair of trekkers struggling up the hill, about thirty minutes away from Namche.
'Keep going – three more hours and you're in Namche,' I smiled as they both physically crumbled before my eyes and sat down disheartened on the side of the trail. I walked on feeling no guilt and blaming my behaviour on the drugs and altitude.
As we carried on down Mingmar told me a story about when he'd gone to work in Japan for a year. He was doing construction work with ten other Sherpas and he now spoke fluent Japanese. When they'd first arrived, however, they didn't speak a word of the language. One of their gang went off to buy some food and came back with various tins of things and they cooked up a very good dinner. The following night they invited some other workers from a nearby camp to share a meal. When these Japanese workers arrived for dinner they saw the tins and were horrified.
'That is bruddy cat food!' they shouted.
Tears rolled down Mingmar's face as he told me the story and the sound of our laughter echoed off the steep valley walls.
After two and a half hours we reached the bottom and crossed the suspension bridge over the river. As we crossed we passed two trekkers walking along happily, seemingly without a care in the world. Little did they know of the fiendish ascent right ahead of them. It was like driving along on the empty side of the motorway having just passed a five-mile traffic jam on the other side and seeing people driving towards it unaware of what was to come. It made you feel good . . . Or maybe that's just me?
We walked along the riverbank until we stopped for lunch. We had vegetable curry, rice and some powerful chillies. The chillies certainly woke me up, and afterwards we made good time and were soon in Monjo. This was where we'd stayed on the first night and I was under the impression that we were stopping there. Mingmar, sadly, had other plans and so we marched on. He wanted us to sleep in Phadking, the village where we had stopped for yak and chips on the first day. It was another two hours' walk and my knee groaned in agony. I slipped on my headphones and listened to music. I started doing a much better pace, no doubt also helped by us having descended more than 3,000 feet. I marched to Bon Iver, Lana Del Rey, the Stranglers, Marianne Faithfull and the Divine Comedy. My legs were like lead and my knees ached with every step but I was managing not to stop for too many breaks. Eventually we crossed the final bridge and it was but a short five-minute walk up to the Green Village Guesthouse. As if on cue, on came 'Glad It's All Over' by Captain Sensible.
Walking through the main drag of the village there were several hectic games of Phadking underway. This is a game not unlike billiards but played with checkers on a flat piece of wood. The player must slide his or her checker across the wooden top and try to knock his or her opponent's checker into a small round hole cut into the corner of the wood. The table is sprinkled with flour to make it slippery. As the game was named after the village, I presumed it originated there – but apparently different versions of it are played all over the world. Either way it was certainly popular.
We climbed the stairs into the Green Village and I threw my bags on the bed and collapsed. I was filthy. A thick film of dust and dirt covered me but I couldn't have cared less. I was lying down and only three hours from Lukla and the flight back to Kathmandu. I remembered the steep walk down on the first day and I already knew that I had taken my last steps in the Khumbu Valley: I needed more horse.
Later that evening, as we sat around the wood stove in the centre of the Green Valley dining room, I broached the subject with Mingmar. He seemed unsurprised by my request and rang someone on his mobile to organize it. The stove was gloriously hot and I could feel the heat seeping into my tired bones. Outside the skies had darkened and the clouds had rolled down the valley, lowering the temperature by ten degrees in a second.
I was sick of beer so I bought a bottle of XXX Rum for us to drink. It did the trick and Mingmar was soon three sheets to the wind and suggesting we move on to the local drink, _rakshi._ This is made from millet and I'd had enough local spirits to know that it was going to be rough. I was wrong. It was utterly delicious, subtle, and really hit the spot. We sat around the Sherpa Aga and talked bollocks for hours.
My horse arrived early the following morning and, to my surprise, it was Tiza, my second day horse. The look of horror in the poor animal's eyes as she spotted me was unmistakable. She backed away, trying to turn round and bolt for home, but Scary Lady now saw me as a total cash cow and grabbed the reins for dear life.
The climb up to Lukla wasn't nearly as bad as I remembered and on the frequent flat bits I felt very embarrassed when passing fresh-faced backpackers heading off on their treks and staring at the lazy bastard on a horse. The last half-hour, however, was reassuringly steep and I was very pleased to have Tiza. I think I can confidently say that she didn't feel the same way. It started to snow quite heavily. We looked back up towards the Holy Mountain above Khumjung and it was white. We'd got out just in time. Mingmar thought otherwise.
'Snow good for Eti tracks,' he said ruefully.
Damn it, he was right – but I wasn't going back. I'd had enough and wanted a warm bath and my legs back. We trudged on up the path until we came to the final slope, where a memorial to the victims of the Yeti Air crash reminds trekkers that these are dangerous mountains. We passed by and crossed under the arch demarking the end of the trail. As we entered the town I quickly hopped off Tiza and walked in before anyone could see me. I paid Scary Lady off and patted Tiza on the nose. Tiza turned her head very slowly to look at me.
Our eyes met and we shared a brief moment and then she said, clear as day, 'Don't ever come back here again, you fat bastard.'
I jumped back in surprise and looked around to see if anybody else had heard her. Nobody appeared to have done so. I have had many such occasions, when I have been convinced that animals have spoken to me. It's either a very special skill or the first signs of severe mental illness. My son, Jackson, claims that every cat he encounters winks at him. It appears to be a family trait.
We said goodbye to Tiza and Scary Lady and trudged through the snow into town to my last guesthouse. I had to get a plane to Kathmandu the following morning and it wasn't looking promising. Given the height and the variable weather conditions, flights had sometimes been cancelled for up to a week. With the snow still falling hard, I had a sneaking suspicion that I might be getting to know Lukla rather well.
I spent the rest of the day writing up my notes and drinking cup after cup of sweet black tea. It was bollock-numbingly cold and, for the first time, I got out my rented down jacket and put it on in the communal room of the guesthouse. As with all Nepalese guesthouses, there was a wood burning stove in the middle of the room – but it wasn't lit. Various members of the guesthouse family came in, turned on a telly and watched an Indian show called _Dance India Dance._
This was a succession of terrible dance acts, one being a woman dancing round her ironing board while another was a man in drag dancing inside a closet . . . Subtle it was not.
The judges all spoke in Hindi/English saying anodyne things like, 'Very sweet act; I wish you the best of luck.' After the panel had spoken a rather creepy man in a leather chair (who reminded me of Cyril from _That's Life!)_ appeared to make the final decision. He slammed anyone male but was incredibly complimentary about any woman performer: 'You have a most fabulous form and such a charming smile . . .'
The Nepalese family _oohed_ and _aahed_ at every act as I desperately hinted that it might be time to light the stove. They ignored me and so I poured green chilli all over my Sherpa stew hoping it might warm me up. I went to bed at seven p.m. and had the coldest night I have ever spent in a bed (and I have slept in two ice hotels).
I woke at five-thirty absolutely certain that all planes would be cancelled. Miraculously however, the day became clear and sunny and the runway had been magically cleared of snow. I had a coffee and walked over to the airstrip. Mingmar came to say goodbye and gave me a white silk scarf for 'safe travels'. He was going to walk all the way back to Khumjung to help his parents build a new house. We said our goodbyes and parted. He was a great guy but I felt totally emasculated by him.
At seven a.m. a siren sounded to indicate the imminent arrival of a plane. Not only was it clear in Lukla, but conditions in Kathmandu were good too. The plane landed and passengers got off while porters hurried to offload baggage and hurl ours on. Meanwhile someone constantly rotated the propellers manually so that they wouldn't freeze up.
The engines roared and we started hurtling down the slope towards the drop. At the last moment the plane went up the ramp and was catapulted into the void. Everyone screamed, as the plane appeared to almost come to a standstill in mid-air. Then, somehow, it got some invisible traction and we were off. People started breathing again and unclenching their fists. The fat woman next to me let go of my arm and I felt blood start to flow to my fingers again. I looked out of the little window at the snowy peaks to my right. As I did I could almost swear that I saw a large hairy creature, about eight feet tall and covered in a reddish-brown hair. As I stared the creature raised its right hand and extended two fingers in the international sign of dismissal. I rubbed my eyes and looked again, but the mountains had disappeared and we were enveloped in soft white clouds that would carry us back to Kathmandu and the drudgery of the known world.
### Nessie
'Whoever fights monsters should see to it that in the process he does not become a monster.'
Friedrich Nietzsche
I couldn't really be an English monster-hunter and not go after the Loch Ness Monster, possibly the most famous monster in the world.
'Nessie' has somehow grabbed the world's imagination: you can mention her anywhere and people know the story. The term 'monster' was first coined to describe Nessie in 1933 by Alex Campbell, a water bailiff and part-time journalist, in an article he wrote for the _Inverness Courier._ This was the year in which the legend really kicked off, with a plethora of sightings of a large creature in the loch that continue to this day. There was apparently always something eerie about the loch, however: tales of something in the water stretch back to the time of St Columba, in the sixth century, who supposedly witnessed a man who was being attacked by a large 'water creature' in the River Ness.
In more recent times the canny locals of the loch have done their very best to use the story to attract the lucrative tourist dollar. A trip to Scotland would not be complete without a visit to Loch Ness. It's something that appeals to the whole family. Dad can be a monster-hunter, Mum can enjoy the scenery and the kids can get a cheap thrill worrying that they might be eaten by this modern-day celebrity dinosaur.
Truth be told, though, the idea of travelling to Scotland after my recent adventures – and as opposed to searching for the Chupacabra in Puerto Rico or the Wild Man of Borneo – was not that appealing. I was a monster-hunter and wanted to hack through jungles and climb mountains, not drive up the M1. Also, on this trip I'd be travelling with my wife and two kids. Don't get me wrong. I adore my family but it's just that when I'm 'working' I like to travel alone. This isn't because I don't miss my family when I'm gone (I do, terribly) but is because, in my opinion, to write a proper book you need to be on your own. This allows you to people-watch, eavesdrop, explore and get into trouble.
I occasionally write for the _Sunday Times_ and they had provided me with a brand-new Mercedes 'family wagon' to review. They suggested that I go off on a trip with my family to somewhere in the UK to test out the vehicle. It was half term and the kids were bored. Everything was pointing in one direction. We were going to Loch Ness to monster-hunt en famille. God help us all.
The Mercedes turned up. It was a lovely, posh new yuppie-mobile and it had been brought over from Germany especially for this story. _No pressure . . ._ I thought to myself as I climbed in. I got in the wrong side. It was a left-hand drive with German number-plates. The kids loved it, playing with all the buttons in the back that turned your seat into a Jacuzzi or some such nonsense.
I punched 'Inverness' into the complicated sat-nav system and it announced in a German/English accent that the trip was about 500 miles. It was also claiming that it would take me eight and a half hours. This was crazy. I thought that Germans had no speed limits on their autobahns? It seemed the moment it got over here it was getting all Little Englander with me.
'How long until we get there, Dad?' The kids were restless and we were still parked outside the house.
'About three hours,' I said.
'Dad is lying. It's about nine hours, kids,' said Stacey.
_'Nine hours!_ OMG! What the hell are we going to do for nine hours?' cried my little monsters.
'We could play I Spy?' I suggested, hoping they'd say no.
'Why did you tell them it was nine hours?' I whispered to Stacey.
'Because that's how long it is,' she replied.
'Now they're pissed off already,' I whispered back.
'Why are you whispering?' asked Stacey.
'Forget it,' I said, trying to find the button that turned the engine on.
'I spy with my little eyes something beginning with L . . . !' cried the back seat.
'It's L for Long Time – too long to be in a car!' They all laughed and for a moment forgot they were unhappy . . . But only for a moment.
I found the button and pressed it. The engine came on but was so 'eco' I had to get out of the car to check, as I couldn't hear anything. I got back in and drove out through the gates. We were off, monsters hunting monsters.
The drive was pretty much as I'd imagined it would be. The kids bickered most of the way up. Someone hit someone; someone denied hitting someone; someone hit someone back.
'Shut up.'
'No, you shut up.'
'Dad, Parker hit me.'
'Mum, the little brat kicked me . . .'
As the soap opera developed in the back seat, things weren't much better up front. Because we were in a left-hand drive, Stacey – who was in the 'exposed' passenger seat – was convinced that every oncoming car was going to hit her. Every single time a car went past us she would flinch and scream: 'You're in the wrong fucking lane, you big bonehead! Aaaaaaaaaarrrghghgh!'
By Burford, just twelve miles away from home, I was fantasizing about murdering her. We had eight hours and forty-five minutes to go and I realized that I was on my very own _National Lampoon's Vacation._
Suddenly a car pulled out in front of me without looking. I slammed on the brakes and just managed to avoid hitting him. I was about to remonstrate with the single-cell organism in the clapped-out Ford when he beat me to it.
'Fucking German bastard. Learn to fucking drive, you Nazi arsehole!'
He sped off, giving the finger to both the entire German nation and me. Stacey and I looked at each other and had to laugh.
Although we were after the Loch Ness Monster, we were not going to be staying beside Loch Ness. Experience has taught me that if a hotel doesn't have a swimming pool, or at the very least a hot tub, then kid trouble lies ahead. I'd found a hotel with a pool in Inverness so the idea was to hang out there and make investigative forays to the nearby loch.
I must have been high on hallucinogens when I agreed to this idea. By the time we eventually rolled into Inverness, ten hours later, nobody was talking to anybody. We simply communicated via a series of sharp elbows and seat kicks.
All I wanted to do was hit the bar, have a drink and get something to eat, but the kids had got the overpowering stench of chlorine in the lobby into their noses and wanted to swim. Reluctantly I got them into swimsuits, found towels and wandered down to the pool.
To my delight there was a hot tub right next to the pool. Unfortunately there was also an enormous tattooed Geordie sitting bang in the middle of it. Without much choice I slipped into the bubbly soothing waters. I tried to look away from the tattooed Geordie but he just sat and stared at me as though I'd spilt his pint. He appeared to be struggling to say something and it took a while but it eventually came out.
'Are youse him off the telly?'
I didn't want to talk to him. I didn't want to talk to anyone. I just wanted to sit there, relax and get over my ten-hour drive from hell. If I did this, though, he would tell everyone he knew in prison that I was a wanker who was 'up' himself and needed a 'good kicking'.
I turned and smiled and looked bashful and said, 'Yes, I'm afraid so.'
He looked at me for a while before speaking again: 'No you're not him. Are youse mocking me?'
I didn't know what to say to this without getting a smack.
'No, no, it's me. I'm Dom Joly from the telly. How you doing?' I tried to sound relaxed and friendly.
'I thought you were that fella from the comedy-house show, wasssiisname?' His red face looked like it was about to explode with exertion.
'I don't know. I really don't know. Sorry, I'm really tired and just want to chill out.' I leant back and closed my eyes and hoped he'd go away.
'Do you think youse better than me?'
I didn't wait for any more. I got up, got out of the hot tub, grabbed the kids out of the pool and started marching them towards the changing rooms. They started protesting and trying to get back into the pool.
The tattooed Geordie shouted, 'Are those your kids?'
I ignored him and we got into the changing rooms to find a man standing naked, with one leg up on the bench, blow-drying his pubic hair. We left straight away and headed for the room.
'Hey, that was quick – how was it?' asked Stacey.
'Fine,' I said, looking for the mini-bar.
After half an hour or so we went downstairs to get something to eat, only to find that the hotel restaurant closed at eight p.m.
We went next door to a 'posh' Italian restaurant. It was 'posh' enough to have nothing for the kids to eat, but 'shit' enough to leave us all very unsatisfied and with a huge bill.
As we walked back to the hotel I had to try to explain to the kids why a man was drunkenly punching another man in the face while a woman hit the same man with a handbag.
Alcohol and weak genes,' I said.
I didn't know whether they understood and I didn't care. I just wanted to go to bed.
The next morning we woke up and everyone seemed to be in better moods. The sun was shining and Inverness looked rather lovely in the dappled morning sunlight.
To get everyone in the mood and get them up to speed on the basic legend, I showed them the _Arthur C. Clarke's Mysterious World_ episode on 'Lake Monsters'. I was a little bit worried that it might freak them out and that they wouldn't want to go to the loch, but they spent the whole time howling with laughter at what people wore in 1980.
Stacey and I laughed along, remembering some of our own eighties faux pas. There was me with my crimped hair and make-up, and her with the Princess Di flick and then the perm, oh God, the perm . . .
Down we went to breakfast for the usual joyless UK-hotel experience. Having passed on the heart-attack buffet, I asked for a cappuccino and was stared at as though I'd just demanded moon rock. Surely we all know about coffee now, even up here? The fifteen-year-old boy in charge of breakfast eventually agreed to go next door to the bar and ask. He was soon back, however.
'The manager says it's impossible.'
I gave up and sat down to nibble on a tough croissant. Parker trapped Jackson's leg between their chairs and gave it a little squeeze. Jackson started screaming. Everyone in the room looked up from their troughs and started staring.
'Is that the guy from _I'm a Celeb_ . . . ?' asked a scrawny mother of four potential burglars on the next-door table, as though I wasn't there.
'No,' replied the multi-earringed father out on a rare bout of parole. 'What would he be doing staying here?' I had to admit that he had a point. Even in the Congo I'd managed to get a cappuccino.
A double-decker bus drove past the window. It had a huge advert for _Celebrity Big Brother_ on Five. This would have been an encouraging sign of modernity had the show not ended five weeks previously . . .
We drove towards Loch Ness. We were on our way monster-hunting and I didn't have the foggiest notion of what we were going to do. We stopped at a place that had a statue of Nessie outside it. We got a couple of photographs and the kids wanted to know whether we were done now we'd found it.
They insisted on going into the shop that was like a Scottish mega-mart. Anything was Nessied up and for sale. There were Nessie hats, Nessie humps, Nessie shirts, Nessie fridge magnets, Nessie posters and . . . Nessie everything. There was also everything Scottish you would never want. The kids went mental and bought silly tartan hats, stickers, stuffed toys. I tried to appeal to Stacey but she had gone all misty-eyed and Scottish and reminded me that both sides of her Canadian family – the MacDougalls and the Johnstones – were from here. She started buying books called things like _Your Clan Guide_ and wanted tartan from each side. I was going to be bankrupted in seconds by monsters and ancestry.
My phone beeped indicating that I had a text. It was the features editor at the _Sun_ , Caroline. She'd read a Tweet that I'd posted asking anyone for help in finding people to talk to about Nessie. She had sent me a list of every nutter – sorry, 'specialist' – in the area. I looked down the list.
The first one was a guy called Steve. He was the man who had jacked in his job as a burglar-alarm installer and now lived on the shores of the loch in a former mobile library. He earned his living making little figurines of Nessie while keeping a permanent watch on the loch for the 'beastie'. I felt a little guilty when I saw his name.
About twelve years ago, just before we started filming _Trigger Happy TV_ , Sam Cadman and I made a recce trip up to Loch Ness. We were looking into doing some filming in the area and were driving around thinking of ideas. We heard about this guy in his little mobile home and, after a rather long session in the local pub, came up with a plan. Using driftwood, we made the shapes of an enormous pair of clawed feet. We attached these to two poles and then we waited for the lights in the beach hut to go out. When they did we approached stealthily and wandered all around the thing making 'Nessie prints'. We never hung around to see the excitement the next morning, but I'd always felt a little guilty. I rang the number and a relaxed voice said 'Hello' after about ten rings.
'Hi, is that Steve?' I asked.
'Yes, it is. Who's wanting to know?' asked the relaxed voice.
'Hi, my name is Dom and I'm writing a book on monsters. I'm currently up in the Loch Ness area and was wondering if I could pop by and have a chat with you?' I tried to sound like a friendly non-judgemental type of guy.
'Well, that would have been great but there's just one little problem. I'm currently lying on a beach in Thailand.'
This was a turn up for the books. It seemed that there was decent money to be made in Nessie-figurine making. I thanked Steve and neglected to mention our earlier half-meeting in the shape of fake footprints.
I rang the next name on the list. It was a man called Tony Harmsworth. He answered straight away.
'Hello.'
'Hi, is that Tony Harmsworth?'
'It is he. Who are you?'
'My name is Dom and I'm writing a book about monsters and I was given your name by a journalist as someone I should talk to about Nessie.'
'Ah, well, yes, I'm definitely someone you should talk to. Unfortunately I'm currently laid up in bed with a bad back. How long are you up here for?'
'Not long I'm afraid.'
'Then it's not going to be possible as I'm totally immobile at the moment.'
Things were not going well with my investigations. Then Tony had a suggestion.
'You should go and see Adrian Shriner at the Loch Ness Exhibition in Drumnadrochit.' I thanked Tony and hung up because the family had just come out of the shop laden down with their Scottish booty. We tried to cram it all into the Mercedes' boot before heading off down the loch to Drumnadrochit.
The Loch Ness Centre & Exhibition was not hard to miss. We parked up next to a pond containing another mock-up statue of Nessie and wandered inside past a small yellow submarine that looked like it was straight out of the adventures of Tintin. Tintin did in fact come to Scotland, although not to Loch Ness. He headed up to the 'Black Island', where he ended up discovering that the Bigfoot-type animal that so terrified locals was a gorilla used by counterfeiters to keep nosey people away.
We entered the exhibition: six rooms featuring slideshows, video clips and exhibits, all lit in quite a slick manner. It's pretty professional. The problem was that I had come here for some monster stories. Although the whole place is sold on 'Nessie', the entire exhibition does its very best to deconstruct the myth and leave you at the exit wondering why you bothered to come to Loch Ness. It seemed that Adrian Shriner was not a believer – not any more, anyhow.
I met him in the vast shop through which you're channelled to get back to your car. He certainly looked the part – a crazy long beard and sporting the full tweed. He was the epitome of a mad professor.
I'd left my 'Dom Joly Monster Hunter' card at the entrance and it seemed to have done the trick. He was happy to tell me about how he got started.
'Essentially I'm a lazy man and I saw monster-hunting as a quick path to glory. People might feel that going down into the depths of the loch in a tiny submarine was brave, but it's a lot easier than hiking to the North Pole.' I liked Adrian. He had a twinkle in his eye and was clearly a smart guy. When he'd first come to Loch Ness, in the late sixties, the world had pretty much been explored and explained. Monster-hunting was a way to have an adventure while also cocking a snook at established science.
He admitted that he had started off as a very keen Nessie enthusiast but was now more interested in working out what the famous sightings actually were. He had become a sceptic.
Age is a great rationalizer,' he said, chuckling.
Looking around the shop, however, it was clear that he was making a great living from Nessie, whether or not it existed.
The family was getting restless again and complaining that they were hungry. I sighed and said goodbye to Adrian before getting back into the family wagon. We drove past Urquhart Castle, a beautiful ruin that sits on the shores of the loch. This was the setting of the famous 'humps' photo taken by Peter MacNab in 1955. I'd read a book called _The Loch Ness Story_ , written – surprisingly – by BBC reporter and Nessie enthusiast Nicholas Witchell. He's probably best known for remarks Prince Charles made about him, having forgotten he had a microphone on: 'I can't bear that man. I mean, he's so awful, he really is . . .'
Witchell was and is firmly convinced of Nessie's existence. He had MacNab's account of the moment the photo was taken.
I was returning from a holiday in the north with my son and pulled the car up on the road just above Urquhart Castle. It was a calm, warm, hazy afternoon. I was all ready to take a shot of Urquhart Castle when my attention was held by a movement in the calm water over to the left. Naturally I thought of the 'Monster' and hurriedly changed over the standard lens of my Exacta (127) camera to a six-inch telephoto.
As I was doing so a quick glance showed that some black or dark enormous water creature was cruising on the surface. Without a tripod and in a great hurry I took the shot. I also took a very quick shot with another camera, a fixed-focus Kodak, before the creature submerged.
I remembered seeing that photograph when I was a kid. I tried to tell my kids about it but they needed food and were not in the slightest bit interested. We continued on until we reached the end of the loch at Fort Augustus. We parked up and went to the Bothy for lunch. I ordered haggis and felt like a bit of a tourist but I didn't care. I genuinely love haggis and have it as often as I can. My kids asked me what was in it and I started to try to explain but they looked properly ill so I stopped. Do Scots really eat haggis? From what I saw in Inverness, the national diet seems to be chips and curry sauce washed down with a deep-fried Mars Bar and a fag. (This is the moment when, if you're Scottish, you get all angry and put the book down to Tweet some abuse at me – but why do you guys do this? If someone abuses the English, we just laugh it off or invade you . . .)
We left Fort Augustus and drove back to Inverness. The kids were annoyed that we hadn't seen Nessie and wanted to hit the pool. I took a quick look to check that the tattooed Geordie wasn't in the hot tub. It was all clear. I got the kids ready again and took them down. The smell of chlorine was particularly strong but at least there was nobody about. I hopped into the hot tub only to find that it had become a cold tub overnight. Meanwhile the kids jumped into the pool but then got out quickly rubbing their eyes in pain and crying. Somebody had just dumped a vat of chlorine in the water and it was completely un-swimmable. We retreated to our room, giving the brain-dead mullet behind the poolside front desk an evil look that didn't even register.
I tried to book us a restaurant for supper but everywhere was full. I just couldn't understand it: surely Inverness isn't _that_ popular? Then I found out that it was Valentine's Day. I was slightly mortified and considered bluffing my way through, as Stacey appeared not to know either. In the end I came clean. Fortunately neither of us are really Valentine's obsessives so it wasn't too much of a disaster. We did, however, have to eat downstairs, alone in the bar. I munched on my distinctly unromantic carb fest of chicken balti (mostly potato) with chips and rice. Stuffed, we staggered up to our twin beds that the hotel had so kindly provided for our romantic evening. We ended our Valentine's night propped up in separate beds playing Scrabble together on iPads.
When we'd been driving about I'd spotted a sign advertising 'Nessie Cruises' on the loch, and I'd booked us a passage for the following day. The kids were a little bit worried that Nessie might attack us but everyone eventually agreed to the trip.
The next morning we left Inverness again and headed off towards the loch. On the way Parker asked Siri, the 'brainbox' who lives in my iPhone, whether Nessie exists. Siri was pretty adamant: 'No. The Loch Ness Monster is a mixture of misidentification and hoaxes.'
We caught the eleven o'clock cruise. We'd had two alternatives – a two-hour trip that included a stop-off at Urquhart Castle, or the one-hour 'basic'. We opted for the two-hour version. I was astonished to find that the boat was packed. It was low season but there were maybe a hundred people paying twelve pounds a pop, and there were five trips a day. Somebody was making a hell of a lot of money out of something that probably didn't exist. I looked around us – there were Italians, Russians, Indians, Cockneys, Martians . . .
The cruise kicked off with some God-awful Scottish 'folk' music.
I presumed that this had to be the 'magic' of Bruce Macgregor, as his CD was for sale to anybody suffering from a spot of tone-deafitis. When Bruce had finished, on came a rather unexcited recorded commentary. I wanted to hear all about Nessie and where various sighting had taken place. This, after all, was the only reason people were here. Nobody in Japan woke up in the morning and planned their dream trip to Loch Lomond . . . Well, maybe they did, because they like their whiskey, but you get my point. Everyone round the world has heard of the Loch Ness Monster and this is why they were here.
While the actor hired to read the script told us about what birds we could see on the loch, I drifted off and looked at the boat's depth finder. The water was currently 727 feet deep: that's seriously deep. Parts of Loch Ness are deeper than the North Sea. I drifted back into consciousness in time to hear the actor get scientific and start telling us about the 'Great Glen', the geological fault that tore right across Scotland. Loch Ness, like the Okanagan, was a glacial trough. Then Bruce Macgregor came back on and several people on board looked close to suicide as they realized that they would be stuck on this hell boat for some time longer.
We went up top on to the open roof, where we listened to the actor, clearly struggling with the dullness of the text, tell us that badgers could be found in the surroundings. Parker and I looked around for lifeboats. Just as it couldn't get any worse, it did. Rod Stewart's 'Sailing' came on.
The only real mention of the monster that had brought us to this loch in the first place was in a series of 'monster toys' available to purchase on board. These monsters, however, were known as 'Jessy' not 'Nessie'. I presumed that someone had bought the copyright to Nessie, but Jessy was not the best alternative. Essentially everybody on board was there looking for a 'Big Jessy'.
As we approached the castle the boat went past it before swinging round to starboard to dock. I looked out of the window and spotted the wake: a curious corkscrew-type affair that looked just like the 'humps' in the Peter MacNab photograph. Curiously it was in almost exactly the same position as it had been in the photo. The rest of the loch was flat calm and our wake did make a very effective set of humps. There was no boat in the Peter MacNab photo but maybe it had disappeared behind the castle, as ours had done? Like almost everything else in the world of monsters . . . I hadn't the foggiest.
We docked at Urquhart Castle and the foreign hordes, who'd never managed its conquest when it was a 'working' castle, offloaded and swamped the place.
I spoke to a man who looked like he was a monster-hunter. I judged this by the fact that he was alone, clearly felt personal hygiene was for scientists and was wearing a T-shirt proclaiming that 'Nessie Exists!'
He told me that that there was an enormous underwater cave under the rock shelf supporting the castle. This, he told me, was where the creatures lived and how they avoided the extensive sonar scans that had raked the loch. Ogopogo was also supposed to live in a cave, under Rattlesnake Island. I told my new friend about this and he whipped out a notebook and started asking me a series of questions. He was writing down my answers in a fairly unintelligible scrawl, his tongue hanging out of his mouth in concentration. He told me that he had several photographs of Nessie that 'nobody has seen'. I asked him why he hadn't shown them to anybody. He told me that he wasn't ready: he was getting all his facts and putting them together into a damning exposé that would blow the story wide open.
'Did you know that, after the first photograph was taken in 1933, the police were ordered by the government to make sure that they stopped anyone attacking the creature?' He looked me straight in the eyes. I admitted that I had not been aware of this fact.
The man smiled. 'I've got lots of facts, I have.'
I asked him if I could have his email to ask him some further questions. He stopped smiling and looked petrified.
'Email? You don't want to use that . . . Ever.' He looked around the boat suspiciously.
'Why ever not?' I asked.
'They track everything I do. They came to my house once and broke in. I fought a man off with a stick. They were from the government.' He was whispering now, clearly concerned that the small Japanese man sitting next to him was one of MI5's top agents. I thanked him for the chat and backed away to the mental safety of my family and we spent the rest of the cruise looking for badgers.
In the car I looked down my list of Nessie contacts. There was one for a guy called Miko who was supposedly the head of the 'Nessie Fan Club'. Not expecting too much from this one, I rang the number and spoke to him. He was very chipper and suggested that we should meet up in the car park of Drumnadrochit.
My family had tired of monster-hunting and wanted to go to the cinema. I dropped them off in Inverness and then drove back to the loch for my meeting with Miko.
I sat in the car park waiting for my contact like some curious Cold War contact. A couple of locals wandered past and stared at me hard, trying to work out if I was a German dogger. Finally a blue car with a Finnish flag on the back pulled in and circled the car park before parking up alongside me. I lowered my window, as did the driver of the blue car. There were two people inside, a man and an elderly woman. The elderly woman looked at me and then asked me if I was Dom. She sounded like Arnold Schwarzenegger's grandmother.
'You var Dum?'
I nodded to indicate that I was indeed 'Dum'.
'Vollow uss please.' The window closed and the blue car slid out of the car park and turned right. I followed and we left the village and then turned right again on to a tiny track that wound its way up the hill that rose above Urquhart Castle. After five minutes or so we came to a stop beside a little white cottage with a fabulous view over the castle and the loch. Miko jumped out of the car and introduced himself. He was Finnish and Arnie's granny was, in fact, his mother. They were both extreme Nessie enthusiasts and happy to talk about anything. A woman called Terry, who was originally from South Carolina, owned the cottage that we were outside. Miko used her place as a location for a couple of webcams that some company in America had provided him with.
The idea was that anyone in the world could control the cameras and scan the loch at any time, creating a non-stop, international team of monster watchers. Unfortunately, despite the webcams having been blessed by a pair of white witches, it seemed that not everybody was as diligent as they should be. Terry had two rather fine-looking sheep that grazed in the field in front of her cottage. These sheep had become something of an online hit with hundreds of people going on the website to watch them do very little. Some people emailed to ask questions about the sheep, while one couple had come all the way over from Germany to meet them.
Miko told me that there was intense rivalry between the different groups of monster-hunters around the loch. Adrian Shriner, the mad professor I'd met earlier, was considered to be the don of the Loch Ness mafia. Shriner's first exhibition was originally called the 'Official Exhibition' and then became 'Loch Ness 2000' before becoming the current 'Loch Ness Experience'. Meanwhile a guy called Donald Skinner opened up another attraction just next door, called the 'Original Monster Exhibition'. This was now called 'Nessieland' after legal threats had been issued for 'passing himself off'.
A monster-hunter who lived on the other side of the Loch had rowed over and daubed 'Shriner is a madman' in orange paint on Urquhart Castle. There had also been incidents of padlocks being put on people's gates and boats being burnt.
'It was a bit of a closed shop when I first got here,' said Miko, looking down dolefully towards Drumnadrochit. He'd actually worked at Shriner's place for a while before breaking off to go solo. I asked him whether they'd ever got any famous Loch Ness enthusiasts while he worked at the exhibition.
'We had Kylie Minogue . . . And the President of Botswana.' Sadly these visits weren't at the same time and history did not record Kylie's views on the monster story.
I drove away from the cottage, having made loose plans to organize some sort of boat hunt during the following couple of days. There was talk of sonar equipment and it all sounded rather exciting. Sadly, when I returned to the hotel it was to find my whole family in full mutiny. They had had their fill of Inverness and wanted to head back south and go home. This put a serious dent in the time I'd wanted to spend around the loch. It was a valuable lesson. Monster-hunting is a lonely business and not the sort of thing you do with a young family. I told them that I would think about their request while I had a relaxing soak in the hot tub.
I had it to myself for about five minutes and was just starting to enjoy it when a young guy and his very fat girlfriend got in. They must have been about seventeen and they totally ignored me; they had eyes only for each other. They sat close together with expressionless faces and it was only after two minutes or so that I noticed that she appeared to be giving him a hand job under the bubbles. That was it. Enough was enough. Nessie would have to wait for another day. It was time to hang up my monster-hunting boots for a while.
#### Epilogue
The one thing everyone asks me on hearing about my monster-hunting activities is: 'So, do they exist?' The answer, I'm afraid, is still a rather disappointing 'I don't know'. I suppose that this is better than definite proof that they don't?
I think that there is definitely something of quite some size in Lake Okanagan. I know this because I think I saw it with my own eyes. Whether it is Ogopogo or some sort of sturgeon, I know not. The fact that the sightings there go back so far and have been recorded by so many people makes me feel that there must be something 'unknown' in the murky depths. My only hope is that someone will get a photo or some footage while using a decent camera and not suffering from Parkinson's disease. Just thinking about the amount of shaky footage I have viewed online in the last year makes me rather sick.
The Hibagon was always going to be the most 'dodgy' of my hunts and I rather think that this was probably some escaped monkey who happened to be seen at a time coinciding with sightings of creatures like the Yeti and Bigfoot around the world. For some reason 1965-1975 seems to have been an extraordinarily productive time for monster sightings. Maybe it was fashionable? Maybe people had Super 8 cameras for the first time? Who knows? But I am not that convinced about the Hibagon. I was so thrilled, however, to be able to visit Hiroshima and Nagasaki. Before I went to Japan, whenever I thought about these places I pictured scenes of massive destruction and horror. Now I think of them as bustling cities full of life and friendliness, a true testament to the human spirit of survival.
The Congo was the most difficult place that I have ever travelled in, and there is certainly no question that if the Mokèlé-mbèmbé exists then it has chosen one of the most remote and unvisited areas of the world to do it in. The real problem in Africa is to distinguish between reality and mysticism. Tribes talk about things in the spiritual world in exactly the same way as they would something real and earthbound. I think it is likely that something unknown to science exists in these vast wetlands but think it highly unlikely that it is a dinosaur. It seems to me that the creature might be more like a manatee, a sort of hybrid hippo/rhino thing. Who knows what lies in Lake Tele and the surrounding wetlands? Certainly not me after my disastrous trip there. But I hope that someone else makes the trip and can tell me more about it. One piece of advice: don't make a schedule; just go with the flow . . .
I also believe that there is some sort of creature roaming the thick woods of Pacific Coast America. There have been sightings from way back, but what really convinced me was the fact that so many of the people who have seen things are not keen to talk to anyone about them. I'm still not sure about the Patterson/Gimlin film. If it's authentic, then it's the most convincing and astounding piece of footage in cryptozoology There's something fishy about it, however. Before my trip I hadn't realized that Patterson was an avid Bigfoot hunter. I just thought he was someone who'd stumbled on the creature and happened to have a camera. It just seems too lucky. On the other hand, the location was very remote and you wouldn't have needed to go that far to fake some footage. Sometimes I look at the film and spot a zip down the back. Other times I'm convinced. You'll have to decide for yourselves. I still think about those footprints Richard and I saw, though . . . What made them?
I think the Yeti is possibly the most believable of all the 'monsters' I looked into. The type of people who have reported sightings of footprints or the actual creature tend to be credible, serious types – adventurers and climbers who have no real gain in allowing people to think that they are crazy. I met so many Sherpas who all had the same belief in this creature and expressed zero surprise in regularly hearing the cries or having yaks killed by one. The mundane way in which Mingmar's brother showed me the photos he'd recently taken of footprints was compelling in itself. There's also the sheer inaccessibility of parts of the Himalayas, which makes it very possible, in my view, that something unknown lives there. The skull at the monastery in Khumjung is a puzzle. I have no way of ascertaining whether or not it's real but it certainly looked the part and I like to think that I got within a pane of glass from touching a Yeti.
And, finally, the Loch Ness Monster. Sadly, I don't believe that there is anything in there. I think the loch is too small and has been host to too much proper investigation for something to have remained unrevealed were it there. I'm not just saying this so that I don't have to go back there, promise.
It's been fun being a monster-hunter. I adore travelling anyway but to travel with a sense of purpose, however spurious, is so much more exhilarating. I've met some wonderful people and been to some amazing places. I've also met some real creeps and been to a couple of places I won't ever return to.
Whatever, life is short and the world is wide. Just get out there and go have your own adventures. Me, I fancy a bit of a lie down by a pool somewhere hot with a good book. I wonder what Borneo is like at this time of year . . .
|
{
"redpajama_set_name": "RedPajamaBook"
}
| 1,446
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\section{Introduction}
This paper explores the range and valence of real Smirnov functions. Smirnov functions, a well studied class of functions \cite{Duren}, are analytic functions on the open unit disk $\mathbb{D}$ which can be written as the quotient of two bounded analytic functions where the denominator is an outer function. Real Smirnov functions, studied in \cite{GMR, MR2021044, Helson, Helson2, MR1889082}, are those Smirnov functions which have real boundary values almost everywhere. In a nutshell, we will characterize all possible ranges of such functions and all possible finite valences on their range. The two main theorems of this paper (terminology, motivation, and plenty of examples to be reviewed below) are the following:
\begin{Theorem*}
If $\varphi$ is a non-constant real Smirnov function, then $\varphi(\mathbb{D})$ is either
$$\varphi(\mathbb{D}) = \mathbb{C}_{+} \setminus F \; \; \mbox{or} \; \; \varphi(\mathbb{D}) = \mathbb{C}_{-} \setminus G \; \; \mbox{or} \; \; \varphi(\mathbb{D}) = \mathbb{C} \setminus (F \cup G \cup E),$$
where $E \subsetneq \mathbb{R}$ and closed, $F \subseteq \mathbb{C}_{+}$ is relatively closed and has
capacity zero, and $G \subseteq \mathbb{C}_{-}$ is relatively closed and has capacity zero.
Moreover, given any closed $E \subsetneq \mathbb{R}$, any relatively closed $F \subseteq \mathbb{C}_{+}$ of capacity zero, and any relatively closed
$G \subseteq \mathbb{C}_{-}$ of capacity zero, there are real Smirnov functions
with ranges $\mathbb{C}_{+} \setminus F$, $\mathbb{C}_{-} \setminus G$,
and $\mathbb{C} \setminus (E \cup F \cup G)$.
\end{Theorem*}
\begin{Theorem*}
The valence of every real Smirnov function with finite valence is given
by the valence of a plane valence tree, and any valence arising from a
plane valence tree is the valence of a real Smirnov function.
\end{Theorem*}
The inspiration for this paper, and what informs our results, comes from the study of unbounded Toeplitz operators on the Hardy space $H^2$ of the open unit disk $\mathbb{D}$ (see \cite{MR3004956, MR2418122} and below). Here, for a general analytic function $\varphi$ on $\mathbb{D}$, one can define the Toeplitz operator
$$T_{\varphi}: \mathcal{D}(T_{\varphi}) \to H^2, \quad T_{\varphi} f = \varphi f,$$
where $\mathcal{D}(T_{\varphi})$, the domain of $T_{\varphi}$, is defined by
$$\mathcal{D}(T_{\varphi}) = \{f \in H^2: \varphi f \in H^2\}.$$ Sarason \cite{MR2418122} showed that $\mathcal{D}(T_{\varphi}) \not = \{0\}$ if and only if
\begin{equation}\label{99sdhsfh}
\varphi = \frac{b}{a},
\end{equation} where $b$ and $a$ are bounded analytic functions on $\mathbb{D}$ and $a$ has no zeros. Such $\varphi$ comprise the well-known {\em Nevanlinna class} $N$ \cite{Duren}. It can also be arranged so that
$$a(0) > 0 \; \; \mbox{and} \; \; |a|^2 + |b|^2 = 1$$ almost everywhere on the unit circle $\mathbb{T}$. With these normalizing conditions, this representation is unique. In the same paper, Sarason also showed that $\mathcal{D}(T_{\varphi})$ is dense in $H^2$ if and only $a$ in \eqref{99sdhsfh} is an outer function. These $\varphi$ comprise the {\em Smirnov class} $N^{+}$. Observe from \cite[Ch.~2]{Duren} that
\begin{equation}\label{bcvcdf}
\bigcup_{p > 0} H^p \subseteq N^{+},
\end{equation}
where $H^p$, the {\em Hardy classes}, are the analytic functions $f$ on $\mathbb{D}$ for which the $p$-integral means
\begin{equation}\label{Mp}
M_{p}(r, f) := \left(\int_{0}^{2 \pi} |f(r e^{i t})|^p \frac{d t}{2 \pi}\right)^{1/p}
\end{equation} are uniformly bounded for $r \in [0, 1)$.
Classical theorems of Fatou and Riesz \cite[Ch.1, 2]{Duren} say that for each $\varphi \in N^{+}$ the radial limit
\begin{equation}\label{radiallimit}
\varphi(e^{i t}) := \lim_{r \to 1^{-}} \varphi(r e^{i t})
\end{equation}
exists (and is non-zero) for almost every $t \in [0, 2 \pi]$.
We say $\varphi \in N^{+}$ belongs to the {\em real Smirnov class} $N^{+}_{\mathbb{R}}$ if
$$\varphi(e^{i t}) \in \mathbb{R}$$
for almost every $t$ (see some examples below).
These real Smirnov functions have been studied in \cite{GMR, MR2021044, Helson, Helson2, MR1889082} and a full characterization of them was given by Helson \cite{Helson, Helson2} as
\begin{equation}\label{sdhfjsd;gfgee2}
\varphi \in N^{+}_{\mathbb{R}} \iff \varphi = i \frac{u + v}{u - v},
\end{equation}
where $u$ and $v$ are inner functions and $u - v$ is an outer function.
When $\varphi \in N^{+}_{\mathbb{R}}$ and
$$\langle f, g \rangle = \int_{0}^{2 \pi} f(e^{i t}) \overline{g(e^{i t})} \frac{d t}{2 \pi}$$
denotes the usual inner product
on $H^2$ (considered in the usual way, via radial limit functions, as a closed subspace of $L^2$), one can use
the fact that $\varphi(e^{i t})$ is real for almost every $t$ to see that $$\langle T_{\varphi} f, g\rangle = \langle f, T_{\varphi} g\rangle, \quad f, g \in \mathcal{D}(T_{\varphi}).$$ In other words, $T_{\varphi}$ is a densely defined symmetric operator on $H^2$.
Standard results from the theory of unbounded symmetric operators \cite[Vol. II, Ch.~VII]{A-G} show that when $\varphi \in N^{+}_{\mathbb{R}}$ and $\lambda \not \in \mathbb{R}$, the densely defined operator $T_{\varphi} - \lambda I$ has closed range and the deficiency numbers
\begin{equation}\label{668wyeuhfjw}
d_{\varphi}(\lambda) := \operatorname{dim}(\operatorname{Rng}(T_{\varphi} - \lambda I))^{\perp},
\end{equation}
where $\operatorname{Rng}$ denotes the range of an operator,
are constant on each of the half planes
$$\mathbb{C}_{+} = \{z: \Im z > 0\}, \; \; \mathbb{C}_{-} = \{z: \Im z < 0\}.$$ Moreover, given a pair $(m, n)$, where $m, n \in \mathbb{N}_{0} \cup \{\infty\}$, there is a $\varphi \in N^{+}_{\mathbb{R}}$ with
$$(d_{\varphi}(i), d_{\varphi}(-i)) = (m, n).$$ It is also the case that both deficiency numbers are finite if and only if $\varphi \in N^{+}_{\mathbb{R}}$ is a rational function.
To get to our discussion of the range of $\varphi$, the focus of this paper, we unpack this a bit further as was done in \cite{Helson}. Observe that $\operatorname{Rng}(T_{\varphi} - \lambda I)$ is not only a closed subspace of $H^2$ but it is also invariant under the shift operator $Sf = z f$ on $H^2$ and thus, by Beurling's theorem \cite[Ch.~7]{Duren},
$$\operatorname{Rng}(T_{\varphi} - \lambda I) = \Theta H^2$$
for some inner function $\Theta$.
Let $\Theta_{\lambda}$ denote the inner factor of $\varphi - \lambda$.
All functions in
$\Theta H^2$ have $\Theta_{\lambda}$ as a divisor.
Moreover, since we can write $\varphi - \lambda = b/a$ where $b$ and $a$ are in
$H^\infty$ and $a$ is outer, it follows that the inner factor of
$(\varphi - \lambda)$ is the inner factor of $b$, and thus the
inner factor of $(\varphi - \lambda)a$ is precisely $\Theta_{\lambda}$.
Thus $\Theta = \Theta_\lambda$.
This means that
$$(\operatorname{Rng}(T_{\varphi} - \lambda I))^{\perp} = (\Theta_{\lambda} H^2)^{\perp}$$ which is a model space \cite{MR1761913, MR0270196, MR3526203, MR1895624}, a typical invariant subspace for the backward shift operator $S^{*}$. Moreover, the model space $(\Theta_{\lambda} H^2)^{\perp}$ has finite dimension $n$ if and only if $\Theta_{\lambda}$ is a finite Blaschke product of degree $n$. Since
$$\ker(T^{*}_{\varphi} - \overline{\lambda} I) = (\operatorname{Rng}(T_{\varphi} - \lambda I))^{\perp} = (\Theta_{\lambda} H^2)^{\perp},$$
and for each $w \in \mathbb{D}$,
$$T^{*}_{\varphi} k_{w} = \overline{\varphi(w)} k_{w}, \quad k_{w}(z) = \frac{1}{1 - \overline{w} z},$$
we see that
$$\bigvee \{k_{w}: \varphi(w) = \lambda\} \subseteq \ker(T^{*}_{\varphi} - \overline{\lambda} I) = (\Theta_{\lambda} H^2)^{\perp}.$$
In the above, $\bigvee$ denotes the closed linear span.
Hence, using the fact that $\Theta_{\lambda}$ is the inner factor for $\varphi - \lambda$, we see that for $\lambda \in \mathbb{C} \setminus \mathbb{R}$,
$$\lambda \in \varphi(\mathbb{D}) \iff \Theta_{\lambda}(w) = 0 \; \; \mbox{for some $w \in \mathbb{D}$}.$$
Furthermore, the valence
\begin{equation}\label{bbbbbbx}
v_{\varphi}(\lambda) := \operatorname{card}\{w \in \mathbb{D}: \varphi(w) = \lambda\}
\end{equation} will be the degree of the Blaschke factor of $\Theta_{\lambda}$.
For example, if the inner factor of $\Theta_{\lambda}$ is either a unimodular constant or a singular inner function (which will have no zeros in $\mathbb{D}$), then $\lambda \not \in \varphi(\mathbb{D})$, i.e., $v_{\varphi}(\lambda) = 0$. On the other hand, if $\Theta_{\lambda}$ has is an infinite Blaschke factor, then $v_{\varphi}(\lambda) = \infty$. Note that
$$v_{\varphi}(\lambda) \leqslant d_{\varphi}(\lambda), \quad \lambda \not \in \mathbb{R},$$ but equality does not always hold. For example, $\Theta_{\lambda}$ might be the product of a finite Blaschke product of degree $n$ and a singular inner function. In this case $v_{\varphi}(\lambda) = n$ while $d_{\varphi}(\lambda) = \infty$.
Thus, characterizing
the range of $\varphi$ will involve a discussion of the $\lambda \in \mathbb{C} \setminus \mathbb{R}$ such that $\varphi - \lambda$ has a non-trivial Blaschke factor.
Rudin \cite{MR0235151}, generalizing a classical theorem of Frostman \cite[p.~37]{C-L}, showed that for nearly all $\lambda \in \mathbb{C} \setminus \mathbb{R}$, the inner factor of $\varphi - \lambda$ is a Blaschke product. Here ``nearly all'' means that this property holds with the possible exceptional set of logarithmic capacity (capacity for short) zero. See \cite{Fisher, MR1334766} for basic facts about logarithmic capacity and see \cite{MR779463, GRM, MR2986324} for more on Blaschke products.
In the above, we are allowing a unimodular constant to count as a Blaschke factor (of order zero). In this degenerate case we see that $\Theta_{\lambda} \equiv \xi$ for some $\xi \in \mathbb{T}$ and so
$$(\operatorname{Rng}(T_{\varphi} - \lambda I))^{\perp} = (\xi H^2)^{\perp} = \{0\}.$$ Thus if $\Theta_{\lambda}$ is a unimodular constant function for one $\lambda \in \mathbb{C}_{+}$ (or one $\lambda \in \mathbb{C}_{-})$ then, since $d_{\varphi}$ is constant on each of $\mathbb{C}_{+}$ or $\mathbb{C}_{-}$, it follows that $\Theta_{\lambda}$ is a constant unimodular function for all $\lambda \in \mathbb{C}_{+}$ (or all $\lambda \in \mathbb{C}_{-}$). If $\Theta_{\lambda}$ is a constant unimodular function for one $\lambda \in \mathbb{C}_{+}$, then $\varphi(\mathbb{D}) \cap \mathbb{C}_{+} = \varnothing$ (similarly for some $\lambda \in \mathbb{C}_{-}$ and hence $\varphi(\mathbb{D}) \cap \mathbb{C}_{-} = \varnothing$). Hence, for example, if $\varphi(\mathbb{D}) \cap \mathbb{C}_{+}$ omits an open disk about $\lambda \in \mathbb{C}_{+}$, then $\varphi - \lambda$ is an outer function, i.e., $\Theta_{\lambda}$ is a constant unimodular function. In this case the above discussion implies that $\varphi(\mathbb{D}) \cap \mathbb{C}_{+} = \varnothing$.
Thus, using the above analysis, along with the fact that $\varphi(\mathbb{D})$ is an open connected subset of $\mathbb{C}$ (open mapping theorem), we have the following possibilities for the range of $\varphi \in N^{+}_{\mathbb{R}}$:
\begin{equation}\label{789ruwoiepdws}
\varphi(\mathbb{D}) = \mathbb{C}_{+} \setminus F \; \; \mbox{or} \; \; \varphi(\mathbb{D}) = \mathbb{C}_{-} \setminus G \; \; \mbox{or} \; \; \varphi(\mathbb{D}) = \mathbb{C} \setminus (F \cup G \cup E),
\end{equation}
where $F \subseteq \mathbb{C}_{+}$ and $G \subseteq \mathbb{C}_{-}$ are relatively closed subsets of capacity zero and $E \subsetneq \mathbb{R}$ is closed. The question we ask and answer in this paper is whether or not we can actually obtain all of these possibilities.
In addition, we also discuss the valance on these ranges.
To give the reader a feel for where we are heading, let us consider a few simple examples of ranges of $\varphi \in N_{\mathbb{R}}^{+}$.
\begin{Example}
If
$$\varphi_1(z) = i \frac{1 + z}{1 - z},$$
then $\varphi_1 \in N^{+}$ (since it is the
quotient of two bounded analytic functions
and the denominator $1 - z$ is outer) and
$$\varphi_1(e^{i t}) = - \cot(t/2) \in \mathbb{R}, $$
which says that $\varphi_1 \in N_{\mathbb{R}}^{+}$.
Furthermore, $\varphi_1(\mathbb{D}) = \mathbb{C}_{+}$.
In a similar way, we see that if
$$\varphi_2(z) = -i \frac{1 + z}{1 - z},$$ then $\varphi_2 \in N_{\mathbb{R}}^{+}$ and $\varphi_2(\mathbb{D}) = \mathbb{C}_{-}$.
If $\theta$ is the singular inner function
$$\theta(z) = \exp\left(\frac{1 + z}{1 - z}\right),$$
then $\theta(\mathbb{D}) = \mathbb{D} \setminus \{0\}$ and thus if $\psi_{1} := \varphi_1 \circ \theta$, then $\varphi_1 \in N^{+}_{\mathbb{R}}$ and
$\psi_{1}(\mathbb{D}) = \mathbb{C}_{+} \setminus \{i\}$. Observe that the singleton $\{i\}$ has capacity zero \cite[p.~140]{MR1334766}.
Given a relatively closed subset $W \subseteq \mathbb{D}$ of capacity zero, there is an inner function $\sigma$ such that $\sigma(\mathbb{D}) = \mathbb{D} \setminus W$ \cite{C-L}. Then $\psi_2 = \varphi_1 \circ \sigma \in N^{+}_{\mathbb{R}}$ and $\psi_{2}(\mathbb{D}) = \mathbb{C}_{+} \setminus F$, where $F = \psi_1(W)$ has capacity zero.
\end{Example}
\begin{Example}
If
$$\varphi_3(z) = \left(\frac{1 + z}{1 - z}\right)^4$$ then
$$\varphi_{3}(e^{i t}) = \cot^{4}(t/2) \in \mathbb{R}$$ and,
since
$$z \mapsto \frac{1 + z}{1 - z}$$
maps $\mathbb{D}$ onto the right-half plane $\{z: \Re z > 0\}$, then $\varphi_{3}(\mathbb{D}) = \mathbb{C} \setminus \{0\}$.
\end{Example}
\begin{Example}
If
\begin{equation*}%
\varphi_4(z) = \frac{z}{(1 - z)^2},
\end{equation*}
the well-known Koebe function, then
$$\varphi_4(e^{i t}) = -\frac{1}{2} \frac{1}{1 - \cos t} \in \mathbb{R}$$ and
$\varphi_4(\mathbb{D})$ is the single slit domain $\mathbb{C} \setminus (-\infty, -\tfrac{1}{4}]$.
\end{Example}
\begin{Example}
If
$$\varphi_5(z) = \frac{i z}{1 - z^2},$$
then
$$\varphi_5(e^{i t}) = - \frac{1}{2} \csc t \in \mathbb{R}$$ and $\varphi_5(\mathbb{D})$ turns out to be the double slit domain
$$\mathbb{C} \setminus ((-\infty, -\tfrac{1}{2}] \cup [\tfrac{1}{2}, \infty)).$$
\end{Example}
Of course one can compose any of the functions $\varphi_j$ from these examples with interesting inner functions, like was done in the first example, to obtain ranges taking the form $\mathbb{C}_{+} \setminus F$, $\mathbb{C}_{-} \setminus G$, and $\mathbb{C} \setminus (F \cup G \cup E)$ for relatively closed sets $F$ and $G$ of capacity zero and a closed set $E \subsetneq \mathbb{R}$. Can we obtain all of these possibilities as ranges for given $E, F, G$?
\section{The main range result}
Our main result about the range of $\varphi \in N_{\mathbb{R}}^{+}$ is the following:
\begin{Theorem}\label{MT}
If $\varphi \in N^{+}_{\mathbb{R}}$ and non-constant, then $\varphi(\mathbb{D})$ is either
$$\varphi(\mathbb{D}) = \mathbb{C}_{+} \setminus F \; \; \mbox{or} \; \; \varphi(\mathbb{D}) = \mathbb{C}_{-} \setminus G \; \; \mbox{or} \; \; \varphi(\mathbb{D}) = \mathbb{C} \setminus (F \cup G \cup E),$$
where $E \subsetneq \mathbb{R}$ and closed, $F \subseteq \mathbb{C}_{+}$ is relatively closed and has
capacity zero, and $G \subseteq \mathbb{C}_{-}$ is relatively closed and has capacity zero.
Moreover, given any closed $E \subsetneq \mathbb{R}$, any relatively closed $F \subseteq \mathbb{C}_{+}$ of capacity zero, and any relatively closed
$G \subseteq \mathbb{C}_{-}$ of capacity zero, there are functions in $ N^{+}_{\mathbb{R}}$
with ranges $\mathbb{C}_{+} \setminus F$, $\mathbb{C}_{-} \setminus G$,
and $\mathbb{C} \setminus (E \cup F \cup G)$.
\end{Theorem}
Our proof needs a variation of a result from \cite[p.~119]{Fisher}.
\begin{Lemma}\label{99w8w8w8w8w+}
Suppose $f$ is a non-constant function belonging to $H^p$ for some $p \in (0, \infty)$ and $\mathcal{E} \subseteq \mathbb{T}$ of positive Lebesgue measure for which
$$\lim_{r \to 1^{-}} f(r \xi) =: f(\xi)$$ exists for
each $\xi \in \mathcal{E}$ . If
$E = \{f(\xi): \xi \in \mathcal{E}\}$, then $E$ has positive capacity.
\end{Lemma}
\begin{proof}
We follow the original proof from \cite[p.~119]{Fisher} with a slight variation.
First, note that since the almost everywhere defined boundary $\xi \mapsto f(\xi)$ function on $\mathbb{T}$ is a limit of
continuous functions (the dilations $f_r(\xi) := f(r \xi)$) on $\mathbb{T}$,
by Egorov's theorem there is a set of positive measure
that is a subset of $\mathcal{E}$ on which the boundary function
$f$ is continuous. By the
inner regularity of Lebesgue measure, this set has a compact
subset of positive measure. Without loss of generality we may
take $\mathcal{E}$ to be this set. Then $f(\mathcal{E})$ is compact,
and
$|f| \leqslant K$ on $\mathcal{E}$ for some $K > 0$.
Suppose towards a contradiction that $E$ has zero logarithmic capacity. Then
by Evan's theorem
\cite[p.~33]{Fisher}, for some probability measure $\sigma$ on $\mathbb{T}$, the logarithmic potential
$$p(z) = - \int_{E} \log |z - \zeta| d\sigma(\zeta)$$
satisfies
$$\lim_{z \to \xi} p(z) = +\infty, \quad \xi \in E.$$
The function $u = p(f)$ is harmonic on $\mathbb{D}$ and satisfies the condition
$$\lim_{r \to 1^{-}} u(r \xi) = + \infty, \quad \xi \in \mathcal{E}.$$
Let $v$ be the harmonic conjugate of $u$ on $\mathbb{D}$ and define
$F = e^{-u - i v}$.
Assuming that $F \in H^p$ (a fact we will prove momentarily), we see that
$$\lim_{r \to 1^{-}} |F(r \xi)| = 0, \quad \xi \in \mathcal{E}.$$
But since $\mathcal{E}$ has positive Lebesgue measure, this would mean that $f \equiv 0$ \cite[p.~17]{Duren}, a contradiction.
To finish, we now show that $F \in H^p$. Observe that a use of Jensen's inequality \cite[p.~43]{MR1334766} and the integral means definition of $H^p$ from \eqref{Mp}
shows that
\begin{align*}
\int_{0}^{2 \pi} |F(r e^{i t})|^p \frac{d t}{2 \pi} & = \int_{0}^{2 \pi} \exp\left(\int_{E} \log|f(r e^{i t}) - \zeta|^p d \sigma(\zeta)\right)\frac{d t}{2 \pi}\\
& \leqslant \int_{0}^{2 \pi} \left(\int_{E} |f(r e^{i t}) - \zeta|^p d \sigma(\zeta)\right) \frac{d t}{2 \pi}\\
& = \int_{E}\left( \int_{0}^{2 \pi} |f(r e^{i t}) - \zeta|^p \frac{dt}{2 \pi}\right) d \sigma(\zeta)\\
& \leqslant 2^p (\sup_{0 < r < 1} M(r, f)^{p} + K^p).
\end{align*}
Since the last quantity above is independent of $r \in (0, 1)$, this shows that $F \in H^p$ and thus completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem \ref{MT}]
The first part of the theorem follows from the discussion preceding \eqref{789ruwoiepdws}. For the second part (obtaining all possible types of ranges), we proceed as follows.
First consider the case where the range contains points in both the
upper and lower half planes.
Since $E$ is a proper closed subset of $\mathbb{R}$, it follows that $\mathbb{R} \setminus E$ has at least one component. If it has {\em exactly one} component, then $E$ must be one of the following four options:
\begin{equation}\label{s8dufds}
\varnothing, \quad (-\infty, c], \quad [c, \infty), \quad (-\infty, a] \cup [b, \infty) , \quad (a < b).
\end{equation}
We will first deal with the case where $\mathbb{R} \setminus E$
has {\em at least two} components.
By means of a real translation of our function at the end, we can assume $0 \in E$ and, for some $a <0$ and $0 < b < c$, that
\begin{equation}\label{3094ouitrgjlfksd}
E \subseteq (- \infty, a] \cup [0, b] \cup [c, \infty).
\end{equation}
Define
$$E_{+} = E \cap [0, \infty), \; \; E_{-} = E \cap (-\infty, 0)$$ and the open set
\begin{equation}\label{Omega}
\Omega = \{\Re z > 0\} \setminus (\wt{E}_1 \cup \wt{E}_2 \cup \wt{E}_3 \cup \wt{F} \cup \wt{G}),
\end{equation}
where
$$\wt{E}_1 = \{x^{\frac{1}{4}}: x \in E_{+}\},$$
$$\wt{E}_2 = e^{i \frac{\pi}{4}} \{(-x)^{\frac{1}{4}}: x \in E_{-}\},$$
$$\wt{E}_3 = e^{-i \frac{\pi}{4}} \{(-x)^{\frac{1}{4}}: x \in E_{-}\},$$
and $\wt{F}$ and $\wt{G}$ are the intersections of the right half plane with the
images of $F$ and $G$ respectively under the multivalued map
$z \mapsto z^{1/4}$.
\begin{figure}
\begin{tikzpicture}[scale=2.0]
\draw (-2,0) node{};
\draw[thick] (0,0)--(1,0) node[anchor=west]{$b^{1/4}$};
\draw (.5,0) node[anchor=south]{$\widetilde{E}_1$};
\draw[thick,->] (2,0) node[anchor=east]{$c^{1/4}$} -- (4,0)
node[anchor=west]{$\widetilde{E}_1$};
\draw[thick,->] (1,1) node[anchor=north west]{$|a|^{1/4}e^{\pi i/4}$} -- (2,2) node[anchor = west]{$\widetilde{E}_2$};
\draw[thick,->] (1,-1) node[anchor= south west]{$|a|^{1/4}e^{-\pi i/4}$} -- (2,-2)node[anchor = west]{$\widetilde{E}_3$};
\draw[thick,dashed] (1,.25)--(1.25,.3125) node[above]{$\widetilde{F}$};
\draw[thick,dashed] (.25,-1)--(.3125,-1.25)node[right]{$\widetilde{F}$};
\draw[thick] (0,2) -- (0,-2);
\draw[dashed] (.5,-.25) --( .75,-.25)node[below]{$\widetilde{G}$};
\draw[dashed] (.25,.5) --( .25,.75)node[above]{$\widetilde{G}$};
\draw[fill=black] (1,0) circle(.05cm);
\draw[fill=black] (2,0) circle(.05cm);
\draw[fill=black] (1,1) circle(.05cm);
\draw[fill=black] (1,-1) circle(.05cm);
\draw (2.5,1.5) node{\LARGE $\Omega$};
\end{tikzpicture}
\caption{The region $\Omega$ when $E = (- \infty, a] \cup [0, b] \cup [c, \infty).$}\label{figa}
\end{figure}
See Figure \ref{figa} for an illustration of $\Omega$ when
$$E = (- \infty, a] \cup [0, b] \cup [c, \infty).$$
Since $\Omega$ is contained in
$$\{\Re z > 0\} \setminus \left([0, b^{\frac{1}{4}}] \cup [c^{\frac{1}{4}}, \infty) \cup e^{i \pi/4} [|a|^{\frac{1}{4}}, \infty) \cup e^{- i\pi/4} [|a|^{\frac{1}{4}}, \infty) \right)$$
and this last set is connected, one concludes, also using the containment in \eqref{3094ouitrgjlfksd} along with the fact that extracting a set of capacity zero does not disconnect a domain \cite[p.~68]{MR1334766}, that $\Omega$ is connected.
By \cite[p.~125]{MR1344449} there exists an analytic covering map for $\Omega$, such that $\psi(\mathbb{D}) = \Omega$. Furthermore, since $\psi(\mathbb{D})$ is contained in a half-plane, then $\psi$ belongs to $H^{p}$ for all $p \in (0, 1)$ \cite[p.~109]{Garnett}. Thus, by \eqref{bcvcdf}, $\psi \in N^{+}$.
The map $\psi$ is a covering map from $\mathbb{D}$ to $\Omega$, which
means that each point of $\Omega$ is contained in an open neighborhood $U$ such that
$\psi^{-1}(U)$ consists of disjoint open sets each of which is homeomorphic to $U$
under $\psi$.
We now claim that if the radial limit
\begin{equation}\label{oos9d9}
\lim_{r \to 1^{-}} \psi(r e^{i t})
\end{equation}
exists, which it will for almost every $t$ (see \eqref{radiallimit}), then this value must belong to $\partial \Omega$.
Indeed, if this were not the case, then, for some particular $t$, the limit would be equal to some
$w \in \Omega$. Now choose an open neighborhood $U$ of $w$ such that
$\psi^{-1}(U)$ consists of disjoint sets each of which is homeomorphic to $U$
under $\psi$. Let $W$ be an open neighborhood of $w$ contained in
$U$ and such that
$\overline{W} \cap U$ is compact.
Then
$$\psi^{-1}(W) = \bigcup_{a \in A} V_a,$$ where
$A$ is some index set and the $V_a$ are pairwise disjoint open sets that are each
homeomorphic to $W$ under $\psi$.
Thus each $V_a$ has compact closure in $\psi^{-1}(U)$.
For some $b \in [0,1)$, the curve
$$r \to \psi([r e^{i t},e^{i t})), \quad r \in [b, 1),$$
must lie entirely in $W$ since $\psi$ has radial limit of $w$
at $e^{i t}$. But since the $V_a$ are
disjoint open sets, this means that the ray
$$[r e^{i t},e^{i t}), \quad r \in [b, 1),$$
must lie entirely in one of the $V_a$. But this is impossible because
each of the $V_a$ has compact closure in $\psi^{-1}(U)$ but
$e^{i t} \not \in \psi^{-1}(U)$.
The set $\wt{F}$ is the union of two images of $F$ under maps that are
Lipschitz in any annulus centered at the origin. Since $F$ has
capacity $0$, each of the images has capacity $0$
\cite[p.~137]{MR1334766}, and their union $\wt{F}$ also has capacity
zero \cite[p.~57]{MR1334766}. The same applies to $\wt{G}$, and thus
$\wt{F} \cup \wt{G}$ has capacity zero.
By Lemma \ref{99w8w8w8w8w+}, the values in those sets cannot be
boundary values of $\psi$, except possibly on a set of measure $0$.
Setting $\varphi = \psi^{4}$ we see that $\psi \in H^{p}$
for all $p \in (0, \tfrac{1}{4})$ and thus, again from \eqref{bcvcdf}, $\psi \in N^{+}$. Moreover, since
\[
\lim_{r \to 1^{-}} \psi(r e^{i t}) \in \partial \Omega \setminus (\wt{F} \cup \wt{G})
\]
for almost every $\theta$, we see that
$$\lim_{r \to 1^{-}} \varphi(r e^{i t}) \in \mathbb{R}$$ for almost every $t$. Thus $\varphi \in N_{\mathbb{R}}^{+}$.
The construction of $\Omega$ from \eqref{Omega}, and the fact that $0 \in E$, will show that
\[\varphi(\mathbb{D}) = \{z^4: z \in \Omega\} = \mathbb{C} \setminus (E \cup F \cup G).\]
For the cases where the desired range is $\mathbb{C}_{+} \setminus F$ or $\mathbb{C}_{-} \setminus G$, and for the case where
$E = (-\infty, c]$ or $E = [c, \infty)$ or $E = (-\infty, a] \cup [b, \infty)$ and the desired range is
$\mathbb{C} \setminus (E \cup F \cup G)$, we let $\varphi$ be the covering map from $\mathbb{D}$ onto the desired range. The proof that this map has the required properties is similar to the proof of the
first case above. We use the fact that any mapping from $\mathbb{D}$ into a domain with at least one slit that
goes to $\infty$ is in $H^p$ for all $p \in (0,\tfrac{1}{2})$ \cite[p.~110]{Garnett}.
If the desired range is $\mathbb{C} \setminus (F \cup G)$, we let $\wt{F}$ and $\wt{G}$ be the
images of $F$ and $G$ respectively under the multivalued map $z \mapsto z^{1/2}$ -- which are of capacity zero \cite[p.~137]{MR1334766}. Let
$\psi$ be the covering map from $\mathbb{D}$ onto
$\mathbb{C} \setminus (\wt{F} \cup \wt{G} \cup [1,\infty))$, and let $\varphi = \psi^2$. The proof that mapping
has the required properties is similar to the proofs above.
\end{proof}
\begin{Remark}
\begin{enumerate}
\item The mapping $\varphi$ constructed above is actually outer, as long as $0$ is not in the range.
To see this, observe that, in the first part of the proof,
$\psi(\mathbb{D}) \subseteq \{\Re z > 0\}$.
Such functions are outer \cite[p.~109]{Garnett}. Since the product of outer functions is another outer function,
this means that $\varphi = \psi^4$ is outer.
In the other cases in which $0$ is not in the range, we have that $\varphi$ is a map onto a domain with a linear slit
containing $0$ and going to $\infty$.
But any map onto such a domain is outer, since we can define the square root of such a mapping,
and that mapping will be onto a half plane omitting $0$ and thus outer (and belong to $H^p$ for all $p \in (0, 1)$ \cite[p.~109]{Garnett}).
\item The proof of Theorem \ref{MT} also
shows that we can find a
$\varphi$ with the desired properties that is in $H^p$ for each $p \in (0,\tfrac{1}{4})$.
\end{enumerate}
\end{Remark}
\section{Controlling the valence}
A key step in the construction in Theorem \ref{MT} was the uniformization theorem \cite[p.~125]{MR1344449}. However, it is not clear from our construction how one can control the valence of $\varphi$.
In this regard, one can ask the following question: Suppose we are given an closed set $E \subsetneq \mathbb{R}$ and a pair $(m, n)$, $m, n \in \mathbb{N}_0 \cup \{\infty\}$. Can we find a $\varphi \in N^{+}_{\mathbb{R}}$ such that the valence of $\varphi$ is equal to $m$ on $\mathbb{C}_{+}$, $n$ on $\mathbb{C}_{-}$, and such that $\varphi(\mathbb{D}) = \mathbb{C} \setminus E$? Can we say anything about the valence of $\varphi$ on $\mathbb{R} \setminus E$?
Let us start with a few observations. Extending a standard proof of the open mapping theorem for analytic functions, one can prove the following. Recall that $v_{\varphi}$ is the valence function from \eqref{bbbbbbx} and $d_{\varphi}$ is the deficiency index from \eqref{668wyeuhfjw}.
\begin{Proposition}
For $\varphi \in N_{\mathbb{R}}^{+}$ and $N = 1, 2, \ldots$, the set
$$\{w \in \mathbb{C}: v_{\varphi}(w) \geqslant N\}$$ is an open subset of $\mathbb{C}$.
\end{Proposition}
Note that the above result does not hold when $N = \infty$.
\begin{Example}
Theorem \ref{MT} says that we can find a $\varphi \in N_{\mathbb{R}}^{+}$ such that
$$\varphi(\mathbb{D}) = \mathbb{C} \setminus [0, \infty).$$ However, the previous proposition says that we can't find a $\varphi \in N_{\mathbb{R}}^{+}$ with the same range and that also satisfies
$$v_{\varphi}|_{\mathbb{C}_{+}} = 1, \quad v_{\varphi}|_{\mathbb{C}_{-}} = 2, \quad v_{\varphi}|_{\{x < 0\}} = 3$$
since $v_{\varphi}|_{\mathbb{C}_{+}}$ and $v_{\varphi}|_{\mathbb{C}{-}}$ must be at least $3$.
\end{Example}
Next we explore when the valence is finite.
\begin{Proposition}\label{oriuioewur}
For $\varphi \in N_{\mathbb{R}}^{+}$ the following are equivalent.
\begin{enumerate}
\item[(i)] $\varphi$ is a rational function;
\item[(ii)] $v_{\varphi}|_{\mathbb{C}_{+}}$ and $v_{\varphi}|_{\mathbb{C}_{-}}$ are finite;
\item[(iii)] There are two relatively prime finite Blaschke products $B_1, B_2$ such that $B_1 - B_2$ has no zeros on $\mathbb{D}$ and such that
\begin{equation}\label{pppsd9sd97fd8}
\varphi = i \frac{B_1 + B_2}{B_1 - B_2}.
\end{equation}
\item[(iv)] $d_{\varphi}(i)$ and $d_{\varphi}(-i)$ are finite.
\end{enumerate}
Furthermore,
\begin{enumerate}
\item[(a)] if any of the above equivalent conditions hold, we have
$$v_{\varphi}|_{\mathbb{C}_{+}} = \operatorname{deg}(B_2), \quad v_{\varphi}|_{\mathbb{C}_{-}} = \operatorname{deg}(B_1).$$
\item[(b)] if any of the above conditions do not hold then either $v_{\varphi}|_{\mathbb{C}_{+}}$ or $v_{\varphi}|_{\mathbb{C}_{-}}$ is infinite nearly everywhere.
\end{enumerate}
\end{Proposition}
\begin{proof}
$(i) \iff (iv)$ is from \cite{Helson}.
$(iii) \implies (i)$: One can show directly that a $\varphi$ given by \eqref{pppsd9sd97fd8} is a rational function in $N_{\mathbb{R}}^{+}$.
$(i) \implies (iii)$: Set $g = (\varphi - i)/(\varphi + i)$ and observe that $g$ is meromorphic function on $\mathbb{D}$ that is continuous with unimodular boundary values on $\mathbb{T}$. A classical theorem of Fatou \cite{GRM} says that
$$g = \frac{B_1}{B_2},$$ where $B_1$ and $B_2$ are relatively prime Blaschke products. The result now follows with a simple computation.
$(iii) \implies (ii)$: We follow an argument from \cite{GRM}. The M\"{o}bius transformation
$$\psi(z) = \frac{z - i}{z + i}$$ is injective and maps $\mathbb{C}_{+}$ onto $\mathbb{D}$ and $\mathbb{C}_{-}$ onto $\widehat{\mathbb{C}} \setminus \mathbb{D}^{-}$. If $w \in \mathbb{C}_{+}$, the number of solutions to $\varphi(z) = w$ is the same as the number of solutions to $\psi \circ \varphi (z)= \psi(w) = \eta \in \mathbb{D}$. Writing this out, this is same as the number of solutions to
$$\eta = \frac{\varphi(z) - i}{\varphi(z) + i} = \frac{B_2(z)}{B_1(z)}.$$
To examine the number of zeros in $\mathbb{D}$ of $B_2 - \eta B_1$, observe that on $\mathbb{T}$ we have
$$|\eta B_1| = |\eta| < 1 = |B_2|$$ and so, by Rouche's Theorem, the number of zeros in $\mathbb{D}$ of $B_2$ and $B_2 - \eta B_1$ are the same. This proves that $v_{\varphi}(w) = \operatorname{deg}(B_2)$ whenever $w \in \mathbb{C}_{+}$. The corresponding valence on $\mathbb{C}_{-}$ follows in a similar way. This also verifies $(a)$.
$(ii) \implies (iv)$: Suppose $v_{\varphi}|_{\mathbb{C}_{+}}$ is finite but $d_{\varphi}(i) = \infty$. Since $d_{\varphi}|_{\mathbb{C}_{+}}$ is constant, we see that $d_{\varphi}(\lambda) = \infty$ for all $\lambda \in \mathbb{C}_{+}$. By \cite{MR0235151}, the inner factor $\Theta_{\lambda}$ of $\varphi - \lambda$ is a Blaschke product for all $\lambda \not \in \mathbb{R}$ except possibly for a set of capacity zero. As discussed in \eqref{bbbbbbx},
$$v_{\varphi}(\lambda) = \operatorname{deg}(\Theta_{\lambda}) = \infty$$ for nearly all $\lambda \in \mathbb{C}_{+}$, a contradiction. An analogous argument holds for $v_{\varphi}|_{\mathbb{C}_{-}}$. This also proves $(b)$.
\end{proof}
The techniques in the above proof also show the following.
\begin{Corollary}
If $\varphi \in N_{\mathbb{R}}^{+}$ and $v_{\varphi}|_{\mathbb{C}_{+}}$ is non-constant, then $v_{\varphi}(\lambda) = \infty$ for nearly all $\lambda \in \mathbb{C}_{+}$. An analogous result holds for $v_{\varphi}|_{\mathbb{C}_{-}}$
\end{Corollary}
\begin{Remark}\label{ooookkk}
We point out a paper \cite{MR554397} which gives some information about how one can, under mild technical conditions, define an inner function whose valence can be prescribed on various closed subsets of the $\mathbb{D}$ of zero capacity. Since real Smirnov functions take the form $i (u + v)/(u - v)$ (recall Helson's characterization from \eqref{sdhfjsd;gfgee2}), where $u$ and $v$ are inner, this creates examples of real Smirnov functions with wild valence behavior.
\end{Remark}
We now characterize the possible valences of real Smirnov functions of finite valence. As seen in Proposition \ref{oriuioewur}, these are the rational real Smirnov functions.
A {\itshape disk tree}
is a type of Riemann surface made by welding together copies
of $\mathbb{D}$ and
$$\mathbb{D}^* := \{z \in \widehat{\mathbb{C}}: |z| > 1 \text{ or } z = \infty\}.$$
To each copy of $\mathbb{D}$ and $\mathbb{D}^*$ in the disk tree is associated a positive
integer $m$ which we call the valence.
A {\itshape admissible}
arc on a disk tree is an arc on the boundary of a copy of
$\mathbb{D}$ or $\mathbb{D}^*$ of the form
$\{e^{i\theta} : a < \theta < b\}$ where $(a,b)$ contains no multiple of
$2\pi / m$. The {\itshape image arc} for a given admissible arc
is the arc
$\{e^{i\theta} : am < \theta < bm\}.$ This is the image of the admissible arc
under the function $z^m$.
An admissible arc is called a {\itshape free arc} if it is part of the
boundary of the disk tree, in other words, if it has
{\itshape not}
been welded to
another arc in the disk tree. An admissible arc is a
{\itshape welded arc} if it has been welded to another arc in the
disk tree.
We will now formally define disk trees inductively.
A copy of $\mathbb{D}$ or $\mathbb{D}^*$ with a valence is a disk tree.
Let $X$ be a disk tree. Let $Z$ be a new copy of $\mathbb{D}$ or $\mathbb{D}^*$ with valence
$m$. Let $Y$ be a copy of $\mathbb{D}$ or $\mathbb{D}^*$ in $X$ with valence $n$, where
$Y$ is a copy of $\mathbb{D}$ if $Z$ is a copy of $\mathbb{D}^*$, and vice versa.
Suppose we are given a free arc on $Y$ and a free arc on $Z$
and that both of them have the same image arc.
We may weld $Y$ and $Z$ together on their
free arcs by the map $Y \ni z \mapsto z^{n/m} \in Z$.
See \cite[II.3C]{Ahlfors-Sario} for more on welding Riemann surfaces.
We will also give an explicit example of welding in
Example \ref{ex:twodisctree}.
We say the resulting surface of $X$ welded to $Z$
is a disk tree if it still has a free arc remaining.
Any disk tree is simply connected (van Kampen's theorem \cite[Ch.~1]{MR1867354}). It is also conformally equivalent to the
disk. Indeed, by the uniformization theorem, it is equivalent to the
disk, the plane, or the Riemann sphere.
But we may weld another copy of the disk (or complement of the disk)
to any disk tree and still
obtain a simply connected Riemann surface, since any disk tree has
a free arc. Thus we can obtain the original
disk tree from the new Riemann surface by removing a set with
infinitely many points. But if we take a set with infinitely many points
away from either the Riemann sphere, the plane, or the disk, and we
are left with a simply connected set, that set must be conformally
equivalent to the disk.
For a disk tree $X$, define
$$f_X: X \mapsto \widehat{\mathbb{C}} \setminus \{1\}$$ by
$f_X(z) = z^m$ for $z$ in a copy of $\mathbb{D}$ or $\mathbb{D}^*$ with valence $m$, or for
$z$ in a welded arc that belongs to a copy of $\mathbb{D}$ with valence $m$.
The function $f_X$ is clearly meromorphic in each copy of $\mathbb{D}$ or
$\mathbb{D}^*$. By construction, it is continuous in a neighborhood of each welded
arc. To see this, suppose that some admissible arc $I$ in a copy of
$\mathbb{D}$ or $\mathbb{D}^*$ with valence $n$ (call the copy $Y$)
is welded to an admissible arc $J$ in a copy
of $\mathbb{D}$ or $\mathbb{D}^*$ of valence $m$ (call the copy $Z$). Let
$Y'$ and $Z'$ be sufficiently small neighborhoods of $I$ and $J$ in
$Y \cup I$ and $Z \cup J$, respectively.
Note that by definition of the welding the map
\[
\varphi(z) =
\begin{cases}
z^{n/m} &\text{ if $z \in Y'$}\\
z &\text{ if $z \in Z'$}\\
\end{cases}
\]
is well defined, continuous and even conformal.
But $f_X(z) = \varphi(z)^m$ in $Y' \cup Z'$.
Thus by Morera's theorem, $f_X$ is analytic in a neighborhood of each
welded arc and thus meromorphic on $X$.
When restricted to a copy of
$\mathbb{D}$ or $\mathbb{D}^*$ that has valence $m$
the function $f_X$ has valence $m$
at each point of
$\mathbb{D}$ or $\mathbb{D}^*$ respectively, and valence $0$ elsewhere.
Also, when restricted to a welded
arc, the function $f_X$ has valence $1$ on the image arc
and $0$ elsewhere.
Thus, $f_X$ has valence on $\mathbb{D}$ equal to the sum of the valences of
the copies of $\mathbb{D}$ in the disk tree, and similarly for $\mathbb{D}^*$.
Its valence on a point in $\partial \mathbb{D} \setminus \{1\}$ is
equal to the number of image arcs in which it appears,
where if two arcs are welded together in the disk tree we
count their image arc (which is the same for both the welded arcs)
as appearing only once.
Let
$\varphi : \mathbb{D} \mapsto X$ be a conformal map from $\mathbb{D}$ to $X$. Then
$f_X \circ \varphi$ is a (meromorphic)
map from $\mathbb{D}$ to $\widehat{\mathbb{C}} \setminus \{1\}$.
\begin{Example}\label{ex:twodisctree}
We give an explicit construction of a disk tree.
This disk tree will have valence $1$ on $\mathbb{D}$ and
$2$ on $\mathbb{D}^*$.
See Figure \ref{figchart} for an illustration of some aspects of this
example.
Let
$X$ consist of one copy each of $\mathbb{D}$ and $\mathbb{D}^*$, together with
the boundary of $\partial \mathbb{D} \setminus \{1\}$.
For $0 < \theta < 2 \pi$,
identify the point
$e^{i \theta}$ on $\partial \mathbb{D} \setminus \{1\}$ with
the point $e^{i \theta / 2}$ on $\partial \mathbb{D}^* \setminus \{1\}$.
We will weld along these
identified boundary points. To do this explicitly,
let \[ \overline{\mathbb{D}} \supset U_1 = \{re^{i\theta} : 1/2 < r \leqslant 1 \text{ and }
0 < \theta < 2\pi\}\]
and
\[ \overline{\mathbb{D}^*} \supset U_2 = \{re^{i\theta} : 1 \leqslant r < 2 \text{ and }
0 < \theta < \pi\}\]
and let $U = U_1 \cup U_2$.
Take coordinate charts $\varphi_1: \mathbb{D} \rightarrow \mathbb{C}$ and
$\varphi_2: \mathbb{D}^* \rightarrow \mathbb{C}$ and
$\varphi_3: U \rightarrow \mathbb{C}$, where
$\varphi_1(z) = z$ and $\varphi_2(z) = 1/z$ and
\[
\varphi_3(z) =
\begin{cases}
z^{1/2} &\text{if $z \in U_1,$}\\
z & \text{if $z \in U_2$}
\end{cases}
\]
where we take the branch of $z^{1/2}$ with $(-1)^{1/2} = i$ and branch
cut along the positive real axis.
We can take as a basis for the open sets in $X$ sets that are open in
$\mathbb{D}$ or in $\mathbb{D}^*$ or sets that are the inverse images of
open sets under $\varphi_3$. Thus $U$ is an open set.
(Note that any set that is the inverse image of an open set under
$\varphi_3$ and lies entirely in $\mathbb{D}$ is open in $\mathbb{D}$; the same may be
said for $\mathbb{D}^*$.)
Notice that $\varphi_3$ is continuous, and that
$\varphi_1(\varphi_3^{-1}(z)) = z^2$ and
$\varphi_2(\varphi_3^{-1}(z)) = z^{-1}$. Both of these maps are analytic in
their domains. Thus we have made $X$ into a Riemann surface with the
given charts.
Define \[
f_X(z) =
\begin{cases}
z &\text{ for $z \in \mathbb{D}$ or $z \in \partial \mathbb{D}$}\\
z^2 &\text{ for $z \in \mathbb{D}^*$ or $z \in \partial \mathbb{D}^*$}.
\end{cases}
\]
Then $f_X$ is analytic from $X$ into $\widehat{\mathbb{C}} \setminus \{1\}$.
To see this, note that
$$f_X(\varphi_1^{-1}(z)) = z, \quad f_X(\varphi_2^{-1}(z)) = 1/z^2, \quad f_X(\varphi_3^{-1}(z)) = z^2.$$
Also, $f$ has valence $1$ on $\mathbb{D}$, valence $2$ on $\mathbb{D}^*$, and
valence $1$ on $\partial \mathbb{D} \setminus \{1\}$.
\begin{figure}
\begin{tikzpicture}
\draw[dashed] (0,1) circle (.5cm) (.5,1)--(1,1);
\draw (0,1) circle (1cm) (-1,1) node[left]{$U_1$};
\draw (5,0) arc (0:180:1);
\draw (3,1) node{$U_2$};
\draw[dashed]
(6,0) arc (0:180:2)
(5,0)--(6,0)
(2,0)--(3,0);
\draw[dashed] (6,-4) arc (0:180:1)
(7,-4) arc (0:180:2)
(5.5,-4) arc (0:180:.5)
(3,-4)--(4.5,-4)
(5.5,-4) -- (7,-4);
\draw[dashed] (-2,-3) circle (.5cm)
(-2,-3) circle(2cm)
(-1.5,-3)--(0,-3);
\draw[->] (.75,-.25)--(-1,-1) node[midway,right,below]{$f_X$};
\draw[->] (1,-.25)--(4,-2) node[midway,right,above]{$\varphi_3$};
\draw[->] (2.75,-3.5) -- (.25, -3.5) node[midway,below]{$z^2$};
\draw (5,-1.5) node {$\varphi_3(U_2)$};
\draw (5,-4.5) node {$\varphi_3(U_1)$};
\end{tikzpicture}
\caption{\label{figchart} An illustration of $\varphi_3$ and
$f_X$ on $U$, from Example \ref{ex:twodisctree}.}
\end{figure}
\end{Example}
We now define some types of graphs which we need to state the main
result.
A {\itshape plane valence tree}
is a graph that is a tree. To each node is associated a
label of either $\mathbb{C}_+$ or $\mathbb{C}_-$ and
positive integer $m$, called the valence. A node with label
$\mathbb{C}_+$ may only be adjacent to nodes labeled $\mathbb{C}_-$, and nodes
labeled $\mathbb{C}_-$ may only be adjacent to nodes labeled
$\mathbb{C}_+$. To each edge is associated
an open interval in $\mathbb{R}$. The interval may be all of
$\mathbb{R}$ but may not be empty. We make the requirement that
a disjoint union of all the intervals on edges coming from a node is a
subset of a disjoint union of $m$ copies of $\mathbb{R}$,
where $m$ is the valence of the node.
We require that some node has the property that
a disjoint union of $m$ copies of $\mathbb{R}$,
where $m$ is the valence of the node,
contains a disjoint union of all the intervals on edges coming from the node,
as well as another open interval in $\mathbb{R}$. We say that such a node
has a free interval.
For a plane valence tree, the valence of a point in $\mathbb{C}_+$ is the sum of
the valences of the $\mathbb{C}_+$ nodes, and similarly for $\mathbb{C}_-$. For a point
in $\mathbb{R}$, it is the number of times it appears in an edge of the valence
tree.
Define
$$\psi = -i \frac{z + 1}{z - 1}.$$ Then
$\psi$ maps $\mathbb{D}$ to $\mathbb{C}_+$ and $\mathbb{D}^*$ to $\mathbb{C}_-$ and
$\mathbb{R}$ to $\partial \mathbb{D} \setminus \{1\}$.
We may form a {\itshape disk valence tree}
by mapping $\mathbb{C}_+$ and $\mathbb{C}_-$ and $\mathbb{R}$ in the
labeling of the valence tree
to $\mathbb{D}$ and $\mathbb{D}^*$ and $\partial \mathbb{D} \setminus \{1\}$
under $\psi^{-1}$.
\begin{Example}
\begin{figure}[ht]
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=10em
]
\node[hp] {$\mathbb{C}_+$: 2}
child { node[hp] {$\mathbb{C}_-:5$}
edge from parent
node[left]{$(0,1)\ $}}
child { node[hp] {$\mathbb{C}_- : 2$}
child {node[hp] {$\mathbb{C}_+ : 1$}
child {node[hp] {$\mathbb{C}_- : 1$}
edge from parent
node[right]{$\ (7,8)$}
}
child {node[hp] {$\mathbb{C}_- : 1$}
edge from parent
node[right]{$\ (9,10)$}
}
edge from parent
node[right]{$(-3,5)$}
}
edge from parent
node[right]{$\ (-3,5)$}
}
;
\end{tikzpicture}
\caption{\label{fig:expv}An example plane valence tree.}
\end{figure}
Figure \ref{fig:expv} is an example of a plane valence tree.
By Theorem \ref{thm:VT} (see below),
there is a real Smirnov function with valence $3$ on
$\mathbb{C}_+$, valence $9$ on $\mathbb{C}_-$, valence $3$ on $(0,1)$, valence
$2$ on $(-3,0] \cup [1,5)$, valence $1$ on $(7,8)$ and $(9,10)$,
and valence $0$ elsewhere.
\end{Example}
\begin{Theorem}\label{thm:VT}
The valence of every real Smirnov function with finite valence is given
by the valence of a plane valence tree, and any valence arising from a
plane valence tree is the valence of a real Smirnov function.
\end{Theorem}
The proof of Theorem \ref{thm:VT} needs the following valence result.
It is an exercise in \cite{Hayman} (Example 3.1 of Section 3.3), but
for the sake of completeness we give the proof.
\begin{Lemma}\label{poirt9fdg}
Let $f$ be an analytic function in $\mathbb{D}$ of valence at most $m$.
Then $f \in H^p$ for every $p \in (0, \tfrac{1}{2 m})$.
\end{Lemma}
\begin{proof}
Recall the definition of the $p$-integral means $M_{p}(r, f)$ from \eqref{Mp} and define
$$M_{\infty}(r, f) := \sup_{|z| = r} |f(z)|.$$
By \cite[Thm.~1]{Cartwright} (see also \cite[Sec.~2.3]{Hayman}) we have
\begin{equation}\label{minty}
M_\infty(r,f) = O\left(\frac{1}{(1 - r)^{2 m}}\right).
\end{equation}
From \cite[Theorem 3.2]{Hayman} we see that if $f$ is $m$-valent and
$0 < r_0 < r < 1$ then
\[
M_p(r,f) \leqslant M_\infty(r_0,f)^p + m \max\Big(m,\frac{m^2}{2}\Big)
\int_{r_0}^r \frac{M_\infty(t,f)^p}{t} \, dt.
\]
Applying the estimate in \eqref{minty} for $M_\infty(r,f)$ shows that the function
$$r \mapsto M_p(r,f)$$ is bounded when $p \in (0, \tfrac{1}{2 m})$, i.e., $f \in H^p$ for all $p \in (0, \frac{1}{2m})$.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:VT}]
Given a plane valence tree, form the associated (disk) valence tree
under the mapping $\psi^{-1}$.
Construct a disk tree $X$ where nodes labeled $\mathbb{D}$ with valence $m$ correspond
to disks of valence $m$, and similarly for $\mathbb{D}^*$.
If $Y$ is a copy of $\mathbb{D}$ or $\mathbb{D}^*$ with valence $m$,
consider the arcs labeling edges
connected to the corresponding node.
Call these arcs $I_1, \ldots, I_k$.
The disjoint union of these arcs
are a subset of a disjoint union of $m$ copies of
$\partial \mathbb{D} \setminus \{1\}$.
This means that we can find a disjoint set of admissible arcs on
the boundary of $Y$ whose image arcs are precisely the arcs
$I_1, \ldots, I_k$.
Given copies of $\mathbb{D}$ and $\mathbb{D}^*$ with corresponding nodes connected by an edge,
weld them together
together on arcs with image arcs equal to the arc labeling the
edge between them.
This is possible by the above remarks.
Since the valence tree has the free arc property
and is a tree, $X$
will have a free arc on some copy of $\mathbb{D}$ or $\mathbb{D}^*$ contained
in it and will be a disk tree.
Let $\varphi$ be a conformal map from $\mathbb{D}$ onto $X$.
The map $g = \psi \circ f_X \circ \varphi$ has valence
equal to the valence of the plane valence tree. Since $g$ has finite
valence, it is in the Smirnov class by Lemma \ref{poirt9fdg}.
Every
point in
$\mathbb{C} \setminus (\mathbb{R} \cup \{i,-i\})$ has a neighborhood $U$ such that
$g^{-1}(U)$ consists of disjoint sets that are each homeomorphic to $U$
under $f$.
The reason that we exclude $i$ and $-i$ is that $i=\psi(0)$ and
$-i=\psi(\infty)$, and the points $0$ and $\infty$ may be the image under
$f_X$ of points where $f_X$ has zero derivative.
The fact that $\{i,-i\}$ has zero capacity together with
the same argument
used to prove \eqref{oos9d9} shows that $g$ has real boundary values almost
everywhere.
We mention in passing that, in fact, since $g$ has finite
valence and thus is rational (by the
result from \cite{Helson}), it
has real boundary
values everywhere, except for a finite number of points where it has
$\infty$ as a boundary value.
%
For the other direction, suppose that $f$ is real Smirnov with finite valence.
Then $f$ is rational and thus continuous on $\mathbb{D}$, when viewed as a map
into $\widehat{\mathbb{C}}$.
Consider $f^{-1}(-\infty, \infty)$. This is a set of branched analytic
arcs in $\mathbb{D}$, with endpoints only at $\mathbb{D}$ and branch points only at
points where $f'$ is zero. Each branch point has an even number of analytic
arcs coming from it.
Also, $f^{-1}(\mathbb{C}_+)$ is a finite disjoint union of open sets we will call
upper regions (similarly for $\mathbb{C}_-$ and lower regions).
By the maximum principle for harmonic functions
applied to the real part of $f$,
no upper region can have more than two of
the analytic arcs going into a branch point as boundary arcs.
Also by the maximum principle, no two upper regions can share a boundary curve,
and each upper region is simply connected.
Given an upper region, note that its boundary curve is mapped into
$\mathbb{R}$, and so by the argument principle each point in
$\mathbb{C}_+$ has the same valence $m$ under $f$ restricted to the upper region.
This means the boundary of the upper region must contain some point
that maps to $\infty$.
The same applies to lower regions.
By the orientation preserving properties of analytic functions, each
upper region maps its boundary (considered as positively oriented) to
the real line in a way so that moving along the boundary increases the
value on the real line (except at $\infty$). A lower region has
the opposite property.
Thus, by the argument principle, the boundary of an upper or lower region
must map onto $\mathbb{R} \cup \{\infty\}$
exactly $m$ times.
We give Figures \ref{fig:discregiona} and \ref{figb}
for illustration.
\begin{figure}
\begin{tikzpicture}[scale = 3.0]
\draw (.55,0) circle (.55cm);
\draw (-.7,0) circle (.4cm);
\draw (.85,.1) circle (.2cm);
\draw (0,.75) node{$\mathbb{C}_+$:2};
\draw (-.75,0) node{$\mathbb{C}_-$:1};
\draw (.4,0) node{$\mathbb{C}_-$:1};
\filldraw[color=white, even odd rule] (0,0) circle (1cm)
(-1.5,-1.5) -- (-1.5,1) -- (1.5,1) -- (1.5,-1.5) -- cycle;
\draw (0,0) circle (1cm);
\draw[->] (1.1,.18) node[right]{$\mathbb{C}_+$:1} -- (.9,.1);
\draw (-1,0) node[left]{$(-\infty,-1)$};
\draw (0,-1) node[below]{$(-\infty,0)$};
\draw (1,-.2) node[right]{$(-2,0)$};
\draw (1,.02) node[right]{$(-2,\infty)$};
\draw (1,.35) node[right]{$(3,\infty)$};
\draw (0,1) node[above]{$(3,\infty)\cup(-\infty,-1)$};
\draw (-.3,0) node[left]{$I_1$};
\draw (0,0) node[left]{$I_2$};
\draw (.68,.2) node[left]{$I_3$};
\end{tikzpicture}
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=10em
]
\node[hp]{$\mathbb{C}_+$:2}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[left]{$(-1,\infty)$}}
child { node[hp] {$\mathbb{C}_-$:1}
child { node[hp] {$\mathbb{C}_+$:1}
edge from parent node[right]{$(-\infty,-2)$}
}
edge from parent node[right]{$(0,3)$}
}
;
\end{tikzpicture}
\caption{\label{fig:discregiona}$I_1=(-1,\infty)$, $I_2=(0,3)$, $I_3=(-\infty,-2)$.
Valence is $3$ on $\mathbb{C}_+$, $2$ on $\mathbb{C}_-$, $2$ on $(0,3)$, $1$ on
$(-1,0]$ and $[3,\infty)$,
and $1$ on $(-\infty,-2)$. Note that as we proceed counterclockwise
around the boundary of an upper region and
clockwise around the boundary of a lower region, the
image under $f$ increases on the real line.}
\end{figure}
\begin{figure}
\begin{tikzpicture}[scale = 4.0]
\draw (.55,0) circle (.55cm);
\draw (0,0) -- (-.6,.8);
\draw (0,0) -- (-.6,-.8);
\draw (-.3,.4) -- (0,1);
\draw (-.3,.4) -- (5/13,12/13);
\draw (-.3,.4) -- (-1,0);
\draw (-.3,.4) -- (-5/13,12/13);
\draw (-12/13,-5/13) circle (.2cm);
\draw (-12/13,-5/13) circle (.35cm);
\draw (-5/13,-12/13) -- (3/5,-4/5);
\filldraw[color=white, even odd rule] (0,0) circle (1cm)
(-1.5,-1.5) -- (-1.5,1) -- (1.5,1) -- (1.5,-1.5) -- cycle;
\draw (0,0) circle (1cm);
\draw (.5,0) node{\bfseries $\mathbb{C}_-^E$};
\draw (-.5,0) node{\bfseries $\mathbb{C}_-^H$};
\draw (0,-.5) node{\bfseries $\mathbb{C}_+^A$};
\draw (.1,.5) node{\bfseries $\mathbb{C}_+^B$};
\draw (-.2,.8) node{\bfseries $\mathbb{C}_+^C$};
\draw (0,.75) node{\bfseries $\mathbb{C}_-^F$};
\draw (-.45,.75) node{\bfseries $\mathbb{C}_-^G$};
\draw (-.65,.5) node{\bfseries $\mathbb{C}_+^D$};
\draw (0,-.95) node {\bfseries $\mathbb{C}_-^I$};
\draw (-.865,-.3) node {\bfseries $\mathbb{C}_-^K$};
\draw (-1.3,-.1) node {\bfseries $\mathbb{C}_-^J$};
\draw[->](-1.1,-.1)--(-.95,-.1);
\end{tikzpicture}
\vspace{-.5in}
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=5em
]
\node[hp]{$\mathbb{C}_+$ ABCD}
child { node[hp] {$\mathbb{C}_-$ E } }
child { node[hp] {$\mathbb{C}_-$ F }
child { node[hp] {$\mathbb{C}_+$ J }
child { node[hp] {$\mathbb{C}_-$ K }}}
}
child { node[hp] {$\mathbb{C}_-$ G } }
child { node[hp] {$\mathbb{C}_-$ H } }
child { node[hp] {$\mathbb{C}_-$ I } }
;
\end{tikzpicture}
\caption{\label{figb} The disk with upper and lower regions, and the
corresponding valence tree. Upper regions are labeled $\mathbb{C}_+$ and lower
regions are labeled $\mathbb{C}_-$. Edges and valences are not labeled.
Note that, if
we start with upper region A, we obtain the upper collection with
A, B, C, and D as the first node.}
\end{figure}
We will now give a method that, given a finite valence real Smirnov function,
constructs a plane valence tree such that the valence of the tree
corresponds to the valence of the function.
We proceed by induction on the number of regions.
If there is only one region, and this region has valence $m$, construct
a plane valence tree with only one node of valence $m$.
Suppose there is more than one region.
We may replace upper regions by lower regions in the following argument if
needed.
Take an upper region, and consider all
upper regions sharing common boundary points
{\itshape inside the unit disk} with the given region.
Take the union, including the boundary points and boundary arcs.
Repeat the process for all the new upper regions added, and continue
until it is no longer possible to do so. This forms a finite union of
upper regions (and their boundaries) -
call it $X$, and call it an upper collection.
By the maximum principle for harmonic functions, $X$ is simply
connected.
$X$ is not the whole disk by our assumption.
The
boundary of $X$ must intersect the boundary of the disk since some point in
the boundary of each upper region maps to $\infty$.
Thus, the complement of $X$ consists of a union of simply connected
sets.
For the $j^{th}$ set, let $\varphi_j$ be the conformal
map from the unit disk onto this set. By the induction hypothesis,
we may form a valence tree for the $j^{th}$ set, using the function
$f \circ \varphi_j$ instead of the function $f$, and by starting with
a lower region instead of an upper region.
(We could also apply the above reasoning directly to the $j^{th}$
set without using the conformal map $\varphi_j$.)
Let $T_j$ denote the valence tree for the $j^{th}$ set.
Now suppose the total valence of
all components of $X$ is $M$. Draw a $\mathbb{C}_+$ node with valence $M$.
Draw edges from the
node for $X$ to the nodes in the trees $T_j$
that correspond to lower collections sharing boundary arcs with $X$.
There is at least one edge to each $T_j$ because every $T_j$ shares a
boundary arc with $X$.
We will later see that there is exactly one edge to each $T_j$.
Label each arc's edges with the intervals corresponding to the values of
$f$ on the edges.
We will show the graph formed is a tree.
Consider an analytic arc in $f(-\infty,\infty)$ that approaches the
boundary of the circle and is part of the boundary of the upper collection.
If we start from a point in the arc that is on the unit
circle and follow the arc,
it either terminates at another point of the circle, or at a branch
point. If this is the case, some other arc going from the branch point
must be the boundary of the same lower region as the original arc;
follow the new arc. We may continue until we hit the boundary of the
unit circle, which we must since the arc can never approach the same branch
point again, and there are finitely many branch points.
Call the combination of arcs $\gamma$.
The combination of arcs $\gamma$
will be part of the boundary of some lower region, call it $L$.
The lower region $L$ can only have a common boundary with the
upper collection $X$
along $\gamma$, since the upper collection $X$ is connected
and $\gamma$ intersects the circle at its two ends.
The component of the complement of the upper collection $X$ that
contains $L$ can have common boundary with $X$ only on
$\gamma$, for the same reason.
This shows that the graph we form is a tree.
The arc $\gamma$ (not counting points on the unit circle) maps to
some subset of
$(-\infty,\infty)$ once. This follows since as a point travels along
$\gamma$, the image of the point always increases on the real line,
or always decreases, since $\gamma$ is part of the boundary of a
lower region.
This shows each edge is labeled with an interval in $\mathbb{R}$.
Since the boundary of the upper collection maps onto
$\mathbb{R} \cup \{\infty\}$ exactly $M$ times, the disjoint
union of all the intervals labeling edges coming from the
upper collection is contained in
a disjoint union of $M$ copies of $\mathbb{R}$.
Note that
there is some node
(say of valence $m$) such that m disjoint copies of $\mathbb{R}$ minus
the disjoint union of its edges contains an (open) interval.
This follows from the
fact that some upper or lower region must have a boundary arc in
common with the unit disk.
Thus, the graph we construct is a plane
valence tree.
The valence of the
tree is the same as that of $f$ by construction.
\end{proof}
Note that if a holomorphic function has bounded finite valence on $\mathbb{C}_+$ and
$\mathbb{C}_-$, it must have finite valence on $\mathbb{R}$ by the open mapping theorem.
We now give some examples.
\begin{Example}\label{sdfjfdg}
Let $n \geqslant 1$ and
$$(a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n)$$ be a finite set of open intervals such that none of the intervals is the entire real line and
$(a_j,b_j)$ is disjoint from $(a_{j+1}, b_{j+1})$ for each $j$. Then there
is a function from $N_{\mathbb{R}}^{+}$ whose range is
\[
\bigcup_{j=1}^n (a_j, b_j) \cup \mathbb{C}_+ \cup \mathbb{C}_-.
\]
Moreover, the valence of each point of $\mathbb{C}_+$ is $\lfloor n/2 \rfloor + 1$ and
the valence of each point of $\mathbb{C}_-$ is $\lceil n/2 \rceil$. The valence
of each point in $\mathbb{R}$ is equal to the number of the intervals
$(a_j, b_j)$ in which it lies.
\end{Example}
Clearly we can interchange the roles of $\mathbb{C}_+$ and $\mathbb{C}_-$ in the above
example.
We could deduce this from our previous theorem, but we will first give
an independent proof that is simpler than the proof of the previous
theorem.
\begin{proof}%
Construct a Riemann surface as follows. Weld a copy of
$\mathbb{C}_+$ to $\mathbb{C}_-$ along the interval $(a_1, b_1)$.
Now weld the copy of $\mathbb{C}_-$ to a different copy of
$\mathbb{C}_+$ along the interval $(a_2, b_2)$.
Now weld this copy of $\mathbb{C}_+$ to a different copy of
$\mathbb{C}_-$ along the interval $(a_3, b_3)$.
Proceed in this manner
until all of the intervals are exhausted.
Call this Riemann surface $X$.
Let $\theta$ be the
projection map from $X$ to $\mathbb{C}$ that takes a given point
in the Riemann surface to the corresponding point in either
$\mathbb{C}_+$, $\mathbb{C}_-$, or $\mathbb{R}$.
We now claim that the Riemann surface is conformally equivalent to $\mathbb{D}$. If it were not, then, since it is simply connected, it would be
equivalent to either the Riemann sphere or the complex plane (uniformization theorem). It is not
equivalent to the sphere since it is not compact. %
It is not equivalent to the complex plane since we could weld another
half plane onto the last half plane welded onto
$X$ and still have a simply connected surface, and if
we remove infinitely many points from any simply connected Riemann
surface and are left with a simply connected surface, the new surface
must be equivalent to the disk by the uniformization theorem and
the Riemann mapping theorem.
Let $\varphi$ be the conformal map
from $\mathbb{D}$ to the Riemann surface and
$$f = \theta \circ \varphi.$$ Then
$f$ maps from $\mathbb{D}$ into $\mathbb{C}$. Since $f$ has valence at most
$\lfloor n/2 \rfloor + 1$ at each point, it belongs to some Hardy space (Lemma \ref{poirt9fdg}) and thus
belongs to $N^+$.
We will now show that
$f$ has real (radial) boundary values almost everywhere.
To see this, note that every point in
$\mathbb{C} \setminus \mathbb{R}$ has a neighborhood $U$ such that
$\theta^{-1}(U)$ consists of disjoint sets that are each homeomorphic to $U$
under $\theta$.
Thus
every point in
$\mathbb{C} \setminus \mathbb{R}$ has a neighborhood $U$ such that
$f^{-1}(U)$ consists of disjoint sets that are each homeomorphic to $U$
under $f$.
The same argument as used to prove \eqref{oos9d9} shows that no (radial) boundary values of
$f$ lie in $\mathbb{C} \setminus \mathbb{R}$.
\end{proof}
For example, for $n=3$,
the valence tree for this example is as shown below.
The nodes are shown with their valences.
\begin{tikzpicture}[hp/.style={rectangle,draw},
grow = right, level distance=8em]
\node[hp] {$\mathbb{C}_+ :1$}
child { node[hp] {$\mathbb{C}_- : 1$}
child {node[hp] {$\mathbb{C}_+ : 1$}
child{node[hp] {$\mathbb{C}_- : 1$}
edge from parent
node[below]{$(a_3,b_3)$}
}
edge from parent
node[below]{$(a_2,b_2)$}
}
edge from parent
node[below]{$\ (a_1,b_1)$}
}
;
\end{tikzpicture}
%
\begin{Example}
The only possible valence trees for a real Smirnov function with
valence $1$ on $\mathbb{C}_+$ and valence $1$ on $\mathbb{C}_-$ have a
$\mathbb{C}_+$ node of valence $1$ connected to a $\mathbb{C}_-$ node of
valence $1$. The interval labeling the edge connecting them
can be any nonempty open interval, except $\mathbb{R}$, since if the
edge was labeled $\mathbb{R}$ neither node would have a free interval.
Thus the valence of such a real Smirnov function is $1$ on a
nonempty open proper subinterval of $\mathbb{R}$, and $0$ elsewhere on
$\mathbb{R}$.
\end{Example}
\begin{Example}
Let us find all possible valences for real Smirnov functions with
valence $2$ on $\mathbb{C}_+$ and valence $1$ on $\mathbb{C}_-$.
The valence tree
for such a function has either one $\mathbb{C}_+$ node of valence
$2$, or two $\mathbb{C}_+$ nodes of valence $1$. These nodes cannot be
adjacent. Figure \ref{fig:vt21} shows all possible valence trees.
In tree I, $I_1$ and $I_2$ must be disjoint; both the top and
bottom node automatically have free intervals.
In tree II, $I_1$ can be an arbitrary non-empty open interval.
Because the $\mathbb{C}_+$ node has valence $2$ and only one edge, it
has a free arc automatically.
So the valence at every point in $\mathbb{R}$ is either $1$ or $0$. The range
on $\mathbb{R}$
is either the union of two open intervals, or is one interval.
\begin{figure}
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{I};
\node[hp]{$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:1}
child { node[hp] {$\mathbb{C}_+$:1}
edge from parent node[left]{$I_2$}
}
edge from parent node[left]{$I_1\,$}
}
;
\draw (2,0) node{II};
\draw(3,0) node[hp]{$\mathbb{C}_+$:2}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[left]{$I_1\,$}
}
;
\end{tikzpicture}
\caption{\label{fig:vt21} All possible valence tress with valence
$2$ on $\mathbb{C}_+$ and $1$ on $\mathbb{C}_-$.}
\end{figure}
\end{Example}
\begin{Example}
Let us find all possible valences for real Smirnov functions with
valence $2$ on $\mathbb{C}_+$ and valence $3$ on $\mathbb{C}_-$. The valence tree
for such a function has either one $\mathbb{C}_+$ node of valence
$2$, or two $\mathbb{C}_+$ nodes of valence $1$. These nodes cannot be
adjacent. Figure \ref{fig:vt23} shows all possible valence trees.
Note that if all the other conditions for being a real valence tree
are satisfied, some node in a tree must have a free interval if
there is some node of valence one connected to at least two nodes, one of
which connects to no other nodes.
If a node of valence $m$ has less than $m$ edges connected to it,
the free interval condition is also automatically satisfied.
These remarks apply to all the trees in Figure \ref{fig:vt23} except for
tree VIII, which also automatically has a free interval since two of the
intervals labeling its edges must be disjoint.
In tree I, we require $I_1$, $I_3$, and $I_4$ to be pairwise disjoint,
and $I_1$ and $I_2$ to be disjoint. In tree II, we require $I_1$ and
$I_3$ to be disjoint. The intervals $I_1$ and $I_2$ do not have to
be disjoint, since the node they have in common has valence $2$, and
any two intervals can fit disjointly into two copies of $\mathbb{R}$.
In tree III, we require $I_1$ and $I_3$ to be disjoint, and $I_1$ and
$I_2$ to be disjoint. In tree IV, the intervals $I_1$ and $I_2$ can
be arbitrary.
In tree V, intervals $I_1$ and $I_2$ must be disjoint, intervals
$I_1$ and $I_3$ are disjoint, and intervals $I_3$ and $I_4$ are
disjoint. In tree VI, $I_1$ must be disjoint from
$I_2$ and $I_2$ must be disjoint from $I_3$.
In tree VII, $I_2$ must be disjoint from $I_3$.
In tree VIII, the intervals $I_1$, $I_2$, and $I_3$ must be able to fit into
two disjoint copies of $\mathbb{R}$ without intersecting. This is
equivalent to requiring that two of them be disjoint. In trees
IX and X, there are no requirements on the intervals. Note that
for all cases, the conditions above imply that
some node must have a free interval, and the above intervals are
allowed to be $\mathbb{R}$ if this does not conflict with any of the above
conditions.
Considering all possible cases shows that
the range of the function on $\mathbb{R}$ may be (counting
multiplicity) %
%
%
%
the union of four open intervals
$I_1$, $I_2$, $I_3$ and $I_4$ where $I_1$ and $I_2$ are disjoint,
$I_1$ and $I_3$ are disjoint, and $I_3$ and $I_4$ are disjoint.
The range of the function on $\mathbb{R}$ (counting multiplicity) may also
be the union of three open intervals, at least two of which
are disjoint, or it may be the union of two open intervals; or it will
be one open interval. These are the only cases.
Some of these intervals may be $\mathbb{R}$ if the conditions are satisfied
(although they cannot be empty).
To take a concrete example, from the first case we see that the
range could be (counting multiplicities)
$$(0,1) \cup (2, \infty) \cup (2, 3) \cup (4,5).$$ In other words,
the valence would be one on $(0,1)$, two on $(2,3)$ and $(4,5)$,
one on $[3,4] \cup [5,\infty)$, and zero on the rest
of $\mathbb{R}$.
\begin{figure}[htbp!]
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{I};
\node[hp]{$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:1}
child { node[hp] {$\mathbb{C}_+$:1}
edge from parent node[left]{$I_2$}
}
edge from parent node[left]{$I_1\,$}
}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[left]{$I_3\!$}}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[right]{$\,I_4$}}
;
\end{tikzpicture} \hfill
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{II};
\node[hp]{$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:2}
child { node[hp] {$\mathbb{C}_+$:1}
edge from parent node[left]{$I_2$}
}
edge from parent node[left]{$I_1\,$}
}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[right]{$I_3\!$}}
;
\end{tikzpicture} \hfill
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{III};
\node[hp]{$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:1}
child { node[hp] {$\mathbb{C}_+$:1}
edge from parent node[left]{$I_2$}
}
edge from parent node[left]{$I_1\,$}
}
child { node[hp] {$\mathbb{C}_-$:2}
edge from parent node[right]{$I_3\!$}}
;
\end{tikzpicture}
\hfill
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{IV};
\node[hp]{$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:3}
child { node[hp] {$\mathbb{C}_+$:1}
edge from parent node[left]{$I_2$}
}
edge from parent node[left]{$I_1\,$}
}
;
\end{tikzpicture}
\vspace{1em}
\begin{minipage}{.45\linewidth}
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{V};
\node[hp]{$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:1}
child { node[hp] {$\mathbb{C}_+$:1}
child{ node[hp] {$\mathbb{C}_-$:1}
edge from parent node[left]{$I_4$}}
edge from parent node[left]{$I_3$}
}
edge from parent node[left]{$I_1\,$}
}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[left]{$I_2\!$}}
;
\end{tikzpicture} \hfill
\end{minipage}
\begin{minipage}{.5\linewidth}
\begin{tikzpicture}[hp/.style={rectangle,draw},grow=right,
level distance = 4em,
sibling distance=5em
]
\draw (-1,0) node{VI};
\node[hp]{$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:1}
child { node[hp] {$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:2}
edge from parent node[above]{$I_3$}
}
edge from parent node[above]{$I_2$}
}
edge from parent node[above]{$I_1\,$}
}
;
\end{tikzpicture}
\vspace{1em}
\begin{tikzpicture}[hp/.style={rectangle,draw},grow=right,
level distance = 4em,
sibling distance=5em
]
\draw (-1,0) node{VII};
\node[hp]{$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:2}
child { node[hp] {$\mathbb{C}_+$:1}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[above]{$I_3$}
}
edge from parent node[above]{$I_2$}
}
edge from parent node[above]{$I_1\,$}
}
;
\end{tikzpicture}
\end{minipage}
\hfill
\vspace{1em}
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{VIII};
\node[hp]{$\mathbb{C}_+$:2}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[left]{$I_1\,$}
}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[left]{$I_2\!$}}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[right]{$\,I_3$}}
;
\end{tikzpicture} \hfill
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{IX};
\node[hp]{$\mathbb{C}_+$:2}
child { node[hp] {$\mathbb{C}_-$:2}
edge from parent node[left]{$I_1\,$}
}
child { node[hp] {$\mathbb{C}_-$:1}
edge from parent node[right]{$I_2\!$}}
;
\end{tikzpicture} \hfill
\begin{tikzpicture}[hp/.style={rectangle,draw},
sibling distance=3em
]
\draw (-1,0) node{X};
\node[hp]{$\mathbb{C}_+$:2}
child { node[hp] {$\mathbb{C}_-$:3}
edge from parent node[left]{$I_1\,$}
}
;
\end{tikzpicture}
\caption{\label{fig:vt23} All possible valence tress with valence
$2$ on $\mathbb{C}_+$ and $3$ on $\mathbb{C}_-$.}
\end{figure}
\end{Example}
\def$'$} \def\cprime{$'${$'$} \def$'$} \def\cprime{$'${$'$}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 33
|
The International Conference of Data Protection and Privacy Commissioners (ICDPPC) connects 119 privacy and data protection authorities from across the globe. The ICDPPC first met in 1970 and has since become the leading data protection forum in the world. The 2018 conference, attracted more than 1,500 participants. EPIC President Marc Rotenberg, along with EPIC Board member Professor Anita Allen, EPIC Advisory Board member Malavika Jayaram, and Apple CEO and former EPIC Champion of Freedom Tim Cook, were among the speakers at the conference.
The EPIC Public Voice Fund provided support for NGO participation this year. A record number of NGOs were expected to attend, with over 130 delegates from 85 different NGOs registered.
The Public Voice Coalition organized a symposium on "AI, Ethics, and Fundamental Rights." Professor Anita Allen was the keynote speaker. Andrea Jelinek, Chair of the European Data Protection Board, Elizabeth Denham, UK Information Commissioner, Helen Dixon, Data Protection Commissioner for Ireland, Eduardo Bertoni, Director of the Argentina National Data Protection Authority, and Dr. Alessandro Mantelero, Rapporteur on Artificial Intelligence and Data Protection for Council of Europe (COE) were among the panelists.
The "Universal Guidelines for Artificial Intelligence" (UGAI) were announced at the Public Voice conference. The UGAI is the first human rights framework for AI. More than 150 experts and 40 NGOs, representing 30 countries around the world endorsed the UGAI. Among the signatories are the American Association for the Advancement of Science, the world's largest multidisciplinary scientific society, BEUC (the European Consumer Organization), and human rights advocate and former world chess champion Garry Kasparov.
At the larger conference, Professor Allen (@Lawprofaallen) delivered a dynamic keynote address to the entire body of privacy commissioners focused on ethics as the "basis of character and moral life." She described law and ethics as coexisting – stating that "Ethics are respected as the ideal foundation of law and professional standards." She recently published an essay in New Europe, "Why Ethics Now?," discussing these points. Allen also publicly endorsed the UGAI during her keynote.
Cook reminded us that, "This crisis is real. It is not imagined, or exaggerated, or crazy." Cook endorsed the GDPR and called for comprehensive privacy legislation in the US. See full speech below.
At the closing session, EPIC President Marc Rotenberg (@Marc Rotenberg) described the importance of civil society participation in the annual privacy conference. "This cannot be a conversation between only governments and industry. Democratic legitimacy requires public participation," said Mr. Rotenberg. He thanked Giovanni Buttarelli and the Data Protection Commissioners for their support for the Public Voice and the work of civil society.
Speaking to the conference theme, Mr. Rotenberg emphasized the importance of ethics to emerging challenges in the data protection field, such as AI. He described the development of the Universal Guidelines for AI, which acknowledged current legal rights but also incorporated ethical guidelines from computer science and human rights norms. "Ethics tells us not only what the law is, but also what the law should be," said Mr. Rotenberg.
For more news on the conference, follow EPIC on twitter @EPICprivacy. See more about Tim Cook's, EPIC Champion of Freedom Award in 2015. Learn more about Professor Anita Allen, a member of EPIC's Board of Directors, and author of several books, including Privacy Law and Society.
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{
"redpajama_set_name": "RedPajamaC4"
}
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f/k/a archives . . . real opinions & real haiku
just getting to know you
Filed under: Haiku or Senryu,q.s. quickies,viewpoint — David Giacalone @ 11:16 pm
We wouldn't be the f/k/a Gang, if we weren't frantically scampering to meet a self-imposed deadline on a Saturday night. It's a good thing we can re-use the same words written the first time we closed down this weblog, in October 2003, with only minor changes:
Doing ethicalEsq f/k/a has been a very rewarding experience, whether the correspondents agreed with me or not. Until I started a web journal, [my f/k/a Gang of alter egos] thought the internet might be used to sustain established friendships and relationships (mostly with email), but couldn't possibly create new ones of any significant value. Well, I was wrong.
Comments and e-correspondence sparked by this website have put me in touch with some very good [talented and interesting] human beings, who can scarcely be blamed for being lawyers [or haiku poets]. Although they're a lot busier than I am, I hope to continue to connect with them across cyberspace.
At the end of this posting, I've listed (alphabetically fairly randomly) a number of the web-log related folks who have become more than just pixelated names to me, due to the quality and/or quantity of their communications, insights, inspiration, or assistance.
sweet grapes
the conversation passes
between friends
… by Hilary Tann – The Heron's Nest VIII:1
Far more often than I could have imagined 6 years ago, this weblog has sparked real conversations — the kind that nurture real friendships. Before I list the names of people across the blogisphere who have been the most generous to me and this weblog, I want to share some haiku and senryu about conversations. (more…)
rivers, sunset, metaphors galore
Filed under: Haiku or Senryu,Schenectady Synecdoche — David Giacalone @ 9:07 pm
February thaw
a new patch of orange
… by dagosan
Catching another sunset or two in photos before we "archivize" this weblog on Saturday seemed like a good idea, as the afternoon waned today. [click "more" below to see some of the photographs] Naturally, I managed to dawdle so long at this keyboard that I only caught the last few moments before the sun dipped behind nearby hills. My timing was a metaphor of sorts for much that has happened (and not) lately in my life. Of course, the sunset itself was a too-obvious symbol (along with the promised sunrise after a long dark night) for the ending of an important era in my life. (more…)
Comments Off on rivers, sunset, metaphors galore
all that great haikai
Filed under: Haiga or Haibun,Haiku or Senryu — David Giacalone @ 10:53 am
In this last week of new posting at f/k/a, how can I possibly put together a piece that pays adequate homage to the vast body of haikai — haiku, senryu and related poetic-literary genres — that our Honored Guest Poets have allowed me to share with you? In two words: I can't.
Beginning in late November 2003, with a little feature located in our Sidebar called "haikuesque," this weblog has brought you "one-breath poetry" by some of the finest English-language haiku poets alive (plus hundreds of translations of the work of 19th Century Japanese Haiku Master, Kobayashi Issa, by David G. Lanoue). In total, 27 well-known and respected haijin have generously let me share their poetry with you, in my role as Haiku Missionary, bringing the joys of "real haiku" to lawyers and other folk not familiar with the genre. [The post "Yes, Lawyers and haiku" explains why haiku seems like a perfect art form for lawyers and others in our too-busy society.] Little did I know that rubbing elbows with some of the best haiku poets would inspire me to work hard at the craft myself, and would also result in my making some of my very closest friends.
Other than repeating here my heartfelt, immense gratitude to each of our Honored Guests, there really is no sufficient way to express my thanks or sum up their contribution to the success of this weblog. As suggested here, I have neither the time nor inclination to select my "favorite" haiku by each poet. Happily, their haikai will remain at this site for as long as Weblogs at Harvard Law School exists. So, I hope readers of f/k/a will use our search function or go often to our Honored Guest Poets Index page, and click on links to each poet's f/k/a archive. Then, sample their wares, and let them seduce you with the charms of haiku.
In alphabetical order, and with haiku-like pith, the f/k/a Gang says: "many thanks for all that great haikai; best wishes, and 'auf Wiedersehen' " to our Haiku Family: Roberta Beary, Randy Brooks; Yu Chang; Tom Clausen; Devar Dahl; Alice Frampton; Barry George; Lee Gurga; Carolyn Hall; Gary Hotham; Jim Kacian; David G. Lanoue; Rebecca Lilly; Peggy Willis Lyles; Paul Miller; Ed Markowski; Matt Morden; Pamela Miller Ness; W.F. "Dr. Bill" Owen; Tom Painting; Andrew Riutta; John Stevenson; George Swede; Hilary Tann; Michael Dylan Welch; and Billie Wilson.
alone at sunset
i pick a pair
of faded daylilies
the morning rush—
the whiteness of last night's snow
….. by David Giacalone – Legal Studies Forum (Vol. XXXII, No. 1. 2008)
Instead of further farewell fanfare regarding our Honored Guest Poets, I'm going to do what I would have done in the normal course of events this week: Present more haikai selected as among the very best of their genre for inclusion in "white lies: Red Moon Anthology 2008" (see our prior post for details). (more…)
nostalgic about Blawg Review
Filed under: Haiku or Senryu,q.s. quickies — David Giacalone @ 9:36 am
.. Ed & Edison in Schenectady (Jan. 2009) .. ..
What a strange coincidence: Just as I was announcing that this would be the last week of production for f/k/a, my friend "Ed Post" was putting together this week's version of Blawg Review — #200!! — which opens with a link to Darren Rowse's ProBlogger post, "If your blog died today . . . what would it be remembered for?". Happily, the 200th milestone for Blawg Review is not its last edition. As its anonymous Editor puts it:
"Not to worry; we've come to praise Blawg Review, not to bury it. This moot funeral is not a morbid affair, but a celebration of everything good about Blawg Review."
Like every issue of Blawg Review, this week's puts the spotlight on the best material posted during the prior week at law-related weblogs. As part of the 200th-edition celebration, Ed has structured this issue around an apt Traveling Wilburys metaphor — a musical group composed of rock-n-roll superstars whose collaboration magically "was greater than the sum of its parts."
One of my favorite poems posted here at f/k/a is this senryu by lawyer-poet Barry George:
his quiet funeral—
a man who did
most of the talking
………………. by barry george
The faux funeral of "Ed Post" and his Blawg Review inspired dagosan to pen a new version this morning:
his noisy wake —
the man who let others
do most of the talking
…. by dagosan
Blawg Review, which is to say Ed and many of his hosts, has always been bery-bery good to this weblog — from giving us the Blawg Review "Creative Law Blog Award" in 2005 [see "thanks a lot (for all this pressure)," Dec. 27, 2005], to including f/k/a in Ed's "Simply the Best" Top Ten Blawg lists [see our post, October 5, 2007], letting us host Blawg Review #52 (April 11, 2006), and mentioning our work often in the weekly Review. Behind the scenes, Ed has also often acted as our long-distance proofreader extraordinaire (saving the Gang from many embarrassments), and as cheerleader and moral support when stress and fatigue and Weltschmerz made me want to throw in the towel.
Ed's two stops in Schenectady to visit this cranky blawger — memorialized here and there — were testaments to the ability of the blawgiverse to create and nurture more than virtual friendships.
So, congratulations, Ed, for creating an enduring, high-quality blog carnival. And, heartfelt thanks for all you've done to create and celebrate the blawger community, and done for this little weblog and its humbled Editor.
bookie's funeral
the undertaker pays
his debt
…. by ed markowski
As usual, Ed has also reminded me that I have a lot work to do this week — crafting an auto-obituary and apologia for this weblog. Because we tried to close down this little project once before, after only 19 weeks in busines, I guess the second (and last) time should go a little more smoothly. See "exitedEsq: going dormant (gonna miss ya)" (October 11, 2003) Re-reading that post, I see there were a lot of lessons I never learned and a lot of mistakes repeated since our premature death notice.
On the other hand, we got such nice obits from other bloggers (back before Denise had even coined the word "blawg"), it's a wonder we ever started back up. Living up to our death press was quite daunting. See, e.g., this humble-making post by law-blog supertar Ernie Svenson, a/k/a Ernie the Attorney, "Requiem for a Heavyweight – ethicalEsq? is shutting down" (Oct. 12, 2003). Actually, the blog-obituaries were so generous, it's a wonder I haven't sought even more long before now.
update: And, it's happening again — nice words inspired by our leaving town. See Scott Greenfield's "Phoenix Rising" (Feb. 24, 2009).
Wait, I'm supposed to be concentrating on lawyer fees this week. I am so easy to distract.
Let's close with a few topical poems written by lawyer-poets:
funeral dirge –
we bury the one
who could carry a tune
…. by David Giacalone – Frogpond, Vol. 31:2 (Spring/Summer 2008)
repub. "white lies: Red Moon Anthology 2008"
the seeds she ordered
in today's mail
funeral over
the deadbolt
slides into place
his death notice. . .
the get-well card
still in my briefcase
… by Roberta Beary
"funeral over" – from the haibun "Stranger Danger" – Frogpond XXVIII:2 (2005)
"after" – Shiki Haikusphere 10th Anniversary Anthology (2007)
"his death notice" – New Resonance 2
is prune juice your cup of tea?
Filed under: haijin-haikai news,Haiku or Senryu — David Giacalone @ 7:01 pm
…. Prune Juice Journal …
at last in his coffin
depressed friend
is smiling
… by George Swede – Prune Juice (Issue 1, Winter 2009)
what's left of the cheese
has a bite
…. by Jim Kacian – Prune Juice (Issue 1)
.. Haiku legend Alexis Rotella has uncorked her first distillation of Prune Juice: Journal of Senryu and Kyoka (Issue 1, Winter 2009), which she describes as a biannual print and digital journal "dedicated to publishing and promoting fine senryu and kyoka in English." Issue 1 offers more than 130 poems by about four dozen haijin, many of them very well-known for their well-crafted poems and wry insight into human nature.
Senryu are structured like haiku, and kyoda like tanka, but their focus is different. As Alexis explains:
"Senryu generally emphasize human foibles and frailties, usually satirically, ironically, humorously. Season words are not necessary nor usual in senryu. Kyoka have a different history than senryu; nevertheless, for modern kyoka in English, the definition is similar: a poem in the tanka form but with the satirical, ironic, humorous aspects of senryu.a poem in the tanka form but with the satirical, ironic, humorous aspects of senryu."
Agreeing with the bumper sticker from StickEm2/CafePress, Alexis tells us that senyru "is an outlet, a therapy of sorts." She wants poets and readers to use senryu and kyoka to help reveal and share their real emotions, saying in her introduction to Issue I:
"I hope this issue inspires you to step up, to come and mingle with the rest of us—to make a toast with a glass of prune juice in honor of the plum blossoms who, without that delicious metaphorical elixir that gets things moving, would not exist. And if you are one who hides behind a potted plant, come out come out whoever you are."
Alexis seeks to publish senryu and kyoka that range from "gently humorous to the most wicked satire" — and advises that "Our tastes run towards the wicked end of the scale."
Frankly, the curmudgeons in the f/k/a Gang like to sip, rather than swig, senryu. And, we're a little wary (maybe even weary) of editors and poets trying to give us shocking or "wicked" poems. So, we plan to decant our Prune Juice a little at a time. With Alexis Rotella at the helm, however, we're pretty sure a lot of readers will be filling their cup to the brim with Prune Juice, and asking for refills.
Here are a few more poems by members of our f/k/a family of Honored Guest Poets from Prune Juice: Journal of Senryu and Kyoka (Issue 1, Winter 2009):
the jangle
of handcuffs
Instead of an air conditioner . . .
I return
with popsicles
… by Tom Clausen
new to the group—
sitting in back with
the artificial plants
… by Jim Kacian
reading of the will
cremated mother
rematerializes
the feud continues—
shoveled snow piled high
on the property line
first ice
on mother's gravestone . . .
her tea time
.. click for an annual subscription to the Prune Juice print edition ($32 with S&H) ..
p.s. Seven-Day Countdown: Speaking of feeling our emotions, getting things moving and setting ourselves free, the f/k/a Gang plans to stop adding to this weblog as of March 1, 2009. It will remain online, with thousands of haiku and senryu, and a lot of law-related and cultural punditry. But, the last f/k/a posting will roll off your Editor's fingers no later than Feb. 28, 2009. We'll try to write a few more posts related to lawyer fees before we hang up our blawger sword; then we'll be looking for something more enjoyable and less stressful to do online. Naturally, we'll have a little more to say when we sign off at the end of this week.
afterwords: Many thanks to Scott Greenfield at Simply Justice for his kindly post reacting to my announcement that f/k/a is closing down production. See "Phoenix Rising" (Feb. 24, 2009)
the pond ices over –
impressionist to
cubist overnight
early March –
the weather vane goose
still heading south
small sad face
in the puddle –
last weekend's snowman
…….. by David Giacalone – Simply Haiku (Autumn 2006, Vol. 4 no. 3)
Albany City Court Judge says local sex offender law is pre-empted
In a thoughtful 12-page decision, dated Feb. 18, 2009, Albany [New York] City Court Judge Thomas K. Keefe refused to enforce the City's sex offender residency law, using the Oberlander case as precedent, and refusing to follow a decision by his City Court benchmate, Judge Rachel Kretser. See Peo. v. James Blair (File #08-186882); "Sex offender residency case tossed" (Albany Times Union, Feb. 20, 2009). After citing the recent proposal to ban sex offenders from living near eachother in Colonie (see our prior post), Judge Keefe notes:
"As easily imagined and as was already noted by the Legislature, these 'not in my backyard' local residency restrictions create great difficulties for the Division of Parole, local probation and social service agencies to locate appropriate housing for sex offenders."
The Times Union notes:
"The conflicting decisions from the same court could send mixed messages to city police.
"Attorney Terence Kindlon, whose firm is suing the county pro bono, said he believes it would be 'more intelligent than not to refrain from prosecuting these cases'."
" . . . Detective James Miller, a spokesman for the Albany Department of Public Safety, said officers in the city will keep making arrests.
As the Times Union Politics Blog noted yesterday evening, "Amid all this, state Supreme Court Justice Roger McDonough is still considering a constitutional challenge to county law nearly identical to the one made in Rockland." Justice McDonough has a summary judgment motion before him in the suit mentioned above brought by Terence Kindlon.
It's clear that we need statewide action on sex offenders. However, we also need politicians who will have the courage to oppose counterproductive and ineffective residency bans — like the fear-mongering S.01300, proposed by Senate Majority Leader Malcolm A. Smith — that prevent whole classes of sex offenders from living in most populated areas, rather than allowing professionals to locate housing most appropriate for each individual sex offender. See our prior post "don't let a bad idea go statewide" (Feb. 2, 2009). If courage is lacking, perhaps politically-motivated leaders from rural areas of the state will rise up against S.01300, which will force many sex offenders to live in less-populated areas.
p.s. See the informative Wall Street Journal article "After Prison, Few Places for Sex Offenders to Live: Georgia's Rules That Keep Some Convicted Felons Far From Children Create Challenges for Compliance, Enforcement" (Feb. 19, 2009; via Corey Yung)
Need a more inspiring subject to head you toward the weekend? How about more haiku from the latest issue of Frogpond [Vol. 32:1 (Winter 2009)], written by our Honored Guest Poets?
windowless classroom
we talk about thinking
Appalachian spring
can I still learn
to play the violin
…. by Yu Chang – Frogpond Vol. 32:1 (Winter 2009)
lovers still
a falling petal
catches moonlight
tightly coiled–
the fetus kicks
…. by Peggy Willis Lyles – Frogpond Vol. 32:1 (Winter 2009)
she would have
polished the silver
Mom's memorial
Mom's ashes
lighter than expected
…. by Carolyn Hall – Frogpond Vol. 32:1 (Winter 2009)
late August
eel grass
breaks the surface
…. by Hilary Tann – Frogpond Vol. 32:1 (Winter 2009)
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officer johnson's undercover operation [updated]
Filed under: Haiku or Senryu,Schenectady Synecdoche — David Giacalone @ 11:57 am
.. The tired old fogies at f/k/a want to thank the energetic Scott Greenfield for covering the latest Schenectady cop scandal at his Simple Justice weblog, so we won't have to think too hard this morning. See "Even Cops Need Some Sleep" (Feb. 19, 2009)
Teaser: Schenectady Police Officer Dwayne Johnson made three times his base pay last year, while averaging 75 hours a week on the clock (making him, at $168,000, the highest paid employee in Schenectady's history). However, after several late-night stakeouts, Schenectady Gazette reporter Kathleen Moore reported yesterday that Officer Johnson has been parking his car outside a local apartment that is not his home for a few hours every Tuesday night since November, during his patrol shift. Despite being tracked by a GPS monitor in his unit, no supervisor caught the apparent dereliction of duty. See "Chief: Cop 'stealing time': Johnson, tops in pay, out of car during shift" (by Kathleen Moore, Feb. 18, 2009).
Responding to the question from Schenectady Police Chief Mark Chaires, "how dumb can you get?", Scott points out that "neither Chief Chaires nor anybody else on the force thinks that somebody ought to take the occasional gander at their top earner, the big money man, to make sure they are getting their money's worth?" Scott then muses: "How dumb? Not as dumb as you, Chief."
Follow-ups today (Feb. 19, 2009): "Cop case probed for collusion: Chief wants to know why supervisors didn't notice AWOL officer's absences" (Daily Gazette , Feb. 19, 2009); "Editorial: In Sch'dy, Car 10, where are you?" (Daily Gazette, Feb. 19, 2009; "He deserves to be fired, and anybody but a union officer or lawyer, or perhaps arbitrator, would agree."); "High-paid cop accused of slacking off" (WNYT/CH.13, Feb. 18, 2009, with video); "Did Schenectady's $168G cop spend hours away? Schenectady probes whether highest-paid officer was at apartment while on duty" (Albany Times Union, February 19, 2009); "Tarnishing the badge: A decade of trouble for Schenectady police" (Times Union, by Paul Nelson, Feb. 19, 2009)
update (Feb. 20, 2009): The Gazette tells us this morning that Officer Johnson was "suspended without pay Thursday while the department investigates the extent of his absences during his overnight patrols." He apparently will have to be paid if kept on suspension longer than 30 days. "Absent officer out for month: Bennett begins cop AWOL probe; union issues cited" (Feb. 20, 2009). I'm surprised that Public Safety Commissioner Wayne Bennett believes "it will take well over a month to finish the investigation into Johnson's absences. Also under review are the supervisors who did not notice them and the officers who may have tipped him off when internal affairs attempted to catch him in the act early last Tuesday." I'm not surprised that he expects the police union to argue napping has become a "past practice," approved regularly by lower-level supervisors, that cannot be changed without union approval.
The Gazette notes that "Some officers, who spoke anonymously, say everyone who works long shifts takes naps, beginning at lunchtime. They argued that an unspoken rule in the department allows napping to continue after lunch as long as police get up as soon as they get a call." Bennett says: "If someone had the absolute and unmitigated gall to call [napping] a past practice, well, supervisors do not have that kind of authority to authorize that."
In his update this morning at Simple Justice, Scott Greenfield trumpets "The new frontier for police contracts: Napping Clauses."
Officer Johnson is 49 years old and apparently considers a double shift to be his regular work day. The f/k/a Gang understands the need to nap (although, altogether, we alter egos aren't working 75 hours a week), but we agree with the Gazette that if the conduct is proven, Officer Johnson should be fired. At the very least, some major auditing of his time records is needed, plus more scrutiny of his so-called supervisors.
Undercover? Lawyer Greenfield concludes: "But don't fear that Johnson will go unpunished. My bet is that his wife will have a few questions about what he was doing in that apartment every Tuesday morning."
We don't get paid overtime (nor anytime) here at f/k/a, but we're always workin' hard trying to bring you some of the best haiku around. As promised yesterday, here are poems written by a few of our Honored Guest Poets that were selected for the newest issue of Frogpond [Vol. 32:1, Winter 2009]. We'll have another batch later this week.
turning back on a dead end street —
one odor changes
… by Gary Hotham – Frogpond Vol. 32:1 (Winter 2009)
heat lightning the crooked split in the watermelon
… by w.f. owen – Frogpond Vol. 32:1 (Winter 2009)
a flock of turnstones
skirt the shore
a field sparrow flashes
… by Tom Painting – Frogpond Vol. 32:1 (Winter 2009)
I finally share the secret
with my cat
….. by Alice Frampton – Frogpond Vol. 32:1 (Winter 2009)
the small fir
the barren windbreak sifting a rainy fog
…. by Tom Clausen – Frogpond Vol. 32:1 (Winter 2009)
winter night
the heat comes on
a retinal sun
wanders through
the observation car
… by John Stevenson – Frogpond Vol. 32:1 (Winter 2009)
our long bathtub soak —
a ring around
…. by David Giacalone – Frogpond Vol. 32:1 (Winter 2009)
p.s. Speaking of criminal justice in Schenectady, the print version of the Daily Gazette has an article on p. B3 titled "Imposter [sic] suspect in Regents exam faces lesser charge" (Feb. 19, 2009). In it we learn that District Attorney Robert Carney won't be charging Deandre M. Ellis with burglary [illegally entering a building intending to commit another crime] for entering a Schenectady school to take a Regents exam in disguise for another student. We were doubtful of the arresting officers' legal reasoning in a post on Jan. 29, 2009 (scroll to second story). Instead, Ellis is being charged with misdemeanor criminal impersonation, which he denied at his arraignment yesterday. DA Carney explains that "There has to be some sort of notice or communication to [a] person that 'you're not welcome' to convert [entering a public building like a school] to a trespass," on which to hang a burglary count. According to the Gazette:
"But Carney likened the case to a shoplifter. Anyone is allowed in a store, until they're asked to leave. But a shoplifter isn't charged with burglary, Carney said, even though they may have entered with the intent to steal."
Tonsorial-forensic experts should note a mystery raised in the case: Ellis wore a wig when posing as a female student in January. As you can see above, he has short spiky hair in his mug shot. But, three weeks later, he appeared in court with "long hair, past his shoulders." Neither Ellis nor his public defender were willing to comment on the issue. Could it be Ellis will claim he always goes around in the long wig and therefore was not trying to impersonate the female student?
.. two good ideas from Schenectady County . .
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frogpond brings HSA winners
.. The newest issue of Frogpond [Vol. 32:1, Winter 2009], the journal of the Haiku Society of America, arrived at my door this snowy February afternoon. Frogpond always has a lot of winning haiku, but this issue also announces the winners of HSA's most prestigious annual contests: The 2008 Kanterman Merit Book Awards for best published books in 2007; the 2008 Henderson Award for best haiku; and the 2008 Brady Award for best senryu.
As usual, several of f/k/a's Honored Guest Poets have been honored this year.
John Stevenson received 1st and 3rd place awards in the Harold G. Henderson Haiku Contest for 2008:
fifteen minutes
of mince pie
[1st Place, 2008 Henderson Contest]
my attention
[3rd Place, 2008 Henderson Contest]
Michael Dylan Welch won 2nd Place in the Gerald Brady Memorial Contest for 2008, with this senryu:
busy Italian restaurant–
sung to the wrong table
[2nd Place, 2008 Brady contest]
Among the Mildred Kanterman Memorial Book Awards for 2008:
Roberta Beary's The Unworn Necklace (Snapshot Press 2007) placed third [find poems and discussion at f/k/a here]
While Matt Morden's Stumbles in Clover (Snapshot Press 2007; discussed here at f/k/a) shared Honorable Mention honors with Gary Hotham's Missed Appointment (Lilliput Review 2007; featured at f/k/a in posts here and there)
The Best Anthology award went to Jim Kacian's Big Sky – The Red Moon Anthology of English-Language Haiku 2006 (Red Moon Press 2007; find sample poems at the bottom of this prior post)
The Best Haibun award went to "Dr. Bill" w.f. owen for his book small events (Red Moon Press 2007)
.. In the very near future, we'll share poems from the Winter 2009 issue of Frogpond written by our Honored Guests (update: go here and there). Below the fold, you will find a list of all the winners from the three contests described above (soon, you will be able to find all the winning poems and the comments of the judges by clicking on the link for each contest at the HSA Haiku Contests page):
a preference for congeniality
Filed under: Haiku or Senryu,lawyer news or ethics,viewpoint — David Giacalone @ 12:53 pm
.. .. .. The Jerks vs. the Genial: Law professor Jeff Harrison started an interesting discussion last week in a posting at MoneyLaw titled "Ready, Set, Punt" (Feb. 10, 2009). He notes that likablity is a "Pretty crazy way to pick a football team right? The team would lose every game." Harrison then asks:
"Is there any reason to think the 'like' factor is different for law faculty success. At least in football there will be an objective measure of success and an opportunity to cut players. In law school hiring there are no measures and the initial hiring decisions are for lifetime jobs."
Prof. Harrison concludes by opining that likability "sounds like a great approach if you are deciding who you want to go down to the bar with after school for a drink — which sadly may be the standard by which much hiring is done. It's a disaster for the stakeholders of a law school."
In response, Gabriella Montelle wrote "They Like Me, They Like Me Not" (February 12, 2009) at her On Hiring weblog on the Chronicles of Higher Education website. She invited readers to answer two questions:
"Is likability a reasonable consideration in hiring, firing, and tenure decisions or do some committees place too great an emphasis on it? How does it factor into hiring decisions in your department?"
Montelle's piece attracted a variety of responses, and one Comment by a "humanities doctoral candidate" ["HDC"] impressed Louisville U. law dean Jim Chen so much, he turned it into a separate posting at MoneyLaw called "You like me" (Feb. 13, 2009). [Chen's "Rocket man" post over the weekend about the remarkably valuable yet unselfish play of NBA player Shane Battier may also be related, as part of his ongoing talent versus character debate. via Simple Justice]. Commentor HDC's insights included saying:
"The really good scholars are self-confident, and that confidence allows them to treat everyone else with respect and kindness. They are excited about ideas, and they are willing to share. Most of all, they are willing to collaborate — they are the ones organizing symposia, inviting guest speakers, cultivating graduate students, and just generally creating the kind of atmosphere where good work flourishes and everyone benefits.
Meanwhile, Jeff Harrison wrote "But will you love me tomorrow" (Feb. 13, 2009) in answer to Dean Chen, saying that in the faculty context likability or "niceness" is the code for "are you someone with whom I will be socially and politically comfortable." He insists that "Nice in a faculty meeting is only slightly connected to morality, selflessness, or charity." Going back to the football analogy, Harrison concludes:
"If personal social and political comfort are critical in determining who gets an offer to join your faculty, it's like a team thinking more about getting drunk together than winning games."
An anonymous commentor then told Prof. Harrison that the football analogy was not as apt for a faculty as a comparison to a baseball team. Using Barry Bonds as an example, he states:
"In other words, superstars are worthless if they create a bad vibe in the clubhouse. . . . but the point is, good scholars who aren't good colleagues are not worth having around, and whatever is 'good' about their scholarship will be worthless if they aren't the sort of person who can get along with colleagues, train students, and just generally make their work environment a pleasant place to be."
In my experience, HDC and the anonymous commentor have it right. As Jim Harrison suggests, faculty should not be trying to hire or promote only persons who fit within their personal socio-ideological comfort zone. But, they would do well to look for colleagues who match brilliance with unselfishness and congeniality — or, to be more precise, a person who is "genial" in the sense suggested in Merriam-Webster's definition:
3 a: favorable to growth or comfort . . . b: marked by or diffusing sympathy or friendliness
4: displaying or marked by genius
Naturally (this being the cranky old f/k/a Gang speaking), we do not mean "nice" like the smiley-faced gladhanders with gold stars for every student and colleague. Nor do we mean "nice" in Harrison's sense of "just like me," as sameness is boring and intellectual quicksand. Law school faculties need bright minds willing to challenge individuals and institutions, and debate issues of law and policy — but, there is no reason to accept less than respect for eachother and agreeable disagreement. [You need, of course, to respect colleagues and students enough to ask hard questions and expect rigorous thinking.]
Law faculty jobs are far too desirable and desired for us to believe that faculty or students have to put up with jerks and selfish manipulators in order to assure brilliance in scholarship or in the classroom. Because there are more than enough more-than-capable candidates, there should be a preference for the genial over the jerkish. That preference may in fact turn out to be a wonderful tool for behavior modification.
In his posting 2007 "talent versus character," Jim Chen notes how often others have been enablers, willing to justify the odious conduct of a faculty member by saying "He's a smart guy. Brilliant, even." That echoed my assertion that same year that:
[H]aving a high IQ is never an excuse for having a low EQ; it's a reason to demand that our leaders (and our kids) demonstrate and nurture a robust "Emotional Intelligence."
Daniel Goleman introduced most of us to the notion of EQ, in his 1996 bestseller Emotional Intelligence: Why It Can Matter More Than IQ. (well-reviewed here; click for a quick recap of the "Four Components of Emotional Intelligence") . . . I'm still amazed at how many otherwise-sensible people are willing to overlook or excuse the emotional immaturity and ineptness of a colleague, friend or family member (and the harm it causes other people), if the low-EQ is attached to a significantly high IQ — and, especially, if accompanied by a large bank account or a powerful position. I think having a high IQ makes the failure to appreciate, nurture and develop ones EQ rather inexcusable.
It was two years ago this week that we wrote about Robert I Sutton's then-new book "The No Asshole Rule: Building a Civilized Workplace and Surviving One That Isn't" (Warner Business Books, 2007, and an identically-titled article in American Lawyer/Law.com (Feb. 20, 2007). The article explains:
"According to Bob, an asshole is one who oppresses, humiliates, de-energizes, or belittles his target (generally someone less powerful then himself), causing the target to feel worse about herself following an interaction with the asshole. (And, as his examples prove, this behavior is not by any means limited to male perpetrators or female victims.) These jerks use tactics such as personal insults, sarcasm and teasing as vehicles for insults, shaming, and treating people as if they're invisible to demean others. Sutton distinguishes temporary assholes . . . from certified assholes, who routinely show themselves to be nasty people. The latter, he argues, must go [from the workplace]."
A$$holes surely do not belong in law offices (even though many clients think they want such characters to champion their causes). They're even less appropriate in legal academia — especially, when their nasty little show is turned on "impressionable" law students, the very people paying their salaries.
Sutton's book offers a 24-question self-test to see if you are "a certifiable asshole." You can take Sutton's Asshole Rating Self-Exam (ARSE) at Guy Kawasaki's ElectricPulp website. Search and tenure committees might want to ask themselves how their candidates might fare if they took ARSE and answered honestly.
At her Chronicles of Higher Education weblog, Ms. Mentor advised last week that "They're Out to Get Me: No matter how good you are at your work, your colleagues won't keep you if they don't like you" (Feb. 10, 2009). She says this advice is especially important in perilous times like now, when jobs that once seemed secure seem quite shaky; and she asks whether "your colleagues already avoid you as a sour, combative personality — someone who'll waste department energy on vendettas?". I'd like to think that law schools would insist on basic geniality from each of their faculty members in good times, too. In the long run, their "stakeholders" deserve both brilliance and high EQ from every law professor. There are far too many willing candidates to settle for any less.
p.s. Blawging with EQ: If you have a preference for thoroughness and straight-talk, and also wonder who's been writing good material at lawyer weblogs, check out Mark Bennett's Blawg Review #199, at his Defending People blawg.
We can't promise you consistently high EQ here at f/k/a, but we'll try our best. What we do promise is consistently high-quality haiku. For example, here's another installment in our project presenting poems from past issues of Modern Haiku. They're written by poets who later became members of our f/k/a Honored Guest family. Here are more from Modern Haiku Vol. XXVIII: 1 (Winter-Spring 1997), which have not appeared before here at f/k/a:
almost 200 years of air–
George Washington died
…. by Gary Hotham – Modern Haiku Vol. XXVIII: 1
On the boardwalk
a blind man listens to the sea
finding its way back
… by George Swede – Modern Haiku Vol. XXVIII: 1
water splashing down–
the warmth of the sun
on my eyelids
little waterfall–
they come to see
why we're not speaking
pushing in walnuts
with my heel–
… by Lee Gurga – Modern Haiku Vol. XXVIII: 1
fall rains
of mushrooms
tail tucked,
a collie skirts
the bungee jumpers
.. by John Stevenson – Modern Haiku Vol. XXVIII: 1
the darkness deepest
where the snowy owl was
… by Yu Chang – Modern Haiku Vol. XXVIII: 1
Valentine flamingos return to the Stockade [updated]
.. they're back: Lawrence's Valentine Flamingos, 2009 ..
As we reported in detail this time last year in "Lawrence and the Flamingos – a Stockade Valentine mystery," a flock of pink flamingos (genus "phoenicopteris ruber plasticus" ) has been returning each Valentine's Day to the traffic circle home of Lawrence the Indian, at the intersection of Front, Ferry and Green Streets, in Schenectady's Stockade neighborhood. The f/k/a Gang planned to be up at sunrise on Saturday, February 14th, to see whether our Valentine Flamingo miracle would continue in 2009, and to snap some pictures, if it did.
To our surprise, while strolling the neighborhood at sunset tonight, February 13, the "flamboyance" of fourteen flamingos had already landed at the feet of Lawrence. We don't know if the blustery winds blowing the past two days across the Northeast accounts for their premature arrival, but Valentine romantics will have even more time to enjoy this Stockade Valentine tradition.
.. . . . . . . . . . . . . ..
.. sunset, Feb. 13, 2009, photos by D.A. Giacalone ..
The light wasn't great for this amateur photographer to capture the event this evening, but the photos above surely hint at the joy the big pink birds bring to Valentine lovers and Stockade residents each year (thanks to two flamingo shepherds who want to remain anonymous). We promise to take more photos tomorrow in full daylight and add them below, along with a flock of flamingo haiku and senryu. [follow-up: the tradition continues; see suns along the Mohawk, Flamingo Visitation 2011.]
To whet your appetite, here are two haiku written specifically for this year's Stockade Flamingo event by Roberta Beary, our lawyer friend and much-honored haiku poet:
peeking out
of his daughter's blouse
flamingo tattoo
sober now
dad uprights
… by Roberta Beary for f/k/a's Flamingo Flamboyance 2009
[Click to read the Schenectady Gazette's coverage of the 2008 arrival of the flamingos.]
Valentine stroll
neither lover mentions
the pink flamingos
first warm day
she plants
the pink flamingo
.. by ed markowski – Modern Haiku (2008)
. . . . continued (Saturday morning, February 14, 2009):
two pink flamingos
& a waitress named Sally…
summer begins
… by ed markowski
frost on
the flamingo's beak –
Valentine breakfast alone
Snapping photos with near-frozen fingers around 8 AM this morning, I couldn't help but feel a bit of Valentine empathy for poor old Lawrence, standing there like a prop among the flamboyantly romantic flamingos, and gazing longingly again today at the lovely clientele of Arthur's Market. You may recall that our Lawrence statue was originally a carving done by wood carver Samuel Anderson Robb, about 1860, for cigar-store-Indian vendor William Demuth. In DeMuth's 1872 catalog, Lawrence is listed as "No. 53 Indian Chief." Like the shy and proud Kaw-Lija (lyrics) Lawrence "never got a kiss." As Hank Williams sung in 1952:
Kaw-Lija, was a lonely Indian never went nowhere
His heart was set on the Indian maiden with the coal black hair
Kaw-Lija-A, just stood there and never let it show
So she could never answer "YES" or "NO".
Click for a YouTube clip of "Kaw-Lija" (performed by Johnny Cash and Hank Williams, Jr. at the Grand Olde Opry). Please, don't be an "ol' wooden head" like Kaw-Lija and Lawrence — take a risk and let her know you care. Maybe next Valentine will be a little less lonely, and you'll be viewing the Stockade Flamingos hand-in-hand.
on the pink flamingo
.. by ed markowski
cracked wing
parting her pink robe
…………… by Yu Chang, from A New Resonance (1999)
— hurry: you've only got 'til sunset to catch the Valentine flamingos —
pink envelope
Valentine hugs and kisses
from Mom
update (Feb. 15, 2009): The Sunday Albany Times Union has an entertaining article about the Valentine Flamingos. See "Pink flamingos back in Stockade." Reporter Paul Grondahl says:
"Nobody has claimed credit for spawning this quirky urban mystery. Of course, nobody's trying too hard to crack the case and spoil the suspension of disbelief.
"The sheer audacity and cockeyed romanticism of this random act of oddity inspired the first sing-along in front of the flamingos."
No one told the f/k/a Gang to show up to participate or snap a few shots. Nevertheless, you can click to see a YouTube Stockade 2009 Valentine video, with photos by Mabel Leon and Beverly Elander (produced by Jennifer Wells). Due to a technical malfunction, you won't hear zany Stockadians singing Rogers & Hart's "My Funny Valentine," but will have to settle for a performance by Carly Simon and Frank Sinatra.
Another long-legged-avian Valentine tradition: The Heron's Nest Readers' Choice Awards (f/k/a Valentine Awards): Managing Editor John Stevenson announced this morning (Feb. 14, 2009) the winners of the Ninth Annual Readers' Choice Awards, for the best haiku in The Heron's Nest of 2008 (Vol. X, which is also available in a paper edition). Congratulations to all the winners.
Poem of the Year: Fay Aoyagi had the Poem of the Year, which can be seen here. Runners-up honors for best poem went to Burnell Lippy, Christopher Herold, and Harriot West.
Grand Prize, Poet of the Year, went to Burnell Lippy for his consistently fine haiku. Runners-up honors went to Carolyn Hall, Christopher Herold and Gary Hotham. Carolyn and Gary are, of course, f/k/a Honored Guest poets. Carolyn is a perennial winner of haijin awards, and Gary seems to be more active again writing his much-admired poetry for leading haiku journals. For a little Valentine reflection, here are a pair by each of them from The Heron's Nest Vol. X:
enough sunrise —
a small window
in an old hotel
playground swings —
a strong wind replaces
…. by Gary Hotham – The Heron's Nest X (2008)
an eagle sighting —
the frailty
in my father's hug
needles of rain
the talk show guest
addresses my problem
…. by Carolyn Hall – The Heron's Nest X (2008)
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celebrating Lincoln and Darwin
Filed under: Haiku or Senryu,q.s. quickies — David Giacalone @ 12:43 pm
. . Abe Lincoln . . born February 12, 1809 . . Charles Darwin .
My plan to start spending a lot less time weblogging has run smack up against the intriguing coincidence of today's joint birth bicentennial for Abraham Lincoln and Charles Darwin. Some commentators think it's too much of a stretch trying to link the two great men based merely on their birth date. As with the use of juxtaposition in good haiku, however, I've found the comparison — including contrasts, similarities, and differences — to be an interesting and illuminating way to recall what each man has meant to their own times and to ours, and to look at both men in new ways.
There have been two recent books focusing on Darwin and Lincoln together, as unique human beings and towering historical figures:
"Rebel Giants: The Revolutionary Lives of Abraham Lincoln & Charles Darwin"
by David R. Contosta (Prometheus Books, 2008)
"Angels and Ages: A Short Book About Darwin, Lincoln, and Modern Life" by Adam Gopnik (Knopf, 2009)
If you're too busy Twittering, actually working, or merely napping, to find and read either or both of them, I'd suggest making the time for two articles that do a good job with the task of comparing Lincoln and Darwin:
"Who Was More Important: Lincoln or Darwin?" by Malcolm Jones (Newsweek, July 7, 2009)
Illustration: Bryan Christie Design; photos: Corbis … ..
.. "How Lincoln and Darwin Shaped the Modern World" by Adam Gopnik (Smithsonian Magazine, February 2009; illustration by Joe Ciardiello)
Also see the first 16 minutes of Gopnik on Charlie Rose (Feb. 4, 2009; the remainder of the interview is an Appreciation of John Updike)
Lincoln and Darwin are both known for ideas (emancipation and evolution, respectively) that threatened powerful vested interests and deep-held beliefs. The books and articles mentioned above describe the importance of their theories — and the importance of the personal characteristics and social environment of each man in shaping their life's work. I'm going to focus here on another important similarity. As Adam Gopnik says:
"Darwin and Lincoln helped remake our language and forge a new kind of rhetoric that we still respond to in politics and popular science alike. They particularized in everything, and their general vision rises from the details and the nuance, their big ideas from small sightings. They shared logic as a form of eloquence, argument as a style of virtue, close reasoning as a form of uplift. Each, using a kind of technical language—the fine, detailed language of naturalist science for Darwin; the tedious language of legal reasoning for the American—arrived at a new ideal of liberal speech."
In Newsweek, Malcolm Jones wrote:
"Lincoln united the North behind him with an eloquence so timeless that his words remain fresh no matter how many times you read them. Darwin wrote one of the few scientific treatises, maybe the only one, worth reading as a work of literature. Both of them demand to be read in the original, not in paraphrase, because both men are so much in their prose. To read them is to know these elusive figures a little better. Given their influence on our lives, these are men you want to know."
" . . . The quality of Darwin's mind is in evidence everywhere in this book, but so is his character—generous, open-minded and always respectful of those who he knew would disagree with him, as you might expect of a man who was, after all, married to a creationist."
". . . Lincoln, no less than Mark Twain, forged what we think of today as the American style: forthright, rhythmic, muscular, beautiful but never pretty. As Douglas L. Wilson observes in "Lincoln's Sword," his brilliant analysis of the president's writing, Lincoln was political, not literary, but he was, every bit as much as Melville or Thoreau, "perfecting a prose that expressed a uniquely American way of apprehending and ordering experience." What Lincoln says and how he says it are one. You cannot imagine the Gettysburg Address or the Second Inaugural in words other than those in which they are conveyed.
In addition to the notion of their "beautiful but never pretty" writing style, Jones also describes characteristics of the two men that remind me very much of some of my favorite haiku poets:
" . . . Like Darwin, Lincoln was a compulsive scribbler, forever jotting down phrases, notes and ideas on scraps of paper, then squirreling the notes away in a coat pocket, a desk drawer—or sometimes his hat—where they would collect until he found a use for them in a letter, a speech or a document. He was also a compulsive reviser."
Putting yourself in your writing; valuing logic and clear prose; paying attention to details and being open to new ways of looking at the world; and advocating strongly-held beliefs even when some of your closest kith and kin disagree: these are all characteristics worth emulating, as we think of two famous men who were born two hundred years ago today.
p.s. If you're interested, we've written quite a few times on Lawyer Lincoln.
Our time is up for blogging today. We'll leave you with the next installment in our project presenting poems from past issues of Modern Haiku written by poets who later became members of our f/k/a Honored Guest family.
Here are three from Modern Haiku Vol. XXVIII: 1 (Winter-Spring 1997), written by our haiku friend Tom Clausen. We'll delve further into XXVIII:1 over the next few days.
freed from the cat–
baby meadow lark
all speckles
all the panes broken–
in and out of the mill
pigeons fly
garage door open
at the funeral home—
nothing there
… by Tom Clausen – Modern Haiku Vol. XXVIII: 1
cur-mudgeonly valentine
Filed under: Haiga or Haibun,Haiku or Senryu,q.s. quickies — David Giacalone @ 10:27 am
Valentine's Day –
a new sign says
.. Valentine's Day has often brought out the curmudgeonly side of the f/k/a Gang. [see, e.g., our posts "not really in a Valentine mood" and "off-peak romance"] This year, JC Penney's declaration of Doghouse Prevention Week has turned the secretly-romantic Prof. Yabut into a growling cur, rather than a lapdog. Penney's wants men to know that "No Bad Gift Will Go Unpunished," and its Beware of the Doghouse website allows sweethearts to send their guy a warning or even list him as being In the Doghouse. Naturally, in addition to graphic examples of what happens in the doghouse, there are many (expensive) suggestions on how to avoid or get out of Casa Canine.
We are not impressed. Instead, we repeat our contention from 2005 that "Cherries in the Snow" author Emma Forrest makes a very good point:
"Love is so delicate, you can't afford to risk it on fake holiday." (AP/Nashua Telegraph, "British author had no need for Valentine's Day rubbish," Feb. 20, 2005)
All quips aside about stimulus (or stimulated) packages, our economic crisis seems like a perfect opportunity for Valentine lovers (and even spouses) to let each other know it's the thought not the price tag that counts. Indeed, in today's Schenectady Gazette article "Economy tops love this year: Retailers expect recession to cut into Valentine's Day spending" (February 11, 2009), we learn that "Low-cost items this Valentine's are expected to have greater sway over lovers on the prowl for gifts." For example, folks are buying half-pound boxes of candy rather than the larger heart-shaped offerings at Krause's in Colonie. [Sharing fewer calories has many other advantages of course, in a nation where waistlines and bottoms keep expanding, even when the economy shrinks.]
The Gazette also reports that "The National Retail Federation said American adults are expected to spend an average of $102.50 on Valentine's gifts and merchandise, compared with $122.98 a year earlier." In addition,
"BISWorld Research, a Los Angeles market research firm, earlier this month projected holiday card sales to rise over the year by 1.1 percent and candy sales to increase 0.9 percent. But holiday apparel, dining out and jewelry are forecast to take the biggest hits, declining 6.7 percent, 6.1 percent and 5.1 percent, respectively.
More cards and fewer diamonds sounds like a good trend to us. However, if you're heading for the doghouse, we suggest you click to hear Hank Williams' plaintive request that his good dog "Move It on Over" and let the bad dog squeeze in, too.
If you'd like to tell your beloved how you feel in more than one language, click here for "Valentine's Day phrases in 8 languages."
Now's a great time to reprise Roberta Beary's haibun from Modern Haiku (Vol. 39:1, Winter 2008):
What I Mean Is
everyone knows everything old people know only the good die young and kids know parents don't know it all and teachers know students wait until the day before the project is due and you and i both know that love doesn't conquer anything in fact it doesn't even come close
as if it mattered
i pocket
a red leaf
………………………………… by Roberta Beary, Modern Haiku 39:1 (2008)
And a couple of senryu by Ed Markowski:
we do nothing
the sensous curves
of a snow drift
p.s. National Inventors' Day (February 11): If the love of your life loves creativity and service to humanity, Prof. Yabut suggests you remind her (or him) that February 11th is both Thomas A. Edison's birthday and National Inventors' Day. (via Securing Innovation weblog, which has a familiarly-anonymous editor). If you really want to impress her, bring her to the far-too-little-known Edison Exploratorium in downtown Schenectady. The Exploratorium aims to "preserve, promote and celebrate the unique heritage of Edison and the pioneers who gave birth to the electric age here in 'The Original Electric City'." You might get sent to the doghouse for giving her an electric iron, washing machine or microwave oven, but you'll light up her eyes with exhibits filled with those and other items pioneered in Schenectady.
Can't make it to Schenectady? You can find dozens of YouTube clips from the Edison Exploratorium, including one featuring Charles P. Steinmetz, General Electric's Chief Engineer and Scientist (1865 – 1923), who wanted to use inventions like the production and distribution of energy:
"to develop the most perfect civilization the world has ever seen. The civilization not for a minority depending on the labor of masses of slaves or serfs but a real civilization of benefit to all the members of the human race."
…. finally, our lonely-guy 2008 Valentine haiga (photo Mama G. 1951):
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stein and hull and more white lies
Filed under: Haiku or Senryu,q.s. quickies — David Giacalone @ 8:54 pm
. . We didn't "phone this one in," but we woulda if we coulda.
frigid night
the radiator wakes me,
lulls me to sleep
.. How would you have answered Jacob A. Stein's headline question in his newest Legal Spectator column, "Did You Read the Latest Opinion of the Supreme Court?" (Washington Lawyer, February 2009). The sage columnist lists 37 possible answers, but you should feel free to add 38. Other ______________ .
To give you a taste, here's a dozen replies offered by Stein. (The f/k/a Gang are leaning toward #17.)
1. I read about it in Linda Greenhouse's story in The New York Times.
4. I read the headnotes, or was it the syllabus?
5. I read about it in the newsletter we send to clients.
7. Let's just say I flipped through it.
8. I liked that strong language in the dissent.
11. Is it filled with original intent?
17. I usually take it to bed with me. It must be under the covers.
18. Congress will take care of it, just you wait and see.
19. Was it another punitive damages case, or was it the gun case?
30. Why don't they televise the arguments?
36. How much of these opinions do the clerks write?
37. I am going into the hospital for some minor surgery, and I will read it there.
……. What about Dan Hull? Forget the disparaging remarks of Ron Baker, Kevin O'Keefe and Enrico Schaefer, what am I ever going to do about JD Hull. Over at his What About Clients? weblog, Dan insists on saying really nice things about me — or, at least about someone named David Giacalone at f/k/a who bears little or no resemblance to your cranky Editor. For recent examples, see here ("the blawgosphere's spiritual leader David Giacalone . . . is the only sensitive guy we ever liked even a little bit") and there ("spiritual leader and technical adviser in one person . . . Keep reading him"). When I leave comments asking for retractions, Dan forgets to post them. So, let's get a few of things straight:
if you come here for either spiritual or technical advcse, you need a Plan B
we're trying to keep a low profile these days, and our pundit necks are more likely to be in our shells than stuck out very far trying to make waves, pass judgments, or garner links
we've been unsuccessfully striving for haiku- and zen-like humility at f/k/a for almost six years; Hull's hallelujahs are not helping
we know Dan is not merely fishing for mutual compliments, 'cause he's got one of the most-praised weblogs in the entire blawgiverse (see the left-SideBar on his WAC home page); so, we're starting to get paranoid about hidden agendas and anxious about living up to Dan's high expectations. Humbled and honored. But, mostly humbled.
Despite the above lapse in judgment, Dan seems to be on the mark with his list of "The Big Six must-reading lawyer sites" for quickly and efficiently getting "all the news–and new ideas" from lawyer weblogs. He also does a very good job there limning the legal profession's young Slackoiesie.
dust from a slap
on the horse's rump
… by w.f. owen – white lies: RMA 2008
orig pub. Mainichi Daily News #706
Fibs and White Lies: Dan Hull's praise quite naturally reminds us to continue to bring you selections from "white lies: Red Moon Anthology 2008" (by Jim Kacian and the Red Moon Press Editorial Staff, January 2009; see our prior post). "white lies" contains some of the very best haiku-related poetry published in 2008, and our Honored Guest poets are, as always, well-represented.
checkout line
could talk to anyone
writers' conference–
from a toilet stall I hear
someone quoting me
……….. by John Stevenson – white lies: RMA 2008
"checkout line" – Upstate Dim Sum 2008/I
"writers' conference" – Modern Haiku 39:1
a cushion of pine needles
I recall my past
as pleasant
…. by George Swede – white lies: RMA 2008
orig. pub. Acorn 21
between flights
I summarize my life
for a stranger
…. by Hilary Tann – white lies: RMA 2008
orig. pub. Upstate Dim Sum 2008/II
re-prize: Modern Haiku (Summer 1996)
Filed under: Haiku or Senryu — David Giacalone @ 9:56 am
Modern Haiku XXVII:2 (Summer 1996) was published long before HaikuEsq had read his first real haiku poetry. As we said last month, he's going through back issues to find haiku and senryu written by poets who later became members of our f/k/a Honored Guest family.
The following haiku (plus one senryu) were originally published in the Summer 1996 issue of Modern Haiku. Except for the first one by Jim Kacian, this is the first appearance of each poem at f/k/a . . . . .
first autumn wind
not feeling the knife
slice my finger
..…. by Jim Kacian – Modern Haiku XXVII:2
snow that was part
the work gloves off–
wind slips by
my fingers
rain softens
the paper bag–
softer air
she lifts her hands–
water splashes back
on water
window light in the mirror–
a gray hair
among the gray hairs
…. by Gary Hotham– Modern Haiku XXVII:2
conquers
… by David Lanoue – Modern Haiku XXVII:2
[follow the adventures of David's ferret in his novel Haiku Wars; see our post]
Barber's 'Adagio for Strings'
drove right past
the rest area
spring . . .
'HAIKU 1' parked
down by the pond
…. by Lee Gurga – Modern Haiku XXVII:2
in Mohawk's shallows
–prawn on my toes
the train tracks are silent
…. by Yu Chang – Modern Haiku XXVII:2
and magnifying glass
a coin with his birthdate
old slippers
the comfort
coming apart
calling in sick
her own cheerful-sounding
recorded voice
… by John Stevenson – Modern Haiku XXVII:2
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. . . From 2003 to 2009, f/k/a ["formerly known as"] was the home of "breathless punditry" and "one-breath poetry." It is all here in our Archives. You'll find commentary on lawyers and legal ethics, politics, culture, & more, plus "real" haiku by over two dozen Honored Guest Poets.
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"redpajama_set_name": "RedPajamaCommonCrawl"
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\section{Introduction}
\label{sec:introduction}
Interactions of neutral atoms in their ground states with atoms in highly excited Rydberg states attracted considerable scientific attention (\cite{Beigman-rev,Omont1977} and references therein) since Fermi's development of the theory explaining pressure shifts of atomic Rydberg spectral lines \cite{fermi1934}. \textcite{Greene-prl} discovered that the pair of atoms with one of them in the excited state and the other in the ground state (perturber), can form a long-range Rydberg molecules with large internuclear distances and unusual properties (trilobite states) \cite{Granger,Hamilton-PhD,Khuskivadze}. Later, \textcite{Hamilton2002} in their theoretical work predicted that another class of long-range Rydberg molecular states arises when the electron interaction with the perturber exhibits a shape resonance (butterfly states). These theoretical findings were experimentally verified \cite{Greene2006-prl,Bendkowsky2009} and they have been the focus of numerous theoretical \cite{Liu-Rost,Kurz,Junginger,Anderson} and experimental \cite{Vadla,Bendkowsky-prl,Bellos,Anderson-prl,Krupp-prl} investigations. It was found that the temperature and density of the alkali-metal atoms in a Bose-Einstein condensate are particularly favorable for the laser excitation of these long-range molecular Rydberg states.
Initial theoretical papers \cite{Greene-prl,Granger,Hamilton2002} predicted the existence of the trilobite and butterfly states of Rb$_2$ at internuclear distances in the range $10^3-10^5$\thinspace a.u. with energies corresponding to $n\approx30$. Those works are based on an assumption that the interaction of the Rydberg electron with the neutral perturber can be approximated using the Fermi's pseudopotential \cite{fermi1934}.
However, the measurements \cite{Greene2006-prl,Vadla,Bellos} were performed at lower energies corresponding to $n\approx 8\dots11$ and they show similar bound states of Rb$_2$ in this energy range. Calculations of the model potential energy curves (PECs) in this range of energies based on the Fermi's pseudopotential show local minima \cite{Greene-prl} that can be assigned to the measured spectral lines corresponding to the bound states. However, those minima of the PECs are located at internuclear distances $50-200$\thinspace a.u. Those values are considerably below the range $10^3-10^5$\thinspace a.u. for which the theory by \textcite{Greene-prl} was originally developed. Although the qualitative agreement of the theoretical models with the experimental spectra is encouraging, the quantitative validity of the Fermi's pseudopotential at shorter internuclear distances is not entirely clear. The purpose of this paper is to introduce a computational method that is capable of treating this range of smaller internuclear separations ($20 - 420$\thinspace a.u.) and lower energies corresponding to $n\approx8\dots20$ from the first principles. This will allow us to test the validity of Fermi's pseudopotential under these conditions.
It is necessary to mention that more recent experimental works \cite{Bendkowsky-prl,Anderson-prl,Krupp-prl} focus on the energies corresponding to $n\approx35\dots46$. The measured results are in agreement with the theoretical model by \textcite{Greene-prl} applied to Rb$_2$ at internuclear separations $10^3-3\times10^3$\thinspace a.u.
\textcite{Khuskivadze} developed an \textit{ab initio}\xspace method to calculate the PECs and dipole moments of the long-range excited molecular states. In this work the Rydberg atom is represented by the electron in the Coulomb potential, while the effects of the electronic structure of the atomic core are included in the calculation as a correction of the Coulomb wave functions parametrized by the quantum defects \cite{Davydkin}. The positively charged atomic core is placed at the distance $R$ from the neutral perturber that is represented by the short-range model potential in two lowest partial waves. The one-electron Schr\"odinger equation\xspace with both potentials is solved in the sphere centered around the perturber and with a radius that allows the short-range model potential to vanish outside the sphere. The general solutions and their derivatives are then smoothly matched on the sphere with the wave function in the outer region. That matching is performed in terms of the Coulomb Green's function\xspace \cite{Hostler,Hostler1963}, since the only potential considered in the outer region is due to the positive charge of the Rydberg core. This technique guarantees that the correct bound-state asymptotic boundary conditions are satisfied and also that the complicated structure of the wave function in the outer region is treated analytically. In fact, it is necessary to evaluate the Coulomb Green's function\xspace and its radial derivative only on the matching sphere and the wave function elsewhere in the outer region does not explicitly appear in the calculation. Due to the bound-state boundary conditions, the wave function in the inner region can be smoothly matched with the wave function in the outer region only for discrete values of the energy that are identified as the adiabatic energies of the bound states. The method by \textcite{Khuskivadze} takes into account the finite size of the perturber as well as the variation of the Coulomb potential of the other positively charged atomic center throughout the region of the perturber. This effect becomes more relevant with decreasing internuclear separation.
The two-electron computational method developed in this work is a generalization of the approach introduced by \textcite{Khuskivadze}. Particularly, the representation of the perturber and its interaction with the Rydberg atom is more advanced. Since the experimental studies dealing with the long-range molecular states are mainly focused on the systems involving alkali-metal atoms that can be well represented by single electron in the model potential, the approach developed in this paper is also based on that one-electron representation of the perturber.
The interaction of the Rydberg electron with the valence electron of the perturber is a true two-electron repulsion. The additional degree of freedom due to the valence electron allows us to include the effect of the low excited states of the perturber in the theoretical model. It is particularly practical in the studies of systems, where the molecular Rydberg states associated with the second ionization threshold at energies below the first ionization threshold exist. Rb$_2$ is a good example of such system, since the energy of the dissociation channel $\mathrm{Rb}(5p)+\mathrm{Rb}(5p)$ is between the energies of the dissociation channels $\mathrm{Rb}(5s)+\mathrm{Rb}(6p)$ and $\mathrm{Rb}(5s)+\mathrm{Rb}(7p)$ \cite{Spiegelmann}.
The two-electron formulation of the problem also allows us to treat the interaction of the perturber with the Rydberg electron more accurately. This is because the couplings between the Rydberg electron and the valence electron of the neutral perturber are in the present model treated in \textit{ab initio}\xspace manner. Only the closed-shell core of the perturber is represented by a model potential, while in the reference \cite{Khuskivadze} whole the neutral perturber was approximated by the model potential. That approximation required to introduce different model potentials for different spin states (singlet and triplet).
The parameters of the model potential employed in \cite{Khuskivadze} to represent the perturber are optimized to reproduce its electron affinity along with the energies and widths of the low-energy atomic resonances. That interaction potential is restricted to two lowest partial waves (with respect to the center of the perturber), since the low-energy electron interaction with the neutral perturber in higher partial waves is predominated by the repulsive centrifugal barrier that prevents the electron from "probing" the details of the electronic structure of the perturber. However, the presence of another positively charged atomic core can reduce the strength of the repulsive centrifugal barrier and the Rydberg electron may be able to penetrate deeper into the perturber in higher partial waves.
\begin{figure}
\includegraphics{penetscheme.eps}
\caption{Illustration of the model potential of Rb$(5s)$ in the valence region of the perturber (its core is located at $z=0$) along the inernuclear axis $z$ calculated for different internuclear distances $R$ and for the partial waves $l=1,2,3$. The red curve denotes the potential in the absence of the other positively charged atomic core ($R\to\infty$).\label{fig:penetscheme}}
\end{figure}
This mechanism is illustrated in \figref{fig:penetscheme}. It shows the model potential $V_{\text{mod}}(z)$ of Rb$(5s)$ in the presence of another positively charged atomic core calculated along the internuclear axis for several values of the distance $R$:
\begin{equation}
V_{\text{mod}}(z)=
\begin{dcases}
V_{1S}(z)-\frac{1}{\left|z-R\right|}+\frac{l(l+1)}{2z^2}&l=1\vphantom{\frac{0}{0}}\\
-\frac{\alpha_d}{z^4}-\frac{1}{|z-R|}+\frac{l(l+1)}{2z^2}&l=2,3,
\end{dcases}
\end{equation}
where $V_{1S}(r=z)$ is a model potential taken from the reference \cite{Khuskivadze} [Eq. (17)], $\alpha_d=319.2$\thinspace a.u. is the static dipole polarizability of the rubidium atom \cite{Khuskivadze} and $l$ is the angular momentum with respect to the center of the perturber. For the purpose of this qualitative analysis it is sufficient to assume that the atomic potential beyond the core of Rb is predominated by the polarization potential. \figref{fig:penetscheme} shows that if the other positively charged core is sufficiently close, its Coulomb potential lowers the barrier for $l=1$ as well as for $l=2$ so that the electron in a higher Rydberg state can classically penetrate into the valence region. Therefore, the representation of the perturber in $l=2$ at smaller internculear separations is also necessary. On the other hand, the centrifugal barrier for $l=3$ is so repulsive that even when the other core is as close as $R=25$\thinspace a.u. from the center of ther perturber, it is not able to reduce the repulsive barrier enough and the peak of $V_{\text{mod}}(z)$ has a positive value. Therefore, it is difficult for an Rydberg electron in the $f$-wave with any negative energy to penetrate into the perturber space defined by $r\lesssim 7$\thinspace a.u.
Of course, the importance of this effect depends on the size of the perturber and on its distance from the positively charged Rydberg core. As can be seen in \figref{fig:penetscheme}, the model potential employed by \textcite{Khuskivadze} for $l=0$ and $l=1$ is suitable and sufficient for the range of the internuclear separations considered in the reference \cite{Khuskivadze}. The contributions with the higher values of $l$ become relevant only at considerably smaller internuclear distances that are the focus of the present study. Two-electron approach to the problem developed in this paper treats the mechanism described above without a need to construct different perturber potentials for different partial waves.
Another important technical difference between the approach developed in \cite{Khuskivadze} and the present paper is the method used to solve the Schr\"odinger equation\xspace in the inner region. While \textcite{Khuskivadze} perform a direct propagation of the solutions from the center of the sphere to the boundary with careful treatment of the spin-orbit coupling, present work employs the \textsl{R}-matrix\xspace method, where the logarithmic derivative of the wave function on the sphere is calculated and used to ensure the smooth matching to the solution in the outer region. The spin-orbit effects are not considered in the present work.
This paper is focused on the detailed derivation of the two-electron \textsl{R}-matrix\xspace approach to the Rydberg molecular states. A simple application to H$_2$ is discussed and the results are compared with previously published simple models to show that it yields correct energies. Its application to Rb$_2$ and other alkali-metal dimers will be subject of a forthcoming paper.
The rest of this paper is organized as follows: The two-electron computational method is derived in Section \ref{sec:rmatmethod}, the numerical aspects of the calculations are discussed in Section \ref{sec:numeri}, the results obtained using this method for H$_2$ are presented in Section \ref{sec:secresults} and Section \ref{sec:coclusions} contains the conclusions. Atomic units are employed throughout the paper, unless stated otherwise.
\section{Two-electron R-matrix method}
\label{sec:rmatmethod}
Following \textcite{Khuskivadze} and \textcite{Hamilton-PhD}, let us introduce a coordinate system with the center located on the nucleus of the neutral perturber $A$\xspace (\figref{fig:scheme}). The other positively charged center $B$\xspace of the Rydberg atom is located on the negative part of the axis $z$ and its distance from $A$\xspace is denoted as $R$. Axis $z$ is also the quantization axis. Position vectors of the electrons are denoted $\mathbf{r}_1$ and $\mathbf{r}_2$.
\begin{figure}
\includegraphics[scale=1.1]{scheme.eps}
\caption{\label{fig:scheme}Coordinate system used in this work with the center on the nucleus of the perturber $A$\xspace. The nucleus of the Rydberg core $B$\xspace is located at $\mathbf{R}=(0,0,-R)$. Position vectors $\mathbf{r}_1$ and $\mathbf{r}_2$ centered on the origin describe the coordinates of the first and second electron, respectively. Position vectors with respect to the nucleus of the core $B$\xspace are denoted as $\mathbf{r}'_1$ and $\mathbf{r}'_2$.}
\end{figure}
The atomic cores $A$\xspace and $B$\xspace are represented by the spherical model potentials $V_A(r)$ and $V_B(r')$, respectively. The prime in the coordinate denotes that it is taken relatively to the center $B$\xspace, while the unprimed coordinate is relative to the core $A$\xspace. Both $V_A$ and $V_B$ have attractive Coulomb character beyond certain distance from the corresponding core, typically few atomic units. The short-range part of the potential $V_A$ is optimized
to reproduce the energies of the ground and excited states of the perturber with respect to the atomic ionization threshold. Although $V_A(r)$ can be $l$-dependent (see~\cite{Aymar1996} and references therein), such dependence is not needed in this work.
The short-range part of $V_B$ does not explicitly appear in the calculation. Interaction of the Rydberg electron with core $B$ is parametrized by the quantum defects \cite{Davydkin}.
Electronic Hamiltonian of the whole system is
\begin{eqnarray}
\label{eq:totham}
\hat{H}&=&\hat{K}_1+V_A(r_1)+V_B(\left|\mathbf{r}_1-\mathbf{R}\right|)\nonumber\\*
&+&\hat{K}_2+V_A(r_2)+V_B(\left|\mathbf{r}_2-\mathbf{R}\right|)+\frac{1}{r_{12}},
\end{eqnarray}
where $\hat{K}_{1}=-(1/2)\nabla_{1}^2$ and $\hat{K}_{2}=-(1/2)\nabla_{2}^2$ are the operators of the kinetic energy for electrons 1 and 2, respectively. The last term is the Coulomb repulsion of the electrons and $r_{12}=\left|\mathbf{r}_2-\mathbf{r}_1\right|$.
This work deals with such electronic states of the two-electron system, where at least one electron always appears close to the core $A$\xspace (valence electron) and where the internuclear separation $R$ is large enough so that all the region of the valence electron is located in the Coulomb tail of $V_B$ and does not overlap with the core $B$\xspace. The Rydberg electron is asymptotically bound by $V_B$ and its wave function may reach or even exceed the perturber.
Therefore, it is reasonable to introduce the sphere $\mathcal{S}$ centered on the nucleus of the perturber with such radius $r_0$ that whole the region of the space, where the two electrons can appear simultaneously, is confined inside $\mathcal{S}$ (\figref{fig:scheme}).
The basis for the \textsl{R}-matrix\xspace formulation of the problem is provided by the separation of the space to two regions. The region inside the sphere $\mathcal{S}$ accounts for an antisymmetry of the two-electron wave function and the effects of the electron correlation. The region outside the sphere $\mathcal{S}$ deals with a single Rydberg electron and matching of its wave function with the two-electron wave function in the inner region can be done in terms of the Coulomb Green's function\xspace \cite{Hostler1963}.
This imposes the proper boundary conditions and treats the Coulomb singularity analytically. Therefore, it is not necessary to construct any basis set to represent the complicated spatial structure of the one-electron wave function outside $\mathcal{S}$ that spreads over relatively large region of the space.
The restriction that both electrons can simultaneously appear only inside $\mathcal{S}$ and at most one of them around the core $B$\xspace, leads to an asymmetry in the representation of the studied system. It means that even in the case of the homonuclear molecule the computational method discussed here cannot correctly represent the gerade or ungerade symmetry of the two-electron wave function. That may seem like an important deficiency of this computational approach. However, one should keep in mind that it is designed for sufficiently large internuclear distances $R$ and for such energy range, where the gerade and ungerade states of the homonuclear systems form nearly degenerate pairs \cite{Greene-prl,Hamilton2002}. The states where both electrons can simultaneously appear only in the vicinity of the atomic core $A$\xspace, can be represented by the linear combinations of those degenerate wave functions. The situation is different in case of heteronuclear systems, since it is usually obvious, for the given energy range of the interest, which atom is in the valence state and which is in the Rydberg state.
Let us consider now the low bound states of the diatomic cation system as described above, i.e. in the absence of the Rydberg electron. When $R$ is sufficiently large, the bound state is predominately determined by the potential $V_A(r)$ and the effect of the Coulomb tail of $V_B(\left|\mathbf{r}-\mathbf{R}\right|)$ has only perturbative character. Corresponding one-particle Hamiltonian
\begin{equation}
\hat{H}_c=\hat{K}_1+V_A(r_1)+V_B(\left|\mathbf{r}_1-\mathbf{R}\right|)
\label{eq:chanham}
\end{equation}
commutes with the projection of the angular momentum on the internuclear axis $\hat{l}_{1z}$. Therefore, the eigenvalue $m_1$ of $\hat{l}_{1z}$ is a good quantum number and it can be used to categorize the eigenstates of $\hat{H}_c$. Let us assume that $r_0$ is large enough to confine few lowest solutions $\varphi_{im_1}(\mathbf{r}_1)$ of the corresponding Schr\"odinger equation
\begin{equation}
\hat{H}_c\varphi_{im_1}(\mathbf{r}_1)=\epsilon_{im_1}\varphi_{im_1}(\mathbf{r}_1)\label{eq:perchanfuncs}
\end{equation}
with the bound-state boundary conditions. These eigenstates can represent the scattering channels of the problem. The orthonormality of $\varphi_{im_1}(\mathbf{r}_1)$ is assumed in the rest of this paper. When it is desirable to express the parametric dependence of the channel energies on the internuclear separation $R$, it will be denoted as $\epsilon_{im_1}(R)$ in the remaining part of this paper.
\subsection{Inner region}
\label{sec:innerreg}
In order to calculate the \textsl{R} matrix\xspace on the sphere $\mathcal{S}$ variationally, we expand the solutions $\Psi(\mathbf{r}_1,\mathbf{r}_2)$ of the Schr\"odinger equation\xspace
\begin{equation}
\hat{H}\Psi(\mathbf{r}_1,\mathbf{r}_2)=E\Psi(\mathbf{r}_1,\mathbf{r}_2)\label{eq:2esche}
\end{equation}
in the inner region in terms of the basis functions.
In the present study we choose antisymmetric two-electron basis functions coupled to form a state of definite total angular momentum $L$, total electronic spin $S$ and parity $P$. This choice for basis set may seem detrimental since $L$ is not a good quantum number of the diatomic system. However, the only term in the Hamiltonian \eqref{eq:totham} that violates the rotational symmetry is the potential $V_B$ that is a smooth function inside the sphere $\mathcal{S}$. We will show that $V_B$ generates in the inner region a weak $L$-coupling that can be expressed by a finite series expansion of the multipole type. One-particle version of this approach has been utilized by \textcite{Khuskivadze}, where the off-center Coulomb potential couples the eigenstates of the operator $(\hat{L}+\hat{S})^2$, since the $jj$-coupling scheme has been employed by the authors. The theoretical studies of the photoionization spectroscopy of two-electron Rydberg atoms (see~\cite{Aymar1996} and references therein) are based on similar approach, where the dipole electric field couples different eigenstates of $\hat{L}^2$.
Following \textcite{Aymar1996}, the two-electron basis function is expressed in terms of the one-electron radial orbitals and angular spherical harmonics:
\begin{eqnarray}
y_{n_1l_1n_2l_2}^{(LM)}(\mathbf{r}_1,\mathbf{r}_2)&=&C_{n_1l_1n_2l_2}\left[\vphantom{\mathcal{Y}_{l_2l_1}^{(LM)}}u_{n_1l_1}(r_1)f_{n_2l_2}(r_2)\right.\nonumber\\*
\times\mathcal{Y}_{l_1l_2}^{(LM)}(\mathbf{\Omega}_1,\mathbf{\Omega}_2)&+&(-1)^{l_1+l_2+L-S}f_{n_2l_2}(r_1)u_{n_1l_1}(r_2)\nonumber\\*
&\times&\left.\mathcal{Y}_{l_2l_1}^{(LM)}(\mathbf{\Omega}_1,\mathbf{\Omega}_2)\right]/(r_1r_2),\label{eq:2ebasis}
\end{eqnarray}
where $l_1$ and $l_2$ are the angular momenta of the first and second electron, respectively. Eigenvalue $M$ corresponds to the projection of the total angular momentum on the quantization axis $\hat{L}_z$. The indices $n_1$ and $n_2$ represent the one-particle radial orbitals. The set of indices $\{n_1,l_1,n_2,l_2\}$ will be denoted by a joint index $\gamma$ in the rest of this paper. $\mathbf{\Omega}_1=(\theta_1, \phi_1)$ and $\mathbf{\Omega}_2=(\theta_2, \phi_2)$ are the angular coordinates of the first and the second electron, respectively (see \figref{fig:scheme}). The two-electron angular functions $\mathcal{Y}_{l_1l_2}^{(LM)}(\mathbf{\Omega}_1,\mathbf{\Omega}_2)$ are the spherical harmonics in the angular coordinates of both electrons coupled by the Clebsch-Gordon coefficients to form the eigenstates of $\hat{L}^2$ and $\hat{L}_z$, as it is usual in the addition of two angular momenta~\cite{edmonds,Sobelman1972}. $C_{n_1l_1n_2l_2}=1/\sqrt{2(1+\delta_{n_1n_2}\delta_{l_1l_2})}$ is the normalization factor~\cite{Aymar1996}. Since $M$ is the conserved quantum number of the system studied here, it will be skipped from the notation of the basis functions \eqref{eq:2ebasis} and corresponding matrix elements in the rest of the paper. The desired value of $M$ is chosen at the beginning of the calculation as the parameter and it has the same fixed value in all the equations in this paper.
The radial orbitals $u_{nl}(r)$, $n=1\dots N_c$ represent the bound states of the bare neutral perturber, $N_c$ is the number of functions employed in the basis set \eqref{eq:2ebasis} for each value of the angular momentum $l$. One-electron orbitals $u_{nl}(r)$ are calculated variationally as linear combinations of the elements of the $B$-spline\xspace basis set~\cite{Bachau2001}
\begin{equation}
\label{eq:bexcoefs}
u_{nl}(r)=\sum_{\alpha=1}^{N_b}b_{\alpha nl}B_\alpha(r),
\end{equation}
where $B_\alpha(r)$ are the $B$-spline\xspace basis functions defined on the interval $\left[0,r_0\right]$ and $b_{\alpha nl}$ are the expansion coefficients. Substitution of \eqab{eq:bexcoefs} into the radial Schr\"odinger equation\xspace
\begin{subequations}
\label{eq:batorbs}
\begin{eqnarray}
\left[-\frac{1}{2}\frac{d^2}{dr^2}+\frac{l(l+1)}{2r^2}+V_A(r)\right]u_{nl}(r)&=&\varepsilon_{nl}u_{nl}(r)\label{eq:batsche}\\*
u_{nl}(0)&=&0\label{eq:bartorbsbcin}\\*
u_{nl}(r_0)&=&0\label{eq:bartorbsbcout},
\end{eqnarray}
\end{subequations}
its projection onto the basis functions and solution of the obtained generalized eigenvalue problem for each $l$ (since the $B$-splines\xspace form a non-orthogonal basis set) then yields the eigenvalues $\varepsilon_{nl}$ and coefficients $b_{\alpha nl}$. The eigenvectors are normalized so that
\begin{equation}
\int\limits_{0}^{r_0}dr\,u_{nl}(r)u_{n'l}(r)=\delta_{nn'}.
\end{equation}
The radius $r_0$ of $\mathcal{S}$ is chosen large enough so that the effect of the finite box can be neglected for lowest states. This is the usual requirement of the \textsl{R}-matrix\xspace methods.
The set of the radial orbitals $f_{n_2l_2}(r)$ that appear in \eqab{eq:2ebasis} consists of the orbitals $u_{nl_2}(r)$ and it is completed by the set of $B$-splines\xspace as
\begin{equation}
f_{n_2l_2}(r)=
\begin{dcases}
u_{nl_2}(r) & n=1\dots N_c,\;l_2=0,1,\dots\\
B_{\alpha}(r) & \alpha=1\dots N_b
\end{dcases}
.
\end{equation}
Since all the orbitals $u_{nl}(r)$ vanish on the sphere $\mathcal{S}$, they are denoted as ``closed'' in the following text. On the other hand, the last $B$-spline\xspace $B_{N_b}(r)$ has non-zero value at $r=r_0$. Hence, this subset represents the ``open'' orbitals that are neither orthogonal to one another nor to $u_{nl}(r)$. The two-particle basis function \eqref{eq:2ebasis}, where $f_{nl}$ is $u_{nl}$ ($B_n$), is denoted closed (open). The absence of the two-particle basis functions with two electrons in the open orbitals reflects that at most one electron can escape from the reaction volume.
While the open and closed categories of the radial orbitals and two-particle basis functions introduced above have the same meaning as in the reference \cite{Aymar1996}, the character of the open subset is very different. \textcite{Aymar1996} obtained all the radial orbitals by the numerical integration of \eqab{eq:batsche} on the grid. The closed orbitals $u_{nl}(r)$ were calculated in such way that the boundary conditions \eqref{eq:bartorbsbcin} and \eqref{eq:bartorbsbcout} were imposed. These closed orbitals are similar to those introduced in this work, only the method of integration is different. However, the open orbitals in the reference \cite{Aymar1996} were obtained by the numerical integration of the differential equation \eqref{eq:batsche} with the boundary condition in the center \eqref{eq:bartorbsbcin} for arbitrary energy $\varepsilon_{nl}$ without imposing any boundary condition at $r_0$. Of course, the selection of the energy was physically motivated. In contrast to the work by \textcite{Aymar1996}, the open orbitals introduced in this paper do not obey any differential equation and simply form a complete set of functions on the interval $\left[0,r_0\right]$. As a result, the construction of the open orbitals is less ambiguous than in the reference \cite{Aymar1996}, since it does not involve any other parameter. The numerical aspects of this technique are discussed below. Note that similar \textsl{R}-matrix\xspace method to treat two-electron atoms based on the $B$-spline\xspace functions has been developed by \textcite{Hart1997}. \textcite{Zatsarinny} has employed $B$-splines\xspace as a continuum basis set in his \textsl{R}-matrix\xspace calculations of the electron-atom collisions as well. That method and corresponding computer program is capable of treatment of much more complex atoms with many electrons.
Using the basis set \eqref{eq:2ebasis}, the variational principle for the \textsl{R} matrix\xspace can be formulated. We employ the Wigner-Eisenbud pole expansion of the \textsl{R} matrix\xspace \cite{Robicheaux1991, Aymar1996}, where the Bloch operator \cite{Bloch1957}
\begin{equation}
\hat{L}_B=\frac{1}{2r_1}\delta(r_1-r_0)\frac{d}{dr_1}r_1+\frac{1}{2r_2}\delta(r_2-r_0)\frac{d}{dr_2}r_2
\end{equation}
is added to the Hamiltonian \eqref{eq:totham} and the matrix representation $\underline{H}'$ of this modified operator $\hat{H}'=\hat{H}+\hat{L}_B$ in the basis set \eqref{eq:2ebasis} is diagonalized. The matrix elements
\begin{eqnarray}
O_{\gamma\gamma'}^{(LL')}&=&\Braket{y_\gamma^{(L)}|y_{\gamma'}^{(L')}}\label{eq:2eovermatels}\\
{H'}_{\gamma\gamma'}^{(LL')}&=&\Braket{y_\gamma^{(L)}|\hat{H}'|y_{\gamma'}^{(L')}}\label{eq:2ehammatels},
\end{eqnarray}
involve an integration throughout the inner region in both coordinates $\mathbf{r}_1$ and $\mathbf{r}_2$. Both matrices $\underline{H}'$ and $\underline{O}$ are in our calculations organized into blocks defined by corresponding $L$ and $L'$. Each block has a structure corresponding to groupping the basis functions \eqref{eq:2ebasis} into the open and closed subset (see \figref{fig:omatscheme} and \figref{fig:hamscheme}). Note that the non-orthogonality of the open orbitals $f_{nl}(r)$ is transfered to the two-electron basis functions \eqref{eq:2ebasis}. Therefore, the overlap matrix $\underline{O}$ defined by \eqab{eq:2eovermatels} is block diagonal with $L=L'$. The closed-closed parts of the non-zero blocks are formed by the identity matrices $I$ (\figref{fig:omatscheme}).
\begin{figure}
\includegraphics[scale=0.87]{omatscheme-par.eps}
\caption{Block structure of the overlap matrix $\underline{O}$ (see \eqab{eq:2eovermatels}). All the blocks where $L\neq L'$ are zero. The mutual overlaps of the closed basis functions yield the identity matrices $O_{cc}=I$ in the diagonal blocks.\label{fig:omatscheme}}
\end{figure}
Evaluation of the matrix elements \eqref{eq:2ehammatels} for the one-particle operators $\hat{K}_{1,2}$, $V_A$ and $\hat{L}_B$ is straightforward. We first calculate their representation in the $B$-spline\xspace basis and then perform the contraction \eqab{eq:bexcoefs}. The matrix elements in the two-electron basis \eqref{eq:2ebasis} can be expressed in terms of these one-electron elements, since the angular integration is trivial and can be performed analytically. Note that all these operators are diagonal in $L$. Therefore, when matrix $\underline{H}'$ is organized into blocks indexed in rows and columns by $L$ and $L'$, respectively, all the non-zero matrix elements of operators $\hat{K}_{1,2}$, $V_A$ and $\hat{L}_B$ are located only in the diagonal blocks $L=L'$ (see \figref{fig:hamscheme}).
\begin{figure}
\includegraphics[scale=0.87]{hamscheme-par.eps}
\caption{Block structure of the Hamiltonian matrix $\underline{H}'$. All the non-zero matrix elements of the operators $\hat{K}'=\hat{K}_1+\hat{K}_2+\hat{L}_{B}$, $\hat{V}_A=V_A(r_1)+V_A(r_2)$ and $1/r_{12}$ are located only in the diagonal blocks. The non-zero matrix elements of $\hat{V}_B=V_B(r_1')+V_B(r_2')$ are not restricted to any specific blocks. The small scheme on the bottom of the figure shows that every $LL'$-block consists of the sub-matrices corresponding to groupping the basis functions into the open and closed set.\label{fig:hamscheme}}
\end{figure}
The operator of the electron repulsion $1/r_{12}$ is also diagonal in $L$. However, calculation of its matrix elements is computationally more demanding than evaluation of the one-particle terms discussed above. As it is usual in the theory of atomic spectra (for example, see \cite{Sobelman1972,Zatsarinny,Aymar1996} and references therein), the electron repulsion operator is first expressed in terms of the multipole series
\begin{equation}
\frac{1}{\left|\mathbf{r}_1-\mathbf{r}_2\right|}=\sum\limits_{\lambda=0}^{\infty}\frac{4\pi}{2\lambda+1}\frac{r_{<}^\lambda}{r_{>}^{\lambda+1}}\sum\limits_{\mu=-\lambda}^{\lambda}Y^*_{\lambda\mu}(\mathbf{\Omega}_1)Y_{\lambda\mu}(\mathbf{\Omega}_2),
\label{eq:polexpand}
\end{equation}
where $r_<$ ($r_>$) is the smaller (larger) value of $r_1$ and $r_2$, while $Y_{\lambda\mu}(\mathbf{\Omega})$ are the spherical harmonics. The multipole expansion \eqref{eq:polexpand} allows to carry out the angular integration analytically \cite{Sobelman1972}. The radial terms are calculated using the elements
\begin{eqnarray}
R^\lambda(\alpha_1,\alpha_2,\alpha_1',\alpha_2')&=&\int\limits_{0}^{r_0}dr_1\int\limits_{0}^{r_0}dr_2\,B_{\alpha_1}(r_1)B_{\alpha_1'}(r_1)\nonumber\\*
&\times&\frac{r_<^\lambda}{r_>^{\lambda+1}}B_{\alpha_2}(r_2)B_{\alpha_2'}(r_2)
\end{eqnarray}
that are numerically integrated by the cell method proposed by \textcite{Qiu}. These integrals are then contracted with the coefficients $b_{\alpha nl}$ in two, three or four indices, depending on whether the matrix element couples the open-open, open-closed or closed-closed pairs of basis functions \eqref{eq:2ebasis}, respectively. It is this contraction that makes the evaluation of the matrix elements computationally demanding, since in case of the transformation of all four indices the number of necessary multiplications and additions scales with the dimension of the $B$-spline\xspace basis as $N_b^5$ \cite{Diercksen}. Even though the number of orbitals $u_{nl}(r)$ used in construction of the basis functions \eqref{eq:2ebasis} is in practical calculations smaller than the dimension of the $B$-spline\xspace basis ($N_c<N_b$), this step is a bottleneck in the numerical construction of $\underline{H}'$. The same type of transformation is also employed in the \textit{ab-initio} programs for the molecular electronic structure calculations, when the four-index integrals calculated in the atomic basis are transformed to the molecular basis \cite{Diercksen}.
Note that thanks to the independence of the two-electron basis functions \eqref{eq:2ebasis} on $R$, the matrix elements of all the operators discussed above do not depend on the internuclear separation $R$. Therefore, they are calculated only once and used in calculations of the energy spectra at each internuclear separation of the interest.
The only part of the Hamiltonian $\hat{H}$ that depends on $R$, is the potential $V_B(r')$ of the Rydberg atomic core $B$. However, the evaluation of corresponding matrix elements in basis set \eqref{eq:2ebasis} is computationally tractable compared to the matrix elements of the electron-electron repulsion. Assuming that whole the atomic core $B$\xspace is in the outer region, the potential $V_B(r')$ inside the sphere $\mathcal{S}$ has form $V_B(\left|\mathbf{r}-\mathbf{R}\right|)=-1/\left|\mathbf{r}-\mathbf{R}\right|$ and its corresponding matrix elements can be calculated using the pole expansion \eqref{eq:polexpand} as it is explained in Appendix \ref{sec:append}.
Having the Hamiltonian matrix (modified by the Bloch term) $\underline{H}'$ and the overlap matrix $\underline{O}$, the \textsl{R} matrix\xspace can be expressed in terms of the solutions of the generalized eigenvalue problem \cite{Aymar1996, Robicheaux1991, Zatsarinny}
\begin{equation}
\underline{H}'\mathbf{c}_k=E_k\underline{O}\mathbf{c}_k,
\label{eq:geneigen}
\end{equation}
where $E_k$ are the eigenvalues and $\mathbf{c}_k$ are corresponding (column) eigenvectors normalized by the overlap matrix to \cite{Robicheaux1991}
\begin{equation}
\mathbf{c}_k^T\underline{O}\mathbf{c}_{k'}=\delta_{kk'}.
\end{equation}
In order to construct the \textsl{R} matrix\xspace, let us first express the general solutions of the Schr\"odinger equation\xspace \eqref{eq:2esche} in the outer region as a channel expansion
\begin{equation}
\left.\Psi_\beta(\mathbf{r}_1,\mathbf{r}_2)\right|_{r_2\geq r_0}=\sum\limits_{im_1}\varphi_{im_1}(\mathbf{r}_1)Q_{im_1\beta}(\mathbf{r}_2),\label{eq:chanexpand}
\end{equation}
where $\beta$ denotes different independent solutions at the same energy $E$. These solutions usually obey some specific boundary conditions in the outer region (i.e. bound state, linear combination of the incoming and outgoing wave, etc.). When we denote $m_2\equiv M-m_1$ and the scattering wave function $Q_{im_1\beta}(\mathbf{r}_2)$ in the outer region is expressed as
\begin{equation}
\left.Q_{im_1\beta}(\mathbf{r}_2)\right|_{r_2\geq r_0}=\sum\limits_{l_2=\left|m_2\right|}^\infty\frac{1}{r_2}q_{\bar{j}\beta}(r_2)Y_{l_2m_2}(\mathbf{\Omega}_2),
\label{eq:fexpand}
\end{equation}
where $\bar{j}=\{i,m_1,l_2\}$ is the scattering channel, then the \textsl{R} matrix\xspace couples $q_{\bar{j}\beta}(r_0)$ evaluated on the sphere $\mathcal{S}$ with their radial derivatives
\begin{equation}
q_{\bar{j}\beta}(r_0)=\sum\limits_{\bar{j}'}R_{\bar{j}\bar{j}'}q_{\bar{j}'\beta}'(r_0).
\label{eq:rmatcond}
\end{equation}
The notation $q_{\bar{j}'\beta}'(r_0)$ for the derivative of the radial wave function $q_{\bar{j}'\beta}(r)$ evaluated on the sphere $\mathcal{S}$ introduced in \eqab{eq:rmatcond} will be utilized in the remaining part of this paper also for other wave functions.
The elements of the \textsl{R} matrix\xspace can be expressed as \cite{Aymar1996,Robicheaux1991}
\begin{equation}
R_{\bar{j}\bar{j}'}(E)=\frac{1}{2}\sum\limits_{k}\frac{w_{\bar{j}k}w_{\bar{j}'k}}{E_k-E}.
\label{eq:rmatpolex}
\end{equation}
The surface amplitudes $w_{\bar{j}k}$ can be calculated using the eigenvectors $\mathbf{c}_k$. Their explicit form is derived in Appendix \ref{sec:amplider}. The channel index $\bar{j}$ characterizes the valence state of the perturber as well as the partial wave of the electron that can escape from the inner region. Universal validity of the boundary condition \eqref{eq:rmatcond} is the strong side of the \textsl{R}-matrix\xspace method, as the set of solutions $q_{\bar{j}\beta}(r_0)$ in \eqab{eq:rmatcond} is not restricted by any other boundary condition on $\mathcal{S}$ nor in the outer region.
The solution of the generalized eigenvalue problem \eqref{eq:geneigen} is numerically most demanding step of all the calculation. It is necessary to perform a complete diagonalization of the modified Hamiltonian $\underline{H}'$ as all the eigenvalues and eigenvectors are required to construct the \textsl{R} matrix\xspace \eqref{eq:rmatpolex}. Apparently, it is necessary to perform this step for each internuclear distance $R$ of the interest. Note, however, that once having the eigenrepresentation of $\underline{H}'$, the dependence of the pole expansion of the \textsl{R} matrix\xspace on the energy \eqref{eq:rmatpolex} is very simple. This is also the main difference between the present method and another very powerful technique - the eigenchannel \textsl{R} matrix\xspace developed by \textcite{Greene1988}, where the \textsl{R} matrix\xspace can be evaluated by solving a system of linear algebraic equations \cite{Aymar1996}, however, the set of linear equations must be solved for each energy of the interest. Therefore, the method employed here is more convenient, since the search for the bound states is performed on relatively fine grid of energies.
\subsection{Outer region}
\label{sec:outerreg}
One-electron equations (for the coordinate $\mathbf{r}_2$) in the outer region can be obtained by substitution of the solution \eqref{eq:chanexpand} into the Schr\"odinger equation\xspace \eqref{eq:2esche}. Resulting equation is then multiplied by the wave functions $\varphi_{im_1}^*(\mathbf{r}_1)$ and integrated over the coordinate $\mathbf{r}_1$ throughout the inner region. We arrive at the coupled set of differential equations
\begin{eqnarray}
\left[\hat{K}_2\right.&+&\left.\vphantom{\hat{K}_2+}V_B(\left|\mathbf{r}_2-\mathbf{R}\right|)-(E-\epsilon_{im_1})\right]Q_{im_1\beta}(\mathbf{r}_2)\nonumber\\*
&=&\sum_{i'm_1'}U_{im_1i'm_1'}(\mathbf{r}_2)Q_{i'm_1'\beta}(\mathbf{r}_2),\label{eq:outersche}
\end{eqnarray}
where the coupling potential has the following form:
\begin{eqnarray}
U_{im_1i'm_1'}(\mathbf{r}_2)=&-&\int_{\mathcal{V}}dV_1\,\varphi_{im_1}^*(\mathbf{r}_1)\frac{1}{r_{12}}\varphi_{i'm_1'}(\mathbf{r}_1)\nonumber\\*
&+&\frac{1}{r_2}\delta_{ii'}\delta_{m_1m_1'}.\label{eq:outerpoten}
\end{eqnarray}
The domain of integration $\mathcal{V}$ in \eqab{eq:outerpoten} is the volume enclosed by $\mathcal{S}$. The second term reflects the fact the $V_A(r)=-1/r$ for $r>r_0$.
The first term in \eqab{eq:outerpoten} represents the repulsion of the Rydberg electron with the valence electron confined inside the sphere $\mathcal{S}$. The second term describes the attraction between the electron in the outer region with the core $A$\xspace. When we substitute the multipole expansion \eqref{eq:polexpand} into the first term, we immediately see that the first (charge) term of the expansion cancels with the second term of \eqab{eq:outerpoten}. The first non-vanishing term of \eqab{eq:outerpoten} is the dipole potential. It does not vanish because the wave functions $\varphi_{im_1}(\mathbf{r}_1)$ are not typical atomic states. They are eigenstates of the atomic Hamiltonian perturbed by the Coulomb tail of the potential $V_B$ \eqref{eq:chanham}. Hence the non-vanishing dipole term is a consequence of the polarization of the valence electron by the core $B$\xspace. Explicit form of this induced dipole moment is given in Appendix \ref{sec:dipappend}. The induced dipole field was previously employed to model the polarization of the neutral perturber \cite{Khuskivadze} or its core \cite{bottcher, Henriet1984} by other positively charged atomic center. In Section \ref{sec:numeri} we will analyze its magnitude and its dependence on the internuclear distance $R$. Our goal is to provide a numerical evidence that will allow us to neglect the coupling potential on the right-hand side of \eqab{eq:outersche} outside the sphere $\mathcal{S}$.
Provided that the right-hand side of \eqab{eq:outersche} can be neglected, the remaining system of differential equations is not coupled in the indices $i$, $m_1$ and $\beta$. The bound-state energies of the studied system can be found by solving the homogeneous part of \eqab{eq:outersche} for $r_2>r_0$ with the boundary condition \eqref{eq:rmatcond} on the sphere $\mathcal{S}$. Only those solutions that vanish asymptotically are chosen. The two boundary conditions can be satisfied simultaneously only for a discrete set of energies. Since the coordinates of the valence electron do not appear in the rest of this section, the index of the electron position vector in the outer region is dropped and the coordinates are denoted as $\mathbf{r}=(r,\theta,\phi)$. Any solution $X_{im_1}(\mathbf{r})$ that obeys all the desired bound-state boundary conditions can be expressed as a linear combination of the complete set of solutions $Q_{im_1\beta}(\mathbf{r})$ of \eqab{eq:fexpand} as
\begin{equation}
X_{im_1}(\mathbf{r})=\sum_{\beta}A_\beta Q_{im_1\beta}(\mathbf{r})\label{eq:bsradwfacomb}
\end{equation}
and the relation of the radial components $x_{\bar{j}}(r)$ defined by the equation
\begin{equation}
X_{im_1}(\mathbf{r})=\sum_{l_2=\left|m_2\right|}^\infty\frac{x_{\bar{j}}(r)}{r}Y_{l_2m_2}(\mathbf{\Omega})
\label{eq:outerboundradex}
\end{equation}
to the general radial solutions \eqref{eq:fexpand} is
\begin{equation}
x_{\bar{j}}(r)=\sum_{\beta}A_\beta q_{\bar{j}\beta}(r).
\label{eq:aradex}
\end{equation}
The bound-state solutions $X_{im_1}(\mathbf{r})$ satisfy the homogeneous part of \eqab{eq:outersche}, i.e.
\begin{equation}
\left[-\frac{1}{2}\nabla^2+V_B(\left|\mathbf{r}-\mathbf{R}\right|)-(E-\epsilon_{im_1})\right]X_{im_1}(\mathbf{r})=0.
\label{eq:nextoutersche}
\end{equation}
Methods based on the partial-wave expansion of the solution with respect to the center of the perturber $A$\xspace, would be, due to the singularity of $V_B$ at the position of the Rydberg core $B$\xspace, unpractical and computationally very demanding. Instead, we employ a straightforward multichannel generalization of the treatment developed by \textcite{Khuskivadze} that is based on the Green's function\xspace, originally formulated by \textcite{Fabrikant1993}. Consider the Green's function\xspace $G(\mathbf{r},\mathbf{r}'')$ for the potential $V_B$ defined by the equation
\begin{eqnarray}
\left[-\frac{1}{2}\nabla^2\vphantom{V_B(\left|\mathbf{r}-\mathbf{R}\right|)-\frac{k_{im_1}^2}{2}}\right.&+&\left.\vphantom{-\frac{1}{2}\nabla^2}V_B(\left|\mathbf{r}-\mathbf{R}\right|)-\frac{k_{im_1}^2}{2}\right]\nonumber\\*
&\times&G_{im_1}(\mathbf{r},\mathbf{r}'')=-\delta^3(\mathbf{r}-\mathbf{r}''),\label{eq:outergfe}
\end{eqnarray}
where $k_{im_1}=\sqrt{2(E-\epsilon_{im_1})}$. We are interested in such $G_{im_1}(\mathbf{r},\mathbf{r}'')$ that satisfies the asymptotic bound-state boundary conditions. If the Rydberg core $B$\xspace is the hydrogen core, then $G_{im_1}(\mathbf{r},\mathbf{r}'')$ is the Coulomb Green's function\xspace derived in the closed form by \textcite{Hostler1963}. For more complex positive ions, a correction of the Coulomb Green's function\xspace developed by \textcite{Davydkin} is added \cite{Khuskivadze} to account for the non-hydrogen character of the core. That correction is then parametrized by the corresponding atomic quantum defects. When we multiply \eqab{eq:outersche} by $G_{im_1}(\mathbf{r},\mathbf{r}'')$, subtract it from \eqab{eq:outergfe} multiplied by $X_{im_1}(\mathbf{r})$ and then integrate it over $\mathbf{r}$, we arrive at
\begin{eqnarray}
&-&\frac{1}{2}\int_{\mathcal{S}_0}d\Omega\left[G_{im_1}(\mathbf{r},\mathbf{r}'')\nabla^2X_{im_1}(\mathbf{r})-X_{im_1}(\mathbf{r})\right.\nonumber\\*
&\times&\left.\nabla^2G_{im_1}(\mathbf{r},\mathbf{r}'')\right]=\int_{\mathcal{S}_0}d\Omega\,\delta^3(\mathbf{r}-\mathbf{r}'')X_{im_1}(\mathbf{r}),
\label{eq:outvolint}
\end{eqnarray}
where the domain of the integration $\mathcal{S}_0$ is the unit sphere. The Green's function\xspace can be expanded in terms of the spherical harmonics in the angular coordinates as
\begin{eqnarray}
G_{im_1}(\mathbf{r},\mathbf{r}'')&=&\sum_{\lambda=0}^\infty\sum_{\mu=-\lambda}^{\lambda}\sum_{\lambda'=\left|\mu\right|}^\infty\frac{g_{im_1\lambda\lambda'\mu}(r,r'')}{rr''}\nonumber\\*
&\times&Y_{\lambda\mu}^*(\mathbf{\Omega})Y_{\lambda'\mu}(\mathbf{\Omega}''),
\label{eq:gfsphex}
\end{eqnarray}
where $\mathbf{\Omega}''=(\theta'',\phi'')$ are the angular coordinates of $\mathbf{r}''$ and the symmetry with respect to the rotations around the $z$-axis has been utilized \cite{Khuskivadze}. Substitution of this expansion along with \eqab{eq:outerboundradex} into \eqab{eq:outvolint} and integration over $\mathbf{\Omega}$ yields
\begin{eqnarray}
&-&\frac{1}{2}\sum_{\substack{\lambda'\\l_2\in\bar{j}}}\left[g_{\bar{j}\lambda'}(r,r'')\frac{d^2}{dr^2}x_{\bar{j}}(r)-x_{\bar{j}}(r)\frac{\partial^2}{\partial r^2}g_{\bar{j}\lambda'}(r,r'')\right]\nonumber\\*
&\times&Y_{\lambda'm_2}(\mathbf{\Omega}'')=\sum_{l_2}Y_{l_2m_2}(\mathbf{\Omega}'')x_{\bar{j}}(r)\delta(r-r'').\label{eq:angulint}
\end{eqnarray}
The only terms of expansion \eqref{eq:gfsphex} that remain in \eqab{eq:angulint} after the angular integration, are those where $\mu=m_2=M-m_1$ and $\lambda=l_2$. In order to keep the notation simple, index $\mu$ in $g_{im_1\lambda\lambda'\mu}(r,r'')$ was dropped and leading three indices were again merged into the multi-index $\bar{j}=\left\{i,m_1,l_2\right\}$ denoting the scattering channel.
In the outer region ($r>r_0$) the right-hand side of \eqab{eq:angulint} vanishes in the limit $r''\to r_0-$. Both $x_{\bar{j}}(r)$ and $g_{im_1\lambda\lambda'\mu}(r,r'')$ vanish along with their first derivatives for $r\to\infty$ \cite{Hostler1963}. Moreover, the radial integration of \eqab{eq:angulint} can be performed. Subsequent projection on $Y_{\lambda m_2}(\mathbf{\Omega}'')$ then yields
\begin{equation}
\sum\limits_{l_2\in\bar{j}}\left[g_{\bar{j}\lambda}(r_0,r_0)x_{\bar{j}}'(r_0)-x_{\bar{j}}(r_0)g_{\bar{j}\lambda}'(r,r_0)\right]=0,
\end{equation}
where
\begin{equation}
g_{\bar{j}\lambda}'(r_0,r_0)=\lim\limits_{r''\to r_0-}\left.\frac{\partial}{\partial r}\right|_{r=r_0}g_{\bar{j}\lambda}(r,r'').
\end{equation}
Boundary condition \eqref{eq:rmatcond} can be employed at this point to establish the relation between the solutions on the sphere and their derivative in order to smoothly match the solutions in the inner region with the solutions outside $\mathcal{S}$:
\begin{equation}
\sum\limits_{\substack{\bar{j}'\\l_2\in\bar{j}}}\left[g_{\bar{j}\lambda}(r_0,r_0)\delta_{\bar{j}\bar{j}'}-R_{\bar{j}\bar{j}'}g_{\bar{j}\lambda}'(r_0,r_0)\right]
x_{\bar{j}'}'(r_0)=0.\label{eq:bselcond}
\end{equation}
\eqab{eq:bselcond} represents a necessary condition for the set of the radial coefficients $x_{\bar{j}}(r_0)$ to form a bound-state wave function via \eqab{eq:bsradwfacomb} and to possess a fixed logarithmic derivative (or \textsl{R} matrix\xspace $R_{\bar{j}\bar{j'}}$) on the sphere $\mathcal{S}$. The coefficients $x_{\bar{j}}(r_0)$ are in turn expressed as linear combinations of the general solutions $q_{\bar{j}\beta}(r_0)$ (see \eqab{eq:aradex}) that satisfy the boundary condition \eqref{eq:rmatcond} and possess boundary condition of any choice in the outer region. It is convenient to fix the boundary conditions by choosing the derivatives on the sphere $\mathcal{S}$
\begin{equation}
q_{\bar{j}\beta}'(r_0)=\delta_{\bar{j}\beta}\label{eq:outbasdiag}.
\end{equation}
Substitution of \eqab{eq:aradex} and \eqab{eq:outbasdiag} into \eqab{eq:bselcond} yields the system of linear equations for coefficients $A_\beta$
\begin{equation}
\sum\limits_\beta M_{\bar{j}\beta}A_\beta=0,
\end{equation}
where the matrix elements of $\underline{M}$ are the expressions in the square brackets in \eqab{eq:bselcond}. More specifically, introducing the matrices $\underline{\Gamma}$ and $\underline{\Gamma}'$ as
\begin{subequations}
\label{eq:gmatrices}
\begin{eqnarray}
\Gamma_{\bar{j}\bar{j'}}&=&\int_{\mathcal{S}}r_0^3\,d\Omega\int_{\mathcal{S}}r_0^3\,d\Omega''\,G_{im_1}(\mathbf{r},\mathbf{r}'')\nonumber\\*
&\times&Y_{l_2'm_2}^*(\mathbf{\Omega}'')Y_{l_2m_2}(\mathbf{\Omega})\delta_{ii'}\delta_{m_1m_1'}=\nonumber\\*
&=&\delta_{ii'}\delta_{m_1m_1'}g_{\bar{j}l_2'}(r_0,r_0)\\
\Gamma_{\bar{j}\bar{j'}}'&=&\int_{\mathcal{S}}r_0^2\,d\Omega\int_{\mathcal{S}}r_0^3\,d\Omega''\,\lim\limits_{r''\to r_0-}\left.\frac{\partial}{\partial r}\right|_{r_0}\left[rG_{im_1}(\mathbf{r},\mathbf{r}'')\right]\nonumber\\*
&\times&Y_{l_2'm_2}^*(\mathbf{\Omega}'')Y_{l_2m_2}(\mathbf{\Omega})\delta_{ii'}\delta_{m_1m_1'}=\nonumber\\*
&=&\delta_{ii'}\delta_{m_1m_1'}g_{\bar{j}l_2'}'(r_0,r_0),
\end{eqnarray}
\end{subequations}
matrix $\underline{M}$ can be written as
\begin{equation}
\underline{M}=\underline{\Gamma}-\underline{\Gamma'}\,\underline{R}
\end{equation}
and the energies of the bound states can be found by solving the equation $\det(\underline{M})=0$.
Computationally most demanding step in the treatment of the outer region is the construction of matrices $\underline{\Gamma}$ and $\underline{\Gamma}'$. Although the angular integrals in \eqab{eq:gmatrices} can be reduced to three dimensions \cite{Khuskivadze,Hostler1963}, evaluation of the Green's function\xspace along with its normal derivative is the bottleneck of the outer region calculation. It is necessary to evaluate it on the grid of the energies $(E-\epsilon_{im_1})$ for every state of the perturber included in the calculation and for every internuclear distance $R$ of interest. However, these calculations can be efficiently parallelized.
On the contrary to the method discussed here, \textcite{Khuskivadze} do not use the \textsl{R} matrix\xspace to match the wave functions on the sphere $\mathcal{S}$. Instead, they numerically integrated solutions $q_l(r)$ defined by the single-channel version of \eqab{eq:fexpand}. Another approach was presented by \textcite{Hamilton-PhD} who parametrized the solution in the inner region by the phase shifts and introduced the model relation between the local electron momentum and the total energy of the bound state.
\section{Numerical aspects of the calculations}
\label{sec:numeri}
As explained in Section \ref{sec:innerreg}, the two-electron basis set \eqref{eq:2ebasis} consists of the closed and open subsets
\begin{subequations}
\begin{eqnarray}
y_{n_1l_1n_2l_2}^{(L)\mathrm{C}}(\mathbf{r}_1,\mathbf{r}_2)&=&\frac{C_\gamma}{r_1r_2}\hat{A}\left[\vphantom{\mathcal{Y}_{l_1l_2}^{(L)}(\mathbf{\Omega}_1,\mathbf{\Omega}_2)}u_{n_1l_1}(r_1)u_{n_2l_2}(r_2)\right.\nonumber\\*
&\times&\left.\mathcal{Y}_{l_1l_2}^{(L)}(\mathbf{\Omega}_1,\mathbf{\Omega}_2)\right],\label{eq:2eclosedsub}\\
y_{n_1'l_1'n_2'l_2'}^{(L')\mathrm{O}}(\mathbf{r}_1,\mathbf{r}_2)&=&\frac{C_{\gamma'}}{r_1r_2}\hat{A}\left[\vphantom{\mathcal{Y}_{l_1'l_2'}^{(L')}(\mathbf{\Omega}_1,\mathbf{\Omega}_2)}u_{n_1'l_1'}(r_1)B_{\alpha}(r_2)\right.\nonumber\\*
&\times&\left.\mathcal{Y}_{l_1'l_2'}^{(L')}(\mathbf{\Omega}_1,\mathbf{\Omega}_2)\right]\label{eq:2eopensub},
\end{eqnarray}
\end{subequations}
where $\hat{A}$ provides the spin-adapted antisymmetrization of the expression as shown in \eqab{eq:2ebasis}.
If the complete set of the closed orbitals $u_{n_1l_1}(r)$ ($n_1=1\dots N_b-1$ for each $l_1$) was used in the open subset \eqref{eq:2eopensub} of the two-particle basis, the closed basis functions \eqref{eq:2eclosedsub} would be redundant, provided that the set of $B$-splines\xspace is complete. However, such two-electron basis set would be very large and the \textsl{R}-matrix\xspace calculations would not be numerically tractable. In practice, we need only several functions $u_{n_1'l_1'}(r_1)$ to construct the channel functions $\varphi_{im_1}(\mathbf{r}_1)$ defined by \eqab{eq:perchanfuncs}. However, such restricted set of $u_{n_1'l_1'}(r_1)$ combined with the complete set of $B$-splines\xspace in \eqab{eq:2eopensub} leads to an insufficient and asymmetric treatment of the two-electron correlation inside $\mathcal{S}$. We solve this dilemma by introducing the closed part of the basis \eqref{eq:2eclosedsub} in which the one-electron orbitals $u_{n_1l_1}(r_1)$ and $u_{n_2l_2}(r_2)$ span identical sets. The closed part is responsible for the convergence of the correlation interaction inside the sphere $\mathcal{S}$. The open part \eqref{eq:2eopensub} allows one electron to escape the sphere $\mathcal{S}$ and usually requires far fewer one-electron orbitals $u_{n_1'l_1'}(r_1)$.
Apparently, present approach leads to a linear dependence between the open and closed basis function because some of the closed basis elements \eqref{eq:2eclosedsub} can be expressed as proper linear combinations of the open basis functions \eqref{eq:2eopensub}. Some of the molecular \textsl{R}-matrix\xspace methods \cite{tennyson-rev}, based on different one-particle basis sets than the approach presented here, avoid this dependence by prior orthogonalization of the open radial orbitals $B_\alpha(r_2)$ in \eqab{eq:2eopensub} to the set of $u_{n_2l_2}(r_2)$ selected in \eqab{eq:2eclosedsub}. However, the issue of the overcompleteness of the two-electron basis can be easily solved. First, the singular overlap matrix $\underline{O}$ defined by \eqab{eq:2eovermatels} is diagonalized and its kernel is recognized as the set of eigenvectors corresponding to the zero eigenvalues. The orthonormal projection matrix into the regular subspace is constructed using the non-zero eigenvalues and corresponding eigenvectors \cite{Mayer2002}. Afterwards, the projection of the generalized eigenvalue problem \eqref{eq:geneigen} using that matrix is performed. In addition to elimination of the linear dependence from the two-particle basis set, this procedure reduces the overlap matrix to the identity matrix. The reduction of the generalized eigenvalue problem to the standard one makes its computation more memory efficient because it is not necessary to store the overlap matrix.
Thanks to the block diagonal structure of $\underline{O}$ (see \figref{fig:omatscheme}), its diagonalization can be performed block-wise as well as the projection of $\underline{H}'$ (see \figref{fig:hamscheme}). The dimension of the projected Hamiltonian matrix in the calculations presented here is approximately 10\% smaller than the dimension of the basis set before the projection. Removal of the redundant two-electron space is numerically well defined procedure. In our calculations, all the eigenvalues of the subspace to be removed were at least five orders of magnitude smaller than the rest of the spectrum.
The angular component of each basis function \eqref{eq:2ebasis} is characterized by the quantum numbers $l_1$, $l_2$, $l_1'$, $l_2'$, $L$ and $L'$ and their upper limits influence the convergence of the calculation. The range of $l_2'$ in the set of the open functions \eqref{eq:2eopensub} must be large enough to achieve a sufficient angular resolution of the wave function on the sphere $\mathcal{S}$ and our tests showed that partial waves up to $l_2'=4$ is necessary to achieve converged results in case of the hydrogen molecule. The range of $l_1$ and $l_2$ available in the subset of the closed functions \eqref{eq:2eclosedsub} must be large enough to allow a correct representation of the complicated structure of the correlation cusps of the two-electron wave function in the inner region. The functions with $l_{1,2}\leq4$ were used in the calculations discussed below as a result of the requirement that the two-particle basis set must reproduce the electron affinity of the hydrogen atom with accuracy better than 1\thinspace meV. The requirement on $l_1'$ in the open two-particle basis functions is that a sufficient angular basis for the expansion of few lowest wave functions $\varphi_{im}(\mathbf{r}_1)$ in terms of the spherical harmonics must be provided. $l_1'\leq1$ was sufficient in the calculations presented below.
No constraint is applied the total angular momentum $L$. Therefore, for given pair $l_1$, $l_2$, the two-particle functions $y_\gamma^{(L)}$ for each $\left|l_1-l_2\right|\leq L\leq l_1+l_2$ are included in the basis set.
The radial components of the two-electron basis functions are characterized by indices $n_1$, $n_2$, $n_1'$ and $n_2'$. The set of 90 $B$-splines\xspace of the sixth order was used to obtain the results presented below and the radius of the sphere $\mathcal{S}$ was set to $r_0=10\thinspace$ a.u. The range of $n_1$ in the open subset is usually relatively small ($n_1'\leq2$ in the calculations discussed in this work), since usually only few lowest channels $\varphi_{im_1}(\mathbf{r}_1)$ are involved in the calculations. Even those are very weakly perturbed atomic states of the perturber. Therefore, their expansions in terms of $u_{n_1l_1}(r_1)$ have one dominant term for each $l_1$ and few small corrections. Only the ground state of the perturber influenced by the core $B$\xspace was included in the calculations presented here. Considerably larger sets of the radial orbitals are necessary in the closed subsets of $y_\gamma^{(L)}$ to treat the correlation effects of the electron-electron interaction. The converged results presented below were obtained for $n_1,n_2=1\dots20$. All the angular coupling coefficients were calculated using the routines by \textcite{Wei1999} that provide accurate values for higher angular momenta.
A partially parallelized computer implementation of the method discussed in Section \ref{sec:rmatmethod} was developed and executed on the computer cluster with 60 central processing unit (CPU) cores available for this project. Evaluation of every diagonal $L$-block of the modified Hamiltonian matrix $\underline{H}'$ (see \figref{fig:hamscheme}) in the basis set discussed above required approximately 10\thinspace GB of the computer memory, predominately due to the transformation of the matrix elements of the operator $1/r_{12}$ from the $B$-spline\xspace basis to the two-electron basis \eqref{eq:2ebasis}. The amount of the computer memory required for the other steps of the construction of the diagonal blocks of $\underline{H}'$ and $\underline{O}$ is negligible. The total CPU time necessary to calculate all the $R$-independent part of $\underline{H}'$ was approximately 300\thinspace seconds on the Intel Xeon CPU (Ivy Bridge generation). Another step of the calculation that requires considerable computer resources, is the complete diagonalization of $\underline{H}'$ after its projection on the regular subspace. Total dimension of $\underline{H}'$ in the basis set discussed above was 13122, the projection reduced it to 12770. The diagonalization was performed using the parallelized routine dsyev from the program library LAPACK. The diagonalization required 1.2\thinspace GB of the computer memory and 1300\thinspace seconds of the CPU time.
The potential of the perturber polarized by the atomic core $B$\xspace \eqref{eq:outerpoten} is neglected in the outer region. To justify this approximation numerically for the hydrogen perturber, we evaluated the dipole term \eqab{eq:dippoten}, generated by the ground state $\varphi_{10}(\mathbf{r}_1)$, that is the leading term of this potential. Dependence of the corresponding dipole moment $D_{10}$ on the internuclear separation $R$ obtained using \eqab{eq:dipexplic} is shown in \figref{fig:dipdep}. The curve calculated using the same set of the closed orbitals $u_{n_1'l_1'}(r_1)$ (see \eqab{eq:2eopensub}) as discussed above ($n_1'\leq2$, $l_1'\leq1$) is denoted as Basis 1. To study the convergence of the calculated dipole moment, we performed one more calculation of $D_{10}(R)$, where the set of the closed orbitals was extended to three lowest $s$-orbitals, two $p$-orbitals and one $d$-orbital. These results are also plotted in \figref{fig:dipdep} and corresponding curve is denoted as Basis 2. As can be seen, $D_{10}(R)$ calculated in the smaller basis set is considerably lower compared to the one calculated using the more extensive set of orbitals 2, although the two-electron PECs calculated using the smaller basis set 1 are converged (as will be discussed in Section \ref{sec:secresults}). This suggests a higher sensitivity of $D_{10}(R)$ to the set of closed functions $u_{n_1'l_1'}(r_1)$ employed in the open part of the two-electron basis set \eqref{eq:2eopensub}. Our tests (not published here) showed that the curve marked as Basis 2 in \figref{fig:dipdep} does not considerably change with further extension of the set of employed $u_{n_1'l_1'}(r_1)$. When any of the calculated dipole moments is substituted into \eqab{eq:dippoten} for the dipole potential, the value of $U_{D10}(\mathbf{r}_2)$ anywhere in the outer region is more than two orders of magnitude smaller than the Coulomb potential of the core $B$\xspace. Consequently, the potential \eqref{eq:outerpoten} can be neglected outside the sphere $\mathcal{S}$.
\begin{figure}
\includegraphics{diptarg.eps}
\caption{Dipole moment $D_{10}$ of the ground state $\varphi_{10}(\mathbf{r}_1)$ of hydrogen calculated using \eqab{eq:dipexplic} as function of the internuclear separation $R$. The blue dotted curve denoted as Basis 1 was calculated using the same set of the closed orbitals as was used to calculate the energies of the two-electron system. The solid black line denoted as Basis 2 was calculated using the extended set of lowest three $s$-orbitals, lowest two $p$-orbitals and the lowest $d$-orbital.\label{fig:dipdep}}
\end{figure}
It is interesting to analyze the relation between the dependence of $D_{10}$ on the internuclear distance $R$ and the static dipole polarizability of the perturber $\alpha_d$.
As long as $R$ is large enough and the electric field generated by the atomic core $B$\xspace can be within a good approximation considered constant throughout the region of the perturber, the dipole moment $D_{10}$ depends on $R$ as $D_{10}(R)=\alpha_d/R^2$ \cite{Khuskivadze,bottcher,Henriet1984}. Comparison of this dependence with $D_{10}$ calculated using \eqab{eq:dipexplic} for different values of $R$ allows us to evaluate the range of $R$ where this approximation of the constant electric field inside $\mathcal{S}$ is valid.
The values of $D_{10}$ obtained from \eqab{eq:dipexplic} using the smaller set of the closed orbitals (Basis 1 in \figref{fig:dipdep}) can be very accurately fitted to
\begin{equation}
D_{10}(R)=a/R^2,\label{eq:dipfit}
\end{equation}
where $a=1.696$\thinspace a.u. This suggests that the variation of the electric field inside $\mathcal{S}$ does not influence the polarization of the perturber by the core $B$\xspace. The low value of $a$ compared to the static dipole polarizability of the hydrogen atom $\alpha_d=4.5$\thinspace a.u. is the consequence of the insufficient set of the closed orbitals $u_{n_1'l_1'}(r_1)$ for the representation of the dipole moment mentioned above.
The curve denoted as Basis 2 in \figref{fig:dipdep}, calculated using more extensive set of the closed orbitals, also can be very accurately fitted to \eqab{eq:dipfit}, where $a=4.25$\thinspace a.u. This is much closer to the correct value of $\alpha_d$. The reciprocal quadratic dependence of $D_{10}$ on the internuclear separation also shows that the presence of the atomic core $B$\xspace does not considerably affect the static dipole polarizability of the perturber throughout the studied range of $R$.
The computationally most demanding part of the outer region treatment is the numerical evaluation of the matrix elements of the Green's function\xspace and its radial derivative \eqref{eq:gmatrices}. The three-dimensional integration (over the angular coordinates $\theta$, $\theta''$ and $\phi-\phi''$) \cite{Khuskivadze} is performed numerically using the Gauss-Legendre quadrature in each dimension. The convergence tests showed that the results presented below are converged for 20 abscissas in each dimension. The $1/\left|\mathbf{r}-\mathbf{r}''\right|$ singularity is removed from the Green's function\xspace and separately integrated analytically as it is explained in the Appendix of the paper \cite{Khuskivadze}. The Whittaker functions along with their derivatives necessary to evaluate the Coulomb Green's function\xspace \cite{Hostler1963} are calculated using the subroutines by \textcite{Thompson1985}. Note that \textcite{Hamilton-PhD} employed a Taylor expansion of the Green's function\xspace on the sphere $\mathcal{S}$ assuming that $r_0\ll R$, since the authors focused on the limit of large internuclear separations. That approach requires less evaluations of the complicated special functions. Moreover, this work is oriented more towards the situations, when $r_0$ and $R$ are comparable.
The matrix $\underline{M}$ calculated for fixed $R$, as a function of the energy $E$, possesses two kinds of poles. These are \textsl{R}-matrix\xspace poles (see \eqab{eq:rmatpolex}) and the poles of the Green's function\xspace at the bound-state energies of the Rydberg atom in the absence of the perturber. Since all these are known in advance, the energy interval of our interest was split into the subintervals defined by these poles. The situations, where the bound-state of whole the diatomic system has the energy like the Rydberg atom without the perturber, are not the main focus of this computational method and it is less accurate there. The energy grid between each two poles of $\underline{M}$ for the calculations presented below consisted of 3000 energy points and the evaluation of $\underline{M}$ (its dimension is 5) required 600\thinspace seconds of the CPU time.
\section{Results}
\label{sec:secresults}
As a simple demonstration of the presented computational method, we applied it to H$_2$ for calculations of the PECs of its excited $^1\Sigma$ states. The most simple system was selected intentionally, as its treatment does not require any additional parameters that are necessary for other first-row elements and it is free of all the model-related complications. Both model potentials $V_A(r)$ and $V_B(r')$ reduce to pure Coulomb potentials everywhere in the space and the Green's function\xspace does not include any quantum-defect corrections \cite{Davydkin,Khuskivadze,Hamilton-PhD}. Moreover, the results obtained using the method presented here can be compared with those obtained using simple contact-potential models published previously. The technical details of the calculations are discussed above in Section \ref{sec:numeri}.
The PECs for the internuclear separations from 20\thinspace a.u. to 420\thinspace a.u. are presented and they include the nuclear repulsion term $1/R$ that is not part of the electronic Hamiltonian \eqref{eq:totham}. In order to keep the graphs easier to read, the energy -0.5\thinspace a.u. of H($1s$) was subtracted from all the total energies. In the limit $R\to\infty$, every bound-state PEC converges to an energy of the non-interacting system of the perturber in its ground state and the atomic Rydberg state of the other atom $\mathrm{H}(1s)+\mathrm{H}(nl)$. These energies as well as corresponding groups of the PECs will be denoted by $n$ in the rest of this section.
For finite internuclear distances, the Coulomb Green's function\xspace has poles at energies $\epsilon_{10}(R)-1/(2n^2)$, where the channel energy $\epsilon_{10}(R)$ is defined by \eqab{eq:perchanfuncs} and $n\in\mathbb{N}$. Those energies would correspond to eigenstates of a system in which the Rydberg electron does not interact with the neutral atom with the center $A$\xspace. They will be referred to as the non-interacting Rydberg thresholds (NIRTs) in the rest of this section. The energy of the excited channel $\epsilon_{20}(R)$ is so high that all the corresponding NIRTs are above the ionization threshold. Therefore, the excited channels were not included in the outer region treatment. \figref{fig:lowstates} demonstrates (the red dotted curves) that NIRTs exhibit only very weak dependence on $R$ over the studied interval
\begin{figure}
\includegraphics{fig-low.eps}
\caption{Potential energy curves of the excited $^1\Sigma$ states of H$_2$. The nuclear repulsion is included and the energy -0.5\thinspace a.u. of H$(1s)$ was subtracted. The \textsl{R}-matrix\xspace results are plotted with the solid black line, the results of the contact model \cite{Presnyakov} are represented by the solid green line. The dotted red lines are the poles of the Green's function\xspace $\epsilon_{10}-1/(2n^2)+1/R$. The blue dotted line in the inset corresponds to $E_A-1/R$. The points denoted by $\times$ are the energies calculated by \textcite{Corongiu}, the points denoted by $+$ in the lower inset are the energies calculated by \textcite{Staszewska}.\label{fig:lowstates}}
\end{figure}
because the ground state of the perturber $\varphi_{10}(\mathbf{r}_1)$ is only very weakly perturbed by the atomic core $B$\xspace. The first-order perturbation theory yields $\epsilon_{10}(R)=\varepsilon_{10}-1/R$ and addition of the nuclear repulsion cancels the dependence on the internuclear distance. The good agreement of this result with the non-perturbative calculations presented in \figref{fig:lowstates} means that the variation of $V_B$ over the spatial volume, where the wave function of the valence electron is located, does not play any considerable role. This is understandable taking into account that the wave function of H$(1s)$ is very compact.
\begin{figure}
\includegraphics{ellipt.eps}
\caption{Top: PECs of the excited $^1\Sigma$ states of H$_2$. The curves have the same meaning like in \figref{fig:lowstates}, results between NIRTs corresponding to $n=12$ and $n=13$ are presented. Bottom: Electron density of the spheroidal eigenstates $\left|\tau_{i0n}(\mathbf{r}_{\text{pert}})\right|^2$ of H$(n=12)$ located in one focus \cite{Coulson} evaluated at the position of the other focus as a function of the distance between the foci.\label{fig:curdetail}}
\end{figure}
As can be seen in the inset of \figref{fig:curdetail}, the higher-order perturbation becomes more pronounced at the internuclear separations $R\lesssim40$\thinspace a.u.
The PEC on the bottom of \figref{fig:lowstates} is the ion-pair curve. The lower inset of that figure shows that for $R\lesssim40$\thinspace a.u., the ion-pair PEC is slightly diverted from its asymptotic form $-0.5-E_A-1/R$, where $E_A$ is the electron affinity of H. This is another consequence of the increased polarization of the perurber by the core $B$\xspace at smaller $R$. Note that the position and shape of the avoided crossing of the ion-pair PEC with the $4^1\Sigma_u^+$ at $R\approx36$\thinspace a.u. is in excellent agreement with the previously published results of the \textit{ab initio}\xspace quantum chemical calculations by \textcite{Staszewska}. Moreover, the results of \textcite{Corongiu} for higher energies are also in good agreement with the findings presented here. The accuracy of the ion-pair PEC is sensitive predominately to the quality of the representation of the two-electron interaction in the inner region, since the probability density of the Rydberg electron in this state is well localized around the perturber. It is the convergence of this ion-pair PEC in our calculations that requires the set of the closed two-electron basis functions larger than the subset of the open functions.
As can be seen in \figref{fig:assem}, the PECs
\begin{figure}
\includegraphics{fig-assem.eps}
\caption{PECs of the excited $^1\Sigma$ states of H$_2$. The interval of the asymptotic energies of $\mathrm{H}(1s)+\mathrm{H}(nl)$ with $n=12\dots22$ is presented. The colors and the notation has the same meaning as in \figref{fig:lowstates}.\label{fig:assem}}
\end{figure}
(other than the ion-pair curve) can be divided into two categories: The PECs of the first type follow the shape of the $1\sigma_g$ PEC of H$_2^+$ at large internuclear separations and they are located at energies very close below the NIRTs (see also the inset of \figref{fig:curdetail}). These PECs represent the Rydberg states weakly affected by the neutral perturber. For the energy interval of the interest in this work these PECs coincide with the associated NIRTs at $R\gtrsim120\thinspace$a.u. The PECs of the second type are more interesting. They are located at energies between the pairs of subsequent NIRTs at smaller internuclear separations and each of them decreases with increasing $R$ until it coincides with the lower NIRT.
PECs of both types show an oscillatory structure that is better pronounced in the second-type PECs. These undulations can be understood in terms of the simplified model developed by \textcite{Granger} to predict the trilobite-like states of Rb$_2$ \cite{Greene-prl}. In their single-electron approach the perturber is represented by Fermi's contact pseudopotential and the Rydberg center by the pure Coulomb potential (the non-hydrogen character of Rb is neglected). The eigenstates of this model system are expressed as linear combinations of such eigenfunctions $\tau_{nmi}(\mathbf{r})$ of the hydrogen-atom Hamiltonian that are separable in the prolate spheroidal coordinates \cite{Coulson} with the foci located on the atomic cores. The quantum number $m$ is set to zero, since only the $\Sigma$ states are considered. The quantum number $i$ denotes different degenerate eigenstates with the same $n$. \textcite{Granger} applied the first-order perturbation theory to calculate the energies of the model diatomic system as
\begin{equation}
E_n(R)=-\frac{1}{2n^2}+2\pi A\left[k(R)\right]\sum_{i}\left|\tau_{i0n}(\mathbf{r}_{\text{pert}})\right|^2,\label{eq:greenepert}
\end{equation}
where $n$ is the principal quantum number of the atomic Rydberg electron without the perturber, $A\left[k(R)\right]$ is the electron-perturber scattering length and the values of the atomic spheroidal eigenstates are taken at the position of the perturber. \textcite{Granger} showed that in case of Rb$_2$ every minimum in the oscillations of the PEC corresponds to such values of $R$, where the sum in \eqab{eq:greenepert} is dominated by one spheroidal eigenstate of the Rydberg electron. The values of $\left|\tau_{i0n}(\mathbf{r}_{\text{pert}})\right|^2$ for $n=12$ are plotted in the lower part of \figref{fig:curdetail}. As can be clearly seen, the structures in the PECs calculated using the method introduced in this work obviously correspond to the peaks of the spheroidal eigenstates of the hydrogen atom and they are of the same nature like those in Refs. \cite{Greene-prl,Granger}.
More specifically, the peaks of the spheroidal eigenstates correspond to the local maxima of the oscillations in the PECs. This behavior is different from the case of Rb$_2$ discussed by \textcite{Granger}, where the positions of the peaks of the spheroidal atomic eigenstates coincide with the minima of the oscillations. An explanation is quite simple: The scattering length $A(k)$ in \eqab{eq:greenepert} is negative for hydrogen and positive for rubidium \cite{Schwartz,Bahrim2000}.
It is interesting to compare the results of our two-electron \textsl{R}-matrix\xspace calculations with those obtained by the application of the approach developed by \textcite{Greene-prl} and \textcite{Granger} to H$_2$. The generalized energy-dependent scattering length for the interaction of the free electron with the neutral atom is defined as $A(k)=-\tan\delta(k)/k$, where $\delta(k)$ is the $s$-wave scattering phase shift. The effective range theory modified for the presence of the polarization potential \cite{OMalley1961} provides the low-energy expansion
\begin{equation}
-\frac{1}{A(k)}=-\frac{1}{A_0}+\frac{\pi\alpha_d}{3A_0^2}k+\frac{4\alpha_d}{3A_0}k^2\ln\left(\frac{\sqrt{\alpha_d}k}{4}\right)+\mathcal{O}(k^2),\label{eq:mert}
\end{equation}
where $A_0=5.95$ is the singlet zero-energy scattering length \cite{Schwartz} and $\alpha_d=4.5$\thinspace a.u. is the static dipole polarizability of the hydrogen atom. All the terms where $k$ appears in the second and higher order will be neglected here. Following \textcite{Greene-prl}, the local electron momentum $k$ can be for given internuclear separation $R$ calculated as
\begin{equation}
k(R)=\sqrt{2\left(\frac{1}{R}-\frac{1}{2n^2}\right)}.\label{eq:localmom}
\end{equation}
Substitution of \eqab{eq:localmom} and \eqab{eq:mert} into \eqab{eq:greenepert} yields the energies of the model diatomic system. The PEC calculated for $n=12$ is plotted in \figref{fig:curdetail}. It is in encouraging qualitative agreement with the two-electron \textsl{R}-matrix\xspace calculations, including the details of the oscillatory structure. It is not surprising that the agreement is excellent at larger internuclear separations ($R\gtrsim100$\thinspace a.u.) and that the results of the one-electron model increasingly divert from the PEC calculated using the two-electron \textsl{R}-matrix\xspace approach with decreasing $R$. Comparable agreement between the Fermi's model and two-electron \textsl{R}-matrix\xspace calculations was observed for all the energies and internuclear separations $R$ explored in this paper. This is a consequence of the finite size of the perturber that plays more important role at smaller internuclear separations, while the representation by the Fermi's contact pseudopotential reduces the interaction with the perturber to a single spatial point.
Fermi's pseudopotential \cite{fermi1934}
\begin{equation}
V_{Fk}(\mathbf{r})=2\pi A\left[k(R)\right]\delta^3(\mathbf{r}-\mathbf{r}_{\text{pert}}),
\end{equation}
previously applied to Rb$_2$ \cite{Greene-prl,Granger}, is parametrized to treat a collision of a free electron with a neutral target. While this model holds very well for large internuclear separations ($R\sim10^3 - 10^5$\thinspace a.u. \cite{Greene-prl}), some of its assumptions cease to be accurate at smaller distances $R$ explored in this paper. For instance, the dipole moment of the neutral atom gained in the presence of the potential $V_B$ becomes relevant inside the sphere $\mathcal{S}$, although we neglect it in the outer region. Moreover, variation of $V_B$ inside the sphere $\mathcal{S}$ is larger for smaller $R$ and the validity of the free-electron scattering model becomes questionable. We attempt to account for this deficiency by adding the third, $k^2\ln k$, term in the expansion \eqref{eq:mert} that appears to be not included in the original works \cite{Greene-prl,Granger}. This term becomes important because larger variation of the potential $V_B$ around the perturber provides the scattered electron with higher local momenta via \eqab{eq:localmom}. In fact, we also carried out tests (not published here) with only first two terms on the right-hand side of \eqab{eq:mert} and the agreement of such Fermi's model with the two-particle \textsl{R}-matrix\xspace calculations was very poor.
Another simple model worth mentioning here was developed by \textcite{Presnyakov}. It is based on a different Fermi's pseudopotential \cite{blatt-book} and \textcite{Presnyakov} utilized it to study broadening of the excited levels of atoms in an alkali-metal atmosphere. The author solved the Schr\"odinger equation\xspace non-perturbatively using the Green's function\xspace \cite{Hostler1963} and the PECs of the excited bound states $E(R)$ were calculated as the solutions of the transcendental equation
\begin{eqnarray}
-\frac{1}{A_0}&=&\frac{2\Gamma(1-\kappa)}{\kappa}\left\{\left[-\frac{1}{4}+\frac{\kappa}{z}\right]\mathcal{W}_{\kappa,1/2}(z)\mathcal{M}_{\kappa,1/2}(z)\right.\nonumber\\
&+&\left.\mathcal{W}'_{\kappa,1/2}(z)\mathcal{M}'_{\kappa,1/2}(z)\right\},\label{eq:presnyakov}
\end{eqnarray}
where $\kappa=\sqrt{-2E(R)}$, $z=2R/\kappa$, $\mathcal{W}_{\kappa,1/2}(z)$ and $\mathcal{M}_{\kappa,1/2}(z)$ are Whittaker functions \cite{Hostler} and the prime denotes the derivative with respect to the argument. Therefore, the computational method is not based on the perturbation theory. We solved \eqab{eq:presnyakov} numerically to compare the results of this model with the simple models discussed above \cite{Granger,Greene-prl} as well as with our two-electron \textsl{R}-matrix\xspace results. The PECs obtained by solving \eqab{eq:presnyakov} are in excellent agreement with the results of the first-order perturbation calculations \cite{Granger,Greene-prl}, as can be seen in \figref{fig:curdetail} (the agreement is similar throughout all the energy range shown). Such a good agreement indicates that the first-order perturbation theory is very accurate when applied to the states separable in the spheroidal coordinates. Comparison between the solutions of \eqab{eq:presnyakov} and the PECs obtained using our two-electron \textsl{R}-matrix\xspace approach for the lower energies is plotted in \figref{fig:lowstates}. Both calculations are in very good qualitative agreement. They agree very well quantitatively at larger internuclear separations and as $R$ decreases, the energies of the one-electron model are systematically slightly lower than those obtained from the \textsl{R}-matrix\xspace calculations.
\section{Conclusions}
\label{sec:coclusions}
A method for study of long-range Rydberg states of diatomic molecules was introduced in this paper. The method can be viewed as a straightforward extension of the pure one-electron approach of \textcite{Khuskivadze}. In the present work, the two-electron \textsl{R}-matrix\xspace method is employed to calculate the wave function in the region surrounding one of the atomic cores. The wave function smoothly connects to a bound-state solution dictated by a long-range Coulomb field of the second atomic center. The approach presented here does not account for a situation in which both electrons leave the first atomic center and thus the ionization energy of the first core defines the upper energy limit of this approach.
In comparison to the method of \textcite{Khuskivadze}, the present approach brings a few simplifications and some complications. One of the simplifications is the core potential $V_A$ that can be made independent of the total spin state of the system as well as of the total angular momentum with respect to the first atomic core. These properties allow calculations for much shorter internuclear distances $R$ than those explored by \textcite{Khuskivadze}. The shorter internuclear separations $R$ require higher partial waves with respect to the first center and this requirement raises the demands on the accurate representation of the neutral perturber (center $A$\xspace) that was considered different for the partial waves $l=0$ and $l=1$ and every total spin state in \textcite{Khuskivadze}. On the other hand, the two-electron method presented here has to deal with the correlated movement of the electrons in the core potential $V_A$ inside a sphere surrounding the atomic core $A$\xspace. Such complication may also become a gain in situations, where the energies of the molecular Rydberg states associated with the excited states of the perturber are below the ionization threshold of the whole system (for example, Rb$_2$).
The motivation for the development of the presented method is to study the long-range Rydberg states of the alkali-metal diatomics at the internuclear separations $R\approx20-400$\thinspace a.u., corresponding to experiments \cite{Greene2006-prl,Vadla,Bellos}. This work aims to complement previous studies \cite{Greene2006-prl,Bellos} and test the approximation of the contact potential at smaller internuclear distances utilized there to explain the experimental spectra. In order to test the basic mechanisms of the method we decided to test it on the simplest of molecules, H$_2$. Present results accurately reproduce previous \textit{ab initio}\xspace calculations \cite{Corongiu,Staszewska} for lower states ($n=3,4$) and they predict yet undocumented oscillatory character of higher anti-bonding states $\mathrm{H}(1s)+\mathrm{H}(nl)$. While these oscillations are also well reproduced by the two contact-potential models tested in this paper, the absolute energies predicted by the contact-potential models are systematically lower than the more accurate two-particle calculations. Present numerical analysis in the spheroidal coordinates reveals that the physical origin of these oscillations is the same as was explained by \textcite{Granger} for calculated trilobite states, i.e. the oscillations correspond to the peaks of the dominant spheroidal eigenstates.
Our internal numerical tests indicate that the accuracy of the presented energies should be better than 1\thinspace meV. Therefore, we believe that the present method should bring similar level of accuracy to the characterization of the long-range molecular Rydberg states of the alkali-metal molecules. Such studies are planned for the near future.
\begin{acknowledgments}
We are thankful to Chris Greene for the initial impulse for this work. We also acknowledge the support of the Grant Agency of the Czech Republic (Grant No. P208/14-15989P).
\end{acknowledgments}
|
{
"redpajama_set_name": "RedPajamaArXiv"
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Malgrate é uma comuna italiana da região da Lombardia, província de Lecco, com cerca de 4.208 habitantes.
Estende-se por uma área de 2 km², tendo uma densidade populacional de 2104 hab/km². Faz fronteira com Galbiate, Lecco, Valmadrera.
É a cidade natal do atual Arcebispo de Milão, Cardeal Angelo Scola.
Demografia
Comunas de Lecco (província)
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
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Bob has been one of Nashville's Leading contractors and a registered engineer for more than 30 years. He has savored his role developing Nashville and the Middle Tennessee area since he founded Biscan Constructionin 1981. A 1973 graduate of Pennsylvania State University with a B.S. in Civil Engineering, Bob has built a company that continues to be recognized as one of the Greater Nashville 100 companies, handling between 25-30 projects of all sizes simultaneously. Bob spends much of his time working on medical office, office and hotel projects as well as the majority of private jobs the company completes. Many times he is involved in the design process and makes changes to further enhance the owner's vision and budget. Occasionally he partners on a project and handles the developmentfrom inception to completion including all the front end studies and banking.
Born and raised in Middle Tennessee, Jim graduated from Belmont University in 1987 with a BA degree in Finance and Economics. Jim joined Biscan the following year, became a partner 8 years later and currently runs the day to day operations of the company. Along with those duties he also manages Biscan's largest projects, making him a very hands on officer of the company.
Jim married his wife Kate in 1987 and has two sons. Jim also serves on the Board of Trustees for BGA. He is an avid outdoorsman. He enjoys hunting and running his working farms in both Kentucky and Illinois as well as big game hunting out west.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8,955
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{"url":"http:\/\/cms.math.ca\/10.4153\/CMB-1997-041-1","text":"Abstract view\n\n# A New Proof of a Theorem of Magnus\n\nPublished:1997-09-01\nPrinted: Sep 1997\n\u2022 Sal Liriano\nFeatures coming soon:\nCitations \u00a0\u00a0(via CrossRef) Tools: Search Google Scholar:\n Format: HTML LaTeX MathJax PDF PostScript\n\n## Abstract\n\nUsing naive algebraic geometric methods a new proof of the following celebrated theorem of Magnus is given: Let $G$ be a group with a presentation having $n$ generators and $m$ relations. If $G$ also has a presentation on $n-m$ generators, then $G$ is free of rank $n-m$.\n MSC Classifications: 20E05 - Free nonabelian groups 20C99 - None of the above, but in this section 14Q99 - None of the above, but in this section","date":"2013-12-07 10:30:09","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.568164050579071, \"perplexity\": 1070.756971745387}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2013-48\/segments\/1386163053923\/warc\/CC-MAIN-20131204131733-00093-ip-10-33-133-15.ec2.internal.warc.gz\"}"}
| null | null |
package org.tud.inf.st.pccs;
/**
* <!-- begin-user-doc -->
* A representation of the model object '<em><b>Unit</b></em>'.
* <!-- end-user-doc -->
*
*
* @see org.tud.inf.st.pccs.PccsPackage#getUnit()
* @model
* @generated
*/
public interface Unit extends OperatorTableEntry {
} // Unit
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,009
|
Phillip Phillips Steps Outside The Light
Phillip Phillips fans waiting to hear news about possible tour dates in 2015 got some difficult news instead. In an article published January 26th on The Hollywood Reporter's website we learned that Phillip filed a petition with the California Labor Commissioner regarding the contract he signed with 19 Entertainment as part of his American Idol win.
The article reveals that Phillip decided to file the petition based on a number of grievances and issues extending all the way back to his early days as an American Idol winner in the fall of 2012.
From unfair financial compensation, to exploitative and manipulative deals made on Phillip's behalf, the core of the complaints relate to multiple violations by 19 Entertainment of the California Talent Agencies Act. To understand what this means you need to read the petition, which clearly outlines what this "Act" is and how it was violated by 19E and all its affiliated companies. I read the document, and honestly, it was a very difficult read. Yes, the lawsuit is about unfair financial compensation and about actions deemed illegal by the Talent Agencies Act. However, at its core, it really is about an artist's desire for integrity and respect, for true compromise and collaboration. It is about 19 Entertainment failing to care for and act on the best interests of Phillip, one of its most successful artists in recent times. This is no minor detail. It is, in fact, at the heart of the matter.
One of the first words that come to mind when talking about Phillip and his music has always been "authenticity." The quest for it, the importance that Phillip placed on maintaining it, has been evident from day one. It was evident throughout his run on American Idol where he never hid his unease with certain requirements he was expected to fulfill as a show contestant. He's the guy who refused to wear the pretty clothes Tommy Hilfiger suggested he wear on that infamous early Idol episode. He's the guy who, despite being repeatedly compared to Dave Matthews decided to play an obscure Dave Mathews song on the show. He's the guy who, just after winning American Idol, politely declared that "Home" was a beautiful song but one that he would never have written himself. The guy who has refused to play in concert songs he feels don't "best represent" his sound. The guy who showed you can win a huge pop singing competition while remaining completely and absolutely true to yourself. Authenticity is very important for Phillip Phillips.
I remember clearly the day the title of Phillip's second album, Behind The Light, was announced. He was set to play the last show of his Canadian tour in Vancouver, Canada. As he often does, he was also engaged to do an exclusive acoustic performance that morning before the main show. The news about the title broke mid-morning, just before the acoustic show. I had won tickets for that performance and remember thinking what a whirlwind of a day he was going to have. To learn that the title of that album was decided without Phillip's approval and then announced to the press before notifying him is heartbreaking, outrageous. It must have been devastating.
That day Phillip fulfilled every single one of his obligations as an artist on tour. He did the performance and Q&A. Then he did an exclusive photo session with a few people from the audience who had won a second contest. He was then quickly whisked away for sound check at the venue. There were probably interviews with the press in between all of that. He played a heartfelt show for his audience that night. And later, the hour nearing closer to midnight, he came out to greet a big group of fans who had waited for him to say Hi. I was grateful to him but also oddly distraught. It had been, I'm even more certain now, a very long day for him, and yet, he still took the time to come out. Phillip Phillips works very hard. The decision to break out of his contract is not one he must have taken lightly. He has shown–proven beyond any doubt–that he is ready to pay his dues and work himself to exhaustion.
And then there is the matter of "style," that slippery word that means nothing and everything at the same time. In Phillip's case it has found its outlet in the albums, yes, but really and clearly through his brilliant, instrumental-heavy, jam inspired, live performances. Live performances that reveal his love of rock and fusion and experimentation, not really the folk-pop sound that two of his main radio hits, "Home" and "Gone Gone Gone," favoured and that many people expected him to repeat on his second album. Of course, there is nothing wrong with pop and folk, other than these are not the styles that Phillip seems to hold close to his heart. Hence the unease and clear division that one can still hear between the singles and the B-sides on both of Phillip's albums so far. To read that Phillip feels that 19 Entertainment failed to put his interests above their own, financially and creatively, shows again the terrible struggle this must have caused on somebody who values authenticity above all.
The ramifications of this move on Phillip's career are not very clear to me. I have no knowledge what legal consequences this may have on his ability to make music through an established platform, on his ability to perform said music through live shows, or his ability to move, in the short term, to a new management/label that will allow him to continue to work. What is clear to me is that, whatever the consequences, they will have no impact on my admiration and support of Phillip's music and career. I know that the music will continue, and that I, as well as the majority of his fans, will find our way to it, as I'm sure Phillip will find his way to us.
I think back to a comment Phillip used to make after winning the show to explain some of his song choices and decisions. He would say, and I paraphrase, that it was better to fail and know that you did because of something you believed in, did or say, than to fail because of something somebody else told you to believe in, do or say. Stand by your decisions, and, if you fail, own your failures. Your triumphs will then be all the sweeter, for they will have come from the deepest essence of your being. Also, there will be nothing left to hide. Phillip Phillips has nothing left to hide.
19 Entertainment american idol phillip phillips
by Andreina RomeroNews
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|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,884
|
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