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\section{Introduction}
The necessary ingredients to build the late time background evolution of the standard flat cosmological model are the local (or today's) Hubble constant $H_0$ and the amount of pressureless matter, given by the fractionary density parameter $\Omega_{m0}=\rho_m/\rho_{c0}$, where $\rho_{c0}=3H^{2}_0/8\pi G$ is the today's critical density. Another possible degree of freedom is equation of state of dark energy $w_{de}=P_{de}/\rho_{de}$, where $P_{de}$ ($\rho_{de}$) is the dark energy pressure (energy density). This corresponds to the expansion rate $H=\dot{a}/{a}$, where the symbol $``^{.}"$ means derivative with respect to the cosmic time, of the type
\begin{equation}
H = H_0 \left[\frac{\Omega_{m0}}{a^{3}}+(1-\Omega_{m0})a^{-3(1+w_{de})}\right]^{1/2}.
\label{Eq1}
\end{equation}
The function $a(t)$ is the cosmic scale factor. Recent observations including Cosmic Microwave Background (CMB), Baryon acoustic oscillations and Supernovae (SN) type Ia data constrain this scenario close to the preferred parameters values $\Omega_{m0} \simeq 0.3$ and $w_{de}\simeq -1$. The latter result makes (\ref{Eq1}) equivalent to the so called flat $\Lambda$CDM model \cite{Aghanim:2018eyx}.
Since the discovery that the width of the lightcurve peak of Supernovae type Ia (SNIa) is correlated with their peak luminosity (the so called Phillips calibration \cite{Phillips:1993ng}) such objects have been considered reliable standard candles and widely used in observational cosmology. Hence, the SN type Ia Hubble diagram (SNHD) represents one of the observational pillars supporting the cosmological concordance model. However, in terms of the redshift range distribution, even recent SNHD catalogues are limited to redshifts up to $z\sim 2$. Having SN data in the range $1<z<2$ make such samples deep enough to state cosmic acceleration but somehow limited to probe the early background evolution. Then, the search for alternative reliable candles in redshift range beyond the SNHD ($z > 2$) opens the possibility to test cosmology dynamics deep in the universe lifetime at moments between the early Universe and the last scattering surface $z\sim 1100$ where CMB has been released. There are phenomenological attempts in the literature to build a Hubble diagram for objects that are not considered standard candles but are at such high redshifts. This includes Gamma Ray Bursts (GRBs) \cite{Demianski:2016zxi,Wei:2013xx,Izzo:2009bw,Cardone:2009mr,Liang:2008kx,Schaefer:2006pa} and Quasars \cite{Risaliti:2018reu,Lusso:2017hgz, Risaliti:2016nqt, Risaliti:2015zla}. Mostly of these approaches are based on the strategy to find out empirical correlations between observational properties of these objects e.g., fluxes at different wavelengths, emission peaks among others observational quantities in a certain redshift range and then to extrapolate the fitting parameters to the entire sample.
Concerning the Quasars Hubble Diagram (QSOsHD), recently Risaliti \& Lusso presented a robust catalogue containing 1598 inferred luminosity distances of quasars in the redshift range $0.5<z<5.1$ \cite{Risaliti:2018reu}. The final catalogue represents the onset of a gradual refinement on the selection techniques and flux measurements developed along Refs.
\cite{Lusso:2017hgz, Risaliti:2016nqt, Risaliti:2015zla}. The strategy performed in these works is based on a correlation between the X-ray and the optical-ultraviolet (UV) emissions at low redshift i.e., there exists a phenomenological parametrization such that $log (L_X) = \gamma log (L_{UV})+\beta$ where $\gamma$ and $\beta$ are fitting parameters of the luminosities at $2$keV ($L_{X}$) and $2500 \AA$ ($L_{UV}$) \cite{Avni}. Another recent work from this same group has shown that this relation persists to even higher redshifts up to $z\sim 7$ \cite{Salvestrini:2019thn} (see also \cite{Lusso:2019akb}). This corroborates therefore the strategy of Ref.\cite{Risaliti:2018reu}. The latter reference reports that while the low-z subsample (containing objects with redshifts $z<2.1$) of the QSOsHD is consistent with the standard cosmological model, the inclusion of the high-z subsample yields to a $\sim 4 \sigma$ tension with standard cosmology. By fitting the expansion (\ref{Eq1}) to the QSOsHD data it is found a preference for a higher fractionary matter density parameter $\Omega_{m0}>0.4$ and a phantomic dark energy $w_{de}<-1$. It is worth of mentioning that even some SN analysis have reported tensions with relation to the standard cosmology either by stating a marginal evidence for acceleration \cite{Nielsen:2015pga} or inferring a higher $\Omega_{m0}$ values \cite{Shariff:2015yoa}.
Our goal in this work is to detail the analysis of the QSOsHD by searching for the source of such tension in an independent way. Rather than data fitting and performing a parameter estimation via statistical techniques we employ an model-independent estimator for cosmic acceleration proposed in Refs. \cite{Seikel:2007pk,Seikel:2008ms}. This will allow us to answer two basic questions: Is there a tension between the QSOsHD data and the standard cosmological model? Is the QSOsHD a reliable observational source?
In the next section we introduce the method and apply it to the data. We also use SNHD from Pantheon \cite{Scolnic:2017caz} sample as alternative data set of our analysis. This allows us to update the evidence for acceleration available in the SNHD and to promote a proper comparison with the QSOsHD. Our main results are presented in section III and we conclude in the final section.
\section{Evidence for acceleration in Hubble diagram}\label{sect2}
Our analysis is based on the estimator proposed in \cite{Seikel:2007pk, Seikel:2008ms} and revisited recently in Ref. \cite{Velten:2017ire}. Its main goal is to provide a quantitative measurement in terms of a statistical Confidence Level (CL) about the accelerated dynamics of the universe in a model independent way. The construction of such estimator is based on the definition of the deceleration parameter
\begin{equation}\label{qz}
q(z)= \frac{H^{\prime}(z)}{H(z)} (1+z) - 1,
\end{equation}
where a prime $`` \, ^{\prime} \,"$ denotes a derivative with respect to the redshift. An accelerated background expansion at a certain redshift is indicated by $q(z)<0$. From (\ref{qz}) one can obtain
\begin{equation}
{\rm ln}\,\frac{H(z)}{H_0}=\int^z_0 \frac{1+q(\tilde{z})}{1+\tilde{z}} d\tilde{z}.
\label{lnH}
\end{equation}
The null hypothesis proposed in \cite{Seikel:2007pk} is that the universe has never expanded in a accelerated way. This means that $q(z)>0 \, \forall \, z$. See also Refs. \cite{Santos:2007pp,Gong:2007zf} for similar approaches.
By applying the latter inequality to Eq (\ref{lnH}) it turns into
\begin{equation}
{\rm ln}\,\frac{H(z)}{H_0}\geq\int^z_0\frac{1}{1+\tilde{z}}d\tilde{z}={\rm ln}\,(1+z).
\label{ineqLnH}\end{equation}
From the Hubble diagram one obtains the luminosity distance which for a homogeneous, isotropic and expanding background (a Friedman-Lemaitre-Robertson-Walker universe) reads
\begin{equation}
d_L(z)= (1+z) \int^{z}_0 \frac{d\tilde{z}}{H(\tilde{z})}.
\end{equation}
It is possible to recast inequality (\ref{ineqLnH}) via the definition $d_L$ such that it becomes
\begin{equation}
d_L\leq (1+z) \frac{1}{H_0} \int^z_0 \frac{d\tilde{z}}{1+\tilde{z}}=(1+z)\frac{1}{H_0} {\rm ln}(1+z).
\end{equation}
The luminosity distance is essential to compute the observed distance modulus $\mu$, the quantity directly related to the observations,
\begin{equation}
\mu = m-M= 5 log (d_L / Mpc) + 25,
\label{mu}
\end{equation}
where $M$ and $m$ are the absolute and apparent magnitudes respectively.
For each object $i$ at a redshift $z_i$ in the Hubble diagram we can define the quantity
\begin{eqnarray}
&&\Delta \mu_{obs} (z_i) = \mu_{obs}(z_i) - \mu(q=0) \\ \nonumber
&=& \mu_{obs} (z_i)-5{\rm log}\left[\frac{1}{H_0}(1+z_i){\rm ln} (1+z_i)\right]-25,
\label{eqDeltaobs}
\end{eqnarray}
which is the difference between its observed distance modulus $\mu_{obs}^i$ and the distance modulus of a universe with constant deceleration parameter $q=0$ from today until the redshift $z_i$. Then, the estimator has been designed in (8) such that up to this point we have assumed only that light propagates on null geodesics in a homogeneous, isotropic and spatially flat universe.
Clearly, positive $\Delta \mu_{\rm th}$ values indicate acceleration. The face value of $\Delta \mu_{obs}$ is meaningless if its inferred error is not included. Taking into the account the error of the redshift ($\sigma_z$) and peculiar velocities ($\sigma_{v}$) of each SN, the error of the quantity $\Delta \mu_{obs}$ becomes
\begin{equation}
\sigma_i=\left[\sigma^2_{\mu_i}+\left(\frac{5\, ln\left(1+z_i\right)+1}{(1+z_i)\, ln (1+z_i) ln10}\right)^2\left(\sigma^2_z+\sigma^2_v\right)\right]^{1/2}.
\end{equation}
The so called ``single SN analysis" corresponds to computing the quantity $\Delta \mu_{obs}$ for each SN individually. An interesting output of this analysis is that even a few SN indicate that the universe never accelerated (i.e., $\Delta \mu_{obs} <0 $). However, as expected from the statistical nature of such analysis the majority of SN in usual Supernovae samples indicates acceleration.
As already noticed by Refs. \cite{Seikel:2007pk, Seikel:2008ms} the single SN analysis if of limited statistical interest. A more interesting analysis of the estimator $\Delta \mu_{obs}$ is obtained with the so called ``averaged SN analysis". In this analysis we group a number $N$ of objects defining the mean value
\begin{equation}
\overline{\Delta\mu}=\frac{\sum_{i=1}^{N} g_i \,\Delta\mu_{obs}(z_i)}{\sum_{i=1}^{N} g_i},
\label{10}\end{equation}
where the factor $g_i = 1/ \sigma^{2}_i$ makes the data points with smaller errors contribute more to the average. The standard deviation of the mean value is defined by
\begin{equation}
\sigma_{\overline{\Delta\mu}}=\left[\frac{\sum^N_{i=1} g_i \left[\Delta \mu_{obs} \left(z_i\right) - \overline{\Delta\mu}\right]^2}{(N-1)\sum^N_{i=1}g_i}\right]^{1/2}.
\label{11}\end{equation}
The grouping criteria can obey either a fixed redshift range or a fixed object number $N$ per bin. The quantitative value for the evidence of acceleration in each SN bin is given by $\overline{\Delta\mu}$ divided by the error $\sigma_{\overline{\Delta\mu}}$.
One may now wonder what is the expected result for the estimator $\Delta \mu$ for a giving cosmology. In Ref. \cite{Velten:2017ire} we addressed this issue by simulating mock data for certain cosmologies and for specific Hubble diagram distributions. It has been verified that the estimator satisfactorily computes the evidence for acceleration in actual SNHDs like the JLA sample in comparison with simulated catalogues. The full JLA sample provides a $20.40 \sigma$ statistical CL asserting accelerated expansion (see first row of Table \ref{TableI}). This result appeared firstly in Ref. \cite{Velten:2017ire}. Columns in this table represent, from left to right, the sample studied, the evidence for acceleration, the number of objects in the sample and the mean redshift.
We calculate now the evidence for acceleration in the PANTHEON sample. The construction of the HD with the Pantheon SN data \cite{Scolnic:2017caz} employs a different procedure in which the $H_0$ value is degenerated with the absolute magnitude. Then, the PANTHEON sample does not allow to constrain $H_0$. It is also necessary to set a fiducial cosmology to generate the Hubble diagram. This is why one can infer the evidence for expansion for different cosmologies. We present these results in Table \ref{TableI}. It shows that the PANTHEON sample also provides strong evidence favoring acceleration for both the $\Lambda CDM$ model ($+23.73 \sigma$), a $w$CDM model ($+24.23 \sigma$) and a CPL dark energy parameterization ($+23.62 \sigma$). It is also worth noting that when substracting low-z data from such samples i.e., data at $z<0.1$ (as in Ref.\cite{Seikel:2008ms}), the evidence increases.
\begin{table}
\centering
\caption{Averaged evidence (in $\sigma$ of C.L.) for acceleration for different SNHD samples. For the samples indicated with {\it no low-z} data at $z<0.1$ have been not considered.}
{\begin{tabular} {c||c||c||c}
Sample & $\overline{\Delta \mu}$ / $\sigma_{\overline{\Delta\mu}}$ & \# of & Mean \\
& Evidence & Objects & redshift \\
\hline \hline
FULL JLA & +20.40 & 740 & 0.32 \\
JLA no low-z & +22.33 & 588 & 0.40 \\
Pantheon $\Lambda$CDM& +23.73 & 1048 & 0.32 \\
Pantheon $\Lambda$CDM no low-z& +28.25 & 837 & 0.39 \\
Pantheon $w$CDM& +24.23 & 1048 & 0.32 \\
Pantheon $w$CDM no low-z& +28.73 & 837 & 0.39 \\
Pantheon CPL& +23.62 & 1048 & 0.32 \\
Pantheon CPL no low-z& +28.14 & 837 & 0.39 \\
\hline
\end{tabular}}
\label{TableI}
\end{table}
\section{Results for the Hubble diagram of Quasars }
We apply the estimator described in the last section to the QSOsHD developed in Ref. \cite{Risaliti:2018reu}. By using Eqs. (\ref{10}) and (\ref{11}) we obtain the results shown in Table \ref{TableII}. The result for the full QSOsHD sample containing 1598 objects reveals a surprisingly strong lack of evidence favoring acceleration ($-13.60 \sigma$). A similar results is obtained even substracting the low-z sample ($-13.55\sigma$). Indeed, this does not affect QSOs since there are only a few of them at $z<0.1$. We also split the QSOsHD into a subsample with objects at $z<1.3$, the same redshift range of the JLA SN sample. This sub-sample is also not able to provide evidence for acceleration ($-7.95 \sigma$). This splitting is motivated by the procedure adopted in Ref. \cite{Risaliti:2018reu}. The latter assumes a log-linear relation between the rest-frame monochromatic luminosities at 2KeV ($L_X$) and 2500 \AA in this redshift range. The parameters of this relation are found by a joint fit with the JLA SNHD sample. Hence, it can be said that the JLA sample is used to calibrate the QSOsHD at low redshifts. The obtained fitting parameters for the $log (L_X) $ x $ log (L_{UV})$ relation are assumed to be valid to the entire QSOs data sample at higher redshifts. It has been also recently shown that there is no significant evolution of this correlation (fitting parameters) towards high redshifts \cite{Salvestrini:2019thn}. Therefore the method seems robust and pertinent for building a reliable HD of high-z objects. A similar technique has also been used to build the Gamma-Ray Bursts Hubble diagram \cite{Liang:2008kx}.
\begin{table}
\centering\caption{Averaged evidence (in $\sigma$ of C.L.) for acceleration for the QSOsHD. All results adopt $H_0 = 70.0$ km/s/Mpc. }
{\begin{tabular} {c||c||c||c}
Sample & $\overline{\Delta \mu}$ / $\sigma_{\overline{\Delta\mu}}$ & \# of & Mean\\
& Evidence& Objects & redshift \\
\hline \hline
FULL Quasars & -13.60 & 1598 & 1.34 \\
Quasars no low-z & -13.55 & 1587 & 1.35 \\
QSOs $0<z\leq1.3$& -7.95 & 968 & 0.80 \\
QSOs $1.3<z\leq5.1$& -16.21 & 630 & 2.15\\
\hline
\end{tabular}}
\label{TableII}
\end{table}
Results shown in Table \ref{TableII} for the QSOsHD do not support standard cosmology. Rather than inferring deceleration, negative $\overline{\Delta \mu}/ \sigma_{\overline{\Delta \mu}}$ values mean that there is no indication for acceleration. This should be the proper interpretation of such results. Negative $\overline{\Delta \mu}/ \sigma_{\overline{\Delta \mu}}$ are also found even for the low-z subsample though Ref. \cite{Risaliti:2018reu} states that this sub-sample is in agreement with all the main current cosmological probes. This tension is actually caused by the large dispersion of the QSOs at low redshifts and due to the fact some ``far from the $\Lambda$CDM best fit'' data points have very small error bars contributing then more to the estimator.
Fig. \ref{FigCumulative} shows how the cumulative $\Delta \mu$ evolves with the redshift. Black (Blue) dots follow the PANTHEON (QSOs) sample. For the SN data this analysis reveals that indeed high redshift information is demanded in order to establish positive evidence for acceleration. On the other hand, the cumulative $\Delta \mu$ for the QSOsHD clearly evolves to negative values as higher redshifts are taken into account. This confirms that this sample is not able to infer acceleration.
We investigate now on possible sources for such discrepancy i.e., why does not the QSOsHD provide evidence for acceleration?
\begin{figure}[t]
\includegraphics[width=0.48\textwidth]{CumulativePlot2.pdf}
\caption{Cumulative evidence for the SN PANTHEON (black) sample and QSOsHD (blue). }
\label{FigCumulative}
\end{figure}
We show in Fig. \ref{Fig2} the $\Delta \mu$ values for binned (bin width 0.2) data. Black (Blue) data points stands for the binned Pantheon SNHD (QSOsHD). We have performed this with different bin configuration and the general aspect of the binned data remains the same. The two black points in the SNHD with no error bars represent bins with a single SN. We also plot the expected (theoretical) evolution of the estimator (8) which can be computed by replacing $\mu_{obs}$ by $\mu_{th}$ where the latter is calculated with (\ref{mu}) for a given cosmological model.
Solid (Dashed) lines assume $\Omega_{m0}=0.3$ $(\Omega_{m0}=0.45)$. Both curves assume a cosmological constant behavior for dark energy. The Milne universe is represented by the horizontal line at $\Delta \mu =0$. The Einstein-de Sitter universe $\Omega_{m0}=1$ corresponds to the dotted line fully in the region $\Delta \mu <0$.
The inset in Fig. \ref{Fig2} shows the expected $\Delta \mu$ for different cosmologies. It is assumed a dark energy equation of state parametrized by the CPL equation of state (EoS) parameter $w_{de}=p_{de}/\rho_{de}=w_0+w_1(1-a)$, where $p_{de}$ $(\rho_{de})$ is the dark energy pressure (energy density). The quantity $\Delta \mu$ is plotted (in the inset) for various parameters values in the range $-1.2 < w_0 < -0.8$ and $-0.2 < w_1 < +0.2$. Of course the cosmological constant behavior is within this parameter range at $w_0=-1$ and $w_a=0$. For all possible EoS parameter value in the adopted range $\Delta \mu$ remains positive in the interval $0<z<5.1$ for $\Omega_{m0}=0.3$. Therefore one can not expect $\Delta \mu < 0$ in an accelerated cosmology with $\Omega_{m0}=0.3$ when data up to $z\sim 5$ is used. It is therefore worth noting that the results of Table \ref{TableII} demonstrate that the QSOsHD is inconsistent with the standard cosmology $\Omega_{m0}\sim 0.3$ value.
The dashed lines in Fig. \ref{Fig2} show however that $\Delta\mu < 0$ values are expected in a $\Omega_{m0}\sim 0.45$ universe if data in the range $2<z<5$ is available. A $\Lambda$CDM universo with $\Omega_{m0}=0.45$ is indeed accelerated today. In this case the transition redshift from the decelerated phase to the accelerated one is $z^{0.45}_{tr}=0.35$. This fact can therefore yield to the conclusion that there is a hint for $\Omega_{m0} \gtrsim 0.4$ in the QSOsHD as already mentioned in Ref. \cite{Risaliti:2018reu}. However, $\Delta \mu$ has only a weak dependence on the dark energy equation of state parameters. The estimator is indeed more sensitive to $\Omega_{m0}$ than $w_{de}$ values.
\begin{figure}[t]
\includegraphics[width=0.48\textwidth]{FigTheoBinDeltamu.pdf}
\caption{Magnitude $\Delta \mu$ averaged over redshift bins of width $0.2$ for SN PANTHEON (black) sample and QSOsHD (blue). The inset shows the theoretical evolution of the estimator $\Delta \mu$ as a function of the redshift. Solid lines adopt $\Omega_{m0}=0.3$ and remain at positive $\Delta \mu$ values for the interval $0< z < 5$. Cosmologies adopting $\Omega_{m0}=0.45$ are plotted in the dashed lines. The dotted line represents the Einstein de-Sitter cosmology $\Omega_{m0}=1$. Both set of curves plotted in the inset assume parameters values in the range $-1.2 < w_0 < -0.8$ and $-0.2 < w_a < +0.2$. The solid and dashed lines shown with the binned data assumed the $\Lambda$CDM model. }
\label{Fig2}
\end{figure}
\section{Final Remarks}
The Hubble diagram of high redshift objects is a promising tool to increase our understanding the universe's evolution. Risaliti \& Lusso \cite{Risaliti:2018reu} reported recently a compilation of 1598 objects filling the Hubble diagram in the redshift range $0.5<z<5.1$.
We have analysed in this work this sample under the perspective of a model-independent estimator for the background accelerated expansion according the description done in section \ref{sect2}. This estimator provides whether or not there is positive evidence for acceleration in Hubble diagrams data sets. The main result of this work is to show that one can not infer from the Hubble diagram of Quasars that the universe experienced an accelerated phase along its evolution.
We found that the estimator is weakly dependent on the dark energy equation of state. It is therefore not possible to state a tension with standard cosmology as claimed in Ref. \cite{Risaliti:2018reu}. There is indeed a hint for higher $\Omega_{m0}$ values in agreement with obtained in Ref. \cite{Risaliti:2018reu}. But this does not indicate a strong indication for a cosmological tension relating the $\Omega_{m0}$ parameter since the QSOsHD is apparently not self-consistent, via the $\Delta \mu$ test, when QSOs data in the interval $z<2$ is used. Again, even for $z<2$ data the QSOsHD does not provide evidence for acceleration. In this redshift interval, while one expects $\Delta \mu >0$ for a $\Omega_{m0}=0.45$ cosmology (see Fig. \ref{Fig2}) the inferred value from the sample yields to negative $\Delta \mu$ values (as in Table \ref{TableII}). This means that QSOs can not indicate accelerated expansion which is widely confirmed fact. The source of such discrepancy found in this work is related to the fact that the QSOsHD is still very scattered. Further refinements of the sample are necessary to build a trustful Hubble diagram.
{\bf Acknowledgments:} We thank E. Lusso for providing the QSOsHD data used in this work. We appreciated discussion with D. Schwarz. HV thanks CNPq and FAPES for partial financial support. SG thanks CAPES for financial support.
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The Full Wiki
More info on Partick
1911 encyclopedia
Partick: Quiz
Related quizzes
Glasgow quiz
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Categories: Burghs
Question 1: Partick (Partaig in Gaelic) (formerly Perdyc or Perthick) is an area of ________ on the north bank of the River Clyde, just across from Govan.
Liverpool Edinburgh Dundee Glasgow
Question 2: The well known comedian ________ was a Partick resident as a child.
Transatlantic Years Billy Connolly The Secret Policeman's Balls Billy Connolly Bites Yer Bum!
Question 3: Partick railway station is a trunk station serving as an interchange between the local rail, ________ and local bus systems.
Kelvinhall subway station Exhibition Centre railway station Glasgow Subway Glasgow Central station
Question 4: Being within the sphere of influence of the ________ and neighbouring Glasgow's salubrious 'West-End' it has a high student population.
University Marine Biological Station Millport University of Edinburgh University of Birmingham University of Glasgow
Question 5: Partick Thistle Football Club were formed in the area in 1876, but left to play in the ________ area of Glasgow in 1909.
Maryhill Gilshochill Govan Deaconsbank
Question 6: As well as being the fifth busiest train station in ________, it is the only transport hub to connect three different types of public transport.
United Kingdom Scotland England Wales
Question 7: Welsh perth, 'bush or thicket'), adopted into Scottish Gaelic as "Peart(h)aig", giving modern Gaelic "Pearraig" or "Partaig" (the latter form in use on signage at ________).
Anniesland railway station Bridgeton railway station Partick station Argyle Street railway station
Question 8: The Kings of ________ had a residence there, and in 1136 David I (1124-53) granted the lands of Perdyc to the see of Glasgow.
Highland (council area) Strathclyde Local Government (Scotland) Act 1973 Local Government etc. (Scotland) Act 1994
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\section{Introduction}
Graphene based structures offer unique opportunities to engineer
electronic band structure by shape alone.
\cite{novoselov:2004,castroneto:2009} Infinite graphene sheets
exhibit a conic spectrum but finite sized graphene nanostructures
yield a surprisingly broad array of interesting band features. A
subset of graphene nanostructures reveal flat bands. Theoretical
work shows that flat bands can be found, e.g., at the edges of
two-dimensional graphene, \cite{nakada:1996} in one-dimensional
graphene nanoribbons, \cite{nakada:1996,lin:2009,potasz:2010}
hydrogenated graphene nanoribbons, \cite{kusakabe:2003} graphene
dots, \cite{ezawa:2007} and graphene antidots. \cite{vanevic:2009}
Electrons in flat kinetic energy bands pose challenging
theoretical problems. The absence of any dispersion leaves the
Coulomb interaction to govern the low energy physics. Many common
approximations fail in the extreme flat-band limit. A single flat
band cannot lead to intra-band screening as in ordinary Fermi
liquids, e.g., two-dimensional graphene sheets.\cite{dassarma:2007}
Magnetic properties in bulk graphene in particular occur in a regime where large screening effects
(allowed by a dispersive kinetic energy) minimize the impact of the long-range intra-band Coulomb interaction between electrons
(See, e.g., Refs.~\onlinecite{esquinazi:2003,peres:2005,herbut:2006,ohldag:2007,sheehy:2007,huang:2009,cervenka:2009}). Flat
kinetic energy bands, by contrast, do not allow screening and therefore strongly emphasize interaction effects by default.
Furthermore, conventional perturbative
treatments of the interaction (in comparison to the kinetic energy)
fail in flat-bands due to the absence of a small parameter.
Most theoretical studies of interactions in flat bands use the
Hubbard model with an on-site term.\cite{nagaoka:1966,tasaki:1998a,mielke:1999} The on-site
Hubbard model incorporates just the energy penalty for two electrons
to occupy the same site while ignoring the long range part of the
Coulomb interaction. The on-site term leads to surprising ground
states in the flat-band Hubbard model. For example, work by Nagoaka
\cite{nagaoka:1966} finds ferromagnetism in flat bands at specific
fillings, near one particle per site. This is in stark contrast to
antiferromagnetism favored by super exchange in dispersive bands.
Graphene edges, nanoribbons, and dots present physical systems
hosting flat bands. Theoretical modeling typically relies on the
on-site Hubbard model to make predictions. For example, work
studying flat bands in on-site Hubbard models of zig-zag nanoribbons
\cite{fujita:1996,yazyev:2011} uses meanfield theory to argue for
ferromagnetic states along nanoribbon edges but antiferromagnetic
coupling between edges. An ab initio calculation \cite{lee:2005} and
a work using both the weak-coupling renormalization group and the
density-matrix renormalization-group calculation
\cite{hikibara:2003} provide similar results.
\begin{figure}[t]
\centerline{\includegraphics [width=3 in] {fig1}} \caption{(Color
online) Top: Schematic of a zig-zag nanoribbon of carbon atoms.
Bottom: Schematic of a ferromagnetic crystal with one electron for
every three unit cells in one band. The shaded areas correspond to
single unit cells and the arrows indicate aligned electron spins.}
\label{ribbon}
\end{figure}
Motivated by recent experiments on graphene
nanoribbons,\cite{tao:2011} we construct interacting lattice models
of electrons in flat-band nanoribbons. We focus on zig-zag
nanoribbons because here, in contrast to arm-chair ribbons, two flat
bands arise near the Fermi level even in the absence of adsorbates.
\cite{nakada:1996} In the top panel of Fig.~\ref{ribbon}, we
schematically show a zig-zag nanoribbon where $R_0$ ($\sim 2.46
\mathrm{{\AA}}$) labels the width of a unit cell along the ribbon
($x$ direction) and $L_y$ labels the number of zig-zag chains across
the ribbon ($y$ direction). At low densities the absence of
intra-band screening in flat bands suggests that the long-range part
of the Coulomb interaction is relevant. We therefore construct
models that include even the long-range part of the interaction. We choose to
model a very specific regime: flat-bands in zig-zag nanoribbons,
because we expect the absence of conventional screening to cause flat-band electrons to order in a way which is
completely distinct from electrons in bulk graphene.
The goal of our work is to establish a set of working Hamiltonians
of zig-zag nanoribbons. We construct a single-particle basis of
Wannier functions. We use our basis to compute the interaction
matrix elements. We then establish a projection protocol that sets
up approximate flat-band models. Projection into flat bands
delocalizes basis states due to quantum interference. The resulting
flat-band models are highly non-trivial (incorporating two bands,
long-range interactions, and spin) and can lead to many quantum
ground states even in the absence of significant dispersion. We
make simple estimates of the low energy properties of our models at
odd denominator fillings of a single band.
We argue that, at low densities, the long-range part of the Coulomb
interaction supports ferromagnetic quantum crystals (bottom panel of
Fig.~\ref{ribbon}). Crystalline order projected into the flat band
incorporates quantum superpositions because basis states delocalize.
At low fillings direct spin exchange leads to an effective
Heisenberg model. Our simple estimates therefore predict
ferromagnetic crystalline order in certain parameter regimes. Our
work sets the stage for more accurate studies of our models with a
general class of Jastrow-correlated wavefunctions that apply to flat
bands. \cite{wang:2011}
Our protocol differs from conventional band-structure calculations.
Flat bands, in contrast to dispersive bands, are, by default,
strongly interacting. Conventional applications of density
functional theory accurately model the effect of core electrons
while making very local approximations for the Coulomb interaction
between mobile electrons. Flat bands require accurate treatment of
the long-range portion of the unscreened Coulomb interaction between
otherwise mobile electrons.
In Section~\ref{singleparticle} we consider the band structure that
arises from non-interacting tight-binding models of zig-zag
nanoribbons. Two flat bands are identified. In
Section~\ref{wanniersection} we construct localized single-particle
basis states, orthonormal Wannier functions, from carbon $\pi_{z}$
orbitals in the honeycomb lattice model of zig-zag nanoribbons.
Sections~\ref{oneband} and ~\ref{twoband} use the Wannier functions
to explicitly compute Coulomb interaction matrix elements for one
and two flat bands, respectively. Section~\ref{projection} defines a
projection scheme which limits the total many-body model to the
flat-band portion of the single-particle spectrum.
Section~\ref{lowenergy} sorts terms in the many-body model to argue
that, at low fillings, energetics favor ferromagnetic quantum
crystals. Section~\ref{summary} summarizes and looks forward to more accurate studies of the models constructed here.\\
\section{Flat Bands in Zig-Zag Graphene Nanoribbons}
\label{singleparticle}
We consider interacting electrons hopping among carbon sites forming
zig-zag graphene nanoribbons (Fig.~\ref{ribbon}). We first model the
electrons in a simple non-interacting tight-binding picture. The
single-particle tight-binding Hamiltonian is: \cite{castroneto:2009}
\begin{eqnarray}
H_{0}=-t\sum_{\langle n,m\rangle}(\hat{c}_{n}^{\dagger}\hat{c}_{m}^{\vphantom{\dagger}}+\text{h.c.}),
\label{H0}
\end{eqnarray}
where the hopping integral is $t\sim 2.7$ eV for graphene
\cite{castroneto:2009} and the sum is along bonds of the honeycomb
lattice. The second-quantized operator $\hat{c}_{n}^{\dagger}$
creates a fermion at a site $n$. Labels $n$ and $m$ indicate
lattice sites, in contrast to labels for unit cells, $i,j,k,l$, used
in the following.
Two bands near the Fermi level flatten for large ribbon widths.
\cite{nakada:1996} An example band structure for a narrow width,
$L_y=4$, is shown in Fig.~\ref{band}. Near the fermi surface, the
conduction band (upper band, $u$) and valence band (lower band, $d$)
are nearly degenerate for wavevectors $q$ in the region $qR_0 \in
[2\pi/3,4\pi/3]$ and form flat bands. For larger widths the bands
flatten considerably.
We examine the band width with simple ansatz flat-band
single-particle states. \cite{nakada:1996} Considering states in
the region $qR_0 \in [2\pi/3,4\pi/3]$ with even $L_y$:
\begin{eqnarray}
\phi_{\pm}(q,y)&=&(\phi_{A}(q,y),\pm\phi_{B}(q,y))^{T} \nonumber \\
&=&((-u_{q})^{y-1},\pm(-1)^{y-1}(u_{q})^{L_y-y})^{T},
\label{disper1}
\end{eqnarray}
for $y=1,...,L_y$ where $u_{q}\equiv2\mathrm{cos}(qR_0/2)$, the
energy dispersion in band $\Gamma=u,d$ can be computed analytically:
\begin{eqnarray}
|E_{\Gamma}(q)|&\approx&|\phi(q,y)^{T}H_{0}(q)\phi(q,y)|/|\phi(q,y)|^2 \nonumber\\
&=&t(1-u_{q}^2)u_{q}^{L_y}/(1-u_{q}^{2L_y}),
\label{dispersion}
\end{eqnarray}
with
\begin{equation}
H_{0}(q)=t\left(\begin{array}{cc}
0 & Q(q) \\
Q^{\dagger}(q) & 0
\end{array}\right),
Q(q)=\left(\begin{array}{cccc}
u_{q} & 0 & .. & 0 \\
1 & u_{q} & 0 & .. \\
: & : & : & : \\
0 & .. & 1 & u_{q}
\end{array}\right).
\nonumber
\end{equation}
Figure~\ref{band} compares Eq.~(\ref{dispersion}) with the exact results from Eq.~(\ref{H0}).
\begin{figure}[t]
\centerline{\includegraphics [width=3 in] {fig2}} \caption{(Color
online) The dot-dashed lines indicate the energy eigenvalues of
Eq.~(\ref{H0}) versus wavevector for a nanoribbon of width $L_{y}=4$
and a $q$-space mesh of $N=44$. The solid line shows the
approximate expression for the energy, Eq.~(\ref{dispersion}). Two
flat bands form near $qR_{0}=\pi$. In the large $L_{y}$ limit, the
bands flatten for $2\pi/3\leq qR_{0}\leq4\pi/3$. } \label{band}
\end{figure}
Eq.~(\ref{dispersion}) can be used to determine the bandwidth. For
partially filled lattices a narrow range of single-particle basis
states will be occupied. The bandwidth for states in the flat-band
sector vanishes for ribbons with large width:
\begin{eqnarray}
\vert E_{\Gamma}(|q-\pi|\rightarrow\pi/3)\vert\rightarrow \frac{t}{L_{y}}.
\end{eqnarray}
From this estimate we see that band dispersion plays a small role
for dilute ribbons with increasing ribbon widths.
A vanishing bandwidth, due to quantum interference, leaves the
interaction as the dominant term in the many-body Hamiltonian for
electrons. For dilute ribbons we will work in the approximation
that $H_{0}$ adds an overall constant energy shift to the spectrum.
The full Hamiltonian adds the unscreened Coulomb interaction:
\begin{eqnarray}
H_{\text{total}}=H_{0}+H_{V}.
\label{htotal}
\end{eqnarray}
In the following we treat the dispersion as a small correction to
the interacting term. We project the Hamiltonian into the basis of
flat-band states. Our model becomes:
\begin{eqnarray}
H_{\text{total}}&=&\sum_{\mathbf{q}\in \mathrm{BZ},\sigma,\Gamma}E_{\Gamma}(q)\hat{c}_{\mathbf{q}\sigma\Gamma}^{\dag}\hat{c}_{\mathbf{q}\sigma\Gamma}^{\vphantom{\dagger}}
+H_V\nonumber\\
&\rightarrow&\text{constant}+\mathcal{P}^{\dagger}_{\mathrm{FB}}H_V\mathcal{P}_{\mathrm{FB}}^{\vphantom{\dagger}},
\label{eqnprojection}
\end{eqnarray}
where the first equality is written in terms of the creation (annihilation) operators $\hat{c}_{\mathbf{q}\sigma\Gamma}^{\dag}$ ($\hat{c}_{\mathbf{q}\sigma\Gamma}^{\vphantom{\dagger}}$) for Bloch states at wavevector $q$ and band $\Gamma$ in the Brillouin zone (BZ), which are related to the operators for single-particle basis states by a Fourier transform:
\begin{eqnarray}
\hat{c}_{j\sigma\Gamma}^{\dag}=\frac{1}{\sqrt{N}}\sum_{\mathbf{q}\in \mathrm{BZ}}e^{i\mathbf{q}\cdot\mathbf{R}_j}\hat{c}_{\mathbf{q}\sigma\Gamma}^{\dag}.
\label{creator}
\end{eqnarray}
Here $\mathbf{R}_j$ is the lattice vector of the $j$th unit cell,
$N$ defines the number of unit cells and $q$-space mesh, and
$\sigma\in\{\uparrow, \downarrow\}$ denotes spin.
$\mathcal{P}^{\dagger}_{\mathrm{FB}}$ denotes projection into flat
bands such that the many-body eigenstates are constructed from Bloch
states with $qR_0 \in [2\pi/3, 4\pi/3]$. Many-body states
incorporating these values of $q$ will have essentially no kinetic
energy. We consider this model as a centerpiece to understanding the
electronic properties of flat-band nanoribbons at low densities.
To explore possible many-body states in zig-zag nanoribbons we
construct an accurate form for Eq.~(\ref{eqnprojection}) in the
flat-band basis. We note that the absence of any dispersion
excludes intra-band screening as in ordinary Fermi liquids. Thus
many-body eigenstates are determined entirely by the interplay
between various terms in the interaction. It is therefore crucial
to accurately determine the interacting terms in
Eq.~(\ref{eqnprojection}) as prescribed by our choice of
single-particle basis. To construct an accurate single-particle
basis we revisit the underlying simple tight-binding model formed
from overlapping $\pi_{z}$ orbitals. We construct orthonormal
Wannier functions from these orbitals. The Wannier functions will
serve as single-particle basis states, allowing the construction of
competing terms in a many-body model.
\section{Single-Particle Basis States: Flat-Band Wannier Functions}
\label{wanniersection}
In this section we construct a set of single-particle basis states
in nanoribbon flat bands. We superpose carbon $\pi_z$ orbitals to
form orthogonal Wannier functions. The Wannier functions will then,
in later sections, be used to accurately determine interaction
matrix elements.
\begin{figure}[t]
\centerline{\includegraphics [width=3 in] {fig3}} \caption{(Color
online) Two-dimensional Wannier functions plotted as a function of
position in the lattice for a ribbon of width $L_{y}=4$. The
Wannier functions tend to localize near the ribbon edges.}
\label{wannier}
\end{figure}
In an isolated band the Wannier functions are given by:
\begin{eqnarray}
W_j(\mathbf{r})=W_0(\mathbf{r}-\mathbf{R}_j)=\frac{V}{(2\pi)^D}\oint_{\mathrm{BZ}}d\mathbf{q}e^{-i\mathbf{q}\cdot\mathbf{R}_j}\Psi_{\mathbf{q}}(\mathbf{r}),
\label{eq1}
\end{eqnarray}
where $D$ is the dimension, $V$ is the volume of unit cell. The
Bloch functions are
$\Psi_{\mathbf{q}}(\mathbf{r})=\sum_{m=1}^{M}C_{mq}\chi_{mq}(\mathbf{r})$,
with $M$ atomic sites per unit cell.
To make contact with first principles calculations on graphene
nanoribbons \cite{castroneto:2009} we form Bloch functions from
carbon $\pi_z$ orbitals, $\phi(\mathbf{r})=\sqrt{\xi^5/\pi}ze^{-\xi
r}$. The basis states become
$\chi_{mq}(\mathbf{r})=(1/\sqrt{N})\sum_{j=0}^{N-1}e^{i\mathbf{q}\cdot\mathbf{R}_{j}}\phi(\mathbf{r}-\mathbf{R}_{j}-\mathbf{T}_m)$,
where $\mathbf{T}_m$ is the location of the $m$th atom in the unit
cell.
The coefficients $C_{mq}$ and energy eigenvalues $E(q)$ are obtained
from diagonalization of the secular equation:
\begin{eqnarray}
\left[ \tilde{O}^{-1} \tilde{H}(q) \right] {\bf C}_{q}=E(q){\bf C}_{q},
\label{matrixequation}
\end{eqnarray}
where the matrix $\tilde{H}$ follows from the tight-binding
Hamiltonian $H_0$: $\tilde{H}(q)_{mn}=\int
d\mathbf{r}\chi_{mq}^{*}(\mathbf{r}) H_0 \chi_{nq}(\mathbf{r})$ and
the elements of the overlap matrix $\tilde{O}$ are given by
$O_{mn}=\int
d\mathbf{r}\chi_{mq}^{*}(\mathbf{r})\chi_{nq}(\mathbf{r})$. The
eigenvectors ${\bf C}_{q}\equiv\{C_{1q},...,C_{Mq}\}^{T}$ yield the
coefficients used in the definition of the Wannier functions. In
the tight-binding approximation we set $O_{mn}$ proportional to the
elements of the identity matrix, $\delta_{mn}$.
\begin{figure}[t]
\centerline{\includegraphics [width=3 in] {fig4}} \caption{(Color
online) Left panel: Schematic of the energy dispersion for a wide
ribbon with the Fermi level between the degenerate energy bands $u$
and $d$. In this regime low lattice filling allows us to accurately
ignore the finite dispersion near the band edges. Right panel: The
same as the left panel but at larger fillings of the upper $u$ band.
Here the flat-band approximation will only be a good approximation
if the Coulomb interaction is much larger than the band width. }
\label{Ek}
\end{figure}
We solve Eq.~(\ref{matrixequation}) to construct orthonormal Wannier
functions. We consider a one-dimensional lattice of unit cells along
the nanoribbon. The discrete wavevectors become $\mathbf{q}=(2\pi
q/NR_0) \hat{\mathbf{x}}$. The Wannier function located at
$\mathbf{R}_j$ is then:
\begin{eqnarray}
W_j(\mathbf{r})=\frac{1}{N}\sum_{q=0}^{N-1}e^{-i2\pi qj/N}\Psi_{q}(\mathbf{r}).
\label{eq3}
\end{eqnarray}
The Wannier functions defined in this way are unique for a $D=1$
single band model \cite{kohn:1959} but for higher dimensions and
with more bands they are not necessarily unique. \cite{marzari:1997}
We choose a specific set of single-particle basis states by
enforcing $C_{mq}=|C_{mq}|$ at the edge atomic site $m=1$. As a
result we obtain a set of real Wannier functions symmetric about the
$x$ axis.
The above Wannier function can be written as a summation over all
local atomic orbitals $\phi(\mathbf{r})$ located at sites
$\mathbf{r}_{mi}=\mathbf{T}_m+\mathbf{R}_{i}$. Rewriting $W$ at the
origin gives:
\begin{eqnarray}
W_0(\mathbf{r})=N_f \sum_{m=1}^{M}\sum_{i=0}^{N-1}\alpha_{mi}\phi(\mathbf{r}-\mathbf{r}_{mi}),
\label{eq4}
\end{eqnarray}
with weights $\alpha_{mj}=\sum_{q=0}^{N-1}C_{mq}e^{i2\pi qj/N}$ and
normalization constant $N_f$. The coefficients $\alpha$ completely
determine our choice of basis.
We can extend our calculation of the Wannier functions to include
both the upper and lower bands. A denser sampling in momentum space
(i.e., larger $N$) yields more accurate Wannier functions. In
practice, we find that the Wannier function has already converged
when taking $N=44$ for $L_{y}=4$. The Wannier functions of upper and
lower bands for the same sample ribbon are shown in
Fig.~\ref{wannier}. We note that the Wannier functions localize
symmetrically about $x=0$ with an extension of less than four unit
cells. The Wannier functions are also symmetric (antisymmetric)
along $y$ for the upper (lower) band.
The flat-band Wannier functions constructed here correspond to a
specific choice of single-particle basis. By constructing
superpositions of these functions we can equivalently construct a
model using basis states localized on either edge of the ribbon via
a simple rotation in the two-band space. Viewed in this way our
model implicitly includes inter-edge coupling in narrow ribbons
because we work in the basis of $u$ and $d$ bands as opposed to a
two-edge basis.
Our approach can be used to model graphene edges. Our study applies to the edge states
of very wide ribbons provided we superpose our $u$ and $d$ band Wannier functions to construct
left and right edge Wannier functions. Our model can then be used to study edges of very wide
ribbons. But we stress that our model \emph{cannot} apply to the electrons
in the center of graphene because we have considered bands in nanoribbons that carry over only to edge states
in the wide ribbon limit (For a discussion see Ref.~\onlinecite{nakada:1996}). In what follows we focus on
narrow ribbons and only consider Wannier functions in the $u$ and $d$ band
basis.
\section{One-Band Coulomb Model}
\label{oneband}
Interaction effects determine the low energy properties of
Eq.~(\ref{htotal}) in the absence of significant dispersion. When
the chemical potential lies between the nearly flat bands of zig-zag
nanoribbons, the Coulomb interaction sets the dominant energy scale
and mitigates response. Figure~\ref{Ek} shows schematic band
structures for a wide ribbon with the chemical potential at the band
degeneracy (left) and far from the flat-band region (right). In
what follows we focus on dilute systems corresponding to the left
panel. We can, as a first approximation, assume that the valence
band is inert and that only the conduction band, $u$, will be active
under external probes. Projection into the flat $u$ band implies
that the Coulomb interaction alone operates in the massively
degenerate subspace formed from $u$ band single-particle basis
states. In this section we will consider the $u$ band only. In the
following section we will construct a model of both the $u$ and $d$
bands.
We consider an unscreened Coulomb interaction in a single band:
\begin{eqnarray}
\sum_{i,j,k,l,\sigma\sigma^{'}}\mathcal{V}_{ijkl}\hat{c}_{i\sigma}^\dag \hat{c}_{j\sigma^{'}}^\dag \hat{c}_{k\sigma^{'}}^{\vphantom{\dagger}}\hat{c}_{l\sigma}^{\vphantom{\dagger}},
\end{eqnarray}
where the second-quantized operators $\hat{c}_{i\sigma}^{\dag}$
($\hat{c}_{i\sigma}^{\vphantom{\dagger}}$) create (annihilate) a
fermion with spin $\sigma$ in a Wannier state centered at the $i$th
unit cell. The matrix elements $\mathcal{V}$ depend on the basis.
We can rewrite the Coulomb interaction in the $u$ band in a
suggestive form:
\begin{eqnarray}
H_V^{u} &=&V_0\sum_{i} n_{i\uparrow}n_{i\downarrow}+\sum_{i<j}V_{ij}n_in_j-\sum_{i< j}J_{ij}\mathbf{S}_i\cdot\mathbf{S}_j\nonumber\\
&+&\frac{1}{2}\sum_{\{i,j\}\nsubseteq\{k,l\},\sigma\sigma^{'}}V_{ijkl}\hat{c}_{i\sigma}^\dag \hat{c}_{j\sigma^{'}}^\dag \hat{c}_{k\sigma^{'}}^{\vphantom{\dagger}}\hat{c}_{l\sigma}^{\vphantom{\dagger}}.
\label{onebandH}
\end{eqnarray}
Here, the single-component and total density operators are
$n_{i\sigma}=\hat{c}_{i\sigma}^{\dag}\hat{c}_{i\sigma}^{\vphantom{\dagger}}$
and $n_i=n_{i\uparrow}+n_{i\downarrow}$, respectively. The spin
operators
$\mathbf{S}_i=(1/2)\sum_{\sigma\sigma^{'}}\hat{c}_{i\sigma}^{\dag}${\boldmath$\tilde{\sigma}$}$_{\sigma\sigma^{'}}\hat{c}_{i\sigma^{'}}^{\vphantom{\dagger}}$
are defined in terms of the Pauli matrices
{\boldmath$\tilde{\sigma}$}.
Eq.~(\ref{onebandH}) keeps all terms in the full Coulomb
interaction. We compute the matrix elements in the basis of Wannier
functions in the $u$ band. Integral equations for the coefficients
are given in the appendix, Eqs.~(\ref{twobandcoefficients}). The
first term is the ordinary single-site Hubbard term which is the
only term that is commonly used in models of flat-band nanoribbons
(See, e.g., Refs.~\onlinecite{yazyev:2011} and
\onlinecite{fujita:1996}). The second term captures the diagonal
portion of the Coulomb interaction at long range. The absence of a
dispersion implies that these terms can be relevant and must be kept
in accurate models, especially at low fillings. The third term, the
direct exchange term, favors ferromagnetism for $J_{ij} >0$. The
last term represents remaining off-diagonal terms due to the Coulomb
interaction. We find, by direct calculation, that the last terms
are very small compared to the other terms for a single band.
We compute coefficients in Eq.~(\ref{onebandH}) explicitly. We
perform the integrals in Eqs.~(\ref{twobandcoefficients}) by
approximating the exponential part of the $\pi_z$ orbital,
$\phi(\mathbf{r})$, as a linear combination of three Gaussian
functions:
$\sum_{s}\gamma_s(128\beta_s^5/\pi^3)^{1/4}ze^{-\beta_sr^2}$. We
obtain the parameters $\gamma_s$ and $\beta_s$ from the STO-3G
package. \cite{emsl} Data for fitting the $\pi_z$ orbital with
$\xi=1.72$ are listed in Table \ref{tab1}. For numerical results
shown here and in the following sections, we use the Bohr radius,
$a_0=0.53 \mathrm{{\AA}}$, as the unit of length and the Coulomb
energy $e^2/4\pi\epsilon a_0$ ($\sim$ 27.2 eV in vacuum) as the
unit of energy.
\begin{table}[t]
\caption{Fitting parameters for the Gaussian approximation to the
$\pi_z$ orbital with $\xi=1.72$.} \centering
\begin{tabular}{|l|c|c|c|}
\hline
$s$ & 1 & 2 & 3 \\[-.1ex] \hline
$\gamma_s$ & 0.15591627 & 0.60768372 & 0.39195739 \\[-.1ex]
$\beta_s$ & 2.9412494 & 0.6834831 & 0.2222899 \\
\hline
\end{tabular}
\label{tab1}
\end{table}
\begin{table}[t]
\caption{Matrix elements for one-band (u band) case for inter-unit
cell separations of up to $4R_{0}$. } \centering
\begin{tabular}{|c|c|c|c|c|}
\hline
\multicolumn{5}{|l|}{$V_0$=2.24$\times10^{-1}$} \\
\hline
$|i-j|$ & 1 & 2 & 3 & 4 \\
$J_{ij}$ & 2.34$\times10^{-2}$ & 4.68$\times10^{-3}$ & 9.21$\times10^{-4}$ & 1.69$\times10^{-4}$ \\
$V_{ij}$ & 1.43$\times10^{-1}$ & 9.56$\times10^{-2}$ & 6.83$\times10^{-2}$ & 5.23$\times10^{-2}$ \\
\hline
\end{tabular}
\label{tab2}
\end{table}
Table~\ref{tab2} lists the coefficients computed for an $L_y=4$
ribbon. As we see, all coefficients are positive and can be sorted
by $V_0 > V_{ij} > J_{ij} > 0$. The ground state can be determined
by an interplay between leading terms in Eq.~(\ref{onebandH}) and
the chemical potential. These coefficients suggest that partially
filled single bands support the formation of ferromagnetic crystals.
However, the large Coulomb interaction may cause mixing between the
$u$ and $d$ bands. In the next section we construct a two-band
model.
\section{Two-Band Coulomb Model}
\label{twoband}
We now consider a more comprehensive two-band model. The $u$ and
$d$ bands in the flat-band region are essentially degenerate for
wide ribbon widths. The Coulomb interaction can in principle favor
occupancy of both bands or the occupancy of a single band. Accurate
estimates of coefficients in the full two-band model will allow
exploration of the two-band energy landscape to determine the band
occupancy in future work.
We construct Wannier functions in both the $u$ and $d$ bands. The
Hamiltonian is dominated by the following terms:
\begin{eqnarray}
H_V^{ud}&=&\sum_{i,\Gamma}V_{0}^{\Gamma}n_{i\Gamma\uparrow}n_{i\Gamma\downarrow}\nonumber\\
&+&\sum_{i}\left (V_{ii}^{'}n_{iu}n_{id}-J_{ii}^{'}\mathbf{S}_{iu}\cdot\mathbf{S}_{id}\right )\nonumber \\
&+&\sum_{i<j,\Gamma}(V_{ij}^{\Gamma}n_{i\Gamma}n_{j\Gamma}-J_{ij}^{\Gamma}\mathbf{S}_{i\Gamma}\cdot\mathbf{S}_{j\Gamma})\nonumber\\
&+&\sum_{i<j}\sum_{\Gamma \neq \Gamma^{'}}(V_{ij}^{'}n_{i\Gamma}n_{j\Gamma^{'}}-J_{ij}^{'}\mathbf{S}_{i\Gamma}\cdot\mathbf{S}_{j\Gamma^{'}})\nonumber\\
&+&\sum_{i<j}\sum_{\Gamma \neq \Gamma^{'}}\sum_{\sigma\sigma^{'}}(
V_{ij}^{''}\hat{c}_{i\Gamma\sigma}^{\dag}\hat{c}_{j\Gamma^{'}\sigma^{'}}^{\dag}\hat{c}_{j\Gamma\sigma^{'}}^{\vphantom{\dagger}}\hat{c}_{i\Gamma^{'}\sigma}^{\vphantom{\dagger}}\nonumber\\
&+&V_{ij}^{'''}\hat{c}_{i\Gamma\sigma}^{\dag}\hat{c}_{j\Gamma^{'}\sigma^{'}}^{\dag}\hat{c}_{i\Gamma^{'}\sigma^{'}}^{\vphantom{\dagger}}\hat{c}_{j\Gamma\sigma}^{\vphantom{\dagger}}).
\label{fullmodel}
\end{eqnarray}
We have checked, by direct calculation, that other terms involving
three and four centers are much smaller than terms kept in Eq.
(\ref{fullmodel}). Here we see the Hubbard and ferromagnetic terms
as in the one-band case. The last term indicates a non-trivial band
exchange term. The integrals for all coefficients are listed in the
Appendix.
\begin{table}[t!]
\caption{Matrix elements for the two-band case with $L_y=4$ for
inter-unit cell separations of up to $4R_{0}$. } \centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\multicolumn{3}{|l}{$V^{d}_{0}$=2.28$\times10^{-1}$} & \multicolumn{3}{l|}{$V^{u}_{0}$=2.24$\times10^{-1}$}\\
\multicolumn{3}{|l}{$V^{'}_{ii}$=1.91$\times10^{-1}$} & \multicolumn{3}{l|}{$J_{ii}^{'}$=1.32$\times10^{-1}$}\\
\hline
$|i-j|$ & 1 & 2 & 3 & 4 & $D_w$ \\
$V_{ij}^{d}$ & 1.44$\times10^{-1}$ & 9.51$\times10^{-2}$ & 6.79$\times10^{-2}$ & 5.21$\times10^{-2}$ & 1.01 \\
$V_{ij}^{u}$ & 1.43$\times10^{-1}$ & 9.56$\times10^{-2}$ & 6.83$\times10^{-2}$ & 5.23$\times10^{-2}$ & 1.02 \\
$V_{ij}^{'}$ & 1.46$\times10^{-1}$ & 9.56$\times10^{-2}$ & 6.81$\times10^{-2}$ & 5.22$\times10^{-2}$ & 1.02 \\
$J_{ij}^{d}$ & 2.60$\times10^{-2}$ & 3.04$\times10^{-3}$ & 5.75$\times10^{-4}$ & 1.09$\times10^{-4}$ & \\
$J_{ij}^{u}$ & 2.34$\times10^{-2}$ & 4.68$\times10^{-3}$ & 9.21$\times10^{-4}$ & 1.69$\times10^{-4}$ & \\
$J_{ij}^{'}$ & 1.62$\times10^{-2}$ & 3.14$\times10^{-3}$ & 6.54$\times10^{-4}$ & 1.27$\times10^{-4}$ & \\
$V_{ij}^{''}$ & 2.06$\times10^{-2}$ & 7.44$\times10^{-3}$ & 2.95$\times10^{-3}$ & 1.35$\times10^{-3}$ & \\
$V_{ij}^{'''}$ & 1.05$\times10^{-2}$ & 1.82$\times10^{-3}$ & 3.49$\times10^{-4}$ & 6.43$\times10^{-5}$ & \\
\hline
\end{tabular}
\label{tablely4}
\end{table}
Eq.~(\ref{fullmodel}) presents a central result of our work. The
two-band model must be studied for different fillings and different
widths to determine expected ground states. Tables~\ref{tablely4}
and ~\ref{tablely10} show numerically computed coefficients for two
example widths, $L_{y}=4$ and $10$.
The tables show that the electron configurations are determined
primarily by the diagonal components of the Coulomb interaction
(rows 1-3). These rows are nearly equal indicating a band symmetry,
as expected. These rows govern the charge degrees of freedom. Rows
4-6 govern the spin degrees of freedom. The positive elements
support ferromagnetism. The last two rows give rise to band
exchange effects.
We construct a simple fitting form for the first three rows. We
note that the coefficients $V_{ij}^{\Gamma}$ and $V_{ij}^{'}$ can be
thought of as a softened Coulomb interaction between smeared charges
located at separate unit cells $i$ and $j$. For large separations
the charges appear as point charges and interact through the Coulomb
interaction but at short ranges our basis states smear the electron
charge over the width of the ribbon. We approximate
$V_{ij}^{\Gamma}$ and $V_{ij}^{'}$ with a convenient analytic form:
\begin{equation}
V_{i\neq j}\approx \left (\frac{e^2}{4 \pi \epsilon a_0}\right)\frac{a_{0}/R_{0}}{\sqrt{|i-j|^2+D_w^2}},
\label{fitting}
\end{equation}
where the fitting parameter $D_w$ is dependent on the width of the
ribbon and can be determined with a numerical fitting as shown in
Figs.~\ref{fittingw4} and \ref{fittingw10}. The last column of
Tables~\ref{tablely4} and ~\ref{tablely10} shows $D_{w}$ obtained by
fitting.
Eq.~(\ref{fitting}) can be used to approximate the coefficients in
Eq.~(\ref{fullmodel}) at low filling. At low filling
$V_{ij}^{\Gamma}$ and $V_{ij}^{'}$ determine the configuration of
charges. It then suffices to consider spin exchange terms at the
separations fixed by $V_{ij}^{\Gamma}$ and $V_{ij}^{'}$. We use
this procedure to suggest possible low energy solutions to
Eq.~(\ref{fullmodel}).
\section{Flat-Band Projection}
\label{projection}
The flat-band limit, Eq.~(\ref{eqnprojection}), establishes a
unique set of non-perturbative models. In this section we construct
a set of operators that allow flat-band projection of models
constructed in the previous sections. In the following section we
will then use the projected models in simple estimates of the low
energy physics.
\begin{table}[t!]
\caption{The same as Table~\ref{tablely4} but for $L_y=10$.}
\centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\multicolumn{3}{|l}{$V^{d}_{0}$=1.21$\times10^{-1}$} & \multicolumn{3}{l|}{$V^{u}_{0}$=1.17$\times10^{-1}$}\\
\multicolumn{3}{|l}{$V_{ii}^{'}$=1.03$\times10^{-1}$} & \multicolumn{3}{l|}{$J_{ii}^{'}$=5.90$\times10^{-2}$}\\
\hline
$|i-j|$ & 1 & 2 & 3 & 4 & $D_w$ \\
$V_{ij}^{d}$ & 8.93$\times10^{-2}$ & 6.96$\times10^{-2}$ & 5.47$\times10^{-2}$ & 4.45$\times10^{-2}$ & 2.09 \\
$V_{ij}^{u}$ & 8.74$\times10^{-2}$ & 6.84$\times10^{-2}$ & 5.41$\times10^{-2}$ & 4.41$\times10^{-2}$ & 2.15 \\
$V_{ij}^{'}$ & 9.08$\times10^{-2}$ & 6.93$\times10^{-2}$ & 5.44$\times10^{-2}$ & 4.43$\times10^{-2}$ & 2.12 \\
$J_{ij}^{d}$ & 2.82$\times10^{-2}$ & 4.80$\times10^{-3}$ & 1.10$\times10^{-3}$ & 3.36$\times10^{-4}$ & \\
$J_{ij}^{u}$ & 2.65$\times10^{-2}$ & 6.99$\times10^{-3}$ & 1.51$\times10^{-3}$ & 4.30$\times10^{-4}$ & \\
$J_{ij}^{'}$ & 1.65$\times10^{-2}$ & 4.68$\times10^{-3}$ & 1.15$\times10^{-3}$ & 3.10$\times10^{-4}$ & \\
$V_{ij}^{''}$ & 1.75$\times10^{-2}$ & 1.00$\times10^{-2}$ & 5.82$\times10^{-3}$ & 3.52$\times10^{-3}$ & \\
$V_{ij}^{'''}$ & 1.28$\times10^{-2}$ & 2.85$\times10^{-3}$ & 6.30$\times10^{-4}$ & 1.85$\times10^{-4}$ & \\
\hline
\end{tabular}
\label{tablely10}
\end{table}
To enforce flat-band projection we limit all $q$-space sums to the
flat-band region (FBR) $qR_0 \in [2\pi/3,4\pi/3]$. We can therefore
project into a single band by considering a flat-band operator that
limits itself to the FBR:
\begin{eqnarray}
\hat{b}_{j\sigma}^{\dag}\equiv\frac{1}{N}\sum_{l}\sum_{\mathbf{q}\in \mathrm{FBR}}e^{i\mathbf{q}\cdot(\mathbf{R}_j-\mathbf{R}_l)}\hat{c}_{l\sigma}^{\dag}.
\label{proj-creator}
\end{eqnarray}
This operator creates states centered around the unit cell at
$\mathbf{R}_j$. We note that the states created by this operator
have finite overlap with neighbors at $\mathbf{R}_{j+1}$ when the
flat-band region does not encompass the entire Brillouin zone. In
the limit that the flat band encompasses the entire Brillouin zone
the overlap between neighboring states vanishes and we have
$\hat{b}_{j\sigma}^{\dag}\rightarrow\hat{c}_{j\sigma}^{\dag}$. Thus,
the projection into a flat band that incorporates only a fraction of
the Brillouin zone delocalizes basis states.
We can rewrite our model in terms of projected density and spin
operators. The single-component and total projected density
operators are $\rho_{i\sigma}\equiv
\hat{b}_{i\sigma}^{\dag}\hat{b}_{i\sigma}^{\vphantom{\dagger}}$ and
$\rho_i\equiv\rho_{i\uparrow}+\rho_{i\downarrow}$, respectively. The
projected spin operators are defined as:
\begin{eqnarray}
\slashed {\mathbf S }_{j} \equiv \frac{1}{2N}\sum_{\sigma\sigma^{'}}\sum_{\mathbf{q},\mathbf{q'}\in \mathrm{FBR}}e^{i(\mathbf{q}-\mathbf{q'})\cdot\mathbf{R}_j}\hat{c}_{\mathbf{q}\sigma}^{\dag}{\boldsymbol {\tilde{\sigma}}}_{\sigma\sigma^{'}}\hat{c}_{\mathbf{q'}\sigma^{'}}^{\vphantom{\dagger}}.
\label{proj-spin}
\end{eqnarray}
We stress that the projected operators do \emph{not} exhibit
ordinary commutation relations because the underlying operators
create overlapping states, i.e., $\langle 0 \vert \hat{b}_{j+1}
\hat{b}_{j}^{\dag}|0\rangle\neq0$.
\begin{figure}[t!]
\centerline{\includegraphics [width=3 in] {fig5}} \caption{The
diagonal component of the inter-band Coulomb interaction
($V_{ij}^{'}$) for a zigzag graphene nanoribbon with $L_{y}=4$ and
$N=44$. Circles are from numerical evaluation of
Eqs.~(\ref{twobandcoefficients}). The solid line is a fit with
Eq.~(\ref{fitting}) and $D_w=1.02$.} \label{fittingw4}
\end{figure}
\begin{figure}[t !]
\centerline{\includegraphics [width=3 in] {fig6}} \caption{The same
as Fig.~\ref{fittingw4} but for $L_{y}=10$ with $D_w=2.12$.}
\label{fittingw10}
\end{figure}
The projected Hamiltonian can be rewritten entirely in terms of the
above projected operators. Starting from an unprojected model, we
impose projection using the following replacements: $c\rightarrow
b,n\rightarrow\rho,$ and ${\mathbf S}\rightarrow\slashed {\mathbf
S}$. For example, the flat-band projected Coulomb interaction in
the $u$ band becomes:
\begin{eqnarray}
\mathcal{P}^{\dagger}_{u}H^{u}_V\mathcal{P}_{u}^{\vphantom{\dagger}}&=&V_0\sum_{i} \rho_{i\uparrow}\rho_{i\downarrow}+\sum_{i<j}V_{ij}\rho_i\rho_j-\sum_{i< j}J_{ij}\slashed {\mathbf S}_i\cdot\slashed {\mathbf S}_j\nonumber\\
&+&\frac{1}{2}\sum_{\{i,j\}\nsubseteq\{k,l\},\sigma\sigma^{'}}V_{ijkl}\hat{b}_{i\sigma}^\dag \hat{b}_{j\sigma^{'}}^\dag \hat{b}_{k\sigma^{'}}^{\vphantom{\dagger}}\hat{b}_{l\sigma}^{\vphantom{\dagger}}.
\label{proj-onebandH}
\end{eqnarray}
The projected two-band model can also be obtained with a similar
replacement applied to $H^{ud}_V$.
\section{Low Energy Properties}
\label{lowenergy}
We use flat-band projection to discuss possible low energy states
of Eq.~(\ref{eqnprojection}) based on simple energetic arguments. A
detailed quantitative analysis of low energy states is beyond the
scope of the present work. We make progress by ordering terms
according to dominant energy scales. We then focus on example
lattice fillings.
To consider low energy solutions of Eq.~(\ref{eqnprojection}) we
first examine the kinetic term. The kinetic term enforces a
flat-band projection provided the chemical potentials lies near the
flat band, i.e., Fig.~\ref{Ek}a. It is then sufficient to require
that many-body eigenstates of $H_{V}$ utilize Bloch states with
$qR_0 \in [2\pi/3, 4\pi/3]$. We can analyze Eq.~(\ref{fullmodel})
with this $q$-space restriction by using projected operators
constructed in the previous section.
We first point out an intrinsic energetic ordering to each of the
terms in Eq.~(\ref{fullmodel}). We rewrite each of the terms
according to an approximate ordering by energy and in the projected
space:
\begin{eqnarray}
\mathcal{P}^{\dagger}_{ud}H_V^{ud}\mathcal{P}^{\vphantom{\dagger}}_{ud}&=&\sum_{i,\Gamma}V_{0}^{\Gamma}\rho_{i\Gamma\uparrow}\rho_{i\Gamma\downarrow} \nonumber \\
&+&\sum_{i, j,\Gamma,\Gamma'}\left(\overline{V}_{ij}^{\Gamma,\Gamma'}\rho_{i\Gamma}\rho_{j\Gamma'}-\overline{J}_{ij}^{\Gamma,\Gamma'}\slashed {\mathbf S}_{i\Gamma}\cdot\slashed {\mathbf S}_{j\Gamma'}\right)\nonumber\\
&+&H_{\text{Band-exch}},
\label{Vsimplified}
\end{eqnarray}
where we have redefined the diagonal Coulomb terms:
$\overline{V}_{i<j}^{\Gamma\neq\Gamma'}\equiv V'_{ij}$,
$\overline{V}_{ii}^{\Gamma=d,\Gamma'=u}\equiv V'_{ii}$, and
$\overline{V}_{i<j}^{\Gamma=\Gamma'}\equiv V^{\Gamma}_{ij}$,
otherwise $\overline{V}_{ij}^{\Gamma,\Gamma'}=0$. (Note that our
direct calculations find
$\overline{V}_{ij}^{\Gamma\neq\Gamma'}\approx
\overline{V}_{ij}^{\Gamma=\Gamma'}$.) We have also redefined the
off-diagonal exchange terms:
$\overline{J}_{i<j}^{\Gamma\neq\Gamma'}\equiv J'_{ij}$,
$\overline{J}_{ii}^{\Gamma=d,\Gamma'=u}\equiv J'_{ii}$, and
$\overline{J}_{i<j}^{\Gamma=\Gamma'}\equiv J^{\Gamma}_{ij}$,
otherwise $\overline{J}_{ij}^{\Gamma,\Gamma'}=0$. The last term in
Eq.~(\ref{Vsimplified}) corresponds to the last term in
Eq.~(\ref{fullmodel}).
We can understand the low energy properties of the first three
terms in Eq.~(\ref{Vsimplified}) at a few specific fillings.
Considering an inert $d$ band, we assume that the $u$ band is
partially filled at odd denominators, $\nu_{u}=1/(2p+1)$, where
$p=1,2,...$. ($\nu$ indicates the number of particles per basis
state.) Ignoring $H_{\text{Band-exch}}$ allows a decomposition of
basis states into the $u$ and $d$ bands. An inert $d$ band implies
that the inter-band interaction leads to an overall shift of the
chemical potential. A strong external gate bias canceling this
shift should be able to maintain the $u$-band filling
$\nu_{u}=1/(2p+1)$.
In the limit of commuting projected density operators it is well
known \cite{hubbard:1978} that the first terms in
Eq.~(\ref{Vsimplified}) lead to a charge order, i.e.,
one-dimensional Wigner crystals with lattice spacing $2p+1$. We
therefore expect that the $u$-band electrons form a classical Wigner
crystal in the limit that the flat band encompasses the entire
Brillouin zone. The bottom panel of Fig.~\ref{ribbon} depicts a
classical crystal configuration in a single spin state.
In the limit that the projected density operators do not commute,
the case for zig-zag nanoribbons, we predict quantum crystals in
partially filled bands. Quantum crystals arise, in direct analogy
to Wigner crystals, as eigenstates of the projected density
operators. For example, a trial quantum crystal state at
$\nu_{u}=1/(2p+1)$ in spin state $\sigma$ is given by:
\begin{eqnarray}
\prod_{j=0}\hat{b}_{2pj+j,\sigma u}^{\dag}\vert 0\rangle.
\end{eqnarray}
This trial state appears to minimize the energy of the first two
terms in Eq.~(\ref{Vsimplified}) by separating flat-band charges by
an average of $2p$ unit cells. Thus the first two terms in
Eq.~(\ref{Vsimplified}) impose a rigid charge order in the $u$ band.
However, the charges are significantly delocalized. A finite
overlap among neighbors implies that the charges exist in a
superposition of several different unit cells at once: a quantum
crystal.
Provided a rigid charge ordering we consider the next lowest energy
scale: low energy spin properties of Eq.~(\ref{Vsimplified}). We
approximate the spin-spin coupling with an effective Heisenberg
model for the u-band particles at $\nu_{u}=1/(2p+1)$:
\begin{eqnarray}
H_{\text{eff}}^{p}=-J^{u}_{0,2p+1}\sum_{i}\slashed{\mathbf S}_{i,u}\cdot \slashed{\mathbf S}_{i+2p+1,u}.
\label{Heff}
\end{eqnarray}
Eq.~(\ref{Heff}) applies to the case of a single band at odd
denominator filling.
The ground states of Eq.~(\ref{Heff}) are ferromagnetic quantum
crystals. The low energy spin excitations are ferromagnetic magnons.
The underlying rigid charge order enforces a large magnon
wavelength. At $\nu_{u}=1/(2p+1)$ spin wave theory yields
excitation energies:
\begin{eqnarray}
\hbar \omega_{q}=2J^{u}_{0,2p+1} \left[1-\cos((2p+1)R_{0}q) \right].
\end{eqnarray}
This dispersion offers a clear indicator of ferromagnetic crystals
in the spin degrees of freedom.
At finite temperatures the Mermin-Wagner theorem asserts that
spin-spin correlations decay with a finite length scale in the
one-dimensional Heisenberg model. \cite{mermin:1966} Thus,
ferromagnetic ordering holds only up to small length scales. The
spin-spin correlation length at non-zero temperatures, $T$, for the
Heisenberg chain with exchange coupling $J$ is: \cite{kopietz:1989}
\begin{eqnarray}
\frac{\xi_{T}}{(2p+1)R_{0}}=\frac{AJ}{4T}\left[1+B(8T/J)^{1/2}/\pi+\mathcal{O}\left(\frac{T}{J}\right)\right],
\end{eqnarray}
where $A\approx1.1$ and $B\approx0.65$. Our results suggest that
for $L_{y}=10$ at $T=1K$ with $J^{u}_{i,i+3}\approx 1.5\times10^{-3}
(e^{2}/4\pi\epsilon_{0}a_{0})\approx 473K$ the correlation length is
$\xi_{T} /(2p+1)R_{0}\approx 133$. Thus about 390 unit cells
containing 130 $u$-band electrons are included in the formation of a
fully magnetized domain at $\nu_{u}=1/3$ for these parameters.
\section{Summary and Outlook}
\label{summary}
We constructed interacting flat-band lattice models of zig-zag
nanoribbons. A single-particle basis of orthonormal Wannier
functions were built from carbon $\pi_{z}$ orbitals in a
honeycomb-ribbon lattice. The single-particle basis was used to
explicitly compute the Coulomb matrix elements for two ribbon
widths, $L_{y}=4$ and $10$. The total model, Eqs.~(\ref{htotal})
and (\ref{fullmodel}), was then projected into the flat bands of the
single-particle spectrum. The projected flat-band model,
Eq.~(\ref{Vsimplified}), suggests ferromagnetic quantum crystal
ground states.
Our flat-band model, Eq.~(\ref{Vsimplified}), sets the stage for
more accurate analyses with a combination of numerics and many-body
wavefunctions. The absence of a small parameter calls for a
combination of variational studies and diagonalization to verify
proposed ground and excited states.\cite{wang:2011} In addition to
crystals discussed here, uniform quantum liquids are also
possible.\cite{wang:2011}
The models constructed here focus on key physics of interacting
flat bands but exclude several realistic effects. In experiments on
graphene nanostructures many corrections may be required before
making a detailed comparison with experiment. For example, edge
roughness, defects, and substrate disorder can destroy the flat-band
approximation. Furthermore, inter-band screening has also been
ignored in the current study. While intra-band screening was
implicitly incorporated in our model, screening from nearby bands
could lead to corrections to the pure Coulomb model studied here,
e.g., RKKY-type interactions. \cite{brey:2007,saremi:2007}
\vspace{0.2cm}
\section{Acknowledgements}
We thank the Thomas F. Jeffress and Kate Miller Jeffress Memorial Trust, Grant No. J-992, for support.\\
\section{Appendix}
The coefficients in Eqs.~\ref{onebandH} and \ref{fullmodel} are given by:
\begin{eqnarray}
V_{0}^{\Gamma}&=&\int\frac{d^2\mathbf{r}d^2\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|}|W_{0\Gamma}(\mathbf{r})W_{0\Gamma}(\mathbf{r'})|^2, \nonumber\\
J_{ij}^{\Gamma}&=&2\int\frac{d^2\mathbf{r}d^2\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|}W_{i\Gamma}^{*}(\mathbf{r})W_{j\Gamma}(\mathbf{r})W_{i\Gamma}(\mathbf{r'})W_{j\Gamma}^{*}(\mathbf{r'}), \nonumber\\
V_{ij}^{\Gamma}&=&\int\frac{d^2\mathbf{r}d^2\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|}|W_{i\Gamma}(\mathbf{r})W_{j\Gamma}(\mathbf{r'})|^2-\frac{1}{4}J_{ij}^{\Gamma}, \nonumber\\
J_{ij}^{'}&=&2\int\frac{d^2\mathbf{r}d^2\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|}W_{iu}^{*}(\mathbf{r})W_{jd}(\mathbf{r})W_{iu}(\mathbf{r'})W_{jd}^{*}(\mathbf{r'}), \nonumber\\
V_{ij}^{'}&=&\int\frac{d^2\mathbf{r}d^2\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|}|W_{iu}(\mathbf{r})W_{jd}(\mathbf{r'})|^2-\frac{1}{4}J_{ij}^{'}, \nonumber\\
V_{ij}^{''}&=&\int\frac{d^2\mathbf{r}d^2\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|}W_{iu}^{*}(\mathbf{r})W_{id}(\mathbf{r})W_{ju}(\mathbf{r'})W_{jd}^{*}(\mathbf{r'}), \nonumber\\
V_{ij}^{'''}&=&\int\frac{d^2\mathbf{r}d^2\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|}W_{iu}^{*}(\mathbf{r})W_{ju}(\mathbf{r})W_{id}(\mathbf{r'})W_{jd}^{*}(\mathbf{r'}), \nonumber\\
V_{ijkl}&=&\int\frac{d^2\mathbf{r}d^2{\mathbf{r}}^{'}}{|\mathbf{r}-{\mathbf{r}}^{'}|}W_{iu}^{*}(\mathbf{r})W_{lu}(\mathbf{r})W_{ju}^{*}({\mathbf{r}}^{'})W_{ku}({\mathbf{r}}^{'}).
\label{twobandcoefficients}
\end{eqnarray}
The last term is used only in Eq.~(\ref{onebandH}).
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,567 |
{"url":"https:\/\/searchworks.stanford.edu\/catalog?q=%22Van+Roy%2C+Benjamin%2C+%22&search_field=search_author","text":"# Search results\n\n## 32 results\n\nView results as:\nNumber of results to display per page\n\n### 1. Transaction-Cost-Conscious Pairs Trading via Approximate Dynamic Programming[2006]Online\n\nCollection\nIn this paper, we develop an algorithm that optimizes logarithmic utility in pairs trading. We assume price processes for two assets, with transaction cost linear with respect to the rate of change in portfolio weights. We then solve the optimization problem via a linear programming approach to approximate dynamic programming. Our simulation results show that when asset price volatility and transaction cost are sufficiently high, our ADP strategy offers significant benefits over the chosen baseline strategy. Our baseline strategy is an optimized version of a pairs trading heuristic studied in the literature.\n\n### 2. Design of large scale nudge engines [electronic resource][2015]Online\n\nBook\n1 online resource.\nCongestion is a widespread problem in modern urban transportation networks; hundreds of billions of dollars are lost each year due to wasted time, extra fuel consumption, traffic accidents, etc. A significant fraction of these losses is due to peak-hour congestion on road networks, which occurs when the demand for transport exceeds capacity over a short time period each day. In this thesis, we explore the feasibility of reducing peak-hour road traffic congestion using incentives to shift drivers' commute times. We first discuss a practical implementation of such an incentive program \u2014 CAPRI (Congestion And Parking Relief Incentives). This program aimed to reduce peak-hour vehicular traffic into and out of Stanford University. Commuters who sign up for CAPRI earn points for the \"good trips\" they make, and these points can be redeemed for rewards (both monetary and in-kind). CAPRI also includes the capability to personalize incentives based on users' historical behavior. To complement the implementation, we develop a theoretical model for optimally reducing the cost of peak-hour congestion with targeted incentives. We study the evolution of congestion on a highway under time-varying demands using fluid models of traffic flow. We then examine the effect of shifting users' commute times on congestion. We show that ideas from the theory of optimal transport of measures can be used to develop cost-effective incentive schemes to reduce congestion. Specifically, we show that the \"cost of congestion\" and the \"cost of nudging\" are closely related to the Wasserstein distance between measures. We use this relationship to formulate linear programming problems to compute personalized recommendations and incentives to nudge drivers to the off-peak hour. We find that the resultant reduction in the cost of congestion is significant and that it helps to prioritize commuters for nudging based on their origin and destination.\n\n### 3. Efficient learning in sequential optimization [electronic resource][2015]Online\n\nBook\n1 online resource.\nWe consider a broad class of online optimization problems in which a decision-maker must balance between exploration and exploitation while learning from partial feedback. In these problems, the decision-maker repeatedly chooses among a set of possible actions, observes an outcome, and receives a reward representing the utility derived from this outcome. She is uncertain about the underlying system and is therefore initially unsure of which action is best. However, as outcomes are observed, she is able to learn over time to make increasingly effective decisions. Her objective is to choose actions sequentially so as to maximize the expected cumulative reward. We focus on three algorithmic approaches that accommodate flexible statistical modeling, and are capable of experimenting efficiently in broad classes of problems. The first part of the thesis focuses on a design principle known as optimism in the face of uncertainty, which underlies many of the most effective exploration algorithms. We provide a regret bound for an optimistic algorithm that applies broadly and can be specialized to many specific model classes. Our bound depends on a new notion of dimension that measures the degree of dependence among actions. We compare our notion to the Vapnik-Chervonenkis dimension, and explain why that and other measures of dimension used in the supervised literature do not suffice when it comes to analyzing optimistic algorithms. We then turn our attention to Thompson sampling, an elegant algorithm for learning in online optimization problems with partial feedback. We derive a close theoretical connection between Thompson sampling and optimistic algorithms. Due to the connection we derive, existing analysis available for specific optimistic algorithms immediately translates to expected regret bounds for Thompson sampling. The second part of the thesis pushes beyond the optimistic principle, and offers a fresh, information-theoretic, perspective on the exploration\/exploitation tradeoff. We first revisit Thompson sampling from this perspective and provide novel regret bounds that scale with the entropy of the optimal action distribution. Then, we propose a new algorithm--information-directed sampling (IDS)--and study its performance. IDS quantifies the amount learned by selecting an action through an information theoretic measure, and selects actions by optimizing an objective that explicitly balances attaining a high immediate reward and selecting informative actions. We provide a general regret bound for IDS, demonstrate strong performance in simulation, and show through simple analytic examples that it can dramatically outperform Thompson sampling due to the way it quantifies information.\n\n### 4. Generalized low rank models [electronic resource][2015]Online\n\nBook\n1 online resource.\nPrincipal components analysis (PCA) is a well-known technique for approximating a tabular data set by a low rank matrix. This dissertation extends the idea of PCA to handle arbitrary data sets consisting of numerical, Boolean, categorical, ordinal, and other data types. This framework encompasses many well known techniques in data analysis, such as nonnegative matrix factorization, matrix completion, sparse and robust PCA, k-means, k-SVD, and maximum margin matrix factorization. The method handles heterogeneous data sets, and leads to coherent schemes for compressing, denoising, and imputing missing entries across all data types simultaneously. It also admits a number of interesting interpretations of the low rank factors, which allow clustering of examples or of features. We propose several parallel algorithms for fitting generalized low rank models, and describe implementations and numerical results.\nSpecial Collections\n\n### 5. Medical image similarity perception for content-based image retrieval [electronic resource][2014]Online\n\nBook\n1 online resource.\nContent-based image retrieval (CBIR) of perceptually similar images for medical decision support may improve the accuracy and efficiency of radiological diagnosis. A robust reference standard for perceptual similarity in medical images is vital to CBIR applications, but is challenging to create due to the large amount of observer data necessary as well as high inter-observer variability. The aims of this thesis are (1) to develop techniques for creating reference standards based on perceptual similarity that are scalable for large databases, and (2) to model inter-reader variability in observer assessments of image similarity. This thesis first discusses a novel technique for predicting visual similarity for pairs of images in a database from perceptual data obtained by viewing each image individually. In an observer study using 19 CT liver lesions in the portal venous phase, three radiologists provided point-wise ratings for visual attributes of images containing liver lesions displayed individually and also for perceptual similarity in all pair-wise combinations of these images. Using the data generated from viewing images individually, this work develops a scheme to generate ratings of pair-wise similarity using linear fitting, and demonstrates that this is a scalable technique for developing a reference standard. This work also develops a statistical model from these data that may be used to predict inter-reader variability and accuracy of similarity estimates in larger sets of data. Finally, this thesis discusses leveraging compressive sensing techniques, which may be used to complete highly sparse matrices, to impute similarity matrices from a small subset of observer ratings. By being scalable for large databases, this technique may be a reliable and efficient method for creating similarity reference standards. The results of this work thus provide a better understanding of perception of medical image similarity, which may in turn be used to train and validate medical decision support systems.\nSpecial Collections\n\n### 6. Multi-armed bandits with side information [electronic resource][2014]Online\n\nBook\n1 online resource.\nDevelopment of personalized strategies has attracted much attention in recent years. Advances in Information Technology has led to the inundation of information, which has provided the impetus for the development of personalized strategies in diverse fields such as medicine, marketing, finance, and education. A methodological framework for the development of these personalized strategies is the theory of multi-armed bandits with side information, also called \"contextual\" bandits theory, which is an extension of classical (i.e., context-free) multi-armed bandits. This thesis represents an attempt to develop this theory To begin with, we consider the design of clinical trials for developing and testing biomarker-guided personalized therapies. Biomarker-guided personalized therapies offer great promise to improve drug development and patient care, but also pose difficult challenges in designing clinical trials for the development and validation of these therapies. After a review of the existing approaches, we describe new adaptive designs to address these challenges, first for clinical trials in new drug development and then for comparative effectiveness trials involving approved treatments. With the insight from the real-world application, we next consider general multi-armed bandits with side information, and categorizes the approaches to address the multi-armed bandit problems with side information into two groups. One approach uses parametric regression model and the other uses nonparametric regression for the relationship between the reward function of the side information for each arm. We review the existing literature and describe new treatment allocation policies inspired by the design of the biomarker-guided personalized therapy problem. We show that the new treatment allocation policies are applicable to a much more general class of problems than the policies proposed in the literature, and show the improvement of the new policies in simulation studies.\nSpecial Collections\n\n### 7. Problems, models, and algorithms in data-driven energy demand management [electronic resource][2014]Online\n\nBook\n1 online resource.\nA compelling vision for the electricity grid of the 21st century is that of a highly-instrumented system that integrates distributed generation from renewable and conventional sources where superior monitoring allows a targeted, localized, dynamic matching of demand and supply while maintaining a high degree of overall stability. To better monitor demand, utilities have recently deployed massive advanced sensing infrastructure (smart meters) to collect energy consumption data at fine (sub-hourly) time scales from large consumer populations; thus, there is urgent need formalize the new problems and develop the appropriate models, scalable algorithms, and methodologies that can leverage this new information to improve grid operations. The key tension in shaping demand is that while benefits from demand-side management programs are relevant in the aggregate (over many consumers), consumption change happens at the level of the indivdual consumer. As such, incentive schemes (e.g., dynamic pricing) that aim to change certain aspects of the average consumer's consumption may not be optimal for any particular} real consumer. Thus, the perspective this thesis takes is that of data-driven energy program targeting, i.e., using smart meter readings for identifying high-potential types of consumers for certain demand-response and energy-efficiency programs, and designing tailored controls and incentives to improve their usage behavior. This is as much a computational and engineering problem as a management and marketing one. The central contribution of this thesis is on methodology for quantifying uncertainty in individual energy consumption, and relating it to the potential for flexibility for the design and operation of certain demand-side programs. In particular, three algorithmic and modeling contributions are presented that are motivated by the question of comparing and benchmarking the impact and potential of individual consumers to providing flexibility for demand-side management. First, it is noted that individual consumption is empirically observed to be highly volatile; as such no matter how good a predictive model, part of consumption will remain uncertain. Here, this variability is shown to be related to the stress each consumer places on the grid (through their respective cost-of-service); moreover a scalable clustering algorithm is proposed to uncover patterns in variability as encoded in typical distribution functions of consumption. Second, a model of individual consumption is proposed that interprets smart meter readings as the observed outcome of latent, temperature-driven decisions to use either heating, air conditioning, or no HVAC at all; algorithms for learning such response models are introduced that are based on the maximum likelihood estimation framework. The dynamic consumption model is validated experimentally by emphasizing the intended end-use of statistical modeling when comparing with ground-truth data. A third methodological contribution leverages the statistical description of individual consumer response to weather to derive normative, tailored control schedules for thermally-sensitive appliances. These actions are optimal in the sense that they both satisfy individual effort constraints, and contribute to reducing uncertainty in the aggregate over a large population. In addition to the algorithmic and modeling contributions, this thesis presents at great length the application of the methods developed here to realistic situations of segmentation and targeting large populations of consumers for demand-side programs. We illustrate our models and algorithms on a variety of data sets consisting of heterogeneous sources - electricity usage, weather information, consumer attributes - and of various sizes, from a few hundred households in Austin, TX to 120,000 households in Northern California. We validate our dynamic consumption model experimentally, emphasizing the end purpose of decisions made using the outcome of the statistical representation of consumption. Finally, we discuss the two sides of the data coin - increased effectiveness in program management vs potential loss of consumer privacy - in an experimental study in which we argue that certain patterns in consumption as extracted from smart meter data may in some cases aid in predicting relevant consumer attributes (such as the presence of large appliances and lifestyles such as employment or children), but not many others. This, in turn, can enable the the program administrator or marketer to target those consumers whose actual data indicates that they might respond to the program, and may contribute to the debate on what consumers unwillingly reveal about themselves when using energy.\nSpecial Collections\n\n### 8. Scheduling, revenue sharing, and user behavior for aggregated demand response [electronic resource][2014]Online\n\nBook\n1 online resource.\nThe modern electric grid can trace its origins back over 100 years. Naturally, many changes and improvements have occurred during this time as new policies and new technologies are incorporated. Although these changes have, in the main, made distributing electrical energy more reliable and economical, the essential role of the grid has not changed significantly - generate sufficient power to match the quantity of power consumed. It has long been acknowledged that the ability for the operator to dynamically adjust both the quantity of power consumed as well as the quantity of power generated would be a major aid in ensuring the electric grid continues to operate economically. Attempting to adjust the consumption of energy in response to challenges faced by the operator is an approach known as demand response and it is the focus of this thesis. Although the concept of demand response is not new, substantial questions remain to be answered. Questions such as how demand response programs could be implemented for residential users, how participants in these programs should be compensated, and how the actions of such residential users can be modeled. This dissertation seeks to answer these questions by establishing and analyzing analytical models. We analyze demand response programs from the point of view of a Load Serving Entity, '' or aggregator, '' which engages in the energy markets by coordinating and controlling the participants in the program. We describe a novel procedure for direct control of non-preemptive loads by an aggregator. This program is based on a greedy algorithm which schedules the underlying loads in a manner that seeks to reduce the error between the final aggregate load profile and a predefined target profile. Algorithms are presented for two scenarios -- firstly, the case where the loads are known in advance, and secondly where the load demands are stochastic. The aggregator will realize revenue from participating in the energy markets, a portion of which must be returned to the individual users as compensation for being a part of the program. The problem of fairly compensating these users is challenging, and we propose a solution by modeling the scenario in a game theoretic setting and utilizing the concept of the Shapley Value in order to calculate a fair compensation. Finally, we analyze data from an existing demand response program, and model user behavior using a discrete choice modeling framework in order to determine what factors influence the participants response to a program.\nSpecial Collections\n\n### 9. Heuristics for large scale dynamic programming [electronic resource][2013]Online\n\nBook\n1 online resource.\nDynamic programming algorithms are used to solve problems in a diverse variety of fields, including Decision Analysis, Control Systems, Communication Systems, Dynamic Pricing and Bidding, Networking, Supply Chain Management, and Planning. These applications can be challenging, with exponential complexity, so exact solutions are rarely tractable. Therefore, improvements in the efficiency and quality of approximate dynamic programming algorithms can provide significant benefit. One promising approach is on-line Reinforcement Learning, algorithms which balance exploration and exploitation on Markov decision process applications. In practice, however, these algorithms are usually applied off-line to generate a control policy which is then fixed and used on-line. This work focuses on improving off-line policy generation on both discrete and continuous state-space processes, without the distraction of exploitation. It introduces a variety of modifications to existing algorithms and replacements for those algorithms including backtracking, exact and approximate Kalman filter value modeling, variance feedback, visitation allocation, and particle filter optimization. Backtracking re-orders the update calculations to accelerate the learning process for value iteration. Exact Kalman filter value modeling develops a Bayesian dynamic linear Gaussian representation of value using basis functions. Approximate Kalman filter value modeling is a simplified version that appears to be both higher-performance and more robust than exact Kalman filter value modeling. Variance feedback is an adaptive method for updating a value model that exploits prior knowledge while avoiding overconfidence. Visitation allocation is a heuristic to better manage the exploration process. Particle filter optimization is a stochastic optimization technique that can be applied to many policy search problems, including those that are non-convex. These techniques are demonstrated on a variety of benchmark problems, including both discrete and continuous, discounted and undiscounted, and finite and infinite horizon processes.\nSpecial Collections\n\n### 10. Modeling and analysis of the role of energy storage for renewable integration [electronic resource][2013]Online\n\nBook\n1 online resource.\nSpecial Collections\n\n### 11. Sufficient statistics for team decision problems [electronic resource][2013]Online\n\nBook\n1 online resource.\nDecentralized control problems involve multiple controllers, each having access to different measurements but working together to optimize a common objective. Despite being extremely difficult to solve in general, a common thread behind the more tractable cases is the identification of sufficient statistics for each controller, i.e. reductions of the measurements for each controller that do not sacrifice optimal performance. These sufficient statistics serve to greatly reduce the controller search space, thus making the problem easier to solve. In this dissertation, we develop for the first time a general theory of sufficient statistics for team decision problems, a fundamental type of decentralized control problem. We give rigorous definitions for team decisions and team sufficient statistics, and show how team decisions based only on these sufficient statistics do not affect optimal performance. In a similar spirit to the Kalman filter, we also show how to gracefully update the team sufficient statistics as the state evolves and additional measurements are collected. Finally, we show how to compute team sufficient statistics for partially nested problems, a large class of team decision problems that tend to have easier solutions. These team sufficient statistics have intuitive and compelling interpretations. We also show general conditions when these team sufficient statistics can be updated without the state growing in size. To illustrate the results, we give examples for finite-state systems and systems whose variables are jointly Gaussian.\nSpecial Collections\n\n### 12. Three problems in high-dimensional statistical parameter estimation [electronic resource][2013]Online\n\nBook\n1 online resource.\nStatistical parameter estimation is the process of estimating parameters of a system from direct or indirect observations. As noted by Bradly Efron in his recent book [29], at its inception parameter estimation was about simple questions using large datasets, e.g., estimating mean and variance of the age of a population using survey data. Later it evolved into problems involving still simple questions but now small amount of data. Examples included hypothesis testing and estimating parameters of a communication channel using short training sequences. This setting resulted in a fascinating literature and promotion of ideas like statistical eciency and power. In recent years, a new class of parameter estimation has been emerging which is best exemplified by the Netflix challenge. These problems are characterized by vast amount of input data but at the same time complex and large set of parameters to be estimated. In the case of the Netflix challenge, the data consist of about 10^8 known movie ratings and the challenge was to estimate about 10^10 unknown ratings. In this work, we consider three problems of this type, namely, learning and controlling high- dimensional dynamical systems, and statistical signal processing for time of flight mass spectrometry. We present both theoretical and experimental results. First, we consider the problem of learning high-dimensional dynamical systems. A continuous time autonomous dynamical system is described by a stochastic differential equation while a discrete time system is described by a state evolution equation. We consider the problem of learning the drift coefficient of a p-dimensional stochastic differential equation from a sample path of length T. We assume that the drift is parametrized by a high-dimensional vector, and study the support recovery problem in the limit in which both p and T tend to infinity. In particular, we prove a general lower iv bound on the sample-complexity T by using a characterization of mutual information as a time integral of conditional variance, due to Kadota, Zakai, and Ziv. For linear stochastic differential equations, the drift coefficient is parametrized by a p-\u0333>\u0333 p matrix which describes which degrees of freedom interact under the dynamics. In this case, we analyze an 1-regularized least squares estimator and prove an upper bound on T that nearly matches the lower bound on specific classes of sparse matrices. Linear dynamical systems are important modeling tools for many applications, ranging from robotics and control to biology and chemistry. However, in some applications the ultimate goal is to efficiently control an unknown or poorly known system, i.e., the reinforcement learning of the underlying dynamical system. A linear quadratic dynamical system is describe by a state evolution equation and a cost function. The dynamics is similar to that of an autonomous linear dynamical system with the addition of a control term. The cost at each time is a quadratic function of the state and the control exerted at that step. In the second part of this thesis we study the problem of adaptive control of a high dimensional linear quadratic system. Previous work established the asymptotic convergence to an optimal controller for various adaptive control schemes. More recently, for the average cost LQ problem, a regret bound of O(\\sqrt{T}) was shown, apart form logarithmic factors. However, this bound scales exponentially with p, the dimension of the state space. In this work we consider the case where the matrices describing the dynamic of the LQ system are sparse and their dimensions are large. We present an adaptive control scheme that achieves a regret bound of O(p\\sqrt{T}), apart from logarithmic factors. In particular, our algorithm has an average cost of (1+\\epsilon) times the optimum cost after T = polylog(p)O(1\/\\epsilon^2). This is in comparison to previous work on the dense dynamics where the algorithm requires time that scales exponentially with dimension in order to achieve regret of e\u0333 times the optimal cost. We describe the applications of this result in the emerging area of computational advertising, in particular targeted online advertising and advertising in social networks. In the third part of this thesis we tend to the problem of statistical signal processing for time of flight mass spectrometry. Time of flight mass spectrometry is a technique for measuring the mass of ionized chemical species. Traditionally, because of the high data rate and the requirement for the online processing of the signal by the instrument, time of fight mass spectrometers employed only rudimentary signal processing. We employ a simple modification to the conventional time of flight mass spectrometry where a variable and (pseudo)-random pulsing rate is used which allows for traces from different pulses to overlap. This modification requires little alteration to the currently employed hardware. However, it requires a reconstruction method to recover the spectrum from highly aliased traces. We propose and demonstrate an efficient algorithm that can process massive time of flight mass spectrometry data using computational resources that can be considered modest with today's standards. We demonstrate how our method can be employed to increase the throughput of an instrument by an order of magnitude. We expect this to extend the applicability of TOFMS to new domains.\nSpecial Collections\n\n### 13. Convex methods for approximate dynamic programming [electronic resource][2012]Online\n\nBook\n1 online resource.\nhis thesis studies the use of convex optimization in stochastic control problems. We propose methods based on convex optimization for approximate dynamic programming. Dynamic programming is a standard approach to many stochastic control problems, which involves decomposing the problem into a sequence of subproblems to solve for a global minimizer, called the value function. Traditional dynamic programming approaches focus on problems with finite state space and action space, and finding a solution gets prohibitively difficult as the state space or action space grows. With the exception of a few special cases, dynamic programming (DP) is difficult to carry out for general problems. In such cases, approximate dynamic programming (ADP) gives a method for finding a good, if not optimal, policy. In this work, we rely on our ability to (numerically) solve convex optimization problems with great speed and reliability. Using custom generated solvers we can speed up computation by orders of magnitude. In this thesis, we use convex optimization in conjunction with approximate dynamic programming to find the optimal (or suboptimal, but good) policies for an array of stochastic control problems. In the first part of the thesis, we consider applications where we have access to a good controller (which may be very complex): this could be a pilot controlling an aircraft, or a model predictive controller with a long lookahead horizon. In such cases, we can observe the actions chosen in different states, but we do not have access to the underlying objective function used by the controller (such as the pilot). We propose methods that use convex optimization to impute the underlying objective function. This results in a controller that can mimic the behavior of the original controller, often with the complexity reduced greatly. In the second part of the thesis, we develop the mathematical framework and present algorithms for carrying out projected value iteration using quadratic approximate value functions. Although there are no theoretical guarantees, we observe that in practice we achieve very good performance in a reasonable number of steps. We will consider problems in a variety of applications, such as consumer behavior modeling, robotics, finance, input-constrained control, and supply chain management.\nSpecial Collections\n\n### 14. Design and implementation of stochastic control policies via convex optimization [electronic resource][2012]Online\n\nBook\n1 online resource.\nIn this dissertation we consider the design and implementation of control policies for stochastic control problems with arbitrary dynamics, objective, and constraints. In some very special cases, these problems can be solved analytically. For instance, when the dynamics are linear, and the objective is quadratic, the optimal control policy is linear state feedback. Another simple case where the optimal policy can be computed exactly is when the state and action spaces are finite, in which case methods such as value iteration or policy iteration can be used. When the state and action spaces are infinite, but low dimensional, the optimal control problem can be solved by gridding or other discretization methods. In general however, the optimal control policy cannot be tractably computed. In such situations, there are many methods for finding suboptimal controllers that hopefully achieve a small objective value. One particular method we will discuss in detail is approximate dynamic programming (ADP), which relies on an expression for the optimal policy in terms of the value function of the stochastic control problem. In ADP we use the same expression as the optimal policy, but replace the true value function with an approximation. Another widely-used suboptimal policy is receding horizon control (RHC), also known as model predictive control (MPC). In MPC, we solve an optimization problem at each time step to determine a plan of action over a fixed time horizon, and then apply the first input from the plan. At the next time step we repeat the planning process, solving a new optimization problem, with the time horizon shifted one step forward. In the design of policies such as these, we must choose parameters, such as approximate value functions, terminal costs, time horizon, to achieve good performance. Ideally, we would like to be able to compare the performance of a suboptimal controller with the optimal performance, which we cannot compute. In this dissertation, we describe a general method for obtaining a function that lower bounds the value function of a stochastic control problem. Our method yields a numerical lower bound on the optimal objective value, as well as a value function underestimator that can be used as a parameter for ADP or MPC. We can then compare our bound to the performance achieved by our policies. If the gap between the two is small, we can conclude that the policies are nearly optimal, and the bound is nearly tight. Thus, our method simultaneously yields suboptimal policy designs, as well as a way to certify their performance. Our underestimator\/bound is non-generic, in the sense that it does not simply depend on problem dimensions and some basic assumptions about the problem data. Instead, they are computed (numerically) for each specific problem instance. We will see that for many problem families, our method is based on solving a convex optimization problem, thus avoiding the curses of dimensionality' usually associated with dynamic programming. One drawback of the suboptimal policies we design is that an optimization problem must be solved at each time step to determine the input to apply to the system. Using conventional optimization solvers this can take seconds, if not minutes. Thus, applications of these policies have been traditionally limited to systems with relatively slow dynamics, with sampling times measured in seconds, minutes, or hours. In the second part of this dissertation, we outline a collection of optimization methods that exploit the particular structure of the control problem. Our custom methods are up to around 1000 times faster compared with generic optimization packages such as SDPT3 or SeDuMi. These advances, combined with ever-increasing computing power, extends the application of optimization based policies to a wide range of applications, including those with millisecond or microsecond sampling periods.\nSpecial Collections\n\n### 15. Directed learning [electronic resource][2012]Online\n\nBook\n1 online resource.\nIn machine learning, it is common to treat estimation of model parameters separately from subsequent use of the model to guide decisions. In particular, the learning process typically aims to maximize goodness of fit'' without consideration of decision objectives. In this dissertation, we propose a new approach -- directed learning -- which factors decision objectives into the model fitting procedure in order to improve decision quality. We develop and analyze directed learning algorithms for three classes of problems. In the first case, we consider a problem where linear regression analysis is used to guide decision making. We propose directed regression, an efficient algorithm that takes into account the decision objective when computing regression coefficients. We demonstrate through a computational study that directed regression can generate significant performance gains, and establish a theoretical result that motivates it. This setting is then extended to a multi-stage decision problem as our second case, and we show that a variation of directed regression, directed time-series regression, improves performance in this context as well. Lastly, we consider a problem that involves estimating a covariance matrix and making a decision based on that estimate. Such problems arise in portfolio management among other areas, and a common approach is to employ principal component analysis (PCA) to estimate a parsimonious factor model. We propose directed PCA, an efficient algorithm that accounts for the decision objective in the selection of components, and demonstrate through experiments that it leads to significant improvement. We also establish through a theoretical result that the possible degree of improvement can be unbounded.\n\n### 16. Efficient algorithms for collaborative filtering [electronic resource][2012]Online\n\nBook\n1 online resource.\nCollaborative filtering is a novel statistical technique to obtain useful information or to make predictions based on data from multiple agents. A large number of such datasets are naturally represented in matrix form. Typically, there exists a matrix M from which we know a (typically sparse) subset of entries M_ij for (i, j) in some set E. The problem then is to predict\/approximate the unseen entries. This framework of matrix completion is extremely general and applications include personalized recommendation systems, sensor positioning, link prediction and so on. Low rank models have traditionally been used to learn useful information from such datasets. Low-dimensional representations simplify the description of the dataset and often yield predictive powers. As an added benefit, it is easier to store and retrieve low dimensional representations. Finally, many computationally intensive operations such as matrix multiplication and inversion are simplified with low dimensional representations. Singular Value Decomposition (SVD) has traditionally been used to find the lowdimensional representation of a fully revealed matrix. There are numerous algorithms for computing the SVD of a matrix including several parallel implementations and implementations for sparse matrices. However, when the matrix is only partially observed, we show that SVD techniques are sub-optimal. In this work, we will develop algorithms to learn a low rank model from a partially revealed matrix. These algorithms are computationally efficient and highly parallelizable. We will show that the proposed algorithms achieve a performance close to the fundamental limit in a number of scenarios. Finally, the algorithms achieve significantly better performance than the state-of-the-art algorithms on many real collaborative filtering datasets.\nSpecial Collections\n\n### 17. Incentive mechanisms for societal networks [electronic resource][2012]Online\n\nBook\n1 online resource.\nA \"societal network\" is an interconnected structure of resources and consumers that drives a societal process. Examples include roads, electricity grids, health-care systems, and waste management networks. Societal networks are plagued by two pressing problems: (i) the demand for resources exceeds their supply, especially at \"peak'' hours, and (ii) inefficiencies due to wastage are highly permeable, i.e., while the loss to an individual is very small, the overall loss to the societal network as a whole can be substantial. This suggests that small changes in the behavior of each user can lead to significant gains in the efficiency of societal networks. More importantly, such a change in each user is necessary to combat systemic problems such as congestion. There is a strong urgency to the problems in societal networks -- the scale of the problems has reached epic proportions. Fortunately, help is at hand. Over the past few decades, there have been great technological inventions such as the creation of the global Internet, the Web, cellular telephony and cloud computing. Now is the right time to address the problems in societal networks making use of the ubiquitous presence of smart sensing technology. This thesis presents a first attempt at that -- making use of a variety of sensing technology to monitor user behavior, informing the users about their actions and incentivizing them to adopt a desirable behavior. This thesis develops a general approach to affect change in the behavior of the users of societal networks. At the first level, it is based on fine-grained sensing of user behavior and communicating this information to a platform, and using incentives of monetary, social and other kinds to change user behavior. At the second level, the approach is concerned with the design of low-cost and accurate sensors, and backend algorithms to inform, engage and incentivize users to affect the desired behavior change. We describe some pilot projects which we have conducted to test and verify the effectiveness of this approach. The first project, INSTANT (Infosys-Stanford Traffic project), was conducted over 6 months in Bangalore, India, to incentivize roughly 14,000 commuters of Infosys Technologies to shift their commute to off-peak hours. The second project is a recycling experiment conducted over one week at Stanford to incentivize the members of the Stanford community to recycle. Third, we outline our contributions to Steptacular, an employee wellness program run at Accenture, USA. Finally, we describe CAPRI (Congestion and Parking Relief Incentives), a research project currently underway at Stanford University to incentivize off-peak travel and \"off-center\" parking.\nSpecial Collections\n\n### 18. Strategic and adaptive execution [electronic resource][2012]Online\n\nBook\n1 online resource.\nSpecial Collections\n\n### 19. Intermediated blind portfolio auctions [electronic resource][2011]Online\n\nBook\n1 online resource.\nAs much as 12% of the daily volume on the New York Stock Exchange, and similar volumes on other major world exchanges, involves the sale of portfolios by institutional investors to brokers through blind portfolio auctions. Such transactions typically take the form of a first-price sealed-bid auction in which the seller engages a few potential brokers and provides limited information about the portfolio being sold. Uncertainty about the portfolio contents reduces bids, effectively increasing the transaction cost paid by the seller. We consider the use of a trusted intermediary or equivalent cryptographic protocol to reduce transaction costs. In particular, we propose a mechanism through which each party provides relevant private information to an intermediary who ultimately reveals only the portfolio contents and price paid, and only to the seller and winning broker. Through analysis of a game-theoretic model, we demonstrate substantial potential benefits to sellers. For example, under reasonable assumptions a seller can reduce expected transaction costs by as much as 10% under a broker valuation model that induces efficient allocations at equilibrium, and as much as 33% under one that does not. We also consider the effects of intermediation on broker payoffs and social welfare as well as its performance in the context of the second-price auction. In addition, we consider the two-stage game in which sellers have the choice of selecting intermediation or not and show that under certain reasonable modeling assumptions that all sellers will select intermediation at equilibrium.\nSpecial Collections\n\n### 20. Multi-camera vision for smart environments [electronic resource][2011]Online\n\nBook\n1 online resource.\nTechnology is blending into every part of our lives. Instead of having them as intruders in the environment, people prefer the computers to go into the background, and automatically help us at the right time, on the right thing, and in the right way. A smart environment system senses the environment and people in it, makes deductions, and then takes actions if necessary. Sensors are the \"eyes\" of smart environments. In this dissertation we consider vision sensors. Image and video data contain rich information, yet they are also challenging to interpret. From obtaining the raw image data from the camera sensors to achieving the application's goal in a smart environment, we need to consider three main components of the system, i.e., the hardware platform, vision analysis of the image data, and high-level reasoning pertaining to the end application. A generic vision-related problem, e.g., human pose estimation, can be approached based on different assumptions. In our approach the system constraints and application's high-level objective are considered and often define boundary conditions in the design of vision-based algorithms. A multi-camera setup requires distributed processing at each camera node and information sharing through the camera network. Therefore, computation capacity of camera nodes and the communication bandwidth can define two hard constraints for algorithm design. We first introduce a smart camera architecture and its specific local computation power and wireless communication constraints. We then examine how the problem of human pose detection can be formulated for implementation on this smart camera, and describe the steps taken to achieve real-time operation for an avatar-based gaming application. We then present a human activity analysis technique based on multi-camera fusion. The method defined a hierarchy of coarse and fine level activity classes and employs different visual features to detect the pose or activity in each class. The camera fusion methods studied in this dissertation include decision fusion and feature fusion. We show the results of experiments in three testbed environments and analyze the performance of the different fusion methods. Although computer vision already involves complicated techniques in interpreting the image data, it still constitutes just the sensing part of the smart environment system, and further application-related high-level reasoning is needed. Modeling such high-level reasoning will depend on the specific nature of the problem. We present a case study in this dissertation to demonstrate how the vision processing and high-level reasoning modules can be interfaced for making semantic inference based on observations. The case study aims to recognize objects in a smart home based on the observed user interactions. We make use of the relationship between objects and user activities to infer the objects when related activities are observed. We apply Markov logic network (MLN) to model such relationships. MLN enables intuitive modeling of relationships, and it also offers the power of graphical models. We show different ways of constructing the knowledge base in MLN, and provide experimental results.\nSpecial Collections\n\n### Looking for different results?\n\nModify your search: Search all fields\n\nSearch elsewhere: Search WorldCat Search library website","date":"2017-05-26 18:59:46","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\": 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\": 0.47008776664733887, \"perplexity\": 709.8121694672475}, \"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-2017-22\/segments\/1495463608676.72\/warc\/CC-MAIN-20170526184113-20170526204113-00446.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.physicsforums.com\/threads\/coulombs-formula-doubt.298623\/","text":"# Coulomb's formula - doubt\n\n1. Mar 10, 2009\n\n### nikromancer\n\nCoulomb's formula states that the force of attraction between two electrically charged bodies is k(q1*q2\/d^2), where k is 9*10^9 Nm^2\/C^2. However, I have a doubt regarding this.\nWe know that like charges repel and unlike charges attract. Let us take the case of like charges first. If both q1 and q2 are like charges, the value of q1*q2 will be positive. As d and k are also positive, the force of attraction will also be positive. However, this means that like charges ATTRACT each other rather than repel. Similarly, if q1 and q2 are unlike charges, then the force of attraction will be negative and this means that unlike charges repel and like charges attract. But this isn't true.\nSo, shouldn't the formula be F= -k(q1*q2\/d^2) or |F|=k(|q1*q2|\/d^2)?? Please clarify.\n\n2. Mar 10, 2009\n\n### CompuChip\n\nThe formula you give only gives the magnitude of the force, not the direction.\n\nThe full (vector) form giving the force on charge 1 is\n$$k {q_1q_2 \\over r^2}\\mathbf{\\hat{r}}_{21}$$\nwhere $\\mathbf{\\hat{r}}_{21}$ is the vector pointing from charge 2 to charge 1. So if the charges are like, then the force vector will point along this vector, meaning it is repulsive.\n\n3. Mar 10, 2009\n\n### Galileo\n\nYou shouldn't interpret F as the force of attraction. The F in Coulomb's law is simply THE force charge 1 exerts on charge 2. If it is negative, it will be attractive (in the direction to make the distance between chages decrease) and if it is positive, it will be repulsive.\n\nThe notation you are using can confuse whether the force is attractive or not. The unambiguous way of writing coulomb's law is with vector notation (3 dimensions):\n\nThe force charge 1 exerts on charge 2 is\n\n$$\\vec F = K \\frac{q_1q_2}{r^2}\\hat r$$\nwhere $\\hat r$ is the unit vector from q1 to q2 (i.e. $\\vec r_2-\\vec r_1$).\n\nIn one dimension, you can work with signs, but you have to take care with the directions.\n\nF = K(q_1q_2\/d^3) * (x2-x1), where (x2-x1) is the signed distance x2-x1. So if x2>x1, then x2-x_1=d and the force is Kq_1q_2\/d^2. This is positive for like charges, to the force is such that x_2 tends to increase (repellant). Figure out the other cases yourself.","date":"2018-02-21 17:41:51","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\": 0.7087327837944031, \"perplexity\": 626.1642279347531}, \"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-2018-09\/segments\/1518891813691.14\/warc\/CC-MAIN-20180221163021-20180221183021-00485.warc.gz\"}"} | null | null |
AP-NORC Poll: Many Support Impeachment Inquiry But It's Not Top Priority
Economist01/11/2019Бізнес, Новини
More Americans approve of the impeachment investigation into President Donald Trump than disapprove of it, though only about a third say the inquiry should be a top priority for Congress, according to a new poll from The Associated Press-NORC Center for Public Affairs Research.
That solid, if measured, support serves as a warning sign for Trump's White House and reelection campaign, which have insisted that pursuing impeachment will end up being a vulnerability for Democrats heading into 2020.
But the findings present some red flags for Democrats, too: More people say House members are motivated mainly by politics rather than by duty as they investigate the Republican president's dealings with Ukraine and whether he abused his office or compromised national security when he tried to pressure the country to dig up dirt on a political rival.
And assessments of the president's performance generally have remained remarkably stable even as the investigation has unfolded at a rapid clip.
Overall, 47% said they support the impeachment inquiry, while 38% disapprove. Like most assessments of Trump and Washington, views of impeachment are starkly polarized.
A vast majority of Democrats approve of the inquiry, including 68% who strongly approve.
Among them is Sandra Shrewsbury, 70, who lives in Greencastle, Indiana. She said that Trump's impeachment is long overdue.
"I am really concerned about our country if this does not stop," she said of Trump's time in office.She voiced concerns that Trump doesn't have the temperament to be the nation's commander in chief and is doing serious damage to the country's standing.
She was relieved, she said, that after months of hemming and hawing, impeachment proceedings were finally underway.
"I was getting very frustrated with Congress and those investigating because I felt like they were just dragging their heels," she said. "I wish they'd stop worrying about getting reelected themselves and get down to the business they're supposed to be doing. … We pay them to do this job.
"They should have done it a long time, a way long time ago," agreed Monica Galindo, 32, who lives in Camilla, Georgia.
It's another story among Republicans, who overwhelmingly disapprove of the inquiry, including 67% who do so strongly.
"I think its garbage," said Sara Palmer, 42, a staunch Trump supporter who lives in Pocatello, Idaho, and accused Democrats of wasting time and money trying to take down Trump when there are far more important things they should be doing for the country.
"I mean come on!" she said. "There's nothing there. … He didn't do anything wrong."
That's a sentiment shared by a majority, 64%, of Republicans.
Yet even among members of Trump's party, a modest share think he did do something wrong. About a quarter, 28%, think he did something unethical, while 8% think he broke the law.
The public overall has mixed views of whether the president committed any wrongdoing. Most say his interactions with the president of Ukraine were at least unethical. That includes about 4 in 10 who think he did something illegal. About another 3 in 10 think what he did was unethical but not illegal.Trump has insisted he did nothing wrong.
But nearly all Democrats think the president crossed a line, including roughly 7 in 10 who say that he broke the law.
Still, not all Democrats think the inquiry should be Congress' top priority. A quarter think it should be an important but lower priority, and 1 in 10 say it should not be an important priority at all. And while most Democrats support the inquiry, 27% think the House is acting mainly on political motivation to challenge Trump's presidency.
Even as Americans express strong opinions about the inquiry, many have mixed assessments of their own understanding of the impeachment process. Just about 3 in 10 say they understand the process very or extremely well, while roughly as many describe their understanding as limited.
Skylar Iske, 22, who voted for Trump in 2016 but has grown weary of him, said it was difficult for him to oppose the process given his limited awareness of the case for impeachment.
"I don't feel like he should be. But then again, I also don't know what they're investigating," said Iske, who lives outside Des Moines, Iowa.
And there are rare areas where Republicans and Democrats actually agree.
Majorities across party lines think it was inappropriate for Hunter Biden, former Vice President Joe Biden's son, to serve on the board of a Ukrainian energy company while his father was vice president, with only about a quarter of Americans saying it was appropriate.
Roughly 7 in 10, including 6 in 10 Democrats, say it wasn't.
Яка різниця між зарплатою міністра освіти Новосад і молодого вчителя? 22/01/2020
Китай подтвердил что коронавирус передается от человека к человеку 22/01/2020 | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,101 |
\section{Introduction}
Stochastic differential equations (SDEs) and the Kolmogorov partial differential equations (PDEs) associated to them have been widely used in models from engineering, finance, and the natural sciences.
In particular, SDEs and Kolmogorov PDEs, respectively, are highly employed in models for the approximative pricing of financial derivatives.
Kolmogorov PDEs and SDEs, respectively, can typically not be solved explicitly and it has been and still is an active topic of research to design and analyze numerical methods which are able to approximately solve Kolmogorov PDEs and SDEs, respectively (see, e.g.,
\cite{Giles_MultilevelMonteCarlo2008},
\cite{TalayGrahamBook},
\cite{Gyoengy_ANoteOnEulersApproximations1998},
\cite{HighamIntroduction},
\cite{Higham2011stochastic},
\cite{HighamMaoStuart_StrongConvergenceOfEuler2002},
\cite{HutzenthalerJentzen2015_Memoirs},
\cite{Kloeden_SystematicDerivation2002},
\cite{KloedenPlaten1992},
\cite{KloedenPlatenSchurz2012}, \cite{Maruyama_ContinuousMarkovProcessesAndStochasticEquations1955},
\cite{Milstein1974},
\cite{Milstein1995},
\cite{MilsteinTretyakovBook},
\cite{GronbachRitterMinimal2008},
\cite{Roessler2009}).
In particular, there are nowadays several different types of numerical approximation methods for Kolmogorov PDEs in the literature including deterministic numerical approximation methods such as finite differences based approximation methods (cf., for example, \cite{BrennanSchwartz1977}, \cite{BrennanSchwartz1978}, \cite{HanWuFD03}, \cite{KUSHNERFD76}, \cite{YhaoDavisonCorlessFD07}, \cite{SCHWARTZ1977}) and finite elements based approximation methods (cf., for example, \cite{brenner2007mathematical}, \cite{ciarlet1991basic}, \cite{zienkiewicz1977finite}) as well as random numerical approximation methods based on Monte Carlo methods (cf., for example,~\cite{Giles_MultilevelMonteCarlo2008}, \cite{TalayGrahamBook}) and discretizations of the underlying SDEs (cf., for example,
\cite{Gyoengy_ANoteOnEulersApproximations1998},
\cite{HighamIntroduction},
\cite{Higham2011stochastic},
\cite{HighamMaoStuart_StrongConvergenceOfEuler2002},
\cite{HutzenthalerJentzen2015_Memoirs},
\cite{Kloeden_SystematicDerivation2002},
\cite{KloedenPlaten1992},
\cite{KloedenPlatenSchurz2012}, \cite{Maruyama_ContinuousMarkovProcessesAndStochasticEquations1955},
\cite{Milstein1974},
\cite{Milstein1995},
\cite{MilsteinTretyakovBook},
\cite{GronbachRitterMinimal2008},
\cite{Roessler2009}).
The above mentioned deterministic approximation methods for PDEs work quite efficiently in one or two space dimensions but cannot be used in the case of high-dimensional PDEs as they suffer from the so-called curse of dimensionality (cf.\ Bellman~\cite{Bellman}) in the sense that the computational effort of the considered approximation algorithm grows exponentially in the PDE dimension. The above mentioned random numerical approximation methods involving Monte Carlo approximations typically overcome this curse of dimensionality but only provide approximations of the Kolmogorov PDE at a single fixed space-time point.
The key contribution of this paper is to derive and propose a numerical approximation method which aims to overcome both of the above mentioned drawbacks and intends to deliver a numerical approximation of the Kolmogorov PDE on an entire region $[a,b]^d$ without suffering from the curse of dimensionality. The numerical scheme, which we propose in this work, is inspired by recently developed deep learning based approximation algorithms for PDEs in the literature (cf., for example, \cite{BeckEJentzen2017}, \cite{DeepStopping}, \cite{EHanJentzen2017a}, \cite{EYu2017}, \cite{FujiiTakahashiTakahashi2017}, \cite{EHanJentzen2017b}, \cite{Labordere2017}, \cite{raissi2018forward}, \cite{SirignanoSpiliopoulos2017}). To derive the proposed approximation scheme we first reformulate the considered Kolmogorov PDE as a suitable infinite dimensional stochastic optimization problem (see items~\eqref{it:exuniqueSolutions}--\eqref{it:UboundaryCond} in Proposition~\ref{proposition:minimizingPropertyApplied_new} below for details). This infinite dimensional stochastic optimization problem is then temporally discretized by means of suitable discretizations of the underlying SDE and it is spatially discretized by means of fully connected deep artificial neural network approximations (see~\eqref{eq:toMinimizeApprox} in Subsection~\ref{subsec:sgd} as well as Subsections~\ref{subsec:discretizationOfX}--\ref{subsec:DNNapproximations} below). The resulting finite dimensional stochastic optimization problem is then solved by means of stochastic gradient descent type optimization algorithms (see~\eqref{eq:plainGradientDescent} in Subsection~\ref{subsec:sgd}, Framework~\ref{algo:special} in Subsection~\ref{subsec:desc_algo}, Framework~\ref{algo:general_algorithm} in Subsection~\ref{subsec:generalOptDescription}, as well as \eqref{eq:Adam1}--\eqref{eq:Adam2} in Subsection~\ref{sec:adam}). We test the proposed approximation method numerically in the case of several examples of SDEs and PDEs, respectively (see Subsections~\ref{sec:paraboliceq}--\ref{sec:heston} below for details). The obtained numerical results indicate that the proposed approximation algorithm is quite effective in high dimensions in terms of both accuracy and speed.
The remainder of this article is organized as follows. In Section~\ref{sec:derivationalgo} we derive the proposed approximation algorithm (see Subsections~\ref{subsec:kolmogorovEq}--\ref{subsec:sgd} below) and we present a detailed description of the proposed approximation algorithm in a special case (see Subsection~\ref{subsec:desc_algo} below) as well as in the general case (see Subsection~\ref{subsec:generalOptDescription} below). In Section~\ref{sec:examples} we test the proposed algorithm numerically in the case of several examples of SDEs and PDEs, respectively. The employed source codes for the numerical simulations in Section~\ref{sec:examples} are postponed to Section~\ref{sec:source_code}.
\section{Derivation and description of the proposed approximation algorithm}
\label{sec:derivationalgo}
In this section we describe the approximation
problem which we intend to solve (see Subsection~\ref{subsec:kolmogorovEq} below)
and we derive (see Subsections~\ref{subsec:connection}--\ref{subsec:sgd} below)
and specify (see Subsections~\ref{subsec:desc_algo}--\ref{subsec:generalOptDescription} below)
the numerical scheme which we suggest to use to solve this approximation problem
(cf., for example, E et al.~\cite{EHanJentzen2017a},
Han et al.~\cite{EHanJentzen2017b},
Sirignano \& Spiliopoulos~\cite{SirignanoSpiliopoulos2017},
Beck et al.~\cite{BeckEJentzen2017},
Fujii, Takahashi, A., \& Takahashi, M.~\cite{FujiiTakahashiTakahashi2017},
and Henry-Labordere~\cite{Labordere2017}
for related derivations and related approximation schemes).
\subsection{Kolmogorov partial differential equations (PDEs)}
\label{subsec:kolmogorovEq}
Let
$ T \in (0,\infty) $,
$ d \in \N $,
let
$\mu\colon \R^d\to\R^d$
and
$\sigma\colon \R^d\to\R^{d\times d}$
be Lipschitz continuous functions,
let
$
\varphi\colon \R^d \to \R
$
be a function,
and
let
$ u = (u(t,x))_{(t,x)\in [0,T]\times\R^d}
\in C^{1,2}([0,T]\times\R^d,\R)$
be a function with at most polynomially growing partial derivatives
which satisfies for every
$t\in [0,T]$, $x\in\R^d$ that
$u(0,x) = \varphi(x)$ and
\begin{equation}\label{eq:kolmogorovPDE}
\tfrac{\partial u}{\partial t}(t,x)
=
\tfrac12 \operatorname{Trace}_{\R^d}\!\big(
\sigma(x)[\sigma(x)]^{*}(\operatorname{Hess}_x u)(t,x)\big)
+
\langle
\mu(x),(\nabla_x u)(t,x)
\rangle_{\R^d}.
\end{equation}
Our goal is to approximately calculate the function
$\R^d\ni x\mapsto u(T,x)\in\R$ on some subset of $\R^d$.
To fix ideas we consider real numbers $a,b\in\R$ with
$a<b$ and we suppose that our goal is to approximately
calculate the function $[a,b]^d\ni x\mapsto u(T,x)\in\R$.
\subsection{On stochastic differential equations and Kolmogorov PDEs}
\label{subsec:connection}
In this subsection we provide a probabilistic representation for the solutions
of the PDE~\eqref{eq:kolmogorovPDE}, that is, we recall
the classical Feynman-Kac formula for the PDE
\eqref{eq:kolmogorovPDE} (cf., for example, {\O}ksendal~\cite[Chapter 8]{Oksendal_SDE2003}).
Let
$(\Omega,{\ensuremath{\mathcal{F}}},\P)$ be a probability space
with a normal filtration $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$,
let
$W\colon [0,T]\times\Omega \to \R^d$
be a standard $(\Omega,{\ensuremath{\mathcal{F}}},\P,({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]})$-Brownian motion,
and for every $x\in\R^d$ let
$X^x=(X^x_t)_{t\in [0,T]}\colon [0,T]\times\Omega\to\R^d$
be an $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted stochastic process with
continuous sample paths which satisfies that for every
$t\in [0,T]$ it holds $\P$-a.s.~that
\begin{equation}\label{eq:SDEforXstartingAtx}
X^x_t = x + \int_0^t \mu(X^x_s)\,ds + \int_0^t \sigma(X^x_s)\,dW_s.
\end{equation}
The Feynman-Kac formula
(cf., for example, Hairer et al.~\cite[Corollary 4.17 and Remark 4.1]
{HairerHutzenthalerJentzen_LossOfRegularity2015}) and \eqref{eq:kolmogorovPDE}
hence yield that for every $x\in\R^d$ it holds that
\begin{equation}\label{eq:feynman-kac}
u(T,x) = \E\big[u(0,X^x_T)\big] = \E\big[\varphi(X^x_T)\big].
\end{equation}
\subsection{Formulation as minimization problem}
In the next step we exploit \eqref{eq:feynman-kac}
to formulate a minimization problem which is uniquely solved by
the function $[a,b]^d\ni x\mapsto u(T,x)\in\R$ (cf.~\eqref{eq:kolmogorovPDE} above).
For this we first recall the $L^2$-minimization property of the expectation
of a real-valued random variable (see Lemma~\ref{lemma:minimizingProperty} below).
Then we extend this minimization result to certain random fields
(see Proposition~\ref{proposition:minimizingProperty} below).
Thereafter, we apply Proposition~\ref{proposition:minimizingProperty}
to random fields in the context of the Feynman-Kac representation~\eqref{eq:feynman-kac}
to obtain Proposition~\ref{proposition:minimizingPropertyApplied_new} below.
Proposition~\ref{proposition:minimizingPropertyApplied_new}
provides a minimization problem
(see, for instance,~\eqref{eq:functionMinimizedByU} below)
which has the function $[a,b]^d\ni x\mapsto u(T,x)\in\R$
as the unique global minimizer.
Our proof of Proposition~\ref{proposition:minimizingPropertyApplied_new} is based on the elementary auxiliary results in Lemmas~\ref{lem:projection_new}--\ref{lemma:equivalence_of_expectation_representations}. For completeness we also present the proofs of Lemmas~\ref{lem:projection_new}--\ref{lemma:equivalence_of_expectation_representations} here. The statement and the proof of Lemma~\ref{lem:projection_new} are based on the proof of Da Prato \& Zabczyk \cite[Lemma 1.1]{DaPratoZabczyk2008}.
\begin{lemma}\label{lemma:minimizingProperty}
Let $(\Omega,{\ensuremath{\mathcal{F}}},\P)$ be a probability space
and let $X\colon \Omega\to\R$ be an ${\ensuremath{\mathcal{F}}}$/$\B(\R)$-measurable random
variable which satisfies $\E[|X|^2] < \infty$. Then
\begin{enumerate}[(i)]
\item\label{it:minimizingProperty_1}
it holds for
every $y\in\R$ that
\begin{equation}\label{eq:lemma_orthogonality_property}
\E\big[ | X - y |^2\big]
=
\E\big[ | X - \E[X] |^2 \big] + |\E[X]-y|^2,
\end{equation}
\item\label{it:minimizingProperty_2}
it holds that there exists a unique real number $z\in\R$
such that
\begin{equation}\label{eq:lemma_minimizer_existence}
\E\big[ | X - z |^2 \big] = \inf_{y\in\R} \E\big[ | X - y |^2 \big],
\end{equation}
and
\item\label{it:minimizingProperty_3}
it holds that
\begin{equation}\label{eq:lemma_minimizer_uniqueness}
\E\big[ | X - \E[X] |^2 \big] = \inf_{y\in\R} \E\big[ | X - y |^2 \big].
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lemma:minimizingProperty}]
Observe that the fact that $\E[|X|]<\infty$ ensures that
for every $y\in\R$ it holds that
\begin{equation}\label{eq:proof_orthogonality}
\begin{split}
\E\big[ | X - y |^2 \big]
& =
\E\big[ | X - \E[X] + \E[X] - y |^2 \big]
\\
& =
\E\big[ | X - \E[X] |^2 + 2 (X-\E[X]) (\E[X] - y) + |\E[X] - y|^2 \big]
\\
& =
\E\big[ | X -\E[X] |^2 \big] + 2(\E[X]-y) \E\big[ X - \E[X]\big] + |\E[X] - y|^2 \\
& =
\E\big[ | X - \E[X] |^2 \big] + |\E[X] - y|^2.
\end{split}
\end{equation}
This establishes item~\eqref{it:minimizingProperty_1}.
Item~\eqref{it:minimizingProperty_2} and item~\eqref{it:minimizingProperty_3}
are immediate consequences of item~\eqref{it:minimizingProperty_1}.
The proof of Lemma~\ref{lemma:minimizingProperty}
is thus completed.
\end{proof}
\begin{proposition}\label{proposition:minimizingProperty}
Let
$a\in\R$, $b\in (a,\infty)$,
let
$(\Omega,{\ensuremath{\mathcal{F}}},\P)$ be a probability space,
let
$ X = ( X_x )_{ x \in [a,b]^d } \colon [a,b]^d \times \Omega \to \R $
be a
$ ( \mathcal{B}( [a,b]^d ) \otimes \mathcal{F} ) $/$ \mathcal{B}( \R ) $-measurable function,
assume
for every $x\in [a,b]^d$ that $\E[|X_x|^2]<\infty$,
and
assume
that the function $[a,b]^d\ni x\mapsto \E[X_x]\in\R$ is continuous.
Then
\begin{enumerate}[(i)]
\item\label{it:minimizingProperty_prop_1}
it holds that there exists a unique continuous function
$u\colon [a,b]^d\to\R$ such that
\begin{equation}\label{eq:existenceAndUniquenessOfMinimizer}
\int_{[a,b]^d} \E\big[ | X_x - u(x) |^2 \big] \,dx
=
\inf_{v\in C([a,b]^d,\R)} \bigg(\int_{[a,b]^d} \E\big[ | X_x - v(x) |^2 \big] \,dx\bigg)
\end{equation}
and
\item\label{it:minimizingProperty_prop_2}
it holds for every $x\in [a,b]^d$ that $u(x)=\E[X_x]$.
\end{enumerate}
\end{proposition}
\begin{proof}[Proof of Proposition~\ref{proposition:minimizingProperty}]
Observe that item~\eqref{it:minimizingProperty_1}
in Lemma~\ref{lemma:minimizingProperty}
and the hypothesis that
$ \forall \, x \in [a,b]^d \colon \E[ |X_x|^2 ] < \infty $
ensure that for every function $ u \colon [a,b]^d \to \R $
and every $x\in [a,b]^d$ it holds that
\begin{equation}
\label{eq:minimizingPropertyQuantified}
\E\big[ | X_x - u(x) |^2\big]
=
\E\big[ | X_x - \E[X_x] |^2 \big] + | \E[X_x]- u(x) |^2 .
\end{equation}
Fubini's theorem (see, e.g., Klenke~\cite[Theorem 14.16]{Klenke_2014})
hence proves that for every continuous function
$ u \colon [a,b]^d \to \R $ it holds that
\begin{equation}
\label{eq:minimizingPropertyQuantified_2}
\int_{ [a,b]^d }
\E\big[ | X_x - u(x) |^2\big] \, dx
=
\int_{ [a,b]^d }
\E\big[ | X_x - \E[X_x] |^2 \big] \, dx
+
\int_{ [a,b]^d }
| \E[X_x]- u(x) |^2 \, dx .
\end{equation}
The hypothesis that the function
$ [a,b]^d \ni x \mapsto \E[ X_x ] \in \R $
is continuous therefore demonstrates that
\begin{align}
\begin{split}
&\int_{[a,b]^d}
\E\big[ | X_x - \E[X_x] |^2 \big] \, dx
\\&\geq
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{[a,b]^d}
\E\big[ | X_x - v(x) |^2 \big] \, dx
\right)
\\&=
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{ [a,b]^d }
\E\big[ | X_x - \E[X_x] |^2 \big] \, dx
+
\int_{ [a,b]^d }
| \E[X_x]- v(x) |^2 \, dx
\right)
\\&\geq
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{ [a,b]^d }
\E\big[ | X_x - \E[X_x] |^2 \big] \, dx
\right)
\\&=
\int_{ [a,b]^d }
\E\big[ | X_x - \E[X_x] |^2 \big] \, dx .
\end{split}
\end{align}
Hence, we obtain that
\begin{equation}
\label{eq:minimizingPropertyQuantified_3}
\int_{[a,b]^d}
\E\big[ | X_x - \E[X_x] |^2 \big] \, dx
=
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{[a,b]^d}
\E\big[ | X_x - v(x) |^2 \big] \, dx
\right) .
\end{equation}
Again the fact that the function
$[a,b]^d\ni x\mapsto \E[X_x]\in\R$ is continuous therefore
proves that there exists a continuous function $u\colon [a,b]^d\to\R$
such that
\begin{equation}
\label{eq:minimizingPropertyQuantified_4}
\int_{[a,b]^d} \E\big[ | X_x - u(x) |^2 \big] \,dx
=
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{[a,b]^d} \E\big[ | X_x - v(x) |^2 \big] \, dx
\right) .
\end{equation}
Next observe that \eqref{eq:minimizingPropertyQuantified_2}
and~\eqref{eq:minimizingPropertyQuantified_3}
yield that for every continuous function $ u \colon [a,b]^d \to \R $
with
\begin{equation}
\int_{[a,b]^d}
\E\big[ | X_x - u(x) |^2 \big] \, dx
=
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{[a,b]^d}\E\big[ | X_x - v(x) |^2 \big] \, dx
\right)
\end{equation}
it holds that
\begin{align}
\begin{split}
&\int_{[a,b]^d}
\E\big[ | X_x - \E[X_x] |^2 \big] \, dx
\\&=
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{[a,b]^d}
\E\big[ | X_x - v(x) |^2 \big] \, dx
\right)
=
\int_{ [a,b]^d] }
\E\big[ | X_x - u(x) |^2\big] \, dx
\\&=
\int_{ [a,b]^d }
\E\big[ | X_x - \E[X_x] |^2 \big] \, dx
+
\int_{ [a,b]^d }
| \E[X_x]- u(x) |^2 \, dx .
\end{split}
\end{align}
Hence, we obtain
that for every continuous function $ u \colon [a,b]^d \to \R $
with
\begin{equation}
\int_{[a,b]^d}
\E\big[ | X_x - u(x) |^2 \big] \, dx
=
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{[a,b]^d}\E\big[ | X_x - v(x) |^2 \big] \, dx
\right)
\end{equation}
it holds that
\begin{equation}
\int_{[a,b]^d} | \E[X_x] - u(x) |^2 \, dx = 0 .
\end{equation}
This and again the hypothesis that the function
$ [a,b]^d \ni x \mapsto \E[ X_x ] \in \R $
is continuous yield that for
every continuous function $ u \colon [a,b]^d \to \R $
with
\begin{equation}
\int_{[a,b]^d}
\E\big[ | X_x - u(x) |^2 \big] \, dx
=
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{[a,b]^d}\E\big[ | X_x - v(x) |^2 \big] \, dx
\right)
\end{equation}
and every $ x \in [a,b]^d $
it holds that
$ u(x) = \E[ X_x ] $.
Combining this with~\eqref{eq:minimizingPropertyQuantified_4}
completes the proof of Proposition~\ref{proposition:minimizingProperty}.
\end{proof}
\begin{lemma}[Projections in metric spaces]
\label{lem:projection_new}
Let $(E,d)$ be a metric space,
let $n \in \N$, $e_1, e_2, \ldots, e_n \in E$,
and let $P \colon E \rightarrow E $
be the function which satisfies for every $ x \in E $ that
\begin{equation}
\label{eq:projection_1_new}
P(x)
=
e_{
\min\{
k \in \{1,2,\ldots,n\}
\colon
d(x,e_k)
=
\min\{
d(x,e_1),
d(x,e_2),
\ldots,
d(x,e_n)
\}
\}
} .
\end{equation}
Then
\begin{enumerate}[(i)]
\item\label{it:projection_2_new}
it holds for every $ x \in E $ that
\begin{equation}
d(x, P(x))
=
\min_{k\in\{1,2,\ldots, n\}} d(x,e_k)
\end{equation}
and
\item\label{it:projection_1_new}
it holds for every $A\subseteq E$ that
$P^{-1}(A)\in \B(E)$.
\end{enumerate}
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lem:projection_new}]
Throughout this proof let
$ D = (D_1, \ldots, D_n) \colon E \rightarrow \R^n $
be the function which satisfies for every $ x \in E $
that
\begin{equation}\label{eq:definition_of_D}
D(x)
=
\left(
D_1(x),
D_2(x),
\ldots,
D_n(x)
\right)
=
\left(
d(x, e_1),
d(x, e_2),
\ldots,
d(x, e_n)
\right) .
\end{equation}
Note that~\eqref{eq:projection_1_new}
ensures that for every $x\in E$ it holds that
\begin{equation}
d(x,P(x))
=
d(x,e_{
\min\{
k \in \{1,2,\ldots,n\}
\colon
d(x,e_k)
=
\min\{
d(x,e_1),
d(x,e_2),
\ldots,
d(x,e_n)
\}
\}
})
=
\min_{k\in\{1,2,\ldots,n\}} d(x,e_k) .
\end{equation}
This establishes item~\eqref{it:projection_2_new}.
It thus remains to prove item~\eqref{it:projection_1_new}. For this observe that the fact that the function
$d\colon E\times E\to [0,\infty)$ is continuous ensures
that the function $D\colon E\to\R^n$ is continuous. Hence,
we obtain that the function $D\colon E\to \R^n$ is
$ \mathcal{B}(E)$/$\mathcal{B}(\R^n)$-measurable.
Next note that item~\eqref{it:projection_2_new}
demonstrates that for every $k\in \{1,2,\ldots,n\}$,
$x\in P^{-1}(\{e_k\})$ it holds that
\begin{equation}
d(x,e_k)
=
d(x,P(x))
=
\min_{l\in\{1,2,\ldots,n\}} d(x,e_l).
\end{equation}
Hence, we obtain that for every $k\in\{1,2,\ldots,n\}$,
$x\in P^{-1}(\{e_k\})$ it holds that
\begin{equation}\label{eq:first_ineq_for_k_and_min}
k\geq \min\{l\in\{1,2,\ldots,n\}\colon d(x,e_l)
= \min\{d(x,e_1),d(x,e_2),\ldots,d(x,e_n)\}\} .
\end{equation}
Moreover, note that~\eqref{eq:projection_1_new}
ensures that for every
$ k \in \{1,2,\ldots,n\}$, $x\in P^{-1}(\{e_k\})$
it holds that
\begin{align}
\begin{split}
&\min\!\left\{l\in\{1,2,\ldots,n\}\colon d(x,e_l)
= \min_{u\in\{1,2,\ldots,n\}} d(x,e_u)\right\} \\
&\in
\big\{ l \in \{1,2,\ldots,n\} \colon e_l = e_k\big\}
\subseteq\big\{k, k+1, \ldots, n\big\}.
\end{split}
\end{align}
Therefore, we obtain that for every $k\in\{1,2,\ldots,n\}$, $x\in P^{-1}(\{e_k\})$
with $e_k\notin (\cup_{l\in \N\cap [0,k)} \{e_l\})$ it holds that
\begin{equation}
\min\!\left\{l\in\{1,2,\ldots,n\}\colon d(x,e_l)
= \min_{u\in\{1,2,\ldots,n\}} d(x,e_u)\right\}
\geq k.
\end{equation}
Combining this with~\eqref{eq:first_ineq_for_k_and_min} yields
that for every $k\in\{1,2,\ldots,n\}$, $x\in P^{-1}(\{e_k\})$ with
$e_k\notin (\cup_{l\in \N\cap [0,k)} \{e_l\})$
it holds that
\begin{equation}
\min\!\left\{l\in\{1,2,\ldots,n\}\colon d(x,e_l)
= \min_{u\in\{1,2,\ldots,n\}} d(x,e_u)\right\}
=
k .
\end{equation}
Hence, we obtain that for every $k\in\{1,2,\ldots,n\}$
with
$e_k\notin (\cup_{l\in \N\cap [0,k)} \{e_l\})$ it holds
that
\begin{equation}
P^{-1}(\{e_k\}) \subseteq
\left\{x\in E\colon
\min\!\left\{l\in\{1,2,\ldots,n\}\colon d(x,e_l)
= \min_{u\in\{1,2,\ldots,n\}} d(x,e_u)\right\}
=
k\right\} .
\end{equation}
This and \eqref{eq:projection_1_new} show that
for every $k\in\{1,2,\ldots,n\}$ with
$e_k\notin (\cup_{l\in \N\cap [0,k)} \{e_l\})$ it holds
that
\begin{equation}
P^{-1}(\{e_k\})
=
\left\{x\in E\colon
\min\!\left\{l\in\{1,2,\ldots,n\}\colon d(x,e_l)
= \min_{u\in\{1,2,\ldots,n\}} d(x,e_u)\right\}
=
k\right\} .
\end{equation}
Combining \eqref{eq:definition_of_D} with
the fact that the function $D\colon E\to\R^n$
is $\B(E)$/$\B(\R^n)$-measurable therefore demonstrates that
for every $ k \in \{1,2,\ldots,n\}$ with
$e_k \notin (\cup_{l \in \N \cap [0,k)} \{e_l\}) $
it holds that
\begin{align}
\begin{split}
&
P^{-1}(\{e_k\})
\\&=
\left\{
x \in E
\colon
\min\!\left\{
l \in \{1, 2, \ldots, n \}
\colon
d(x, e_l)
=
\min_{u\in\{1,2,\ldots,n\}} d( x, e_u )
\right\}
= k
\right\}
\\&=
\left\{
x \in E
\colon
\min\!\left\{
l \in \{1, 2, \ldots, n \}
\colon
D_l(x)
=
\min_{ u \in \{1,2,\ldots,n\} }
D_u(x)
\right\}
=
k
\right\}
\\&=
\left\{
x \in E
\colon
\left(
\begin{array}{c}
\forall \, l \in\N\cap[0,k)
\colon
D_k(x) < D_l(x) \,\, \text{and} \\
\forall \, l \in \{1,2,\ldots,n\}
\colon
D_k(x) \leq D_l(x)
\end{array}
\right)
\right\}
\\&=
\left[
\bigcap_{l=1}^{k-1}
\underbrace{
\{
x \in E
\colon
D_k(x) < D_l(x)
\}
}_{ \in \mathcal{B}(E) }
\right]
\bigcap
\left[
\bigcap_{l=1}^{n}
\underbrace{
\{
x \in E
\colon
D_k(x) \leq D_l(x)
\}
}_{ \in \mathcal{B}(E) }
\right]
\in
\mathcal{B}(E) .
\end{split}
\end{align}
Hence, we obtain that for every
$ f \in \{e_1,e_2,\ldots,e_n\} $
it holds that
\begin{equation}
P^{-1}(\{f\})
\in
\mathcal{B}(E) .
\end{equation}
Therefore, we obtain that
for every $ A \subseteq E $ it holds that
\begin{align}
\begin{split}
P^{-1}(A)
&=
P^{-1}\!\left(
A \cap \{e_1,e_2,\ldots,e_n\}
\right)
\\&=
\cup_{f \in A \cap \{e_1,e_2,\ldots,e_n\}}
\underbrace{
P^{-1}(\{f\})
}_{ \in \mathcal{B}(E) }
\in \mathcal{B}(E) .
\end{split}
\end{align}
This establishes item~\eqref{it:projection_1_new}.
The proof of Lemma~\ref{lem:projection_new} is thus completed.
\end{proof}
\begin{lemma}
\label{lem:productMeasurable_new}
Let $ (E, d) $ be a separable metric space,
let $ (\cE, \delta) $ be a metric space,
let $ (\Omega, {\ensuremath{\mathcal{F}}}) $ be a measurable space,
let $ X \colon E \times \Omega \rightarrow \cE $
be a function,
assume for every
$ e \in E $ that the function
$ \Omega \ni \omega \mapsto X(e, \omega) \in \cE $
is ${\ensuremath{\mathcal{F}}}$/$\mathcal{B}(\cE)$-measurable,
and assume for every $ \omega \in \Omega $
that the function $ E \ni e \mapsto X(e, \omega) \in \cE $
is continuous. Then it holds that the function
$ X \colon E\times \Omega \to \cE $ is $(\mathcal{B}(E) \otimes {\ensuremath{\mathcal{F}}})$/$\mathcal{B}(\cE) $-measurable.
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lem:productMeasurable_new}]
Throughout this proof let
$ (e_m)_{m \in \N} \subseteq E $ be a sequence which satisfies that
$\overline{\{ e_m \colon m \in \N \}} = E $,
let
$ P_n \colon E \rightarrow E $, $ n \in \N $,
be the functions which satisfy for every $ n \in \N $, $ x \in E $ that
\begin{align}
P_n(x)
=
e_{
\min\{
k \in \{1,2,\ldots,n\}
\colon
d(x,e_k)
=
\min\{
d(x,e_1), d(x,e_2), \ldots, d(x,e_n)
\}
\}
} ,
\end{align}
and let $\mathcal{X}_n \colon E \times \Omega \rightarrow \cE $, $ n \in \N $,
be the functions which satisfy for every $ n \in \N $, $ x \in E $, $ \omega \in \Omega $ that
\begin{equation}
\label{eq:curlyXn}
\mathcal{X}_n(x, \omega)
=
X( P_n(x), \omega) .
\end{equation}Note that
\eqref{eq:curlyXn} shows that for all $ n \in \N $, $ B \in \mathcal{B}(\cE) $
it holds that
\begin{align}
\begin{split}
(\mathcal{X}_n)^{-1}(B)
&=
\left\{
(x, \omega)
\in E \times \Omega
\colon
\mathcal{X}_n(x, \omega) \in B
\right\}
\\&=
\bigcup_{ y \in \operatorname{Im}(P_n) }
\Big(
\left[
(\mathcal{X}_n)^{-1}(B)
\right]
\cap
\left[
(P_n)^{-1}(\{y\}) \times \Omega
\right]
\Big)
\\&=
\bigcup_{ y \in \operatorname{Im}(P_n) }
\left\{
(x, \omega)
\in E \times \Omega
\colon
\Big[
\mathcal{X}_n(x, \omega) \in B
\,\, \text{and} \,\,
x \in (P_n)^{-1}(\{y\})
\Big]
\right\}
\\&=
\bigcup_{ y \in \operatorname{Im}(P_n) }
\left\{
(x, \omega)
\in E \times \Omega
\colon
\Big[
X(P_n(x), \omega) \in B
\,\, \text{and} \,\,
x \in (P_n)^{-1}(\{y\})
\Big]
\right\}
.
\end{split}
\end{align}
Item~\eqref{it:projection_1_new} in Lemma~\ref{lem:projection_new}
hence implies that for all $ n \in \N $, $ B \in \mathcal{B}(\cE) $
it holds that
\begin{align}
\begin{split}
(\mathcal{X}_n)^{-1}(B)
&=
\bigcup_{ y \in \operatorname{Im}(P_n) }
\left\{
(x, \omega)
\in E \times \Omega
\colon
\Big[
X(y, \omega) \in B
\,\, \text{and} \,\,
x \in (P_n)^{-1}(\{y\})
\Big]
\right\}
\\&=
\bigcup_{ y \in \operatorname{Im}(P_n) }
\Big(
\left\{
(x, \omega)
\in E \times \Omega
\colon
X(y, \omega) \in B
\right\}
\cap
\left[
(P_n)^{-1}(\{y\}) \times \Omega
\right]
\Big)
\\&=
\bigcup_{ y \in \operatorname{Im}(P_n) }
\Big(
\big[
\underbrace{
E
\times \left(
\left(X(y, \cdot)\right)^{-1}(B)\right)
}_{\in (\mathcal{B}(E) \otimes {\ensuremath{\mathcal{F}}})}
\big]
\cap
\big[
\underbrace{
(P_n)^{-1}(\{y\})
\times
\Omega
}_{\in (\mathcal{B}(E) \otimes {\ensuremath{\mathcal{F}}})}
\big]
\Big)
\in (\mathcal{B}(E) \otimes {\ensuremath{\mathcal{F}}}) .
\end{split}
\end{align}
This proves that for every $ n \in \N $ it holds that
the function
$
\mathcal{X}_n
$
is
$
(\mathcal{B}(E) \otimes {\ensuremath{\mathcal{F}}})
$/$
\mathcal{B}(\cE)
$-measurable.
In addition, note that
item~\eqref{it:projection_2_new} in Lemma~\ref{lem:projection_new}
and the hypothesis that for every $\omega\in\Omega$ it holds that the
function $E\ni x\mapsto X(x,\omega)\in\cE$ is continuous
imply that for every $ x \in E $, $ \omega \in \Omega $
it holds that
\begin{equation}
\label{eq:productMeasurable_5_new}
\lim_{ n \rightarrow \infty }
\mathcal{X}_n(x, \omega)
=
\lim_{ n \rightarrow \infty }
X( P_n(x), \omega)
=
X(x, \omega) .
\end{equation}
Combining this with the fact that for every $n\in\N$ it holds that
the function
${\ensuremath{\mathcal{X}}}_n\colon E\times\Omega\to\cE$ is
$(\B(E)\otimes{\ensuremath{\mathcal{F}}})$/$\B(\cE)$-measurable
shows that the function $X\colon E\times\Omega\to \cE$
is $(\B(E)\otimes{\ensuremath{\mathcal{F}}})$/$\B(\cE)$-measurable.
The proof of Lemma~\ref{lem:productMeasurable_new} is thus
completed.
\end{proof}
\begin{lemma}
\label{lemma:stochastic_convergence_under_continuous_transformations}
Let $ (\Omega, {\ensuremath{\mathcal{F}}}, \P) $ be a probability space,
let $ ( E, d ) $ and $ ( \cE, \delta ) $
be separable metric spaces,
let $ X_n \colon \Omega \to E $, $ n \in \N_0 $,
be random variables which satisfy
for every $\varepsilon \in (0,\infty)$ that
\begin{equation}
\label{eq:weakConvMetric1}
\limsup_{n\to\infty}
\P(d(X_n,X_0)\geq \varepsilon)
= 0,
\end{equation}
and let $\Phi\colon E \to \cE$ be a continuous function.
Then it holds for every $\varepsilon \in (0,\infty)$ that
\begin{equation}
\label{eq:weakConvMetric2}
\limsup_{n\to\infty}
\P(\delta(\Phi(X_n),\Phi(X_0))\geq\varepsilon)
= 0 .
\end{equation}
\end{lemma}
\begin{proof}[Proof of Lemma \ref{lemma:stochastic_convergence_under_continuous_transformations}]
Note that \eqref{eq:weakConvMetric1}, e.g., Cox et al.~\cite[Lemma 2.4]{CoxJentzenKurniawanPusnik}, and, e.g., Hutzenthaler et al.~\cite[Lemma 4.2]{HutzenthalerJentzenSalimova2016} establish \eqref{eq:weakConvMetric2}. The proof of Lemma~\ref{lemma:stochastic_convergence_under_continuous_transformations} is thus completed.
\end{proof}
\begin{lemma}
\label{lemma:equivalence_of_expectation_representations}
Let
$d, m\in\N$,
$T\in (0,\infty)$,
$L, a\in\R$,
$b\in (a,\infty)$,
let
$\mu\colon \R^d\to\R^d$
and
$\sigma\colon \R^d\to\R^{d\times m}$
be functions which satisfy for every $x,y\in\R^d$ that $\max\{\|\mu(x)-\mu(y)\|_{\R^d},\|\sigma(x)-\sigma(y)\|_{HS(\R^m,\R^d)}\}\le L\|x-y\|_{\R^d}$,
let
$\Phi \colon C([0,T],\R^d) \to \R$ be
an at most polynomially growing
continuous function,
let
$(\Omega,{\ensuremath{\mathcal{F}}},\P)$ be a probability space with a normal filtration
$({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$,
let
$\xi\colon\Omega\to [a,b]^d$ be a continuous uniformly distributed
${\ensuremath{\mathbb{F}}}_0$/$\B([a,b]^d)$-measurable random variable,
let
$W\colon [0,T]\times\Omega\to\R^m$ be a standard
$({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-Brownian motion,
for every $x\in [a,b]^d$ let
$X^x=(X^x_t)_{t\in [0,T]}\colon [0,T]\times\Omega\to\R^d$
be an $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted stochastic process with
continuous sample paths which satisfies that for every
$t\in [0,T]$ it holds $\P$-a.s.~that
\begin{equation}\label{eq:lemma_equivalence_sde_for_x}
X^x_t = x + \int_0^t \mu(X^x_s)\,ds + \int_0^t \sigma(X^x_s)\,dW_s,
\end{equation}
and
let
$\bX\colon [0,T]\times\Omega\to\R^d$ be an
$({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted stochastic process with continuous
sample paths which satisfies that for every $t\in [0,T]$ it
holds $\P$-a.s.~that
\begin{equation}
\label{eq:lemma_equivalence_sde_for_bx}
\bX_t
=
\xi
+
\int_0^t \mu(\bX_s)\,ds
+
\int_0^t \sigma(\bX_s)\,dW_s.
\end{equation}
Then
\begin{enumerate}[(i)]
\item\label{it:measurability}
it holds for every
$x\in [a,b]^d$ that the functions $\Omega\ni\omega\mapsto\Phi((X^x_t(\omega))_{t\in [0,T]})\in\R$
and $\Omega\ni\omega\mapsto\Phi((\bX_t(\omega))_{t\in [0,T]})\in\R$
are
${\ensuremath{\mathcal{F}}}$/$\B(\R)$-measurable,
\item\label{it:GWbound}
it holds for every $p\in[2,\infty)$, $x,y\in[a,b]^d$ that
\begin{equation}
\left(\E\left[\sup_{t\in[0,T]}\|X^x_t-X^y_t\|_{\R^d}\right]^p\right)^{\!\nicefrac{1}{p}}\le \sqrt{2}\exp\!\left(L^2T\big[p+\sqrt{T}\big]^2\right)\|x-y\|_{\R^d},
\end{equation}
\item\label{it:rhs_well_defined_for_every_x}
it holds for every
$x\in [a,b]^d$ that
$
\E\!\left[|\Phi((X^x_t)_{t\in [0,T]})|
+
|\Phi((\bX_t)_{t\in [0,T]})|
\right]
<
\infty,
$
\item\label{it:rhs_continuity_wrt_x}
it holds that the function
$[a,b]^d\ni x\mapsto \E[\Phi((X^x_t)_{t\in [0,T]})]\in\R$
is continuous, and
\item\label{it:identity_of_interest}
it holds that
\begin{equation}
\E\big[ \Phi((\bX_t)_{t\in [0,T]}) \big]
=
\tfrac{1}{(b-a)^d}
\left(
\int_{[a,b]^d} \E\big[ \Phi((X^x_t)_{t\in [0,T]}) \big]\,dx
\right).
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lemma:equivalence_of_expectation_representations}]
Throughout this proof let $c\in[1,\infty)$ be a real number which satisfies for every $w\in C([0,T],\R^m)$ that
\begin{equation}
|\Phi(w)| \leq c\left[1+\sup_{t\in [0,T]} \|w_t\|_{\R^d}\right]^c \label{eq:lemma_equivalence_at_most_polynomial_growth},
\end{equation}
let $p_t\colon C([0,T],\R^m) \to \R^m$, $t\in [0,T]$, be the functions which satisfy for every $t\in [0,T]$, $w=(w_s)_{s\in [0,T]}\in C([0,T],\R^m)$ that $p_t(w)=w_t$, and let $\Psi^N_{x,w}\colon[0,T]\to\R^d$, $N\in \N$, $x\in\R^d$, $w\in C([0,T],\R^m),$
be the functions which satisfy for every $w\in C([0,T],\R^m)$, $x\in\R^d$, $N\in \N$, $ n\in\{0,1,\ldots,N-1\}$, $t\in[\frac{nT}{N},\frac{(n+1)T}{N}]$ that $\Psi^N_{x,w}(0) = x$ and
\begin{equation}
\Psi^N_{x,w}(t) = \Psi^N_{x,w}\!\left(\tfrac{nT}{N}\right)+\left(\tfrac{nt}{T}-n\right)\left[\mu\!\left(\Psi^N_{x,w}\!\left(\tfrac{nT}{N}\right)\right)\tfrac{T}{N}+\sigma\!\left(\Psi^N_{x,w}\!\left(\tfrac{nT}{N}\right)\right) \big( w_{\frac{(n+1)T}{N}} - w_{\frac{nT}{N}} \big)\right].\label{eq:PsiDef}
\end{equation}
Observe that the fact that the Borel sigma-algebra $\B(C([0,T],\R^m))$ is generated by the set
$\cup_{t\in[0,T]}\cup_{A\in\B(\R^m)}\{(p_t)^{-1}(A)\}$ (cf., for example, Klenke~\cite[Theorem~21.31]{Klenke_2014}), the hypothesis that for every $x\in[a,b]^d$ it holds that $X^x\colon [0,T]\times\Omega\to\R^d$ is an $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted stochastic process with
continuous sample paths, and the hypothesis that $\bX\colon [0,T]\times\Omega\to\R^d$ is an
$({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted stochastic process with continuous sample paths demonstrate that the functions
\begin{equation}
\Omega\ni\omega\mapsto(X^x_t(\omega))_{t\in[0,T]}\in C([0,T],\R^d)\label{eq:omegaToX}
\end{equation}
and
\begin{equation}
\Omega\ni\omega\mapsto(\bX_t(\omega))_{t\in[0,T]}\in C([0,T],\R^d)\label{eq:omegaTobX}
\end{equation}
are
${\ensuremath{\mathcal{F}}}$/$\B(C([0,T],\R^d))$-measurable.
Combining this with the fact that the function
$\Phi\colon\allowbreak C([0,T],\R^d)\to\R$ is
$\B(C([0,T],\R^d))$/$\B(\R)$-measurable
implies that for every $x\in [a,b]^d$ it holds that
the functions $\Omega\ni\omega\mapsto\Phi((X^x_t(\omega))_{t\in[0,T]})\in\R$ and
$\Omega\ni\omega\mapsto\Phi((\bX_t(\omega))_{t\in[0,T]})\in\R$
are ${\ensuremath{\mathcal{F}}}$/$\B(\R)$-measurable. This proves
item~\eqref{it:measurability}.
Next observe that \eqref{eq:PsiDef}, the hypothesis that $\mu\colon\R^d\to\R^d$ and $\sigma\colon\R^d\to\R^{d\times m}$ are globally Lipschitz continuous, and the fact that for every $p\in(0,\infty)$ it holds that $\E[\|\xi\|_{\R^d}^p]<\infty$ ensure that for every $p\in(0,\infty)$ it holds that
\begin{equation}\label{eq:lemma_equivalence_lp_sup_numerics}
\begin{split}
&\sup_{N\in\N}
\sup_{x\in [a,b]^d}
\left(
\E\!\left[
\sup_{t\in [0,T]} \| \Psi^N_{x, W}(t) \|_{\R^d}^p
\right]
+
\E\!\left[
\sup_{t\in [0,T]} \| \Psi^N_{\xi, W}(t) \|_{\R^d}^p
\right]
\right)\\
&=\sup_{N\in\N}
\sup_{x\in [a,b]^d}
\left(
\E\!\left[
\max_{n\in \{0,1,\ldots,N\}} \left\| \Psi^N_{x, W}\!\left(\tfrac{nT}{N}\right) \right\|_{\R^d}^p
\right]
+
\E\!\left[
\max_{n\in \{0,1,\ldots,N\}} \left\| \Psi^N_{\xi, W}\!\left(\tfrac{nT}{N}\right) \right\|_{\R^d}^p
\right]
\right)<\infty
\end{split}
\end{equation}
(cf., for example, Kloeden \& Platen~\cite[Section 10.6]{KloedenPlaten1992}). Next note that~\eqref{eq:lemma_equivalence_sde_for_x}, \eqref{eq:lemma_equivalence_sde_for_bx}, \eqref{eq:PsiDef}, the fact that $\mu\colon\R^d\to\R^d$ and $\sigma\colon\R^d\to\R^{d\times m}$ are locally Lipschitz continuous functions, and e.g., Hutzenthaler \& Jentzen~\cite[Theorem 3.3]{HutzenthalerJentzen2015_Memoirs} ensure that for every $x\in[a,b]^d$, $\varepsilon\in(0,\infty)$ it holds that
\begin{equation}
\label{eq:supDiscret1}
\limsup_{N\to\infty}\P\!\left(\sup_{t\in[0,T]}\|X^x_t-\Psi^N_{x,W}(t)\|_{\R^d}\ge\varepsilon\right)=0
\end{equation}
and
\begin{equation}
\label{eq:supDiscret2}
\limsup_{N\to\infty}\P\!\left(\sup_{t\in[0,T]}\|\bX_t-\Psi^N_{\xi,W}(t)\|_{\R^d}\ge\varepsilon\right)=0.
\end{equation}
Combining \eqref{eq:omegaToX}, \eqref{eq:omegaTobX}, and, e.g., Hutzenthaler \& Jentzen~\cite[Lemma 3.10]{HutzenthalerJentzen2015_Memoirs} hence demonstrates that for every $x\in[a,b]^d, p\in(0,\infty)$ it holds that
\begin{equation}
\begin{split}
\E\!\left[\sup_{t\in[0,T]}\|X^x_t\|^p\right]&\le \liminf_{N\to\infty} \E\!\left[\sup_{t\in[0,T]}\|\Psi^N_{x,W}(t)\|^p\right]\\
&\le\sup_{N\in\N}\E\!\left[\sup_{t\in[0,T]}\|\Psi^N_{x,W}(t)\|^p\right]
\end{split}
\end{equation}
and
\begin{equation}
\begin{split}
\E\!\left[\sup_{t\in[0,T]}\|\bX_t\|^p\right]&\le \liminf_{N\to\infty} \E\!\left[\sup_{t\in[0,T]}\|\Psi^N_{\xi,W}(t)\|^p\right]\\
&\le\sup_{N\in\N}\E\!\left[\sup_{t\in[0,T]}\|\Psi^N_{\xi,W}(t)\|^p\right].
\end{split}
\end{equation}
This and \eqref{eq:lemma_equivalence_lp_sup_numerics} assure that for every $ p \in (0,\infty) $ it holds that
\begin{equation}\label{eq:lemma_equivalence_lp_sup_exact}
\sup_{x\in [a,b]^d}
\left(
\E\!\left[
\sup_{t\in [0,T]} \| X^x_t \|_{\R^d}^p
\right]
+
\E\!\left[
\sup_{t\in [0,T]} \| \bX_t \|_{\R^d}^p
\right]
\right)
< \infty .
\end{equation}
Combining
\eqref{eq:lemma_equivalence_at_most_polynomial_growth} and the fact that $\forall\,r \in(0,\infty)$, $ a,b\in\R\colon |a+b|^r\le 2^r(|a|^r+|b|^r)$ therefore demonstrates that for every $p\in(0,\infty)$ it holds that
\begin{equation}\label{eq:lemma_equivalence_uniform_lp_boundedness_Phi}
\begin{split}
&
\sup_{x\in [a,b]^d}
\left( \E\!\left[
|\Phi((X^x_t)_{t\in [0,T]})|^p
\right]
+
\E\!\left[
|\Phi((\bX_t)_{t\in [0,T]})|^p
\right]
\right)
\\&\leq
\sup_{x\in [a,b]^d}
\left(
c^p\,\E\!\left[
|1+\sup\nolimits_{t\in [0,T]} \|X^x_t\|_{\R^d}|^{cp}
\right]
+
c^p\,\E\!\left[
|1+\sup\nolimits_{t\in [0,T]} \|\bX_t\|_{\R^d}|^{cp}
\right]
\right)
\\&\leq
2^{cp}\,c^p
\left(
\sup_{x\in [a,b]^d}
\E\!\left[
2
+ \sup\nolimits_{t\in [0,T]} \|X^x_t\|_{\R^d}^{cp}
+ \sup\nolimits_{t\in [0,T]} \|\bX_t\|_{\R^d}^{cp}
\right]
\right)
<\infty .
\end{split}
\end{equation}
This establishes item~\eqref{it:rhs_well_defined_for_every_x}.
In the next step we observe that
\eqref{eq:lemma_equivalence_sde_for_x}
ensures that for every
$ x,y \in [a,b]^d$, $ t \in [0,T]$
it holds $\P$-a.s.~that
\begin{equation}
X^x_t - X^y_t
=
x-y
+
\int_0^t \left(\mu(X^x_s)-\mu(X^y_s)\right) ds
+
\int_0^t \left(\sigma(X^x_s)-\sigma(X^y_s)\right) dW_s.
\end{equation}
The triangle inequality hence ensures that for every
$ x,y \in [a,b]^d$, $t\in[0,T]$
it holds $\P$-a.s.~that
\begin{equation}
\begin{split}
&\sup_{s\in [0,t]}\|X^x_s - X^y_s \|_{\R^d}
\\
&\leq
\|x-y \|_{\R^d}
+
\sup_{s\in [0,t]} \int_0^s \|\mu(X^x_r)-\mu(X^y_r)\|_{\R^d} \,dr
+
\sup_{s\in [0,t]}\left\|\int_0^t
\left(\sigma(X^x_r)-\sigma(X^y_r)\right) dW_r\right\|_{\R^d}\\
&\leq \|x-y \|_{\R^d}
+
L \left[\sup_{s\in [0,t]} \int_0^s \|X^x_r-X^y_r\|_{\R^d} \,dr \right]
+\sup_{s\in [0,t]}\left\|\int_0^s
\left(\sigma(X^x_r)-\sigma(X^y_r)\right) dW_r\right\|_{\R^d}\\
&= \|x-y \|_{\R^d}
+
L \int_0^t \|X^x_r-X^y_r\|_{\R^d} \,dr
+\sup_{s\in [0,t]}\left\|\int_0^s
\left(\sigma(X^x_r)-\sigma(X^y_r)\right) dW_r\right\|_{\R^d}.
\end{split}
\end{equation}
Therefore, we obtain for every
$ p \in [1,\infty)$, $x,y \in [a,b]^d$, $t\in[0,T]$
that
\begin{equation}\label{eq:supDiffusion}
\begin{split}
\left(\E\!\left[\sup_{s\in [0,t]} \|X^x_s-X^y_s\|^p_{\R^d}\right]\right)^{\!\nicefrac{1}{p}}
& \leq
\|x-y\|_{\R^d}
+
L \int_0^t \left(\E\!\left[\|X^x_r-X^y_r\|^p_{\R^d}\right]\right)^{\!\nicefrac{1}{p}}dr
\\
&
+
\left(\E\!\left[\sup_{s\in [0,t]}
\left\|\int_0^s
\left(\sigma(X^x_r)-\sigma(X^y_r)\right)\,dW_r\right\|^p_{\R^d}
\right]\right)^{\!\nicefrac{1}{p}} .
\end{split}
\end{equation}
The Burkholder-Davis-Gundy type inequality in Da Prato \& Zabczyk~\cite[Lemma 7.2]{DaPratoZabczyk2008} hence shows that for every
$ p \in [2,\infty)$, $ x,y \in [a,b]^d$, $t\in[0,T]$
it holds that
\begin{equation}
\begin{split}
\left(\E\!\left[\sup_{s\in [0,t]} \|X^x_s-X^y_s\|^p_{\R^d}\right]\right)^{\!\nicefrac{1}{p}}
& \leq
\|x-y\|_{\R^d}
+
L \int_0^t \left(\E\big[\|X^x_r-X^y_r\|^p_{\R^d}\big]\right)^{\!\nicefrac{1}{p}}dr
\\
&
\quad +p
\left[\int_0^t
\left(\E\big[\|\sigma(X^x_r)-\sigma(X^y_r)\|_{HS(\R^d,\R^m)}^p\big]\right)^{\!\nicefrac{2}{p}}
dr \right]^{\nicefrac{1}{2}} .
\end{split}
\end{equation}
This demonstrates that for every $p\in[2,\infty)$, $x,y\in[a,b]^d$, $t\in[0,T]$ it holds that
\begin{equation}
\begin{split}
\left(\E\!\left[\sup_{s\in [0,t]} \|X^x_s-X^y_s\|^p_{\R^d}\right]\right)^{\!\nicefrac{1}{p}}
& \le
\|x-y\|_{\R^d}
+
L \int_0^t \left(\E\big[\|X^x_r-X^y_r\|^p_{\R^d}\big]\right)^{\!\nicefrac{1}{p}} dr
\\
&
\quad + Lp
\left[\int_0^t
\left(\E\big[\|X^x_r-X^y_r\|_{\R^d}^p\big]\right)^{\!\nicefrac{2}{p}}
dr \right]^{\nicefrac{1}{2}}.
\end{split}
\end{equation}
H{\"o}lder's inequality hence proves that for every $p\in[2,\infty)$, $x,y\in[a,b]^d$, $t\in[0,T]$ it holds that
\begin{equation}
\begin{split}
\left(\E\!\left[\sup_{s\in [0,t]} \|X^x_s-X^y_s\|^p_{\R^d}\right]\right)^{\!\nicefrac{1}{p}}
& \le
\|x-y\|_{\R^d}
+
L\sqrt{t}
\left[\int_0^t
\left(\E\big[\|X^x_r-X^y_r\|_{\R^d}^p\big]\right)^{\!\nicefrac{2}{p}}
dr \right]^{\nicefrac{1}{2}}
\\
&
\quad + Lp
\left[\int_0^t
\left(\E\big[\|X^x_r-X^y_r\|_{\R^d}^p\big]\right)^{\!\nicefrac{2}{p}}
dr \right]^{\nicefrac{1}{2}}
\\
&\le
\|x-y\|_{\R^d}
+
L\big[p+\sqrt{T}\big]
\left[\int_0^t
\left(\E\big[\|X^x_r-X^y_r\|_{\R^d}^p\big]\right)^{\!\nicefrac{2}{p}}
dr \right]^{\nicefrac{1}{2}}.
\end{split}
\end{equation}
The fact that $\forall \, v,w\in\R\colon|v+w|^2\le 2v^2+2w^2$ therefore shows that for every
$ p \in [2,\infty)$,
$ x,y \in [a,b]^d$,
$t\in[0,T]$ it holds that
\begin{equation}
\begin{split}
&\left(\E\!\left[\sup_{s\in [0,t]} \|X^x_s-X^y_s\|^p_{\R^d}\right]\right)^{\!\nicefrac{2}{p}}\\
&\leq
2\|x-y\|_{\R^d}^2
+
2L^2\big[p+\sqrt{T}\big]^2\int_0^t
\left(\E\big[\|X^x_r-X^y_r\|_{\R^d}^p\big]\right)^{\nicefrac{2}{p}}
\,dr\\
&\leq 2\|x-y\|_{\R^d}^2 +
2L^2\big[p+\sqrt{T}\big]^2\int_0^t \left(\E\!\left[\sup_{s\in [0,r]}\|X^x_s-X^y_s\|^p_{\R^d}\right]\right)^{\!\nicefrac{2}{p}}dr
.
\end{split}
\end{equation}
Combining the Gronwall inequality (cf., e.g., Andersson et al.~\cite[Lemma 2.6]{Andersson2015} (with $\alpha=0$, $\beta=0$, $a=2\|x-y\|_{\R^d}^2$, $b=3L^2\big[p+\sqrt{T}\big]$, $e=([0,T]\ni t \mapsto \left(\E\!\left[\sup_{s\in [0,t]} \|X^x_s-X^y_s\|^p_{\R^d}\right]\right)^{2/p}\in[0,\infty])$ in the notation of Lemma 2.6)) and \eqref{eq:lemma_equivalence_lp_sup_exact} hence establishes that for every
$ p \in [2,\infty)$, $ x,y \in [a,b]^d$, $t\in[0,T]$ it holds that
\begin{equation}
\left(\E\!\left[\sup_{s\in [0,t]} \|X^x_s-X^y_s\|^p_{\R^d}\right]\right)^{\!\nicefrac{2}{p}}
\leq
2\|x-y\|_{\R^d}^2 \exp\!\left(2L^2t\big[p+\sqrt{T}\big]^2\right).
\end{equation}
Therefore, we obtain that for every $p\in[2,\infty)$, $x,y\in[a,b]^d$ it holds that
\begin{equation}
\left(\E\left[\sup_{t\in[0,T]}\|X^x_t-X^y_t\|^p\right]\right)^{\!\nicefrac{1}{p}}\le \sqrt{2}\exp\!\left(L^2T\big[p+\sqrt{T}\big]^2\right)\|x-y\|_{\R^d}.
\end{equation}
This establishes item \eqref{it:GWbound}. Next observe that item \eqref{it:GWbound} and Jensen's inequality imply that for every
$ p \in (0,\infty)$ it holds that
\begin{equation}
\label{eq:uniform_integrability}
\sup_{x,y\in [a,b]^d, x\neq y}
\left(
\frac{\left(\E\!\left[\sup_{t\in [0,T]} \|X^x_t-X^y_t\|_{\R^d}^p\right]\right)^{\!\nicefrac{1}{p}}}
{\|x-y\|_{\R^d}}
\right)
< \infty .
\end{equation}
The hypothesis that the function
$\Phi\colon C([0,T],\R^d)\to\R$ is continuous and Lemma
\ref{lemma:stochastic_convergence_under_continuous_transformations}
hence ensure that for every $\varepsilon\in (0,\infty)$, $(x_n)_{n\in \N_0}\subseteq [a,b]^d$ with $\limsup_{n\to \infty} \|x_0-x_n\|_{\R^d}=0$ it holds that
\begin{equation}
\limsup_{n\to \infty}
\P\!
\left(
|\Phi((X^{x_0}_t)_{t\in [0,T]}) - \Phi((X^{x_n}_t)_{t\in [0,T]})| \geq \varepsilon
\right)
=
0 .
\end{equation}
Combining \eqref{eq:lemma_equivalence_uniform_lp_boundedness_Phi} with, e.g., Hutzenthaler et al.~\cite[Proposition 4.5]{HutzenthalerJentzenSalimova2016} therefore implies that for every $(x_n)_{n\in \N_0}\subseteq [a,b]^d$ with $\limsup_{n\to \infty} \|x_0-x_n\|_{\R^d}=0$ it holds that
\begin{equation}
\limsup_{n\to \infty}
\E\big[
|\Phi((X^{x_0}_t)_{t\in [0,T]}) - \Phi((X^{x_n}_t)_{t\in [0,T]})|
\big]
= 0.
\end{equation}
This establishes item~\eqref{it:rhs_continuity_wrt_x}. In the next step we observe that \eqref{eq:supDiscret1} and \eqref{eq:supDiscret2} ensure that for every
$ \varepsilon \in (0,\infty) $,
$ x\in [a,b]^d $
it holds that
\begin{equation}\label{eq:euler_scheme_convergence_in_probability}
\limsup_{N\to\infty} \left[
\P\!\left(
\sup_{t\in [0,T]} \|
\Psi^N_{x,W}(t)
-
X^x_t \|_{\R^d}
\geq \varepsilon \right)
+
\P\!\left(
\sup_{t\in [0,T]} \|
\Psi^N_{\xi,W}(t)
-
\bX_t \|_{\R^d}
\geq \varepsilon \right)
\right]
= 0.
\end{equation}
The hypothesis that the function $\Phi$ is continuous and
Lemma~\ref{lemma:stochastic_convergence_under_continuous_transformations}
therefore demonstrate that for every
$ \varepsilon\in (0,\infty) $, $ x \in [a,b]^d $ it holds that
\begin{equation}\label{eq:stochastic_convergence_combined}
\limsup_{N\to\infty}
\left[
\P\!
\left(
|\Phi(\Psi^N_{x,W}) - \Phi((X^x_t)_{t\in [0,T]})|
+
|\Phi(\Psi^N_{\xi,W})-\Phi((\bX_t)_{t\in [0,T]})|
\geq \varepsilon
\right)
\right]
= 0 .
\end{equation}
Next observe that
\eqref{eq:lemma_equivalence_at_most_polynomial_growth}
assures that for every
$ N \in \N$,
$ x \in [a,b]^d $,
$ p \in (0,\infty) $
it holds that
\begin{equation}\label{eq:lemma_equivalence_p_integrability}
\begin{split}
& \E\!\left[
|\Phi(\Psi^N_{ x, W} )
-
\Phi( (X^x_t)_{t\in [0,T]} ) |^p
+
|\Phi( \Psi^N_{ \xi, W} )
-
\Phi( (\bX_t)_{t\in [0,T]} ) |^p
\right]
\\[1.5 ex]
&\leq
2^p\,\E\!\left[
| \Phi( \Psi^N_{ x, W } ) |^p
+
|\Phi( (X^x_t)_{t\in [0,T]} ) |^p
\right]
+
2^p\,\E\!\left[| \Phi(\Psi^N_{ \xi, W} ) |^p
+
| \Phi( (\bX_t)_{t\in [0,T]} ) |^p
\right]
\\[1.5 ex]
&\leq
2^pc^p \,
\E\Big[
\left|
1+\sup\nolimits_{t\in [0,T]}
\| \Psi^N_{ x, W} ( t ) \|_{\R^d}
\right|^{cp}
+
\left|
1+\sup\nolimits_{t\in [0,T]} \|X^x_t\|_{\R^d}
\right|^{cp}
\Big]
\\[1.5 ex]
&+
2^pc^p \,
\E\Big[
\left|
1+\sup\nolimits_{t\in [0,T]}
\| \Psi^N_{ \xi, W} (t) \|_{\R^d}
\right|^{cp}
+
\left|
1+\sup\nolimits_{t\in [0,T]} \|\bX_t\|_{\R^d}
\right|^{cp}
\Big]
\\[1.5 ex]
&\leq
4^pc^p \,
\E\!\left[
2
+
\sup\nolimits_{t\in [0,T]} \| \Psi^N_{ x, W} ( t ) \|_{\R^d}^{cp}
+
\sup\nolimits_{t\in [0,T]} \|X^x_t\|_{\R^d}^{cp}
\right] \\[1.5 ex]
&+
4^pc^p \,
\E\!\left[
2
+
\sup\nolimits_{t\in [0,T]} \| \Psi^N_{ \xi, W} ( t ) \|_{\R^d}^{cp}
+
\sup\nolimits_{t\in [0,T]} \|\bX_t\|_{\R^d}^{cp}
\right].
\end{split}
\end{equation}
Combining \eqref{eq:lemma_equivalence_lp_sup_numerics} and
\eqref{eq:lemma_equivalence_lp_sup_exact}
hence shows that for every $ p \in (0,\infty) $ it holds that
\begin{equation}
\label{eq:lemma_equivalence_l2_boundedness}
\sup_{N\in\N}\sup_{x\in [a,b]^d}
\left(
\E\!\left[
|\Phi(\Psi^N_{ x, W} )
-
\Phi( (X^x_t)_{t\in [0,T]} ) |^p
+
|\Phi(\Psi^N_{ \xi, W})
-
\Phi( (\bX_t)_{t\in [0,T]} ) |^p
\right]
\right)
< \infty .
\end{equation}
This, \eqref{eq:stochastic_convergence_combined}, and, e.g., Hutzenthaler et al.~\cite[Proposition 4.5]{HutzenthalerJentzenSalimova2016} imply that for every $ x \in [a,b]^d $
it holds that
\begin{equation}\label{eq:lemma_equivalence_first_l1_convergence}
\limsup_{N\to\infty}
\left(
\E\big[|\Phi(\Psi^N_{x,W})
-
\Phi( (X^x_t)_{t\in [0,T]})|
\big]
+
\E\big[|\Phi(\Psi^{N}_{\xi, W})
-
\Phi( (\bX_t)_{t\in [0,T]})|\big]
\right) = 0.
\end{equation}
Combining \eqref{eq:lemma_equivalence_l2_boundedness}
with Lebesgue's dominated convergence theorem therefore demonstrates that
\begin{equation}\label{eq:lemma_equivalence_second_l1_convergence}
\limsup_{N\to\infty}
\left(\int_{[a,b]^d}
\E\big[| \Phi( \Psi^{N}_{ x, W} )
-
\Phi( (X^x_t)_{t\in [0,T]})|\big]\,dx
\right)
= 0.
\end{equation}
In addition, observe that \eqref{eq:lemma_equivalence_p_integrability}, \eqref{eq:lemma_equivalence_lp_sup_numerics}, and \eqref{eq:lemma_equivalence_lp_sup_exact} prove that for all $p\in(0,\infty)$ it holds that
\begin{equation}\label{eq:Phi_Psi_integrability}
\sup_{N\in\N} \sup_{x\in[a,b]^d} \E\big[|\Phi(\Psi^N_{x,W})|^p\big]<\infty.
\end{equation}
Next observe that~\eqref{eq:lemma_equivalence_first_l1_convergence}
and the fact that $\xi$ and $W$ are independent imply that
\begin{align}
\label{eq:relationBetweenExpectationsOfbXAndXx_1_new}
\begin{split}
\E\!\left[ \Phi( (\bX_t)_{t\in [0,T]} ) \right]
&=
\lim_{ N \rightarrow \infty }
\E\Big[ \Phi\big(\Psi^{N}_{\xi, W}\big) \Big]
\\&=
\lim_{ N \rightarrow \infty }\left[
\int_{ \Omega }
\Phi\!\left(\Psi^{N}_{\xi(\omega), (W_t(\omega))_{t\in[0,T]}}\right) \P(d\omega)\right]
\\&=
\lim_{ N \rightarrow \infty }\left[
\int_{ [a,b]^d \times C([0,T],\R^d) }
\Phi\big(\Psi^{N}_{x, W}\big) \, \big( (\xi, W)(\P) \big)(dx,dw)\right]
\\&=
\lim_{ N \rightarrow \infty }\left[
\int_{ [a,b]^d \times C([0,T],\R^d) }
\Phi\big(\Psi^{N}_{x, w}\big) \, \big( (\xi(\P)) \otimes (W(\P)) \big)(dx,dw)\right].
\end{split}
\end{align}
Combining Fubini's theorem, \eqref{eq:lemma_equivalence_second_l1_convergence}, and \eqref{eq:Phi_Psi_integrability} with Lebesgue's dominated convergence theorem
therefore assures that
\begin{align}
\label{eq:relationBetweenExpectationsOfbXAndXx_2_new}
\begin{split}
\E\!\left[ \Phi((\bX_t)_{t\in [0,T]}) \right]
&=
\lim_{ N \rightarrow \infty }\left[
\int_{ [a,b]^d }
\left(
\int_{ C([0,T],\R^d) }
\Phi\big(\Psi^{N}_{x, w}\big) \, (W(\P))(dw)
\right) (\xi(\P))(dx)\right]
\\&=
\lim_{ N \rightarrow \infty }\left[
\int_{ [a,b]^d }
\left(
\int_{ \Omega }
\Phi\big(\Psi^{N}_{x, w}\big) \, \P(dw)
\right) (\xi(\P))(dx)\right]
\\&=
\lim_{ N \rightarrow \infty }\left[
\int_{ [a,b]^d }
\E\Big[ \Phi\big(\Psi^{N}_{x, W}\big) \Big] \, (\xi(\P))(dx)\right]
\\&=
\int_{ [a,b]^d }
\lim_{ N \rightarrow \infty }
\E\Big[ \Phi\big(\Psi^{N}_{x, W}\big) \Big] \, (\xi(\P))(dx)
\\&=
\int_{[a,b]^d}
\E\!\left[ \Phi( (X^x_t)_{t\in [0,T]} ) \right]
(\xi(\P))(dx)
\\&
=
\frac{1}{(b-a)^d} \left(\int_{[a,b]^d} \E\!\left[ \Phi( (X^x_t)_{t\in [0,T]} ) \right] dx\right).
\end{split}
\end{align}
This establishes item \eqref{it:identity_of_interest}. The proof of Lemma~\ref{lemma:equivalence_of_expectation_representations}
is thus completed.
\end{proof}
\begin{proposition}
\label{proposition:minimizingPropertyApplied_new}
Let
$d, m\in\N$,
$T\in (0,\infty)$,
$a\in\R$,
$b\in (a,\infty)$,
let
$\mu\colon \R^d\to\R^d$ and
$\sigma\colon \R^d\to\R^{d\times m}$
be globally Lipschitz continuous functions,
let
$\varphi\colon\R^d\to\R$ be a function,
let
$u=(u(t,x))_{(t,x)\in [0,T]\times\R^d}\in C^{1,2}([0,T]\times\R^d,\R)$
be a function with at most polynomially growing partial derivatives which satisfies
for every $t\in [0,T]$, $x\in\R^d$ that
$u(0,x) = \varphi(x)$ and
\begin{equation}
\label{eq:differentialu}
\tfrac{\partial u}{\partial t}(t,x)
=
\tfrac12 \operatorname{Trace}_{\R^d}\!\big(
\sigma(x)[\sigma(x)]^{*}(\operatorname{Hess}_x u)(t,x)\big)
+
\langle
\mu(x),(\nabla_x u)(t,x)
\rangle_{\R^d},
\end{equation}
let
$(\Omega,{\ensuremath{\mathcal{F}}},\P)$ be a probability space with a normal filtration $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$,
let
$W\colon [0,T]\times\Omega\to\R^m$
be a standard $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-Brownian motion,
let
$\xi\colon\Omega\to [a,b]^d$ be
a continuous uniformly distributed
${\ensuremath{\mathbb{F}}}_0$/$\B([a,b]^d)$-measurable random variable,
and let
$\bX=(\bX_t)_{t\in [0,T]}\colon [0,T]\times\Omega\to\R^d$
be an $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted stochastic process with
continuous sample paths which satisfies that for every
$t\in [0,T]$ it holds $\P$-a.s.~that
\begin{equation}\label{eq:SDEforXstartingAtXi_new}
\bX_t = \xi + \int_0^t \mu(\bX_s)\,ds + \int_0^t \sigma(\bX_s)\,dW_s.
\end{equation}
Then
\begin{enumerate}[(i)]
\item\label{it:varphiContinuous_new} it holds that the function
$\varphi \colon \R^d \rightarrow \R$ is twice continuously differentiable
with at most polynomially growing derivatives,
\item \label{it:exuniqueSolutions} it holds that there exists a unique continuous function
$U\colon [a,b]^d\to\R$ such that
\begin{equation}
\E\big[ | \varphi(\bX_T) - U(\xi) |^2 \big]
=
\inf_{v\in C([a,b]^d,\R)} \E\big[ | \varphi(\bX_T) - v(\xi) |^2 \big] ,
\end{equation}
and
\item \label{it:UboundaryCond}it holds for every $x\in [a,b]^d$ that $U(x)=u(T,x)$.
\end{enumerate}
\end{proposition}
\begin{proof}[Proof of Proposition~\ref{proposition:minimizingPropertyApplied_new}]
Throughout this proof let
$X^x=(X^x_t)_{t\in [0,T]}\colon [0,T]\times\Omega\to\R^d$, $x\in [a,b]^d$,
be $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted stochastic processes with continuous
sample paths
\begin{enumerate}[a)]
\item which satisfy that for every $t\in [0,T]$, $x\in [a,b]^d$
it holds $\P$-a.s.~that
\begin{equation}
\label{eq:SDEPropertyAppliedForx}
X^x_t = x + \int_0^t \mu(X^x_s)\,ds + \int_0^t \sigma(X^x_s)\,dW_s
\end{equation}
and
\item which satisfy that for every
$\omega\in\Omega$ it holds that the function
$[a,b]^d \ni x \mapsto X^x_T(\omega) \in \R^d$ is continuous (cf., for example, Cox et al.~\cite[Theorem 3.5]{CoxJentzenHutzenthaler2013} and item~\eqref{it:GWbound} in Lemma~\ref{lemma:equivalence_of_expectation_representations}).
\end{enumerate}
Note that the assumption that $ \forall \, x \in \R^d \colon u(0,x) = \varphi(x) $
and the assumption that $u\in C^{1,2}([0,T]\times\R^d,\R)$ has at most polynomially
growing partial derivatives establish item~\eqref{it:varphiContinuous_new}.
Next note that item \eqref{it:varphiContinuous_new} and the fact that for every
$ p \in (0,\infty)$, $x\in[a,b]^d$ it holds that
\begin{equation}
\sup_{t\in [0,T]}
\E[\|X^x_t\|_{\R^d}^p]
< \infty
\end{equation}
assure that for every $x\in [a,b]^d$ it holds that
\begin{equation}
\label{eq:minimizingPropertyApplied_1_new}
\E\big[|\varphi(X^x_T)|^2] < \infty .
\end{equation}
Item~\eqref{it:varphiContinuous_new}, the
assumption that
for every $\omega \in \Omega$
it holds that the function
$
[a,b]^d \ni x
\mapsto X^x_T(\omega)
\in \R^d
$
is continuous, and, e.g., Hutzenthaler et al.~\cite[Proposition 4.5]{HutzenthalerJentzenSalimova2016} hence ensure that the function
\begin{equation}
\label{eq:minimizingPropertyApplied_2_new}
[a,b]^d \ni x
\mapsto
\E[ \varphi(X^x_T) ]
\in \R
\end{equation}
is continuous. In the next step we combine the fact that
for every
$
x \in [a,b]^d
$
it holds that the function
$
\Omega \ni \omega
\mapsto
\varphi(X^x_T(\omega))
\in \R
$
is ${\ensuremath{\mathcal{F}}}$/$\mathcal{B}(\R)$-measurable, the
fact that for every
$
\omega \in \Omega
$
it holds that the function
$
[a,b]^d \ni x
\mapsto \varphi(X^x_T(\omega))
\in \R
$
is continuous, and Lemma~\ref{lem:productMeasurable_new}
to obtain that the function
\begin{equation}
\label{eq:minimizingPropertyApplied_3_new}
[a,b]^d \times \Omega
\ni
(x, \omega)
\mapsto
\varphi(X^x_T(\omega))
\in \R
\end{equation}
is $(\mathcal{B}([a,b]^d) \otimes {\ensuremath{\mathcal{F}}})$/$\mathcal{B}(\R)$-measurable.
Combining this,
\eqref{eq:minimizingPropertyApplied_1_new},
\eqref{eq:minimizingPropertyApplied_2_new}, and
Proposition~\ref{proposition:minimizingProperty}
demonstrates
\begin{enumerate}[A)]
\item that there exists a unique continuous function
$U\colon [a,b]^d\to\R$ which satisfies that
\begin{equation}
\label{eq:minimizingPropertyEqApl_new}
\int_{[a,b]^d} \E\big[ | \varphi(X^x_T) - U(x) |^2 \big] \,dx
=
\inf_{v\in C([a,b]^d,\R)}
\left(
\int_{ [a,b]^d }
\E\big[ | \varphi(X^x_T) - v(x) |^2 \big] \, dx
\right)
\end{equation}
and
\item that it holds for every $x\in[a,b]^d$ that
\begin{equation}
\label{eq:feynmkac2}
U(x)=\E[\varphi(X^x_T)].
\end{equation}
\end{enumerate}
Next note that for every continuous function $V\colon[a,b]^d\to\R$ it holds that
\begin{equation}
\sup_{x\in[a,b]^d}|V(x)|<\infty.
\end{equation}
Item \eqref{it:varphiContinuous_new} hence implies that for every continuous function $V\colon[a,b]^d\to\R$ it holds that
\begin{equation}
C([0,T],\R^d) \ni (z_t)_{t\in [0,T]} \mapsto |\varphi(z_T)-V(z_0)|^2 \in \R
\end{equation}
is an at most polynomially growing continuous function. Combining Lemma~\ref{lemma:equivalence_of_expectation_representations} (with $\Phi=(C([0,T],\R^d) \ni (z_t)_{t\in [0,T]} \mapsto |\varphi(z_T)-V(z_0)|^2 \in \R)$ for $V\in C([a,b]^d,\R^d)$ in the notation of Lemma~\ref{lemma:equivalence_of_expectation_representations}), \eqref{eq:SDEforXstartingAtXi_new}, item~\eqref{it:varphiContinuous_new}, and \eqref{eq:SDEPropertyAppliedForx} hence ensures that for every continuous function $V\colon[a,b]^d\to\R$ it holds that
\begin{equation}
\label{eq:nrRR}
\E\big[|\varphi(\bX_T)-V(\xi)|^2\big]=\frac{1}{(b-a)^d}\left[\int_{[a,b]^d}\E\big[|\varphi(X^x_T)-V(x)|^2\big]\,dx\right].
\end{equation}
Hence, we obtain that for every continuous function $V\colon[a,b]^d\to\R$ with $\E[|\varphi(\bX_T)-V(\xi)|^2]=\inf_{v\in C([a,b]^d,\R)}\E[|\varphi(\bX_T)-v(\xi)|^2]$ it holds that
\begin{equation}
\begin{split}
&\int_{[a,b]^d}\E\big[|\varphi(X^x_T)-V(x)|^2\big]\,dx\\
&=(b-a)^d\left(\frac{1}{(b-a)^d}\left[\int_{[a,b]^d}\E\big[|\varphi(X^x_T)-V(x)|^2\big]\,dx\right]\right)\\
&=(b-a)^d\left(\E\big[|\varphi(\bX_T)-V(\xi)|^2\big]\right)\\
&=(b-a)^d\left(\inf_{v\in C([a,b]^d,\R)}\E\big[|\varphi(\bX_T)-v(\xi)|^2\big]\right)\\
&=(b-a)^d\left(\inf_{v\in C([a,b]^d,\R)}\left(\frac{1}{(b-a)^d}\left[\int_{[a,b]^d}\E\big[|\varphi(X^x_T)-v(x)|^2\big]\,dx\right]\right)\right)\\
&=\inf_{v\in C([a,b]^d,\R)}\left[\int_{[a,b]^d}\E\big[|\varphi(X^x_T)-v(x)|^2\big]\,dx\right].
\end{split}
\end{equation}
Combining this with \eqref{eq:minimizingPropertyEqApl_new} proves that for every continuous function $V\colon[a,b]^d\to\R$ with $\E[|\varphi(\bX_T)-V(\xi)|^2]=\inf_{v\in C([a,b]^d,\R)}\E[|\varphi(\bX_T)-v(\xi)|^2]$ it holds that
\begin{equation}
\label{eq:uniqueUisV}
U=V.
\end{equation}
Next observe that \eqref{eq:nrRR} and \eqref{eq:minimizingPropertyEqApl_new} demonstrate that
\begin{equation}
\begin{split}
\E\big[ | \varphi(\bX_T) - U(\xi) |^2 \big]
&
=
\frac{1}{(b-a)^d} \left(\int_{[a,b]^d} \E\big[ | \varphi(X^x_T) - U(x) |^2 \big] \,dx \right)
\\
& =
\inf_{v\in C([a,b]^d,\R)} \bigg[ \frac{1}{(b-a)^d} \left( \int_{[a,b]^d} \E\big[ | \varphi(X^x_T) - v(x) |^2 \big]\,dx \right) \bigg]
\\
& =
\inf_{v\in C([a,b]^d,\R)} \E\big[ | \varphi(\bX_T) - v(\xi) |^2 \big].
\end{split}
\end{equation}
Combining this with \eqref{eq:uniqueUisV} proves item~\eqref{it:exuniqueSolutions}.
Next note that \eqref{eq:differentialu}, \eqref{eq:feynmkac2}, and the Feynman-Kac formula (cf., for example, Hairer et al.~\cite[Corollary 4.17]{HairerHutzenthalerJentzen_LossOfRegularity2015}) imply that for every $x\in[a,b]^d$ it holds that $U(x)=\E[\varphi(X^x_T)]=u(T,x)$. This establishes item \eqref{it:UboundaryCond}.
The proof of Proposition~\ref{proposition:minimizingPropertyApplied_new} is thus completed.
\end{proof}
In the next step we use Proposition~\ref{proposition:minimizingPropertyApplied_new}
to obtain a minimization problem which is uniquely solved by the function
$[a,b]^d\ni x\mapsto u(T,x)\in\R$. More specifically, let
$\xi\colon\Omega\to [a,b]^d$ be a continuously
uniformly distributed ${\ensuremath{\mathbb{F}}}_0$/$\B([a,b]^d)$-measurable
random variable,
and
let
$\bX\colon [0,T]\times\Omega\to\R^d$ be an $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted
stochastic process with continuous sample paths which satisfies that for
every $t\in [0,T]$ it holds $\P$-a.s.~that
\begin{equation}\label{eq:SDEforXStartingAtXiInSloppyDerivation}
\bX_t = \xi + \int_0^t \mu(\bX_s)\,ds + \int_0^t \sigma(\bX_s)\,dW_s.
\end{equation}
Proposition~\ref{proposition:minimizingPropertyApplied_new} then guarantees that
the function $[a,b]^d\ni x\mapsto u(T,x)\in\R$ is the unique global
minimizer of the function
\begin{equation}\label{eq:functionMinimizedByU}
C([a,b]^d,\R) \ni v \mapsto \E\big[ | \varphi(\bX_T) - v(\xi) |^2 \big] \in \R.
\end{equation}
In the following two subsections we derive an
approximated minimization problem
by discretizing the stochastic process $\bX\colon [0,T]\times\Omega\to\R^d$
(see Subsection~\ref{subsec:discretizationOfX} below)
and
by employing a deep neural network approximation for the
function $\R^d\ni x\mapsto u(T,x)\in\R$
(see Subsection~\ref{subsec:DNNapproximations} below).
\subsection{Discretization of the stochastic differential equation}
\label{subsec:discretizationOfX}
In this subsection we use the Euler-Maruyama scheme
(cf., for example, Kloeden \& Platen~\cite{KloedenPlaten1992}
and Maruyama~\cite{Maruyama_ContinuousMarkovProcessesAndStochasticEquations1955})
to temporally discretize
the solution process $\bX$ of the SDE~\eqref{eq:SDEforXStartingAtXiInSloppyDerivation}.
More specifically, let $N\in\N$,
let $t_0,t_1,\ldots,t_N\in [0,\infty)$ be real numbers
which satisfy that
\begin{equation}
0 = t_0 < t_1 < \ldots < t_N = T.
\end{equation}
Note that \eqref{eq:SDEforXStartingAtXiInSloppyDerivation} implies that for every $n\in\{0,1,\ldots,N-1\}$
it holds $\P$-a.s.~that
\begin{equation}
\bX_{t_{n+1}} = \bX_{t_n} +
\int_{t_n}^{t_{n+1}} \mu(\bX_s)\,ds +
\int_{t_n}^{t_{n+1}} \sigma(\bX_s)\,dW_s.
\end{equation}
This suggests that for sufficiently small mesh size
$\sup_{n\in\{0,1,\ldots,N-1\}} (t_{n+1}-t_n)$
it holds that
\begin{equation}\label{eq:SDEforXApproximated}
\bX_{t_{n+1}} \approx \bX_{t_n} +
\mu(\bX_{t_n})\,(t_{n+1}-t_n) +
\sigma(\bX_{t_n})\,(W_{t_{n+1}}-W_{t_n}).
\end{equation}
Let ${\ensuremath{\mathcal{X}}}\colon\{0,1,\ldots,N\}\times\Omega\to\R^d$
be the stochastic
process which satisfies for every $n\in\{0,1,\ldots,N-1\}$ that ${\ensuremath{\mathcal{X}}}_0 = \xi$
and
\begin{equation}\label{eq:definitionOfXapproximation}
{\ensuremath{\mathcal{X}}}_{n+1} = {\ensuremath{\mathcal{X}}}_n + \mu({\ensuremath{\mathcal{X}}}_n)\,(t_{n+1}-t_n) + \sigma({\ensuremath{\mathcal{X}}}_n)\,(W_{t_{n+1}}-W_{t_n}).
\end{equation}
Observe that \eqref{eq:SDEforXApproximated} and
\eqref{eq:definitionOfXapproximation}
suggest, in turn, that for every $n\in\{0,1,2,\ldots,N\}$ it holds
that
\begin{equation}
{\ensuremath{\mathcal{X}}}_n \approx \bX_{t_n}
\end{equation}
(cf., for example, Theorem~\ref{thm:strong_convergence_euler_method} below
for a strong convergence result for the Euler-Maruyama scheme).
\begin{theorem}[Strong convergence rate for the Euler-Maruyama scheme]
\label{thm:strong_convergence_euler_method}
Let
$T\in (0,\infty)$, $d\in\N$, $p\in [2,\infty)$,
let
$(\Omega,{\ensuremath{\mathcal{F}}},\P)$ be a probability space
with a normal filtration $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$,
let
$W\colon [0,T]\times\Omega\to\R^d$
be a standard $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-Brownian
motion,
let
$\xi\colon\Omega\to\R^d$ be a random variable which satisfies
that $\E[ \| \xi \|_{\R^d}^p]<\infty$,
let
$\mu\colon\R^d\to\R^d$ and $\sigma\colon\R^d\to\R^{d\times d}$
be Lipschitz continuous functions,
let
$\bX\colon [0,T]\times\Omega\to\R^d$ be an $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-adapted
stochastic process with continuous sample paths which satisfies that
for every $t\in [0,T]$ it holds $\P$-a.s.~that
\begin{equation}
\bX_t = \xi + \int_0^t \mu(\bX_s)\,ds + \int_0^t \sigma(\bX_s)\,dW_s,
\end{equation}
for every $N\in\N$ let
$t^{N}_0, t^{N}_1, \ldots, t^{N}_N \in [0,T]$ be
real numbers which satisfy that
\begin{equation}
0 = t^{N}_0 < t^{N}_1 < \ldots < t^{N}_N = T,
\end{equation}
and for every $N\in\N$ let
${\ensuremath{\mathcal{X}}}^{N}\colon \{0,1,\ldots,N\} \times\Omega\to\R^d$
be the stochastic process which satisfies for every $n\in\{0,1,\ldots,N-1\}$
that ${\ensuremath{\mathcal{X}}}^{N}_0 = \xi_0$ and
\begin{equation}
{\ensuremath{\mathcal{X}}}^{N}_{n+1}
=
{\ensuremath{\mathcal{X}}}^{N}_n
+
\mu\big({\ensuremath{\mathcal{X}}}^{N}_n\big)(t^{N}_{n+1} - t^{N}_n)
+
\sigma\big({\ensuremath{\mathcal{X}}}^{N}_n\big)\big(W_{t^{N}_{n+1}} - W_{t^{N}_n}\big).
\end{equation}
Then there exists a real number $C\in (0,\infty)$ such that
for every $N\in\N$ it holds that
\begin{equation}
\sup_{n \in \{0,1,\ldots,N\}}
\Big( \E\big[ \|\bX_{t_n^N} - {\ensuremath{\mathcal{X}}}^{N}_n \|_{\R^d}^p \big]\Big)^{\!\nicefrac{1}{p}}
\leq
C
\left[
\max_{ n \in \{0,1,\ldots,N-1\} }
| t_{n+1} - t_n |
\right]^{\nicefrac{1}{2}}.
\end{equation}
\end{theorem}
The proof of Theorem~2.4 is well-known in the literature
(cf., for instance, Kloeden \& Platen~\cite{KloedenPlaten1992},
Milstein~\cite{Milstein1995},
Hofmann, M{\"u}ller-Gronbach, \& Ritter~\cite{HofmannGronbachRitter2000},
M{\"u}ller-Gronbach \& Ritter~\cite{GronbachRitterMinimal2008},
and the references mentioned therein).
\subsection{Deep artificial neural network approximations}
\label{subsec:DNNapproximations}
In this subsection we employ suitable approximations for the solution
$
\R^d\ni x\mapsto u(T,x)\in\R
$ of the PDE~\eqref{eq:kolmogorovPDE} at time $T$.
More specifically, let $ \nu \in \N $
and let
$\bU = (\bU(\theta,x))_{(\theta,x)\in\R^{\nu}\times\R^d} \colon \R^{\nu} \times \R^d \to \R$
be a continuous function.
For every
\emph{suitable}
$
\theta\in\R^{\nu}
$
and every $x\in[a,b]^d$
we think of
$\bU(\theta,x)\in\R$
as an appropriate approximation
\begin{equation}\label{eq:VThetaApproxVz}
\bU(\theta,x) \approx u(T,x)
\end{equation}
of $u(T,x)$.
We suggest to choose the function $\bU\colon\R^{\nu}\times\R^d\to\R$
as a deep neural network (cf., for example, Bishop~\cite{Bishop_PatternRecognition2016}).
For instance,
let $ \mathcal{L}_d \colon \R^d \to \R^d $ be the function
which satisfies for every $ x = ( x_1, x_2, \dots, x_d ) \in \R^d $ that
\begin{equation}
\mathcal{L}_d( x )
=
\left(
\frac{\exp(x_1)}{\exp(x_1)+1},
\frac{\exp(x_2)}{\exp(x_2)+1},
\dots
,
\frac{\exp(x_d)}{\exp(x_d)+1}
\right)
\end{equation}
(multidimensional version of the standard logistic function),
for every
$ k, l \in \N $,
$ v \in \N_0 = \{0\} \cup \N $,
$ \theta = ( \theta_1, \dots, \theta_{ \nu } ) \in \R^{ \nu } $
with
$
v + l (k + 1 ) \leq \nu
$
let
$ A^{ \theta, v }_{ k, l } \colon \R^k \to \R^l $
be the function which satisfies for every
$ x = ( x_1, \dots, x_k )\in\R^k $ that
\begin{equation}
A^{ \theta, v }_{ k, l }( x )
=
\left(
\begin{array}{cccc}
\theta_{ v + 1 }
&
\theta_{ v + 2 }
&
\dots
&
\theta_{ v + k }
\\
\theta_{ v + k + 1 }
&
\theta_{ v + k + 2 }
&
\dots
&
\theta_{ v + 2 k }
\\
\theta_{ v + 2 k + 1 }
&
\theta_{ v + 2 k + 2 }
&
\dots
&
\theta_{ v + 3 k }
\\
\vdots
&
\vdots
&
\vdots
&
\vdots
\\
\theta_{ v + ( l - 1 ) k + 1 }
&
\theta_{ v + ( l - 1 ) k + 2 }
&
\dots
&
\theta_{ v + l k }
\end{array}
\right)
\left(
\begin{array}{c}
x_1
\\
x_2
\\
x_3
\\
\vdots
\\
x_k
\end{array}
\right)
+
\left(
\begin{array}{c}
\theta_{ v + k l + 1 }
\\
\theta_{ v + k l + 2 }
\\
\theta_{ v + k l + 3 }
\\
\vdots
\\
\theta_{ v + k l + l }
\end{array}
\right),
\end{equation}
let $ s \in \{ 3, 4, 5, 6, \ldots \} $,
assume that $ (s-1) d (d+1) + d + 1 \leq \nu $,
and let
$\bU\colon\R^{\nu}\times\R^d\to\R$
be the function
which satisfies for every
$\theta\in\R^{\nu}$,
$x\in\R^d$
that
\begin{equation}\label{eq:example_net}
\bU(\theta,x)
=
\big( A^{ \theta, (s-1)d(d+1) }_{ d, 1 }
\circ
\mathcal{L}_d
\circ
A^{ \theta, (s-2)d(d+1) }_{ d, d }
\circ
\ldots
\circ
\mathcal{L}_d
\circ
A^{ \theta, d(d+1) }_{ d, d }
\circ
\mathcal{L}_d
\circ
A^{ \theta, 0 }_{ d, d } \big)(x)
.
\end{equation}
The function $\bU\colon \R^{\nu}\times\R^d \to \R$ in \eqref{eq:example_net}
describes an artificial neural network with
$s+1$ layers (1 input layer with $d$ neurons,
$s-1$ hidden layers with $d$ neurons each, and
$1$ output layer with $d$ neurons)
and standard logistic functions as activation functions
(cf., for instance, Bishop~\cite{Bishop_PatternRecognition2016}).
\subsection{Stochastic gradient descent-type minimization}
\label{subsec:sgd}
As described in Subsection~\ref{subsec:DNNapproximations} for every
\emph{suitable} $\theta\in\R^{\nu}$ and every $x\in [a,b]^d$
we think of $\bU(\theta,x)\in\R$ as an appropriate approximation
of $u(T,x)\in\R$.
In this subsection we intend to find a \emph{suitable}
$\theta\in\R^{\nu}$
as an approximate minimizer of the function
\begin{equation}
\label{eq:toMinimizeApprox}
\R^{\nu} \ni \theta \mapsto
\E\big[| \varphi({\ensuremath{\mathcal{X}}}_N) - \bU(\theta,\xi)|^2\big]\in\R.
\end{equation}
To be more specific, we intend to find an approximate
minimizer of the function in \eqref{eq:toMinimizeApprox}
through a stochastic gradient descent-type minimization algorithm
(cf., for instance, Ruder~\cite[Section~4]{Ruder2016},
Jentzen et al.~\cite{JentzenKuckuckNeufeldWurstemberger2018},
and the references mentioned therein).
For this we approximate the derivative of the function
in \eqref{eq:toMinimizeApprox} by means of the Monte Carlo method.
More precisely,
let
$\xi^{(m)}\colon \Omega\to [a,b]^d$, $m\in\N_0$,
be independent continuously uniformly distributed ${\ensuremath{\mathbb{F}}}_0$/$\B([a,b]^d)$-measurable
random variables,
let
$W^{(m)}\colon [0,T]\times\Omega \to \R^d$, $m\in\N_0$,
be independent standard $({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-Brownian motions,
for every $m\in\N_0$ let
$
{\ensuremath{\mathcal{X}}}^{(m)} = ({\ensuremath{\mathcal{X}}}^{(m)}_n)_{n\in\{0,1,\ldots,N\}}
\colon
\{0,1,\ldots,N\} \times \Omega \to \R^d
$
be the stochastic process which satisfies for
every $n\in\{0,1,\ldots,N-1\}$ that
${\ensuremath{\mathcal{X}}}^{(m)}_0 = \xi^{(m)}$ and
\begin{equation}
{\ensuremath{\mathcal{X}}}^{(m)}_{n+1} = {\ensuremath{\mathcal{X}}}^{(m)}_n
+ \mu({\ensuremath{\mathcal{X}}}^{(m)}_n)\,(t_{n+1}-t_n)
+ \sigma({\ensuremath{\mathcal{X}}}^{(m)}_n)\,(W^{(m)}_{t_{n+1}}-W^{(m)}_{t_n}),
\end{equation}
let $\gamma \in (0,\infty)$,
and let
$\Theta\colon\N_0\times\Omega\to\R^{\nu}$
be a stochastic process which satisfies for
every $m\in\N_0$ that
\begin{equation}
\label{eq:plainGradientDescent}
\Theta_{m+1}
=
\Theta_m
-
2\gamma
\cdot
\big(\bU(\Theta_m, \xi^{(m)})-\varphi({\ensuremath{\mathcal{X}}}^{(m)}_N)\big)
\cdot
(\nabla_{\theta}\bU)(\Theta_m,\xi^{(m)}).
\end{equation}
Under appropriate hypotheses we think
for every sufficiently large $ m \in \N $
of the random variable $ \Theta_m \colon \Omega \rightarrow \R^{\nu} $
as a suitable approximation of a local minimum point of the
function~\eqref{eq:toMinimizeApprox} and we think
for every sufficiently large $ m \in \N $
of the random function
$ [a,b]^d \ni x \mapsto \bU(\Theta_n, x) \in \R $
as a suitable approximation of the function
$ [a,b]^d \ni x \mapsto u(T,x) \in \R $.
\subsection{Description of the algorithm in a special case}
\label{subsec:desc_algo}
In this subsection we give a description of the proposed
approximation method in a special case, that is,
we describe the proposed approximation method in the
specific case where a particular neural network approximation
is chosen and where the plain-vanilla stochastic gradient
descent method with a constant learning rate is the employed stochastic
minimization algorithm (cf.\ \eqref{eq:plainGradientDescent} above).
For a more general description of the proposed approximation
method we refer the reader to Subsection~\ref{subsec:generalOptDescription} below.
\begin{algo}
\label{algo:special}
Let
$T,\gamma\in (0,\infty)$,
$a\in\R$, $b\in (a,\infty)$,
$d,N\in\N$,
$s\in\{3,4,5,\ldots\}$,
let
$\nu = sd(d+1)$,
let
$t_0,t_1,\ldots,t_N\in [0,T]$
be real numbers with
\begin{equation}
0 = t_0 < t_1 < \ldots < t_N = T,
\end{equation}
let
$
\mu \colon \R^d\to\R^d
$
and
$
\sigma \colon \R^d\to\R^{d\times d}
$
be continuous functions,
let
$(\Omega,{\ensuremath{\mathcal{F}}},\P,({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]})$
be a filtered probability space,
let
$\xi^{(m)}\colon\Omega\to[a,b]^d$,
$m\in\N_0$,
be independent continuously uniformly distributed
${\ensuremath{\mathbb{F}}}_0$/$\B([a,b]^d)$-measurable random variables,
let
$W^{(m)}\colon [0,T]\times\Omega\to\R^d$,
$m\in\N_0$, be i.i.d.~standard
$({\ensuremath{\mathbb{F}}}_t)_{t\in [0,T]}$-Brownian motions,
for every
$m\in\N_0$
let
${\ensuremath{\mathcal{X}}}^{(m)}\colon \{0,1,\ldots,N\}\times\Omega\to\R^d$
be the stochastic process which satisfies for every $n\in\{0,1,\ldots,N-1\}$
that
${\ensuremath{\mathcal{X}}}^{(m)}_0 = \xi^{(m)}$ and
\begin{equation}
{\ensuremath{\mathcal{X}}}^{(m)}_{n+1}
=
{\ensuremath{\mathcal{X}}}^{(m)}_n
+
\mu({\ensuremath{\mathcal{X}}}^{(m)}_n)\,(t_{n+1}-t_{n})
+
\sigma({\ensuremath{\mathcal{X}}}^{(m)}_n)\,(W^{(m)}_{t_{n+1}}-W^{(m)}_{t_{n}}),
\end{equation}
let $ \mathcal{L}_d \colon \R^d \to \R^d $ be the function
which satisfies for every $ x = ( x_1, x_2, \dots, x_d ) \in \R^d $ that
\begin{equation}
\label{eq:activation}
\mathcal{L}_d( x )
=
\left(
\frac{\exp(x_1)}{\exp(x_1)+1},
\frac{\exp(x_2)}{\exp(x_2)+1},
\dots
,
\frac{\exp(x_d)}{\exp(x_d)+1}
\right)
,
\end{equation}
for every
$ k, l \in \N $,
$ v \in \N_0 = \{0\} \cup \N $,
$ \theta = ( \theta_1, \dots, \theta_{ \nu } ) \in \R^{ \nu } $
with
$
v + l (k + 1 ) \leq \nu
$
let
$ A^{ \theta, v }_{ k, l } \colon \R^k \to \R^l $
be the function which satisfies for every
$ x = ( x_1, \dots, x_k )\in\R^k $ that
\begin{equation}
A^{ \theta, v }_{ k, l }( x )
=
\left(
\begin{array}{cccc}
\theta_{ v + 1 }
&
\theta_{ v + 2 }
&
\dots
&
\theta_{ v + k }
\\
\theta_{ v + k + 1 }
&
\theta_{ v + k + 2 }
&
\dots
&
\theta_{ v + 2 k }
\\
\theta_{ v + 2 k + 1 }
&
\theta_{ v + 2 k + 2 }
&
\dots
&
\theta_{ v + 3 k }
\\
\vdots
&
\vdots
&
\vdots
&
\vdots
\\
\theta_{ v + ( l - 1 ) k + 1 }
&
\theta_{ v + ( l - 1 ) k + 2 }
&
\dots
&
\theta_{ v + l k }
\end{array}
\right)
\left(
\begin{array}{c}
x_1
\\
x_2
\\
x_3
\\
\vdots
\\
x_k
\end{array}
\right)
+
\left(
\begin{array}{c}
\theta_{ v + k l + 1 }
\\
\theta_{ v + k l + 2 }
\\
\theta_{ v + k l + 3 }
\\
\vdots
\\
\theta_{ v + k l + l }
\end{array}
\right),
\end{equation}
let
$\bU\colon\R^{\nu}\times\R^d\to\R$
be the function
which satisfies for every
$\theta\in\R^{\nu}$,
$x\in\R^d$
that
\begin{equation}
\label{eq:neural_network_for_a_generalCase}
\bU(\theta,x)
=
\big( A^{ \theta, (s-1)d(d+1) }_{ d, 1 }
\circ
\mathcal{L}_d
\circ
A^{ \theta, (s-2)d(d+1) }_{ d, d }
\circ
\ldots
\circ
\mathcal{L}_d
\circ
A^{ \theta, d(d+1) }_{ d, d }
\circ
\mathcal{L}_d
\circ
A^{ \theta, 0 }_{ d, d } \big)(x)
,
\end{equation}
and let
$\Theta\colon \N_0\times\Omega\to\R^{\nu}$
be a stochastic process which satisfies for
every $m\in\N_0$
that
\begin{equation}
\Theta_{m+1}
=
\Theta_m
-
2\gamma \cdot
\big(\bU(\Theta_m,\xi^{(m)})-\varphi({\ensuremath{\mathcal{X}}}_N^{(m)})\big) \cdot (\nabla_{\theta} \bU)(\Theta_m,\xi^{(m)})
\end{equation}
\end{algo}
Under appropriate hypotheses we think for every
sufficiently large $ m \in \N $ and every $ x \in [a,b]^d $
of the random variable $\bU(\Theta_m,x) \colon \Omega \rightarrow \R $ in Framework~\ref{algo:special}
as a suitable approximation $\bU(\Theta_m,x)\approx u(T,x)$
of $u(T,x)\in\R$ where $u=u(t,x)_{(t,x)\in[0,T]\times\R^d}\in C^{1,2}([0,T]\times\R^d,\R)$ is a function with at most polynomially growing partial derivatives which satisfies for every $t\in[0,T]$, $x\in \R^d$ that $u(0,x)=\varphi(x)$ and
\begin{equation}
\tfrac{\partial u}{\partial t}(t,x)
=
\tfrac12 \operatorname{Trace}_{\R^d}\!\big(
\sigma(x)[\sigma(x)]^{*}(\operatorname{Hess}_x u)(t,x)\big)
+
\langle
\mu(x),(\nabla_x u)(t,x)
\rangle_{\R^d}
\end{equation}
(cf.~\eqref{eq:kolmogorovPDE} above).
\subsection{Description of the algorithm in the general case}
\label{subsec:generalOptDescription}
In this subsection we provide a general framework
which covers the approximation method derived in
Subsections
\ref{subsec:kolmogorovEq}--\ref{subsec:sgd} above and
which allows, in addition, to incorporate other
minimization algorithms (cf., for example,
Kingma \& Ba~\cite{KingmaBa2015}, Ruder~\cite{Ruder2016},
E et al.~\cite{EHanJentzen2017a}, Han et al.~\cite{EHanJentzen2017b},
and Beck et al.~\cite{BeckEJentzen2017}) than just the plain vanilla
stochastic gradient descent method. The proposed approximation
algorithm is an extension of the approximation algorithm
in E et al.~\cite{EHanJentzen2017a},
Han et al.~\cite{EHanJentzen2017b}, and Beck et al.~\cite{BeckEJentzen2017}
in the special case of linear
Kolmogorov partial differential equations.
\begin{algo}
\label{algo:general_algorithm}
Let
$T \in (0,\infty)$,
$N, d, \varrho, \nu, \varsigma \in \N$,
let
$H\colon [0,T]^2\times\R^d\times\R^d\to\R^d$, $\varphi\colon \R^d\to\R$
be functions,
let
$
( \Omega, {\ensuremath{\mathcal{F}}}, \P, ( {\ensuremath{\mathbb{F}}}_t )_{ t \in [0,T] } )
$
be a filtered probability space,
let
$
W^{m,j} \colon [0,T] \times \Omega \to \R^d
$,
$
m\in\N_0
$,
$
j\in \N
$,
be independent standard $ ( {\ensuremath{\mathbb{F}}}_t )_{ t \in [0,T] } $-Brownian motions on
$(\Omega,{\ensuremath{\mathcal{F}}},\P)$,
let
$
\xi^{m,j}\colon\Omega\to\R^d
$,
$
m \in \N_0
$,
$ j \in \N $,
be i.i.d.\ $ {\ensuremath{\mathbb{F}}}_0 $/$ \B(\R^d) $-measurable random variables,
let
$t_0, t_1, \ldots, t_N\in [0,T]$ be
real numbers with
$0 = t_0 < t_1 < \ldots < t_N = T$,
for every
$\theta\in\R^{\nu}$,
$j\in\N$,
${\bf s}\in\R^{\varsigma}$
let
$ \bU^{\theta,j,{\bf s}}\colon\R^d \to\R $
be a function,
for every
$ m \in \N_0 $,
$ j \in \N $
let
${\ensuremath{\mathcal{X}}}^{m,j} = ({\ensuremath{\mathcal{X}}}^{m,j}_n)_{n\in\{0,1,\ldots,N\}}\colon \{0,1,\ldots,N\}\times\Omega\to\R^d$
be a stochastic process which satisfies for every
$n\in\{0,1,\ldots,N-1\}$ that
${\ensuremath{\mathcal{X}}}^{m,j}_0 = \xi^{ m, j } $ and
\begin{equation}\label{eq:FormalXapprox}
{\ensuremath{\mathcal{X}}}^{m,j}_{n+1}
=
H(t_n,t_{n+1},{\ensuremath{\mathcal{X}}}^{m,j}_n,W^{m,j}_{t_{n+1}}-W^{m,j}_{t_{n}}),
\end{equation}
let $(J_m)_{m\in\N_0}\subseteq\N$ be a sequence,
for every
$m\in\N_0$,
${\bf s}\in\R^{\varsigma}$
let
$\phi^{m,{\bf s}}\colon\R^{\nu}\times\Omega\to\R$ be
the function which satisfies for every
$(\theta,\omega)\in\R^{\nu}\times\Omega$ that
\begin{equation}
\phi^{m,{\bf s}}(\theta,\omega)
=
\frac{1}{J_m}\sum_{j=1}^{J_m}
\left[
\bU^{\theta,j,{\bf s}}\big(\xi^{m,j}(\omega))
-
\varphi\big({\ensuremath{\mathcal{X}}}^{m,j}_N(\omega)\big)
\right]^2,
\end{equation}
for every
$m\in\N_0$,
${\bf s}\in\R^{\varsigma}$
let
$\Psi^{m,{\bf s}}\colon\R^{\nu}\times\Omega\to\R^{\nu}$ be a function
which satisfies for every
$\omega\in\Omega$,
$\theta\in\{\eta\in\R^{\nu}\colon \phi^{m,{\bf s}}(\cdot,\omega)\colon\R^{\nu}\to\R~\text{is differentiable at}~\eta\}$
that
\begin{align}
\Psi^{m,{\bf s}}(\theta,\omega) = (\nabla_{\theta}\phi^{m,{\bf s}})(\theta,\omega),
\end{align}
let
$
\S
\colon
\R^{\varsigma}
\times
\R^{\nu}
\times
(\R^d)^{\N}\to\R^{\varsigma}
$
be a function,
for every
$m\in\N_0$
let $\Phi_m\colon\R^{\varrho}\to\R^{\nu}$
and $\Psi_m\colon\R^{\varrho}\times\R^{\nu}\to\R^{\varrho}$
be functions,
let
$\Theta\colon\N_0\times\Omega\to\R^{\nu}$,
$\bS\colon\N_0\times\Omega\to\R^{\varsigma}$,
and
$\Xi\colon\N_0\times\Omega\to\R^{\varrho}$
be stochastic processes
which satisfy for every $m\in\N_0$ that
\begin{equation}\label{eq:general_batch_normalization}
\bS_{m+1} = \S\bigl(\bS_m, \Theta_{m},
({\ensuremath{\mathcal{X}}}_N^{m,i})_{i\in\N}\bigr),
\qquad
\Xi_{m+1} = \Psi_{m}(\Xi_{m},\Psi^{m,\bS_{m+1}}(\Theta_m)),
\end{equation}
\begin{equation}
\text{and}
\qquad
\Theta_{m+1} = \Theta_{m} - \Phi_{m}(\Xi_{m+1})
\label{eq:general_gradient_step}.
\end{equation}
\end{algo}
Under appropriate hypotheses we think for every
sufficiently large $ m \in \N $ and every $ x \in [a,b]^d $
of the random variable $\bU^{\Theta_m,1,\bS_m}(x) \colon \Omega \rightarrow \R $ in Framework~\ref{algo:general_algorithm}
as a suitable approximation $\bU^{\Theta_m,1,\bS_m}(x)\approx u(T,x)$
of $u(T,x)\in\R$ where $u=u(t,x)_{(t,x)\in[0,T]\times\R^d}\in C^{1,2}([0,T]\times\R^d,\R)$ is a function with at most polynomially growing partial derivatives which satisfies for every $t\in[0,T]$, $x\in \R^d$ that $u(0,x)=\varphi(x)$ and
\begin{equation}
\tfrac{\partial u}{\partial t}(t,x)
=
\tfrac12 \operatorname{Trace}_{\R^d}\!\big(
\sigma(x)[\sigma(x)]^{*}(\operatorname{Hess}_x u)(t,x)\big)
+
\langle
\mu(x),(\nabla_x u)(t,x)
\rangle_{\R^d} ,
\end{equation}where $\mu\colon \R^d\to\R^d$ and $\sigma\colon \R^d\to\R^{d\times d}$ are sufficiently regular functions
(cf.~\eqref{eq:kolmogorovPDE} above).
\section{Examples}
\label{sec:examples}
In this section we test the proposed approximation algorithm (see Section~\ref{sec:derivationalgo} above)
in the case of several examples of SDEs and Kolmogorov PDEs, respectively. In particular, in this section we apply the proposed approximation algorithm to the heat equation (cf.\ Subsection~\ref{sec:paraboliceq} below), to independent geometric Brownian motions (cf.\ Subsection~\ref{sec:geometric} below), to the Black-Scholes model (cf.\ Subsection~\ref{sec:black_scholes} below), to stochastic Lorenz equations (cf.\ Subsection~\ref{sec:lorenz} below), and to the Heston model (cf.\ Subsection~\ref{sec:heston} below). In the case of each of the examples below we employ the general approximation algorithm in Framework~\ref{algo:general_algorithm} above
in conjunction with the Adam optimizer (cf.\ Kingma \& Ba~\cite{KingmaBa2015}) with mini-batches of size 8192 in each iteration step (see Subsection~\ref{sec:adam} below for a precise description).
Moreover, we employ a fully-connected feedforward neural network with one input layer, two hidden layers, and one one-dimensional output layer in our implementations in the case of each of these examples.
We also use batch normalization (cf.\ Ioffe \& Szegedy~\cite{IoffeSzegedy}) just before the first linear transformation, just before each of the two nonlinear activation functions in front of the hidden layers as well as just after the last linear transformation.
For the two nonlinear activation functions we employ the multidimensional version of the function $\R\ni x\mapsto \tanh(x)\in (-1,1)$. All weights in the neural network are initialized by means of the Xavier initialization (cf.\ Glorot \& Bengio~\cite{GlorotBengio10}).
All computations were performed in single precision (float32) on a NVIDIA GeForce GTX 1080 GPU with 1974 MHz core clock and 8 GB GDDR5X memory with 1809.5 MHz clock rate. The underlying system consisted of an
Intel Core i7-6800K CPU with 64 GB DDR4-2133 memory running Tensorflow 1.5 on Ubuntu 16.04.
\subsection{Setting}
\label{sec:adam}
\begin{framework}
\label{framework:adam}
Assume Framework~\ref{algo:general_algorithm}, let $\varepsilon=10^{-8}$, $\beta_1=\frac{9}{10}$, $\beta_1=\frac{999}{1000}$, $(\gamma_m)_{m\in\N_0}\subseteq (0,\infty)$, let $\operatorname{Pow}_r\colon \R^\nu \to \R^\nu$, $r\in(0,\infty)$, be the functions which satisfy for every $r\in(0,\infty)$, $x=(x_1,\ldots,x_\nu)\in\R^\nu$ that
\begin{equation}
\operatorname{Pow}_r(x)=(|x_1|^r, \ldots, |x_\nu|^r),
\end{equation}
assume for every $m\in\N_0$, $i\in\{0,1,\ldots,N\}$ that $J_m=8192$, $t_i=\frac{iT}{N}$, $\varrho=2\nu$, $ T = 1$, $\gamma_m =10^{-3}\mathbbm{1}_{[0,250000]}(m)+10^{-4}\mathbbm{1}_{(250000,500000]}(m)+10^{-5}\mathbbm{1}_{(500000,\infty)}(m)$, assume for every $m\in\N_0$, $x=(x_1,\ldots,x_\nu)$, $y=(y_1,\ldots,y_\nu)$, $\eta=(\eta_1,\ldots,\eta_\nu)\in\R^\nu$ that
\begin{equation}\label{eq:Adam1}
\Psi_m(x,y,\eta) = (\beta_1 x + (1-\beta_1)\eta, \beta_2 y + (1-\beta_2)\operatorname{Pow}_2(\eta))
\end{equation}
and
\begin{equation}\label{eq:Adam2}
\psi_m(x,y) = \left(\left[\sqrt{\tfrac{|y_1|}{1-(\beta_2)^m}}+\varepsilon\right]^{-1}\frac{\gamma_mx_1}{1-(\beta_1)^m},\ldots,\left[\sqrt{\tfrac{|y_\nu|}{1-(\beta_2)^m}}+\varepsilon\right]^{-1}\frac{\gamma_mx_\nu}{1-(\beta_1)^m}\right).
\end{equation}
\end{framework}
Equations~\eqref{eq:Adam1} and \eqref{eq:Adam2} in Framework~\ref{framework:adam} describe the Adam optimizer (cf.\ Kingma \& Ba~\cite{KingmaBa2015}, e.g., E et al.~\cite[(32)--(33) in Section 4.2 and (90)--(91) in Section 5.2]{EHanJentzen2017b}, and line 84 in {\sc{Python}} code \ref{code:common} in Section~\ref{sec:source_code} below).
\subsection{Heat equation}
\label{sec:paraboliceq}
In this subsection we apply the proposed approximation algorithm
to the heat equation (see \eqref{eq:example_constant_coeffs_kolmogorovPDE} below).
Assume
Framework~\ref{algo:general_algorithm}, assume for every
$s,t\in [0,T]$,
$x,w\in\R^d$,
$m\in\N_0$
that
$ N = 1$,
$ d = 100$,
$ \nu = d(2d)+(2d)^2+2d= 2d(3d+1)$,
$ \varphi(x) = \|x\|_{\R^d}^2 $,
$ H(s,t,x,w) = x + \sqrt{2}\operatorname{Id}_{\R^d}w $,
assume that
$
\xi^{0,1}\colon\Omega\to\R^d
$
is continuous uniformly distributed on $[0,1]^d$,
and let
$u = (u(t,x))_{(t,x)\in [0,T]\times\R^d}
\in C^{1,2}([0,T]\times\R^d,\R)$
be an at most polynomially growing function which satisfies
for every $t\in [0,T]$, $x\in\R^d$ that
$u(0,x) = \varphi(x)$ and
\begin{equation}\label{eq:example_constant_coeffs_kolmogorovPDE}
(\tfrac{\partial u}{\partial t})(t,x)
=
(\Delta_x u)(t,x).
\end{equation}
Combining,
e.g., Lemma~\ref{lemma:example_constant_coeffs_exact_solution} below with,
e.g., Hairer et al.~\cite[Corollary 4.17 and Remark 4.1]
{HairerHutzenthalerJentzen_LossOfRegularity2015} shows that for every
$t\in [0,T]$, $x\in\R^d$ it holds that
\begin{equation}
\label{eq:constant_coeffs_concrete_PDE}
u(t,x) = \|x\|_{\R^d}^2 + t\,d.
\end{equation}
Table~\ref{tab:constant_coeffs_linfty_l1_l2} approximately presents
the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL1fehler} below),
the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL2fehler} below),
and
the relative $L^{\infty}(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relLinftyfehler} below) against
$ m \in \{0,10000, \allowbreak50000,\allowbreak 100000, 150000, 200000, 500000, \allowbreak 750000\}$ (cf.\ \textsc{Python} code~\ref{code:paraboliceq} in Subsection~\ref{sec:code_paraboliceq} below).
Figure~\ref{fig:constant_coeffs} approximately depicts
the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL1fehler} below),
the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL2fehler} below),
and
the relative $L^{\infty}(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relLinftyfehler} below) against
$ m \in \{0,100, 200, 300,\ldots,\allowbreak 299800, 299900, 300000\}$ (cf.\ \textsc{Python} code~\ref{code:paraboliceq} in Subsection~\ref{sec:code_paraboliceq} below).
In our numerical simulations for Table~\ref{tab:constant_coeffs_linfty_l1_l2} and Figure \ref{fig:constant_coeffs} we calculated the exact solution
of the PDE~\eqref{eq:example_constant_coeffs_kolmogorovPDE}
by means of Lemma~\ref{lemma:example_constant_coeffs_exact_solution}
below (see \eqref{eq:constant_coeffs_concrete_PDE} above), we approximately calculated the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL1fehler}
\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|dx
\end{equation}
for $ m \in \{0,10000, 50000, 100000, 150000, 200000, 500000, 750000\}$ by means of Monte Carlo approximations with 10240000 samples, we approximately calculated the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL2fehler}
\sqrt{\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|^2dx}
\end{equation}
for $ m \in \{0,10000, 50000, 100000, 150000, 200000, 500000, 750000\}$ by means of Monte Carlo approximations with 10240000 samples, and we approximately calculated the relative $L^\infty(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relLinftyfehler}
\sup_{x\in [0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|
\end{equation}
for $ m \in \{0,10000, 50000, 100000, 150000, 200000, 500000, 750000\}$ by means of Monte Carlo approximations with 10240000 samples (see Lemma~\ref{lemma:MonteCarloSupEstimate} below).
Table~\ref{tab:constant_coeffs_linfty_l1_l2_2}
approximately presents
the relative $L^1(\P;L^1(\lambda_{[0,1]^d};\R))$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m(x)})_{x\in[0,1]^d}$ (see \eqref{eq:relL1fehler2} below),
the relative $L^2(\P;L^2(\lambda_{[0,1]^d};\R))$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m(x)})_{x\in[0,1]^d}$ (see \eqref{eq:relL2fehler2} below),
and
the relative $L^2(\P;L^\infty(\lambda_{[0,1]^d};\R))$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m(x)})_{x\in[0,1]^d}$ (see \eqref{eq:relLinftyfehler2} below), against
$ m \in \{0, 10000, 50000,\allowbreak 100000,\allowbreak 150000, 200000,\allowbreak 500000,\allowbreak 750000\}$ (cf.\ \textsc{Python} code~\ref{code:paraboliceq} in Subsection~\ref{sec:code_paraboliceq} below).
In our numerical simulations for Table~\ref{tab:constant_coeffs_linfty_l1_l2_2} we calculated the exact solution
of the PDE~\eqref{eq:example_constant_coeffs_kolmogorovPDE}
by means of Lemma~\ref{lemma:example_constant_coeffs_exact_solution}
below (see \eqref{eq:constant_coeffs_concrete_PDE} above), we approximately calculated the relative $L^1(\P;L^1(\lambda_{[0,1]^d};\R))$-approximation error
\begin{equation}
\label{eq:relL1fehler2}
\E\Bigg[\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|dx\Bigg]
\end{equation}
for $ m \in \{0, 10000, 50000, 100000, 150000, 200000, 500000, 750000\}$ by means of Monte Carlo approximations with 10240000 samples for the Lebesgue integral and 5 samples for the expectation, we approximately calculated the relative $L^2(\P;L^2(\lambda_{[0,1]^d};\R))$-approximation error
\begin{equation}
\label{eq:relL2fehler2}
\left(\E\!\left[\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|^2dx\right]\right)^{\!1/2}
\end{equation}
for $ m \in \{0, 10000, 50000, 100000, 150000, 200000, 500000, 750000\}$ by means of Monte Carlo approximations with 10240000 samples for the Lebesgue integral and 5 samples for the expectation, and we approximately calculated the relative $L^2(\P;L^\infty(\lambda_{[0,1]^d};\R))$-approximation error
\begin{equation}
\label{eq:relLinftyfehler2}
\left(\E\!\left[\sup_{x\in [0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|^2\right]\right)^{\!1/2}
\end{equation}
for $ m \in \{0, 10000, 50000, 100000, 150000, 200000, 500000, 750000\}$ by means of Monte Carlo approximations with 10240000 samples for the supremum (see Lemma~\ref{lemma:MonteCarloSupEstimate} below) and 5 samples for the expectation. The following elementary result, Lemma~\ref{lemma:example_constant_coeffs_exact_solution} below, specifies the explicit solution of the PDE~\eqref{eq:example_constant_coeffs_kolmogorovPDE} above (cf.\ \eqref{eq:constant_coeffs_concrete_PDE} above). For completeness we also provide here a proof for Lemma~\ref{lemma:example_constant_coeffs_exact_solution}.
\begin{table}[H] \begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
{\begin{tabular}{@{}c@{}}Number \\ of steps\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^1(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^2(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^\infty(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Runtime \\ in seconds\end{tabular}}\\
\hline
0 & 0.998253 & 0.998254 & 1.003524 & 0.5\\
\hline
10000 & 0.957464 & 0.957536 & 0.993083 & 44.6\\
\hline
50000 & 0.786743 & 0.786806 & 0.828184 & 220.8 \\
\hline
100000 & 0.574013 & 0.574060 & 0.605283 & 440.8\\
\hline
150000 & 0.361564 & 0.361594 & 0.384105 & 661.0 \\
\hline
200000 & 0.150346 & 0.150362 & 0.164140 & 880.8 \\
\hline
500000 & 0.000882 & 0.001112 & 0.007360 & 2200.7 \\
\hline
750000 & 0.000822 & 0.001036 & 0.007423 & 3300.6 \\
\hline
\end{tabular}
\caption{Approximative presentations of the relative approximation errors in \eqref{eq:relL1fehler}--\eqref{eq:relLinftyfehler} for the heat equation in \eqref{eq:example_constant_coeffs_kolmogorovPDE}.}
\label{tab:constant_coeffs_linfty_l1_l2}
\end{center}
\end{table}
\begin{figure}
\centering
\includegraphics[scale=0.65]{./errors2.pdf}
\caption{Approximative plots of the relative approximation errors in \eqref{eq:relL1fehler}--\eqref{eq:relLinftyfehler} for the heat equation in \eqref{eq:example_constant_coeffs_kolmogorovPDE}.}
\label{fig:constant_coeffs}
\end{figure}
\begin{table}[H] \begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
{\begin{tabular}{@{}c@{}}Number \\ of \\steps\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^1(\P;L^1(\lambda_{[0,1]^d};\R))$-\\error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^2(\P;L^2(\lambda_{[0,1]^d};\R))$-\\error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^2(\P;L^\infty(\lambda_{[0,1]^d};\R))$-\\error\end{tabular}} & {\begin{tabular}{@{}c@{}}Mean\\ runtime\\ in\\ seconds\end{tabular}}\\
\hline
0 & 1.000310 & 1.000311 & 1.005674 & 0.6\\
\hline
10000 & 0.957481 & 0.957554 & 0.993097 & 44.7\\
\hline
50000 & 0.786628 & 0.786690 & 0.828816 & 220.4 \\
\hline
100000 & 0.573867 & 0.573914 & 0.605587 & 440.5\\
\hline
150000 & 0.361338 & 0.361369 & 0.382967 & 660.8 \\
\hline
200000 & 0.001387 & 0.001741 & 0.010896 & 880.9 \\
\hline
500000 & 0.000883 & 0.001112 & 0.008017 & 2201.0 \\
\hline
750000 & 0.000822 & 0.001038 & 0.007547 & 3300.4 \\
\hline
\end{tabular}
\caption{Approximative presentations of the relative approximation errors in \eqref{eq:relL1fehler2}--\eqref{eq:relLinftyfehler2} for the heat equation in \eqref{eq:example_constant_coeffs_kolmogorovPDE}.}
\label{tab:constant_coeffs_linfty_l1_l2_2}
\end{center}
\end{table}
\begin{lemma}\label{lemma:example_constant_coeffs_exact_solution}
Let
$T\in (0,\infty)$,
$d\in \N$,
let
$C\in \R^{d\times d}$ be a
strictly positive and symmetric matrix,
and let
$u\colon [0,T]\times\R^d \to \R$
be the function which satisfies for every $(t,x)\in [0,T]\times\R^d$
that
\begin{equation}\begin{split}
u(t,x) = \|x\|_{\R^d}^2 + t\operatorname{Trace}_{\R^d}(C).
\end{split}
\end{equation}
Then
\begin{enumerate}[(i)]
\item\label{it:exact_solution_polynomially} it holds that $u\in C^{\infty}([0,T]\times\R^d,\R)$ is
at most polynomially growing
and
\item \label{it:exact_solution_derivative} it holds for every $t\in [0,T]$, $x\in\R^d$ that
\begin{equation}
(\tfrac{\partial u}{\partial t})(t,x)
=
\tfrac12 \operatorname{Trace}_{\R^d}\!\big(C(\operatorname{Hess}_x u)(t,x)\big).
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lemma:example_constant_coeffs_exact_solution}]
First, note that $u$ is a polynomial. This establishes item~\eqref{it:exact_solution_polynomially}.
Moreover, note that for every $(t,x)\in [0,T]\times\R^d$ it holds that
\begin{equation}
(\tfrac{\partial u}{\partial t})(t,x) = \operatorname{Trace}_{\R^d}(C),
\qquad
(\nabla_x u)(t,x) = 2x,
\end{equation}
\begin{equation}
\text{and}\qquad
(\operatorname{Hess}_x u)(t,x) = \big(\tfrac{\partial }{\partial x}(\nabla_x u)\big)(t,x) = 2\operatorname{Id}_{\R^d}.
\end{equation}
Hence, we obtain for every $(t,x)\in [0,T]\times\R^d$ that
\begin{equation}
(\tfrac{\partial u}{\partial t})(t,x)
-
\tfrac12 \operatorname{Trace}_{\R^d}\!\big(C(\operatorname{Hess}_x u)(t,x)\big)
=
\operatorname{Trace}_{\R^d}(C)
-
\tfrac12 \operatorname{Trace}_{\R^d}(2C)
=
0.
\end{equation}
This proves item \eqref{it:exact_solution_derivative}. The proof of Lemma~\ref{lemma:example_constant_coeffs_exact_solution}
is thus completed.
\end{proof}
Lemma~\ref{lemma:MonteCarloSupEstimate} below discloses the strategy how we approximatively calculate the $L^\infty(\lambda_{[0,1]^d};\R)$-errors in \eqref{eq:relLinftyfehler} and \eqref{eq:relLinftyfehler2} above. Our proof of Lemma~\ref{lemma:MonteCarloSupEstimate} employs the elementary auxiliary results in Lemma~\ref{lem:capone} and Lemma~\ref{lem:almost_sure_convergence} below. For completeness we also include proofs for Lemma~\ref{lem:capone} and Lemma~\ref{lem:almost_sure_convergence} here.
\begin{lemma}
\label{lem:capone}
Let $ (\Omega, {\ensuremath{\mathcal{F}}}, \P) $ be a probability space and let $A, \mathcal{O}\in{\ensuremath{\mathcal{F}}}$ satisfy that $\P(\mathcal{O})=1$. Then it holds that
\begin{equation}
\P(A)=\P(A\cap \mathcal{O}).
\end{equation}
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lem:capone}]\
Observe that the monotonicity of $\P$ ensures that
\begin{equation}
\begin{split}
\P(A\cap\mathcal{O}) \le \P(A) &= \P\big([A\cap \mathcal{O}] \cup [A\backslash(A\cap \mathcal{O})]\big)\\
&=\P(A\cap \mathcal{O})+\P(A\backslash(A\cap\mathcal{O}))\\
&=\P(A\cap \mathcal{O})+\P(A\backslash \mathcal{O})\\
&\le \P(A\cap \mathcal{O})+\P(\Omega\backslash \mathcal{O})\\
&=\P(A\cap \mathcal{O})+\P(\Omega)-\P(\mathcal{O})=\P(A\cap \mathcal{O}).
\end{split}
\end{equation}
Hence, we obtain that for every $A,\mathcal{O}\in{\ensuremath{\mathcal{F}}}$ with $\P(\mathcal{O})=1$ it holds that $\P(A)=\P(A\cap \mathcal{O})$. The Proof of Lemma~\ref{lem:capone} is thus completed.
\end{proof}
\begin{lemma}
\label{lem:almost_sure_convergence}
Let $ (\Omega, {\ensuremath{\mathcal{F}}}, \P) $ be a probability space,
let $ X_n \colon \Omega \to [0,\infty) $, $ n \in \N_0 $,
be random variables, assume for every $n\in\N$ that $\P(X_n\ge X_{n+1})=1$, and assume for every $\varepsilon\in(0,\infty)$ that
\begin{equation}
\label{eq:schwachgg0}
\limsup_{n\to\infty}\P(X_n>\varepsilon)=0.
\end{equation}
Then
\begin{equation}
\label{eq:fastsichergg0}
\P\bigg(\limsup_{n\to\infty} X_n = 0\bigg)=1.
\end{equation}
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lem:almost_sure_convergence}]
Throughout this proof let $\mathcal{O}\subseteq \Omega$ be the set given by
\begin{equation}
\mathcal{O}=\cap_{n=1}^\infty \{X_n\ge X_{n+1}\}=\{\forall\, n\in\N\colon X_n\ge X_{n+1}\}.
\end{equation}
Observe that Lemma~\ref{lem:capone} and the hypothesis that for every $n\in\N$ it holds that $\P(X_n \ge X_{n+1})=1$ assure that for every $N\in\N$ it holds that
\begin{equation} \begin{split}
\P(\cap_{n=1}^N\{X_n \ge X_{n+1}\})&=\P([\cap_{n=1}^{N-1}\{X_n\ge X_{n+1}\}]\cap \{X_N \ge X_{N+1}\})\\
&=\P(\cap_{n=1}^{N-1}\{X_n \ge X_{n+1}\}). \end{split}
\end{equation}
This implies that for every $N\in\N$ it holds that
\begin{equation}\begin{split}
\P(\cap_{n=1}^N\{X_n \ge X_{n+1}\}) &= \P(\cap_{n=1}^0\{X_n \ge X_{n+1}\})\\
&=\P(\Omega)=1.\end{split}
\end{equation}
The fact that the measure $\P$ is continuous from above hence demonstrates that
\begin{equation}
\label{eq:PO}
\begin{split}
\P(\mathcal{O})&=\P(\cap_{n=1}^\infty\{X_n \ge X_{n+1}\})=\P(\cap_{N=1}^\infty[\cap_{n=1}^N\{X_n \ge X_{n+1}\}])\\
&=\lim_{N\to\infty}\P(\cap_{n=1}^N\{X_n \ge X_{n+1}\})=1.
\end{split}
\end{equation}
Next note that
\begin{equation}
\begin{split}
\label{eq:darstellungVonGegenWkeitneu}
&\P\bigg( \limsup_{n\to\infty} X_n > 0 \bigg)
\\
&= \P\bigg(\exists\, k \in \N\colon\bigg[\limsup_{n\to\infty}X_n >\tfrac{1}{k}\bigg]\bigg)\\
&= \P\Big(\exists\, k \in \N\colon\forall\, m\in\N\colon\exists\, n\in\N\cap[m,\infty)\colon\big[X_n>\tfrac{1}{k}\big]\Big)\\
&= \P\Big(\cup_{k\in\N}\cap_{m\in \N}\cup_{n\in\N\cap[M,\infty)}\big\{X_n>\tfrac{1}{k}\big\}\Big)\\
&\le \sum_{k=1}^\infty \P\Big(\cap_{m=1}^\infty \cup_{n=m}^\infty\big\{X_n>\tfrac{1}{k}\big\}\Big)\\
&=\sum_{k=1}^\infty\left[\limsup_{m\to\infty}\P(\cup_{n=m}^\infty\{X_n>\tfrac1k\})\right].
\end{split}
\end{equation}
Lemma \ref{lem:capone} and \eqref{eq:PO} therefore ensure that
\begin{equation}
\begin{split}
&\P\bigg( \limsup_{n\to\infty} X_n > 0 \bigg) \\
&\le \sum_{k=1}^\infty\bigg[ \limsup_{m\to\infty}\P\Big([\cup_{n=m}^\infty\{X_n>\tfrac1k\}]\cap\mathcal{O}\Big)\bigg]\\
&\le \sum_{k=1}^\infty\bigg[ \limsup_{m\to\infty}\P\Big([\cup_{n=m}^\infty\{X_n>\tfrac1k\}]\cap\{\forall\, n\in\N\cap[m,\infty)\colon X_m\ge X_n\}\Big)\bigg]\\
&=\sum_{k=1}^\infty\bigg[ \limsup_{m\to\infty}\P\Big(\{X_m>\tfrac1k\}\cap\{\forall\, n\in\N\cap[m,\infty)\colon X_m\ge X_n\}\Big)\bigg]\\
&\le \sum_{k=1}^\infty\left[\limsup_{m\to\infty}\P\big(X_m>\tfrac1k\big)\right].
\end{split}
\end{equation}
Combining this and \eqref{eq:schwachgg0} establishes \eqref{eq:fastsichergg0}. The proof of Lemma~\ref{lem:almost_sure_convergence} is thus completed.
\end{proof}
\begin{lemma}\label{lemma:MonteCarloSupEstimate}
Let
$d\in\N$, $a\in\R$, $b\in(a,\infty)$,
let
$f\colon [a,b]^d\to \R$
be a continuous function,
let
$(\Omega,{\ensuremath{\mathcal{F}}},\P)$ be a probability space,
let
$X_n\colon \Omega\to [a,b]^d$, $n\in\N$,
be i.i.d.\ random variables,
and assume that
$X_1$ is continuous uniformly distributed on $[a,b]^d$.
Then
\begin{enumerate}[(i)]
\item \label{it:weakapprox} it holds that
\begin{equation}\label{eq:lemma_as_convergence}
\P\!\left(
\limsup_{N\to\infty}
\left|
\left[
\max\limits_{1\leq n\leq N} f(X_n)
\right]
-
\left[
\sup\limits_{x\in [a,b]^d} f(x)
\right]
\right|
=
0
\right)
=
1
\end{equation}
and
\item \label{it:pthmomentapprox}
it holds for every $p\in (0,\infty)$ that
\begin{equation}\label{eq:lemma_lp_convergence}
\limsup_{N\to\infty}
\E\!\left[\rule{0cm}{0.9cm}
\left|
\left[
\max\limits_{1\leq n\leq N} f(X_n)
\right]
-
\left[
\sup\limits_{x\in [a,b]^d} f(x)
\right]
\right|^p
\right] = 0.
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lemma:MonteCarloSupEstimate}]
First, observe that the fact that $f\colon [a,b]^d\to\R$ is a continuous
function and the fact that $[a,b]^d\subseteq\R^d$ is a compact set
demonstrate that there exists $\xi\in [a,b]^d$
which satisfies that $f(\xi) = \sup_{x\in [a,b]^d} f(x)$. Next
note that the fact that for every $N\in\N$, $n\in\{1,2,\ldots,N\}$ it holds that
$f(X_n) \leq \sup_{x\in [a,b]^d} f(x)$ implies that for every $N\in\N$ it holds that
$\max_{1\leq n\leq N} f(X_n) \leq \sup_{x\in [a,b]^d} f(x)$. Hence,
we obtain that for every $N\in\N$ it holds that
\begin{equation}
\left|
\big[
\max\nolimits_{1\leq n\leq N} f(X_n)
\big]
-
\big[
\sup\nolimits_{x\in [a,b]^d} f(x)
\big]
\right|
=
\big[
\sup\nolimits_{x\in [a,b]^d} f(x)
\big]
-
\big[
\max\nolimits_{1\leq n\leq N} f(X_n)
\big] .
\end{equation}
Combining this with the fact that $f(\xi) = \sup_{x\in\R^d} f(x)$
ensures that for every $\varepsilon\in (0,\infty)$, $N\in\N$ it holds that
\begin{equation} \label{eq:set_relations}
\begin{split}
&
\big\{|\!\max\nolimits_{1\leq n\leq N} f(X_n) - \sup\nolimits_{x\in [a,b]^d} f(x)|\leq \varepsilon\big\}\\
&= \big\{\max\nolimits_{1\leq n\leq N} f(X_n) \geq \sup\nolimits_{x\in [a,b]^d} f(x) - \varepsilon\big\}
\\
&=
\mathop{\cup}_{n=1}^N \big\{f(X_n)\geq\sup\nolimits_{x\in [a,b]^d} f(x)-\varepsilon\big\}
= \mathop{\cup}_{n=1}^N \big\{|f(X_n) - \sup\nolimits_{x\in [a,b]^d} f(x)| \leq \varepsilon\big\}
\\
&=
\mathop{\cup}_{n=1}^N \big\{|f(X_n) - f(\xi)|\leq\varepsilon\big\}.
\end{split}
\end{equation}
In the next step we observe that the fact that $f\colon [a,b]^d\to\R$ is continuous
ensures that for every $\varepsilon\in (0,\infty)$ there exists $\delta\in (0,\infty)$ such that for
every $x\in [a,b]^d$ with $\|x-\xi\|_{\R^d} \leq \delta$ it
holds that $|f(x)-f(\xi)| \leq \varepsilon$.
Combining this and \eqref{eq:set_relations} shows that for every $\varepsilon \in (0,\infty)$ there
exists $\delta \in (0,\infty)$ such that for every $N\in\N$ it holds that
\begin{equation}\label{eq:estimatingProbabilites}
\begin{split}
&
\P\big(|\!\max\nolimits_{1\leq n\leq N} f(X_n) - \sup\nolimits_{x\in [a,b]^d} f(x)| \leq \varepsilon \big)
=
\P\big( \cup_{n=1}^N \{|f(X_n) - f(\xi)| \leq \varepsilon \}\big)
\\
&\geq
\P\big( \cup_{n=1}^N \{\|X_n-\xi\|_{\R^d} \leq \delta\}\big)
=
1 - \P\big( \cap_{n=1}^N \{\|X_n-\xi\|_{\R^d} > \delta\}\big).
\end{split}
\end{equation}
Hence, we obtain that for every $\varepsilon\in(0,\infty)$ there exists $\delta\in(0,\infty)$ such that
\begin{equation}
\label{eq:144b}
\begin{split}
&\liminf_{N\to \infty} \P\big(|\!\max\nolimits_{1\le n \le N}f(X_n)-\sup\nolimits_{x\in[a,b]}f(x)|\le \varepsilon\big)\\
&\ge 1- \liminf_{N\to\infty}\P\big(\cap_{n=1}^N\{\|X_n-\xi\|_{\R^d}>\delta\}\big).
\end{split}
\end{equation}
Next observe that the fact that the random variables $X_n\colon \Omega\to [a,b]^d$, $n\in\{1,2,\ldots,N\}$,
are i.i.d.\ ensures that for every $\delta\in (0,\infty)$, $N\in\N$ it holds that
\begin{equation}
\label{eq:independentproduct}
\begin{split}
\P\big( \cap_{n=1}^N \{\|X_n-\xi\|_{\R^d} > \delta\}\big)
&=
\prod_{n=1}^N \P(\|X_n-\xi\|_{\R^d} > \delta) \\
&=
\big[\P(\|X_1-\xi\|_{\R^d} > \delta)\big]^N.
\end{split}
\end{equation}
In addition, note that the fact that for every $\delta\in(0,\infty)$ it holds that the set $\{x\in[a,b]^d\colon\|x-\xi\|_{\R^d}\le\delta\}\subseteq \R^d$ has strictly positive $d$-dimensional Lebesgue measure and the fact that $X_1$ is continuous uniformly distributed on $[a,b]^d$ ensure that for every $\delta\in(0,\infty)$ it holds that
\begin{equation}
\P\!\left(\|X_1-\xi\|_{\R^d}>\delta\right)=1-\P\!\left(\|X_1-\xi\|_{\R^d}\le\delta\right)<1.
\end{equation}
Hence, we obtain that for every $\delta\in(0,\infty)$ it holds that
\begin{equation}
\limsup_{N\to\infty}\Big(\big[\P(\|X_1-\xi\|_{\R^d}>\delta)\big]^N\Big)=0.
\end{equation}
Combining this with \eqref{eq:independentproduct} demonstrates that for every $\delta\in(0,\infty)$ it holds that
\begin{equation}
\limsup_{N\to\infty}\P\big( \cap_{n=1}^N \{\|X_n-\xi\|_{\R^d} > \delta\}\big) =0.
\end{equation}
This and \eqref{eq:144b} assure that for every $\varepsilon\in(0,\infty)$ it holds that
\begin{equation}
\liminf_{N\to\infty}\P\big(|\!\max\nolimits_{1\leq n\leq N} f(X_n) - \sup\nolimits_{x\in [a,b]^d} f(x) | \leq \varepsilon \big) =1.
\end{equation}
Therefore, we obtain that for every $\varepsilon\in(0,\infty)$ it holds that
\begin{equation}
\limsup_{N\to\infty}\P\big( |\!\max\nolimits_{1\leq n\leq N} f(X_n) - \sup\nolimits_{x\in [a,b]^d} f(x) | >\varepsilon\big) =0.
\end{equation}
Combining this with with Lemma~\ref{lem:almost_sure_convergence} establishes item~\eqref{it:weakapprox}. It thus remains to prove item~\eqref{it:pthmomentapprox}. For this note that the fact that $f\colon [a,b]^d\to\R$ is globally bounded, item~\eqref{it:weakapprox}, and Lebesgue's dominated convergence theorem
ensure that for every $p\in (0,\infty)$ it holds that
\begin{equation}
\limsup_{N\to\infty} \E\Big[ \big|\!\max\nolimits_{1\leq i\leq N} f(X_i) - \sup\nolimits_{x\in [a,b]^d} f(x) \big|^p\Big] = 0.
\end{equation} This establishes item~\eqref{it:pthmomentapprox}.
The proof of Lemma~\ref{lemma:MonteCarloSupEstimate} is thus completed.
\end{proof}
\subsection{Geometric Brownian motions}
\label{sec:geometric}
In this subsection we apply the proposed approximation algorithm
to a Black-Scholes PDE with independent underlying geometric Brownian
motions.
Assume
Framework \ref{algo:general_algorithm}, let $r=\tfrac{1}{20}$, $\mu=r-\tfrac{1}{10}=-\tfrac{1}{20}$, $\sigma_1=\tfrac{1}{10}+\tfrac{1}{200}$, $\sigma_2=\tfrac{1}{10}+\tfrac{2}{200}$,\ldots, $\sigma_{100}=\tfrac{1}{10}+\tfrac{100}{200}$,
assume for every $s,t \in[0,T]$, $x=(x_1,x_2,\ldots,x_d)$, $w=(w_1,w_2,\ldots,w_d)\in\R^d$, $m\in\N_0$ that
$ N = 1$,
$ d = 100$,
$ \varphi(x) = \exp(-r T)\max\!\big\{[\max_{i\in\{1,2,\ldots,d\}} x_i]-100,0\big\} $, and
\begin{equation}
\begin{split}
&H(s,t,x,w) = \\
&\Big( x_1 \exp\!\big(\big(\mu_1-\tfrac{|\sigma_1|^2}{2}\big)(t-s) + \sigma_1w_1\big), \ldots, x_d \exp\!\big(\big(\mu_d-\tfrac{|\sigma_d|^2}{2}\big)(t-s) + \sigma_dw_d\big)\Big),
\end{split}
\end{equation}assume that
$ \xi^{0,1}\colon\Omega\to\R^d$ is continuous uniformly distributed on $[90,110]^d$,
and
let $u = (u(t,x))_{(t,x)\in [0,T]\times\R^d} \in C^{1,2}([0,T]\times\R^d,\R)$
be an at most polynomially growing function which satisfies
for every $t\in [0,T]$, $x\in\R^d$ that
$u(0,x) = \varphi(x)$ and
\begin{equation}\label{eq:example_geometric_brownian_motion_kolmogorovPDE}
(\tfrac{\partial u}{\partial t})(t,x)
=
\tfrac12 \sum_{i=1}^d
|\sigma_i x_i|^2 (\tfrac{\partial^2 u}{\partial x_i^2})(t,x)
+
\sum_{i=1}^d \mu_i x_i(\tfrac{\partial u}{\partial x_i})(t,x).
\end{equation}
The Feynman-Kac formula
(cf., for example, Hairer et al.~\cite[Corollary 4.17]{HairerHutzenthalerJentzen_LossOfRegularity2015})
shows that for every standard Brownian motion $\mathcal{W} =(\mathcal{W}^{(1)},\ldots,\mathcal{W}^{(d)})\colon[0,T]\times \Omega\to\R^d$ and every $t\in [0,T]$, $x=(x_1,\ldots,x_d)\in\R^d$ it holds that
\begin{equation}\label{eq:example_geometric_brownian_motion_stochastic_representation}
\begin{split}
&u(t,x)
= \\
& \E\!\left[
\varphi\!\left(x_1 \exp\!\left(\sigma_1 \mathcal{W}^{(1)}_t
+
\left(\mu_1 - \tfrac{|\sigma_1|^2}{2}\right)t\right), \ldots, x_d \exp\!\left(\sigma_d \mathcal{W}^{(d)}_t
+
\left(\mu_d - \tfrac{|\sigma_d|^2}{2}\right)t\right)\right)
\right].
\end{split}
\end{equation}
Table~\ref{tab:geometric_linfty_l1_l2}
approximately presents the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL1fehler_bm} below),
the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL2fehler_bm} below),
and
the relative $L^{\infty}(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relLinftyfehler_bm} below) against
$ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$ (cf.\ \textsc{Python} code~\ref{code:geometric} in Subsection~\ref{sec:code_geometric} below).
In our numerical simulations for Table~\ref{tab:geometric_linfty_l1_l2} we approximately calculated the exact
solution of the PDE~\eqref{eq:example_geometric_brownian_motion_kolmogorovPDE} by means of~\eqref{eq:example_geometric_brownian_motion_stochastic_representation}
and Monte Carlo approximations with 1048576 samples, we approximately calculated the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL1fehler_bm}
\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|dx
\end{equation}
for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 81920 samples, we approximately calculated the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL2fehler_bm}
\sqrt{\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|^2dx}
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 81920 samples, and we approximately calculated the relative $L^\infty(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relLinftyfehler_bm}
\sup_{x\in [0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 81920 samples (see Lemma~\ref{lemma:MonteCarloSupEstimate} above).
\begin{table}[H] \begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
{\begin{tabular}{@{}c@{}}Number \\ of steps\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^1(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^2(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^\infty(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Runtime \\ in seconds\end{tabular}}\\
\hline
0 & 1.004285&1.004286&1.009524 & 1 \\
\hline
25000 & 0.842938&0.843021&0.87884& 110.2 \\
\hline
50000 & 0.684955&0.685021&0.719826 & 219.5\\
\hline
100000 & 0.371515&0.371551&0.387978 & 437.9\\
\hline
150000 & 0.064605&0.064628&0.072259 & 656.2\\
\hline
250000 & 0.001220&0.001538&0.010039 & 1092.6\\
\hline
500000 & 0.000949&0.001187&0.005105& 2183.8 \\
\hline
750000 & 0.000902&0.001129&0.006028 & 3275.1\\
\hline
\end{tabular}
\caption{Approximative presentations of the relative approximation errors in \eqref{eq:relL1fehler_bm}--\eqref{eq:relLinftyfehler_bm} for the Black-Scholes PDE with independent underlying geometric Brownian motions in \eqref{eq:example_geometric_brownian_motion_kolmogorovPDE}.}
\label{tab:geometric_linfty_l1_l2}\end{center}
\end{table}
\subsection{Black-Scholes model with correlated noise}
\label{sec:black_scholes}
In this subsection we apply the proposed approximation algorithm
to a Black-Scholes PDE with correlated noise.
Assume
Framework~\ref{algo:general_algorithm},
let $r=\tfrac{1}{20}$, $\mu=r-\tfrac{1}{10}=-\frac{1}{20}$, $\beta_1=\tfrac{1}{10}+\tfrac{1}{200}$, $\beta_2=\tfrac{1}{10}+\tfrac{2}{200}$,\ldots, $\beta_{100}=\tfrac{1}{10}+\tfrac{100}{200}$, $Q=(Q_{i,j})_{(i,j)\in\{1,2,\ldots,100\}}$, $ \Sigma=(\Sigma_{i,j})_{(i,j)\in\{1,2,\ldots,100\}}\in\R^{100\times100}$, $\varsigma_1,\varsigma_2,\ldots,\varsigma_{100}\in\R^{100}$, assume for every $s,t \in[0,T]$, $x=(x_1,x_2,\ldots,x_d)$, $w=(w_1,w_2,\ldots,w_d)\in\R^d$, $m\in\N_0$, $i,j,k\in\{1,2,\ldots,100\}$ with $i<j$
that
$ N = 1$,
$ d = 100$,
$ \nu = d(2d)+(2d)^2+2d=2d(3d+1)$, $Q_{k,k}=1$, $Q_{i,j}=Q_{j,i}=\tfrac12$, $\Sigma_{i,j}=0$, $\Sigma_{k,k}> 0$, $\Sigma\Sigma^\ast=Q$ (cf., for example, Golub \& Van Loan~\cite[Theorem 4.2.5]{GolubVanLoan}), $\varsigma_k=(\Sigma_{k,1},\ldots,\Sigma_{k,100})$, $
\varphi(x) = \exp(-\mu T)\max\!\big\{110-[\min_{i\in\{1,2,\ldots,d\}} x_i],0\big\}
$, and
\begin{multline}
H(s,t,x,w) =\Big(x_1 \exp\!\left((\mu-\tfrac12{\|\beta_1\varsigma_1\|_{\R^d}^2})(t-s) + \langle \varsigma_1,w\rangle_{\R^d}\right),
\ldots,\\
x_d\exp\!\left((\mu-\tfrac12{\|\beta_d\varsigma_d\|_{\R^d}^2})(t-s) + \langle \varsigma_d,w\rangle_{\R^d}\right)\Big),
\end{multline}
assume that
$ \xi^{0,1}\colon\Omega\to\R^d$ is continuous uniformly distributed on $[90,110]^d$,
and let $u=(u(t,x))_{t\in [0,T],x\in\R^d}\in C^{1,2}([0,T]\times\R^d,\R)$
be an at most polynomially growing continuous function which satisfies
for every $t\in [0,T]$, $x\in\R^d$ that $u(0,x)=\varphi(x)$ and
\begin{equation}
\label{eq:example_correlated_geometric_brownian_motion_kolmogorovPDE}
(\tfrac{\partial u}{\partial t})(t,x)
=
\tfrac12 \sum_{i,j=1}^d
x_i x_j \beta_i \beta_j \langle \varsigma_i,\varsigma_j \rangle_{\R^d}
(\tfrac{\partial^2 u}{\partial x_i^2})(t,x)
+
\sum_{i=1}^d \mu_i x_i (\tfrac{\partial u}{\partial x_i})(t,x).
\end{equation}
The Feynman-Kac formula
(cf., for example, Hairer et al.~\cite[Corollary 4.17]{HairerHutzenthalerJentzen_LossOfRegularity2015})
shows that for every standard Brownian motion $\mathcal{W} =(\mathcal{W}^{(1)},\ldots,\mathcal{W}^{(d)})\colon[0,T]\times \Omega\to\R^d$ and every $t\in [0,T]$, $x=(x_1,\ldots,x_d)\in\R^d$ it holds that
\begin{multline}\label{eq:example_correlated_geometric_brownian_motion_stochastic_representation}
u(t,x)
= \E\!\left[
\varphi\!\left(x_1 \exp\!\left(\big\langle \varsigma_1,\mathcal{W}^{(1)}_t\big\rangle_{\R^d}
+
\Big(\mu_1 - \tfrac{\|\beta_1\varsigma_1\|_{\R^d}^2}{2}\Big)t\right),\ldots,\right.\right.\\
\qquad~\left. \left. x_d \exp\!\left(\big\langle \varsigma_d,\mathcal{W}^{(d)}_t\big\rangle_{\R^d}
+
\Big(\mu_d - \tfrac{\|\beta_d\varsigma_d\|_{\R^d}^2}{2}\Big)t\right)\right)
\right].
\end{multline}
Table~\ref{tab:brownian_linfty_l1_l2} approximately presents
the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL1fehler_bs} below),
the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL2fehler_bs} below),
and
the relative $L^{\infty}(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relLinftyfehler_bs} below) against
$ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$ (cf.\ \textsc{Python} code~\ref{code:black_scholes} in Subsection~\ref{sec:black_scholes} below). In our numerical simulations for Table~\ref{tab:brownian_linfty_l1_l2} we approximately calculated the exact
solution of the PDE~\eqref{eq:example_correlated_geometric_brownian_motion_kolmogorovPDE} by means of \eqref{eq:example_correlated_geometric_brownian_motion_stochastic_representation} and
Monte Carlo approximations with 1048576 samples, we approximately calculated the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL1fehler_bs}
\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|dx
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 81920 samples, we approximately calculated the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL2fehler_bs}
\sqrt{\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|^2dx}
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$ by means of Monte Carlo approximations with 81920 samples, and we approximately calculated the relative $L^\infty(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relLinftyfehler_bs}
\sup_{x\in [0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 81920 samples (see Lemma~\ref{lemma:MonteCarloSupEstimate} above).
\begin{table}[H] \begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
{\begin{tabular}{@{}c@{}}Number \\ of steps\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^1(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^2(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^\infty(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Runtime \\ in seconds\end{tabular}}\\
\hline
0 & 1.003383&1.003385&1.011662 & 0.8 \\
\hline
25000 & 0.631420&0.631429&0.640633 & 112.1 \\
\hline
50000 & 0.269053&0.269058&0.275114& 223.3\\
\hline
100000 & 0.000752&0.000948&0.00553 & 445.8\\
\hline
150000 & 0.000694&0.00087&0.004662 & 668.2\\
\hline
250000 & 0.000604&0.000758&0.006483 & 1119.3\\
\hline
500000 & 0.000493&0.000615&0.002774 & 2292.8 \\
\hline
750000 & 0.000471&0.00059&0.002862 & 3466.8\\
\hline
\end{tabular}
\caption{Approximative presentations of the relative approximation errors in \eqref{eq:relL1fehler_bs}--\eqref{eq:relLinftyfehler_bs} for the Black-Scholes PDE with correlated noise in \eqref{eq:example_correlated_geometric_brownian_motion_kolmogorovPDE}.}
\label{tab:brownian_linfty_l1_l2}\end{center}
\end{table}
\subsection{Stochastic Lorenz equations}
\label{sec:lorenz}
In this subsection we apply the proposed approximation algorithm
to the stochastic Lorenz equation.
Assume
Framework~\ref{algo:general_algorithm}, let $\alpha_1=10$, $\alpha_2=14$, $\alpha_3=\tfrac{8}{3}$, $\beta=\tfrac{3}{20}$, let $\mu\colon \R^d\to\R^d$ be a function, assume for every $s,t \in[0,T]$, $x=(x_1,x_2,\ldots,x_d)$, $w=(w_1,w_2,\ldots,w_d)\in\R^d$, $m\in\N_0$
that
$ N = 100 $,
$ d = 3 $,
$ \nu = (d+20)d + (d+20)^2 +(d+20)= (d+20)(2d+21)$, $\mu(x) = (\alpha_1 (x_2-x_1), \alpha_2 x_1 - x_2 - x_1 x_3, x_1 x_2 - \alpha_3 x_3)
$, $\varphi(x) = \|x\|^2_{\R^d}
$, and
\begin{equation}
\label{eq:lorenz_verfahren}
H ( s , t , x , w )
=
x +
\mu(x)(t-s)\mathbbm{1}_{[0,N/T]}(\|\mu(x)\|_{\R^d}) + \beta w
\end{equation}
(cf., for example, Hutzenthaler et al.~\cite{HutzenthalerJentzen2015_Memoirs}, Hutzenthaler et al.~\cite{HutzenthalerJentzenKloeden_StrongConvergenceExplicit2012}, Hutzenthaler et al.~\cite{HutzenthalerJentzenWang}, Milstein \& Tretyakov~\cite{MilsteinTretyakovNonglobal}, Sabanis~\cite{Sabanis13, Sabanis16}, and the references mentioned therein for related temporal numerical approximation schemes for SDEs), assume that
$ \xi^{0,1}\colon\Omega\to\R^d$ is continuous uniformly distributed on $[\tfrac12,\tfrac32]\times[8,10]\times[10,12]$,
and let $u=(u(t,x))_{(t,x)\in [0,T]\times\R^d}\in C^{1,2}([0,T]\times\R^d,\R)$
be an at most polynomially growing function (cf., for example, Hairer et al.~\cite[Corollary 4.17]{HairerHutzenthalerJentzen_LossOfRegularity2015} and H\"ormander~\cite[Theorem 1.1]{hörmander1967}) which satisfies
for every $t\in [0,T]$, $x\in\R^d$ that
$u(0,x) = \varphi(x)$ and
\begin{equation}
\begin{split}
\label{eq:example_lorenz_kolmogorovPDE}
&(\tfrac{\partial u}{\partial t})(t,x)
=
\tfrac{\beta^2}{2}(\Delta_x u)(t,x)
+
\alpha_1(x_2-x_1) (\tfrac{\partial u}{\partial x_1})(t,x) \\
&+
(\alpha_2 x_1 - x_2 - x_1 x_3)(\tfrac{\partial u}{\partial x_2})(t,x)
+
(x_1 x_2 - \alpha_3 x_3)(\tfrac{\partial u}{\partial x_3})(t,x).
\end{split}
\end{equation}
Table~\ref{tab:lorenz_linfty_l1_l2}
approximately presents the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL1fehler_lz} below),
the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL2fehler_lz} below),
and
the relative $L^{\infty}(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relLinftyfehler_lz} below) against
$ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$ (cf.\ \textsc{Python} code~\ref{code:lorenz} in Subsection~\ref{sec:code_lorenz} below). In our numerical simulations for Table~\ref{tab:lorenz_linfty_l1_l2} we approximately calculated the exact
solution of the
PDE~\eqref{eq:example_lorenz_kolmogorovPDE} by means of
Monte Carlo approximations with 104857 samples and temporal SDE-discretizations based on \eqref{eq:lorenz_verfahren} and 100 equidistant time steps, we approximately calculated the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL1fehler_lz}
\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|dx
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 20480 samples, we approximately calculated the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL2fehler_lz}
\sqrt{\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|^2dx}
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 20480 samples, and we approximately calculated the relative $L^\infty(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relLinftyfehler_lz}
\sup_{x\in [0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 20480 samples (see Lemma~\ref{lemma:MonteCarloSupEstimate} above).
\begin{table}[H] \begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
{\begin{tabular}{@{}c@{}}Number \\ of steps\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^1(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^2(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^\infty(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Runtime \\ in seconds\end{tabular}}\\
\hline
0 & 0.995732& 0.995732& 0.996454& 1.0 \\
\hline
25000 & 0.905267& 0.909422& 1.247772& 750.1 \\
\hline
50000 &0.801935& 0.805497& 1.115690& 1461.7 \\
\hline
100000 &0.599847& 0.602630& 0.823042& 2932.1 \\
\hline
150000 &0.392394& 0.394204& 0.542209& 4423.3 \\
\hline
250000 &0.000732& 0.000811& 0.002865& 7327.9 \\
\hline
500000 &0.000312& 0.000365& 0.003158& 14753.0 \\
\hline
750000 &0.000187& 0.000229& 0.001264& 21987.4 \\
\hline
\end{tabular}
\caption{Approximative presentations of the relative approximation errors in \eqref{eq:relL1fehler_lz}--\eqref{eq:relLinftyfehler_lz} for the stochastic Lorenz equation in \eqref{eq:example_lorenz_kolmogorovPDE}.}
\label{tab:lorenz_linfty_l1_l2}\end{center}
\end{table}
\subsection{Heston model}
\label{sec:heston}
In this subsection we apply the proposed approximation algorithm
to the Heston model in \eqref{eq:example_heston_kolmogorovPDE} below.
Assume
Framework \ref{algo:general_algorithm},
let
$ \delta =25$, $\alpha=\tfrac{1}{20}$, $\kappa=\tfrac{6}{10}$,
$ \theta =\tfrac{1}{25} $,
$ \beta=\tfrac{1}{5}$, $\varrho =-\tfrac{4}{5}$,
let $e_{i}\in\R^{50}$, $i\in\{1,2,\ldots,{50}\}$, be the vectors which satisfy that
$e_{1} = (1,0,0,\ldots,0,0)\in\R^{50}$, $e_{2} = (0,1,0,\ldots,0,0)\in\R^{50}$,
\ldots,
$e_{{50}} = (0,0,0,\ldots,0,1)\in\R^{50}$,
assume for every $s,t \in[0,T]$, $x=(x_1,x_2,\ldots,x_d)$, $w=(w_1,w_2,\ldots,w_d)\in\R^d$ that
$ N = 100$,
$ d = 2\delta = 50 $,
$ \nu = (d+50)d + (d+50)^2 +(d+50)= (d+50)(2d+51)$, $
\varphi(x) = \exp(-\alpha T)\max\!\big\{110 - [\textstyle\sum_{i=1}^\delta\tfrac{x_{2i-1}}{\delta}],0\big\}
$, and
\begin{multline}
\label{eq:heston_verfahren}
H( s , t , x , w )= \sum_{i=1}^{\delta}\Bigg( \bigg[x_{2i-1}\exp\!\Big((\alpha-\tfrac{x_{2i}}{2})(t-s)+w_{2i-1}\sqrt{x_{2i}}\Big)\bigg]e_{2i-1}\Bigg.\\
\left.+\bigg[\max\!\bigg\{\Big[
\max\!\Big\{\tfrac\beta2\sqrt{t-s},\max\!\big\{\tfrac{\beta}{2}\sqrt{t-s},\sqrt{x_{2i}}\big\}+\tfrac{\beta}{2}\big(\rho w_{2i-1}+[{1-\rho^2}]^{1/2}w_{2i}\big)\Big\}\Big]^2\right.\\
\Bigg.+\big(\kappa\theta-\tfrac{\beta^2}{4}-\kappa x_{2i}\big)(t-s),0\bigg\}\bigg]e_{2i}\Bigg)
\end{multline}
(cf.\ Hefter \& Herzwurm~\cite[Section 1]{HefterHerzwurm}), assume that
$ \xi^{0,1}\colon\Omega\to\R^d$
is continuous uniformly distributed on $\times_{i=1}^\delta \big([90,110]\times[0.02,0.2]\big)$,
and let $u=(u(t,x))_{(t,x)\in [0,T]\times\R^d}\in C^{1,2}([0,T]\times\R^d,\R)$
be an at most polynomially growing function (cf., for example, Alfonsi~\cite[Proposition 4.1]{Alfonsi2005}) which satisfies
for every $t\in [0,T]$, $x\in\R^d$ that
$u(0,x) = \varphi(x)$ and
\begin{multline}
\label{eq:example_heston_kolmogorovPDE}
(\tfrac{\partial u}{\partial t})(t,x) =
\Bigg[\sum_{i=1}^{\delta}
\Big(
\alpha x_{2i-1} (\tfrac{\partial u}{\partial x_{2i-1}})(t,x)
+
\kappa (\theta - x_{2i}) (\tfrac{\partial u}{\partial x_{2i}})(t,x)
\Big)\Bigg]
\\
+ \Bigg[
\sum_{i=1}^{\delta} \frac{|x_{2i}|}{2}
\Big(
|x_{2i-1}|^2 (\tfrac{\partial^2 u}{\partial x_{2i-1}^2})(t,x) + 2 x_{2i-1} \beta\varrho (\tfrac{\partial^2 u}{\partial x_{2i-1}\partial x_{2i}})(t,x) + \beta^2 (\tfrac{\partial^2 u}{\partial x_{2i}^2})(t,x)
\Big)\Bigg].
\end{multline}
Table~\ref{tab:heston_linfty_l1_l2}
approximately presents the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL1fehler_hs} below),
the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relL2fehler_hs} below),
and
the relative $L^{\infty}(\lambda_{[0,1]^d};\R)$-approximation error associated to
$(\bU^{\Theta_m,1,\bS_m}(x))_{x\in[0,1]^d}$ (see \eqref{eq:relLinftyfehler_hs} below) against
$ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$ (cf.\ \textsc{Python} code~\ref{code:heston} in Subsection~\ref{sec:code_heston} below). In our numerical simulations for Table~\ref{tab:heston_linfty_l1_l2} we approximately calculated the exact
solution of the PDE~\eqref{eq:example_heston_kolmogorovPDE} by means of
Monte Carlo approximations with 1048576 samples and temporal SDE-discretizations based on \eqref{eq:heston_verfahren} and 100 equidistant time steps, we approximately calculated the relative $L^1(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL1fehler_hs}
\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|dx
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 10240 samples, we approximately calculated the relative $L^2(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relL2fehler_hs}
\sqrt{\int_{[0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|^2dx}
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 10240 samples, and we approximately calculated the relative $L^\infty(\lambda_{[0,1]^d};\R)$-approximation error
\begin{equation}
\label{eq:relLinftyfehler_hs}
\sup_{x\in [0,1]^d} \left|\frac{u(T,x) - \bU^{\Theta_m,1,\bS_m}(x)}{u(T,x)}\right|
\end{equation} for $ m \in \{0, 25000, \allowbreak50000,\allowbreak 100000, 150000, 250000, 500000, \allowbreak 750000\}$
by means of Monte Carlo approximations with 10240 samples (see Lemma~\ref{lemma:MonteCarloSupEstimate} above).
\begin{table}[H] \begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
{\begin{tabular}{@{}c@{}}Number \\ of steps\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^1(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^2(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Relative \\ $L^\infty(\lambda_{[0,1]^d};\R)$-error\end{tabular}} & {\begin{tabular}{@{}c@{}}Runtime \\ in seconds\end{tabular}}\\
\hline
0 & 1.038045&1.038686&1.210235& 1.0 \\
\hline
25000 & 0.005691&0.007215&0.053298& 688.4 \\
\hline
50000 & 0.005115&0.006553&0.036513& 1375.2 \\
\hline
100000 & 0.004749&0.005954&0.032411& 2746.8 \\
\hline
150000 & 0.006465&0.008581&0.051907& 4120.2 \\
\hline
250000 & 0.005075&0.006378&0.024458& 6867.5 \\
\hline
500000 & 0.002082&0.002704&0.019604& 13763.7 \\
\hline
750000 & 0.00174&0.002233&0.012466& 20758.8 \\
\hline
\end{tabular}
\caption{Approximative presentations of the relative approximation errors in \eqref{eq:relL1fehler_hs}--\eqref{eq:relLinftyfehler_hs} for the Heston model in \eqref{eq:example_heston_kolmogorovPDE}.}
\label{tab:heston_linfty_l1_l2}\end{center}
\end{table}
\section{\textsc{{\sc Python}} source codes}\label{sec:source_code}
\subsection{{\sc Python} source code for the algorithm}
In Subsections~\ref{sec:code_paraboliceq}--\ref{sec:code_heston} below we present {\sc Python} source codes associated to the numerical simulations in Subsections~\ref{sec:paraboliceq}--\ref{sec:heston} above. The following {\sc Python} source code, {\sc Python} code~\ref{code:common} below, is employed in the case of each of the {\sc Python} source codes in Subsections~\ref{sec:code_paraboliceq}--\ref{sec:code_heston} below.
\lstset{caption={\it common.py}}
\lstinputlisting[label={code:common}]{common.py}
\subsection[A {\sc Python} source code associated to Subsection~\ref{sec:paraboliceq}]{A {\sc Python} source code associated to the numerical simulations in Subsection~\ref{sec:paraboliceq}}
\label{sec:code_paraboliceq}
\lstset{caption={\it example3\_2.py}}
\lstinputlisting[label={code:paraboliceq}]{example3_1.py}
\subsection[A {\sc Python} source code associated to Subsection~\ref{sec:geometric}]{A {\sc Python} source code associated to the numerical simulations in Subsection~\ref{sec:geometric}}
\label{sec:code_geometric}
\lstset{caption={\it example3\_3.py}}
\lstinputlisting[label={code:geometric}]{example3_2.py}
\subsection[A {\sc Python} source code associated to Subsection~\ref{sec:black_scholes}]{A {\sc Python} source code associated to the numerical simulations in Subsection~\ref{sec:black_scholes}}
\label{sec:code_black_scholes}
\lstset{caption={\it example3\_4.py}}
\lstinputlisting[label={code:black_scholes}]{example3_3.py}
\subsection[A {\sc Python} source code associated to Subsection~\ref{sec:lorenz}]{A {\sc Python} source code associated to the numerical simulations in Subsection~\ref{sec:lorenz}}
\label{sec:code_lorenz}
\lstset{caption={\it example3\_5.py}}
\lstinputlisting[label={code:lorenz}]{example3_4.py}
\subsection[A {\sc Python} source code associated to Subsection~\ref{sec:heston}]{A {\sc Python} source code associated to the numerical simulations in Subsection~\ref{sec:heston}}
\label{sec:code_heston}
\lstset{caption={\it example3\_6.py}}
\lstinputlisting[label={code:heston}]{example3_5.py}
\bibliographystyle{acm}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,993 |
Saratow (, ) ist eine Stadt in der gleichnamigen russischen Oblast. Sie hat Einwohner (Stand ) und liegt am rechten, sogenannten "Bergufer" der mittleren Wolga. Saratow ist ein wichtiges Kultur- und Wirtschaftszentrum sowie Universitätsstadt und bildet zusammen mit dem auf der anderen Seite der Wolga gelegenen Engels eine Agglomeration mit rund 1,2 Millionen Einwohnern.
Geographie
Geographische Lage
Saratow befindet sich im europäischen Teil Russlands. Die Stadt liegt an den Rändern des Hügellandes der Wolgaplatte und erstreckt sich bis an das Ufer der Wolga, die hier zum Wolgograder Stausee gestaut wird. Das Stadtzentrum befindet sich unweit des Wolgaufers. Die Wolga ist hier etwa drei Kilometer breit und wird nahe dem Stadtzentrum von der Brücke von Saratow überquert.
Stadtgliederung
Saratow ist Verwaltungssitz der Oblast Saratow, welche den Großraum Saratow umfasst. Während die Oblast Saratow in 38 Rajons (Kreise) unterteilt ist, gliedert sich die Stadt selbst in sechs Stadtteile (ebenfalls Rajon genannt).
Klima
Geschichte
Saratow wurde 1590 als Zarenfestung gegründet. Vorher war die Stadt "Hauptstadt" der Goldenen Horde, des mongolischen Teilreichs in Osteuropa und Westsibirien. Einige Historiker vermuten, dass der Name der Stadt aus dem Tatarischen stammt und so viel wie "Gelber Berg" bedeutet.
Nach einem Ukas des Zaren Fjodor Iwanowitsch begann im Jahre 1590 an der Wolga der Bau einer Bewachungsfestung, nachdem an der Wolga bereits 1586 Samara und 1589 Zarizyn gegründet wurden. Diese Festung sollte die Steppengebiete des Zarentum Russland an der Wolga schützen und die Überfälle der Reitervölker aus den Nachbargebieten eindämmen. Ebenso wichtig war die Sicherung des Wasserweges von Kasan nach Astrachan, der hier verlief. Die Festung wurde unter der Leitung des Heerführers Grigori Sassekin errichtet. Saratow war dabei nicht mehr nur wichtiger Militärstützpunkt, sondern auch großes Handelszentrum als Umschlagplatz des russischen Handels mit dem Orient geworden.
Von der zweiten Hälfte des 18. Jahrhunderts und bis zu den Deportationen im Zusammenhang mit dem Zweiten Weltkrieg lebten in Saratow zahlreiche Wolgadeutsche, deren Tätigkeit auch in der Architektur der Stadt Spuren hinterlassen hat. Sie kamen überwiegend aus Bayern, Baden, Hessen, der Pfalz und dem Rheinland und folgten in den Jahren 1763 bis 1767 der Einladung der deutschstämmigen Zarin Katharina II. Nach dem Überfall des "Dritten Reiches" auf die Sowjetunion im Juni 1941 ließ Josef Stalin das Präsidium des Obersten Sowjets der UdSSR am 28. August 1941 den Erlass "Über die Umsiedlung der im Wolgagebiet ansässigen Deutschen" beschließen. Die Wolgadeutschen wurden der kollektiven Kollaboration beschuldigt, nach Sibirien und Zentralasien deportiert und dort in Arbeitslager der "Arbeitsarmee" (Трудармия) gezwungen, wobei Tausende starben.
Während ihrer Blütezeit im 19. Jahrhundert gehörte die Stadt zu den größten des Russischen Reiches.
Bezogen auf die 1630er Jahre gibt es über Saratow schriftliche Zeugnisse aus einer Reisebeschreibung in deutscher Sprache, die auch eine Abbildung der Stadt enthält.
Im Herbst 1856 wurde in Saratow ein Priesterseminar und 1857 ein Knabenseminar gegründet, das als Vorbereitungsschule für das Priesterseminar dienen sollte. Im Knabenseminar waren jährlich 25 Plätze für Kolonistenkinder reserviert; dabei wurden Waisen und Söhne aus armen Familien bevorzugt. Von den Absolventen des Knabenseminars besuchten nur etwa 20 % das Priesterseminar, die überwiegende Mehrheit ging als Lehrer in die Kolonien. Vier Absolventen des Saratower Priesterseminars, Anton Johannes von Padua Zerr (1849 bis 1932), Josef Aloisius Keßler (1862 bis 1933), Alexander Frison (1873 bis 1937) und Markus Glaser (1880 bis 1950) wurden Diözesanbischöfe.
Im Ersten Weltkrieg war Saratow im Januar 1915 Zielort für Deportationskonvois von Deutschen, Juden, Ungarn, Österreichern und Slawen, die in den westlichen, frontnahen Gebieten des Reiches als mögliche Gefahr eingestuft wurden.
Nach der Auflösung des Gouvernements Saratow 1928 war Saratow in der Russischen SFSR zunächst Zentrum der Oblast Untere Wolga, aus der bereits im selben Jahr der bis 1934 bestehende Krai Untere Wolga wurde. 1932 wurde das Krai-Zentrum nach Stalingrad verlegt. 1934 wurde der Krai Saratov gebildet und schließlich 1936 die bis heute bestehende Oblast Saratow.
Der erste Mensch im Weltall Juri Gagarin studierte an der Technischen Universität in Saratow und landete nach seinem legendären Flug 1961 auch in der Oblast Saratow.
Bevölkerungsentwicklung
Anmerkung: Volkszählungsdaten
Sehenswürdigkeiten
Saratow ist ein bedeutendes Kulturzentrum an der Wolga. Die gesamte Altstadt ist sehr malerisch und hier gibt es zahlreiche prächtige Gebäudeensembles aus vielen Jahrhunderten, so etwa in der ehemaligen Deutschen Straße, heute Uliza Nemezkaja, jetzt Kirow-Prospekt genannt. Auf dem Kirow-Prospekt, heute eine reine Fußgängerzone, oder auf der Promenade am Wolgaufer flanieren Touristen und Einheimische. Prächtige Beispiele des Jugendstils sind in Saratow allerorts zu finden wie etwa die Villa Reineke von Fjodor Schechtel, die er 1912 errichtete. Eine besonders sehenswerte Kirche ist die Anfang des 20. Jahrhunderts erbaute Kirche der Gottesmutter-Ikone Lindere mein Leid. Lohnend ist eine Autofahrt über die drei Kilometer lange alte Wolgabrücke, die das Zentrum von Saratow mit der am gegenüberliegenden Wolgaufer liegenden Stadt Engels verbindet.
Radischtschew-Kunstmuseum
Eine der größten Touristenattraktionen der Stadt ist das am 29. Juli 1885 eröffnete Radischtschew-Kunstmuseum mit über 16.000 Exponaten. Es wurde von Alexei Bogoljubow (1824–1896), dem Enkel von Alexander Radischtschew, gegründet. Es ist eines der größten Museen der Welt für russische Kunst.
Saratower Völkerkundemuseum
Das 1886 gegründete Museum mit einer großen archäologischen Sammlung befindet sich in einem prächtigen klassizistischen Gebäude.
Saratower Oper
Die Saratower Oper wurde 1803 gegründet, nicht nur wegen der prächtigen Architektur ist sie einzigartig und genießt schon lange internationalen Ruhm.
Drama-Theater Saratow
Das Saratower Dramatische Theater ist eins der ältesten Theater Russlands. Gegründet 1803, kann es auf eine über 200-jährige Geschichte zurückblicken.
Wirtschaft
Im Raum Saratow wird vor allem im Wolgatal umfangreich Landwirtschaft betrieben. Im Industriebereich befinden sich in Saratow vor allem Maschinenbau-, chemische und erdölverarbeitende Unternehmen. Auch aufgrund der deutschen Traditionen der Stadt sind zahlreiche deutsch-russische Joint Ventures im Raum Saratow tätig.
Verkehr
Saratow besaß mit dem Flughafen Saratow einen internationalen Flughafen mit vereinzelten Direktverbindungen ins Ausland sowie regelmäßigen Flügen in andere russische Metropolen sowie den Flugplatz Saratow-Dubki. Im August 2019 wurde der neue Flughafen Saratow-Gagarin in Betrieb genommen. Daneben spielen Eisenbahn-Fernverkehrsstrecken sowie nicht zuletzt auch die Wolga als stark befahrene Binnenschifffahrtsstraße eine große Rolle. Der öffentliche Personennahverkehr wird von Buslinien, Trolleybussen und Straßenbahnen getragen. Die Stadt ist Verwaltungssitz der Wolga Regionaldirektion der Russischen Staatsbahn. Die Direktion betreibt nicht nur alle Eisenbahnlinien samt zugehöriger Infrastruktur im Großraum Saratow, sondern auch ein über 4237 Kilometer langes Schienennetz.
Autobahnen und für russische Verhältnisse gut ausgebaute Fernstraßen führen von der Stadt aus direkt in die umliegende Großstädte. Über eine Zweigstrecke der russischen Fernstraße R22 Kaspi ist Saratow mit dem Umland der russischen Hauptstadt Moskau verbunden. Auf dieser Strecke verläuft auch die Europastraße 38 (von Hluchiw, Ukraine nach Qysylorda, Kasachstan). Die Stadt ist über die Fernstraße R158 mit Nischni Nowgorod, Saransk und Pensa verbunden. Hier wird sie von der R228 gekreuzt, die von Sysran nach Wolgograd führt.
Über die Wolga führen drei Brücken:
die Saratower Eisenbahnbrücke (Baujahr 1935),
die Brücke von Saratow (Baujahr 1965) und
die Neue Brücke von Saratow (Baujahre 2000 und 2009).
Kultur und Bildung
Kultur
Saratow gilt im Wolgaraum als Stadt der Kultur, zum einen wegen der langen und bedeutenden Tradition der Stadt, zum anderen natürlich wegen der umfangreichen Kunstsammlung, die mit 16.000 Exponaten und mit Werken alter Meister von globaler Bedeutung ist. Daneben gibt es aber auch zahlreiche Theater, Kinos und weitere große Kultur- und Vergnügungsstätten. Zum Kulturaustausch gibt es in der Stadt eine Vertretung des Goethe-Instituts und ein Deutsches Haus. In Saratow gibt es eine Oper, die schon längst über die russischen Grenzen hinaus von sich reden macht.
Religion
Christentum
Russisch-orthodoxe Kirche
1906 wurde die orthodoxe Kirche der Gottesmutter-Ikone "Lindere mein Leid" errichtet.
Römisch-katholische Kirche
In Saratow entstand am 11. Februar 2002 aus der "Römisch-katholischen Administratur für die Katholiken des lateinischen Ritus im Süden des europäischen Russlands" das katholische Bistum St. Clemens. 2004 wurde das Bistum staatlich anerkannt und ist dadurch eine Art "eingetragener Verein". Die Bischofskirche ist St. Peter und Paul.
Evangelisch-lutherische Kirche
Im Jahr 1793 wurde in Saratow die evangelisch-lutherische St.-Marien-Kirche eingeweiht. Sie lag direkt im Stadtzentrum und wurde 1971 durch Sprengung zerstört. Die Gemeinde versammelte sich danach an verschiedenen Orten und begann im Jahr 2008 mit dem Bau einer neuen Kirche. Es entstand ein Gebäude, das über 200 Personen fassen kann und über zahlreiche kleine und große Räume verfügt. Am 13. Mai 2018 fand die Einweihung der neuen St.-Marien-Kirche statt. An dem Festgottesdienst beteiligte sich auch der Botschafter der Bundesrepublik Deutschland, Rüdiger von Fritsch, der – als ordinierter lutherischer Prediger – die Festpredigt hielt.
Bildung
Weiterführende Bildungseinrichtungen in Saratow sind:
Staatliche Universität Saratow (Tschernyschewski-Universität)
Baukolleg Saratow
Saratower Hochschule des Innenministeriums Russlands
Saratower Militärinstitut für biologischen, chemischen und Strahlenschutz
Saratower Militärisches Medizininstitut
Saratower Staatliche Agrotechnische Universität
Saratower Staatliche Akademie für Veterinärmedizin und Biotechnologie
Saratower Staatliche Medizinuniversität
Saratower Staatliche Rechtsakademie
Saratower Staatliche Wawilow-Agraruniversität
Saratower Staatliche Sozialökonomische Universität
Saratower Staatliche Technische Universität
Saratower Staatliches Konservatorium
Saratower Staatliches Pädagogisches Fedin-Institut
Stolypin-Akademie des Wolgagebietes für den Staatsdienst
Fakultät der Russischen Staatlichen Öffentlichen Technischen Universität für Verkehrswesen
Filiale der Militärischen Artillerieuniversität
Filiale der Russischen Staatlichen Universität für Handel und Wirtschaft Moskau
Filiale der Russischen Akademie für Recht des Wolgagebiets
Die Saratower Theaterschule ist weit über die Grenzen Russlands hinaus bekannt. Hier wurden viele bekannte Schauspieler ausgebildet.
Sport
Zu den bekanntesten Sportvereinen der Stadt gehört der Eishockey-Club HK Kristall Saratow, der am Spielbetrieb Wysschaja Hockey-Liga (WHL) teilnimmt. Seine Heimspielstätte ist die 1969 eröffnete Mehrzweckhalle Sport Palast Kristall, die knapp 5.000 Zuschauerplätze hat und außer für Eishockeyspiele auch für Veranstaltungen in diversen Hallensportarten sowie für Konzerte und Ausstellungen genutzt wird. Die Halle wird ebenfalls Basketballklub BK Awtodor Saratow genutzt, der in der VTB United League spielt.
Im Fußball ist bzw. war die Stadt durch den Verein PFK Sokol Saratow vertreten, der für mehrere Saisons in der höchsten russischen Klasse, der Premjer-Liga, spielte, gegenwärtig jedoch zwischen dem eine Klasse tieferen Perwenstwo FNL und dem drittklassigen Perwenstwo PFL pendelt. Die Stadt besitzt mit dem 1962 eröffneten und bis zu 15.000 Zuschauer fassenden Lokomotiv-Stadion eine eigene Spielstätte.
Kriegsgefangenenlager des Zweiten Weltkrieges
In der Stadt bestand das Kriegsgefangenenlager 238 für deutsche Kriegsgefangene des Zweiten Weltkriegs, das im Herbst 1947 als Lager 368 nach Engels verlegt wurde. Schwer Erkrankte wurden in den beiden Kriegsgefangenenhospitälern 3631 und 5138 versorgt. Südlich der Stadt befand sich ein Friedhof, auf dem 70.000 Kriegsgefangene bestattet wurden; ein weiterer Friedhof mit 2.000 Toten war auf einem der umliegenden Hügel angelegt.
Söhne und Töchter der Stadt
Zu den Söhnen und Töchtern der Stadt Saratow gehören u. a. die Schriftsteller Nikolai Tschernyschewski (1828–1889), Konstantin Fedin (1892–1977), Nadeschda Mandelstam (1899–1980), Alexander Bek (1903–1972) und Erika Müller-Hennig (1908–1985), Pastor Johannes Schleuning (1879–1961), die Politiker Alexei Rykow (1881–1938), Georgi Oppokow (1888–1938), Roman Abramowitsch (* 1966) und Olga Batalina (* 1975), die Mathematiker Alexander Weinstein (1897–1979), Alexander Norden (1904–1993), Wiktor Wagner (1908–1981) und Alexander Olschanski (* 1946), der Physikochemiker und Nobelpreisträger Nikolai Semjonow (1896–1986), die Schauspieler Boris Babotschkin (1904–1975), Boris Andrejew (1915–1982), Oleg Tabakow (1935–2018), Jewgeni Mironow (* 1966) und Filipp Jankowski (* 1968), die Eishockeyspieler Anatoli Fedotow (* 1966), Sergei Nikolajew (* 1972), Denis Platonow (* 1981) und Generaloberst Boris Gromow (* 1943).
Weblinks
Kurzportrait (deutsch)
Bildergalerie von Saratow (russisch)
Website der Stadtverwaltung (russisch)
Die Kultur der Stadt Saratow (russisch)
Kunst aus Saratow (deutsch)
Saratow: Heimat der Wolgadeutschen bei Russia Beyond the Headlines: Aktuelles aus Russland
Einzelnachweise
Ort in der Oblast Saratow
Ort an der Wolga
Ort mit Binnenhafen
Wolgadeutsche
Hochschul- oder Universitätsstadt in Russland
Hauptstadt eines Föderationssubjekts Russlands
Gegründet 1590 | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 9,574 |
\section{Introduction}
Inflation driven by a real single scalar field (inflaton) slowly rolling on a smooth self-interaction
potential represents the minimal class of models in General Relativity (GR) which are in agreement with observations.
Not only are the details of the fundamental nature of the inflaton and of its interaction with other
fields needed to study the stage of reheating after inflation, but also they can be important for
theoretical and phenomenological aspects of its evolution.
Axion inflation, and more generally, inflation driven by a pseudo-scalar field is the archetypal model
to include parity violation during a nearly exponential expansion and it has a rich phenomenology.
An interaction of the pseudo-scalar field with gauge fields of the type
\begin{equation}
\mathcal{L}_{\textup{int}} =- \frac{g\phi}{4} F^{\mu\nu}\tilde{F}_{\mu\nu} \,,
\label{eqn:coupling}
\end{equation}
where $g$ is a coupling constant with a physical dimension of length, or inverse energy (we put
$\hbar=c=1$), leads to decay of the pseudo-scalar field into gauge fields modifying its background
dynamics \cite{Anber:2009ua} and to a wide range of potentially observable signatures including
primordial magnetic fields
\cite{Turner:1987bw,Garretson:1992vt,Finelli:2000sh,Anber:2006xt,Caprini:2014mja,Adshead:2016iae,Caprini:2017vnn,Chowdhury:2018mhj,Sobol:2019xls}, preheating at the end of inflation \cite{Finelli:2000sh,Adshead:2015pva,McDonough:2016xvu} , baryogenesis and leptogenesis \cite{Jimenez:2017cdr,Domcke:2018eki,Domcke:2019mnd},
equilateral non-Gaussianites \cite{Barnaby:2010vf,Barnaby:2011vw,Linde:2012bt}, chiral gravitational
waves in the range of direct detection by gravitational wave antennas \cite{Sorbo:2011rz,Barnaby:2011qe,Ferreira:2014zia,Peloso:2016gqs},
and primordial black holes (PBHs)
\cite{Linde:2012bt,Bugaev:2013fya,Garcia-Bellido:2016dkw,Domcke:2016bkh,Domcke:2017fix}.
The decay of the inflaton into gauge fields due to the coupling in Eq.~\eqref{eqn:coupling}
is a standard problem of amplification of quantum fluctuation (gauge fields) in an external classical
field (the inflaton). Applications of the textbook regularization techniques used for calculation of
quantum effects in curved space-time \cite{Birrell:1982ix,GMM}
have led to interesting novel results in the de Sitter \cite{Finelli:2004bm} and inflationary space-times
\cite{Finelli:2001bn,Finelli:2003bp,Marozzi:2006ky}, also establishing a clear connection between the
stochastic approach \cite{Starobinsky:1986fx,Starobinsky:1994bd} and field theory methods
\cite{Finelli:2008zg,Finelli:2010sh}.
In this paper, we apply the technique of adiabatic regularization \cite{Zeldovich:1971mw,Parker:1974qw}\footnote{It was called 'n-wave regularization' in \cite{Zeldovich:1971mw}.} to
the energy-density and helicity of gauge fields generated through the interaction in Eq.~\eqref{eqn:coupling}.
The evolution equation of gauge fields admits analytical solutions under the assumption of a constant
$\xi\equiv g \dot{\phi}/(2 H)$ \cite{Anber:2006xt}, where $H\equiv \frac{\dot a}{a}$ is the Hubble parameter during inflation,
also considered constant in time. Here we solve in an analytical way for the averaged energy
density and helicity of gauge fields for any value of $\xi$.
Previously only approximate results valid in the strongly coupled regime $\xi \gg 1$ were obtained
in the literature.
Our technique of computing integrals in the Fourier space is based on previous calculations of the
Schwinger effect for a $U(1)$ gauge field in the de Sitter space-time
\cite{Frob:2014zka,Kobayashi:2014zza,Hayashinaka:2016qqn,Domcke:2019qmm},
but now it is applied to a novel problem in which the classical external field is the inflaton.
More recent papers apply similar techniques to calculate backreaction of $SU(2)$ gauge fields
\cite{Lozanov:2018kpk,Maleknejad:2018nxz} and fermions \cite{Adshead:2018oaa,Adshead:2019aac} on the de Sitter space-time.
Our paper is organized as follows. In Section~\ref{sec:setting} and \ref{Energy-momentum tensor} we review
the basic equations and the averaged energy-momentum tensor and helicity of the gauge fields, respectively.
In Section~\ref{Analytical calculation} we present analytical results for the bare averaged quantities,
and we direct the interested reader to Appendix~\ref{Appendix5} and \ref{appendixmass} for more detailed calculations.
In Section~\ref{appendixADI} we outline the adiabatic regularization scheme used (see also \cite{Marozzi:2006ky}).
We also show that counterterms appearing in the adiabatic subtraction method can be naturally interpreted
as coming from renormalization of self-interaction terms of the scalar field either existing in the bare
Lagrangian density, or those which has to be added to it due to the
non-renormalizability of the problem involved (that is clear from $g$ being dimensional).
We then describe the implications of our results to the homogeneous dynamics of inflation with the backreaction
of one-loop quantum effects taken into account in Section~\ref{implicationsdissipative},
particularly focusing on the new regime of validity $|\xi|\lesssim 1$ and commenting on the differences from
previous results existing in the literature. We then conclude in Section~\ref{Conclusions}.
\section{Setting of the problem}
\label{sec:setting}
The Lagrangian density describing a pseudo-scalar inflaton field $\phi$ coupled to a $U(1)$ gauge field is:
\begin{equation}
\mathcal{L} = - \frac{1}{2} (\nabla\phi)^2 - V(\phi)
- \frac{1}{4} (F^{\mu\nu})^2 - \frac{g\phi}{4} F^{\mu\nu}\tilde{F}_{\mu\nu} \,,
\label{eqn:pseudoscalar}
\end{equation}
where $\tilde{F}^{\mu\nu}=\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}/2=\epsilon^{\mu\nu\alpha\beta}
(\partial_\alpha A_\beta-\partial_\beta A_\alpha)/2$ and $\nabla$ is the metric covariant derivative.
We consider the Friedmann-Lema$\rm\hat i$tre-Robertson-Walker (FLRW) metric $\mathrm{d} s^2=-\mathrm{d} t^2+a^2 \mathrm{d} {\bf x}^2$, where $a(t)$ is the scale
factor, and we write the coupling constant $g=\alpha/f$, where $f$ is the axion decay constant.
We consider gauge fields to linear order in a background driven only by a non-zero time-dependent
{\em vev} $\phi(t)$.
In this context, it is convenient to adopt the basis of circular polarization $\bm{\epsilon}_\pm$ transverse
to the direction of propagation defined by the comoving momentum $\mathbf{k}$. In the Fourier space we then have:
\begin{align}
&\mathbf{k}\cdot \bm{\epsilon}_\pm = 0 \,,\\
&\mathbf{k}\times \bm{\epsilon}_\pm = \mp\imath|k|\bm{\epsilon}_\pm \, .
\end{align}
Expanding the second quantized gauge field in terms of creation and annihilation operators for each Fourier mode $\mathbf{k}$, we get:
\begin{equation}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mathbf{A} (t, \mathbf{x})
=\sum_{\lambda=\pm} \int \frac{\mathrm{d}^3k}{(2\pi)^3} \left[ \bm{\epsilon}_\lambda(\mathbf{k})
A_\lambda(\tau,\mathbf{k})a_\lambda^{\mathbf{k}} e^{\imath \mathbf{k}\cdot\mathbf{x}} +H.c.\right] \,,
\end{equation}
where the Fourier mode functions $A_\pm$ for the two circular polarizations satisfy the following equation of motion:
\begin{equation}
\label{eqn:EoM1}
\frac{\mathrm{d}^2}{\mathrm{d}\,\tau^2}A_\pm(\tau,k)+\big( k^2 \bm{\mp} g k \phi' \big) A_\pm(\tau,k)=0\, .
\end{equation}
Here the prime denotes the derivative with respect to the conformal time $\tau$ ($\mathrm{d} \tau = \mathrm{d} t /a$).
The above equation admits a simple analytic solution for a constant
$\dot{\phi}$ $(\equiv d \phi/d t)$ in a nearly de Sitter stage during inflation.
A constant time derivative for the inflaton evolution can be obtained for $V(\phi)=\Lambda^4(1-C|\phi|)$ with $|C\phi| \ll 1$,
or for $V(\phi) \propto m^2 \phi^2$ \cite{St78,Linde:1983gd}.
Natural inflation with $V(\phi) = \Lambda^4 \left[ 1 \mp \cos (\phi/f) \right]$ can be approximated
better and better by $m^2 \phi^2$ for $f \gg M_{pl}$ with $m=\Lambda^2/f$, which is the regime allowed by cosmic microwave background (CMB henceforth) anisotropy measurements
\cite{Savage:2006tr,Akrami:2018odb}.
We thus study the inflationary solution assuming a de Sitter expansion,
i.e. $a(\tau)=-1/(H\tau)$, with $\tau < 0$, $H\simeq\textup{const}$ and $\dot{\phi}\simeq\textup{const}$.
In such a case, we can write $\phi' \simeq -\sqrt{2\epsilon_{\phi}}M_{pl}/\tau$ with $\epsilon_\phi = \dot{\phi}^2/(2 M_{\textup{pl}}^2 H^2)$ being one of the slow-roll parameters.
In this case, the equation of motion for the two circular polarization mode functions becomes
\cite{Anber:2009ua}:
\begin{equation}
\label{eqn:eqbeta}
\frac{\mathrm{d}^2}{\mathrm{d}\,\tau^2}A_\pm(\tau,k)+\Bigl(k^2\pm\frac{2 k \xi}{\tau}\Bigr)A_\pm(\tau,k)=0 \, ,
\end{equation}
where $\xi \equiv g\dot{\phi}/(2H)$.
The above equation reduces to \cite[p.~538]{Abramovitz} with $L=0$ and admit a solution
in terms of the regular and irregular Coulomb wave functions corresponding to the positive frequency for $-k \tau >0$:
\begin{equation}
\label{eqn:solution}
A_\pm(\tau,k) = \frac{\left[G_0(\pm\xi,-k\tau)+\imath F_0(\pm\xi,-k\tau)\right]}{\sqrt{2k}} \,.
\end{equation}
These can be rewritten in the subdomain $-k\tau>0$ in terms of the Whittaker W-functions:
\begin{equation}
\label{eqn:aplusaminus}
A_\pm(\tau,k) = \frac{1}{\sqrt{2k}} e^{\pm\pi\xi/2} W_{\pm\imath\xi,\frac{1}{2}}(-2\imath k\tau) \,.
\end{equation}
Note that the above solutions are symmetric under the change $A_+\to A_-$ and $\xi\to-\xi$ in Eq.~\eqref{eqn:eqbeta}.
\section{Energy-momentum tensor and helicity}
\label{Energy-momentum tensor}
From the Lagrangian density in Eq.~\eqref{eqn:pseudoscalar}, it is easy to derive the
metric energy-momentum tensor (EMT) for the gauge-fields:
\begin{equation}
T_{\mu\nu}^{(F)} = F_{\rho\mu}F^\rho_{\,\,\,\,\nu}+g_{\mu\nu}\frac{\mathbf{E}^2-\mathbf{B}^2}{2}
\end{equation}
with $\mathbf{E}$ and $\mathbf{B}$ being the associated electric and magnetic field.
We then obtain for the energy density and pressure the following expressions:
\begin{align}
\label{T00}
&T^{(F)}_{00} = \frac{\mathbf{E}^2+\mathbf{B}^2}{2} \,, \\
\label{Tij}
&T^{(F)}_{ij} = -E_iE_j-B_iB_j+\delta_{ij}\frac{\mathbf{E}^2+\mathbf{B}^2}{2} \,.
\end{align}
Note that there is no terms in the EMT depending on the pseudo-scalar coupling. Their absence in the $T^{(F)}_{00}$ component follows from impossibility to construct a pseudo-scalar invariant under spatial rotations from $\mathbf{E}$ and $\mathbf{B}$. Then the fact that the EMT trace remains zero in the presence of the interaction of Eq.~\eqref{eqn:coupling}, due to the conformal invariance of the gauge field, leads to the absence of such terms in $T^{(F)}_{ij}$, too.
By using Eqs.~\eqref{T00} and \eqref{Tij}, the Friedmann equations take the form:
\begin{align}
\label{eqn:Friedmann}
&H^2 = \frac{1}{3 M_{pl}^2}\left[ \frac{\dot{\phi}^2}{2}+V(\phi)+\frac{\langle \mathbf{E}^2 + \mathbf{B}^2 \rangle}{2}
\right] \,, \\
\label{eqn:Einstein}
&\dot{H} = -\frac{1}{2M_{pl}^2} \left[\dot{\phi}^2+\frac{2}{3}\langle \mathbf{E}^2+\mathbf{B}^2 \rangle \right]\,.
\end{align}
Using the relations $\Vec{E}=-\Vec{A}'/a^2$ and $\Vec{B}=\Vec{\nabla}\times\Vec{A}'/a^2$,
the averaged energy density is:
\begin{align}
\label{eqn:integral2}
\frac{\langle {\bf E}^2 + {\bf B}^2 \rangle}{2}
& = \int\,\frac{\mathrm{d} k}{(2\pi)^2\,a^4} I(k) \notag \\
& = \int\,\frac{\mathrm{d} k}{(2\pi)^2\,a^4} \,k^2\Bigl[|A_+'|^2 + |A_-'|^2 \notag\\
& \quad \quad + k^2 (|A_+|^2 + |A_-|^2 ) \Bigr] \,.
\end{align}
The electric (magnetic) contribution is given by the first and second (third and fourth) term in
the integrand. It is easy to see that this integral diverges for large momentum $k$. This is a common
behavior for averaged quantities in quantum field theory (QFT henceforth) in curved background, or in external fields, and a renormalization procedure is needed
to remove these ultraviolet (UV) divergences. In Section~\ref{appendixADI} we will use the adiabatic
regularization method \cite{Zeldovich:1971mw,Parker:1974qw} for this purpose, and we will present counterterms needed to renormalize the bare constants in the
Lagrangian in Section~\ref{counterterms}. In the present Section, we identify the UV divergent contributions in the integrands.
Expanding the integrand of Eq.~\eqref{eqn:integral2} for $-k \tau \gg 1$ we obtain quartic, quadratic
and logarithmic UV-divergences:
\begin{equation} \label{eqn:Idiv}
I_{\textup{div}}(k) \sim 2 k^3+\frac{\xi ^2 k}{\tau ^2}+\frac{3 \xi ^2 \left(-1+5 \xi ^2\right)}{4 \tau ^4 k}
+{\cal O} \left(\frac{1}{k}\right)^{3} \,.
\end{equation}
It is interesting to note that the logarithmic divergence changes its sign when $\xi$ crosses
$|\xi| = 1/\sqrt{5}$. On the other hand, expanding in the infrared (IR) limit ($-k \tau \ll 1$)
the integrand of Eq.~\eqref{eqn:integral2} has no IR divergences.
The equation of motion for the inflation $\phi$ is affected by the backreaction of these gauge fields:
\begin{equation}
\ddot \phi + 3 H \dot \phi + V_\phi = g \langle {\bf E} \cdot {\bf B} \rangle \,,
\label{KG}
\end{equation}
where the helicity integral is given by:
\begin{align}
\label{eqn:integral}
\langle {\bf E} \cdot {\bf B} \rangle
= & - \int\,\frac{\mathrm{d} k}{(2\pi )^2\,a^4} J(k) \notag\\
= & - \int\,\frac{\mathrm{d} k\,k^3}{(2\pi )^2\,a^4}
\frac{\partial}{\partial \tau} \Bigl( |A_+|^2 - |A_-|^2 \Bigr) \,.
\end{align}
The integrand in Eq.~\eqref{eqn:integral} has a different divergent behavior compared to the energy
density, since it has only quadratic and logarithmic divergences:
\begin{equation} \label{eqn:Jdiv}
J_{\textup{div}}(k) \sim \frac{\xi k}{\tau^2}-\frac{3\xi(1-5\xi^2)}{2\tau^4 k} + {\cal O} \left(\frac{1}{k}\right)^{5/2} \,.
\end{equation}
Also in this case the integrand in Eq.~\eqref{eqn:integral} does not have any IR pathology.
We point out that, even if we called it 'helicity integral', the above integral in Eq.~\eqref{eqn:integral}
is actually the \emph{derivative} of what is usually called the helicity integral
${\cal H} = \langle {\bf A} \cdot {\bf B} \rangle$ (which is also gauge-invariant for a coupling to a pseudo-scalar).
\section{Analytical calculation of divergent and finite terms}
\label{Analytical calculation}
In order to find an analytical expression for the finite part, we note that the bare integrals in
Eqs.~\eqref{eqn:integral2} and \eqref{eqn:integral} can be solved by using the expression of the mode
functions $A_\pm$ given in Eq.~\eqref{eqn:aplusaminus}.
We identify the divergences by imposing a UV physical cutoff $\Lambda$ in order to avoid time-dependent
coupling constants at low energies: we therefore impose a comoving $k$-cutoff $k_{\textup{UV}}= \Lambda\, a$
\cite{Zeldovich:1971mw,Xue:2011hm} in the integrals in Eqs.~\eqref{eqn:integral2} and \eqref{eqn:integral}.
Note that Eq.~\eqref{eqn:eqbeta} can be solved analytically by Whittaker functions also in presence of
a mass term \cite{Barnaby:2011qe}. We restrict in this section to the results for the massless case,
but we also give the generalization to non-zero mass in Appendix~\ref{appendixmass}.
\subsection{Energy Density}
\label{EnergyAnalytic}
We first compute the energy density (see Appendix~\ref{Appendix5} for details). With the help of the
integral representation of the Whittaker functions and carefully choosing the integration contour we obtain
for the energy density stored in the electric field:
\begin{align}
\label{electric}
&\frac{\langle {\bf E}^2\rangle}{2} =\frac{\Lambda^4}{16\pi^2}-\frac{\xi^2 H^2}{16\pi^2}\Lambda^2
\notag\\
&+\frac{ \xi ^2 \left(19-5 \xi ^2\right) H^4\log (2 \Lambda/H )}{32\pi^2}\notag\\
&+\frac{9 \gamma H^4 \xi ^2 \left(7-5 \xi ^2\right)}{32 \pi ^2 }\notag\\
&+\frac{H^4 \left(148 \xi ^6-163 \xi ^4-221 \xi ^2+18\right)}{96 \pi ^2 \left(\xi ^2+1\right)}\notag\\
&+\frac{H^4 \xi \left(30 \xi ^2-11\right) \sinh (2 \pi \xi )}{128 \pi ^3 }
\notag\\
&-\frac{i H^4 \xi ^2 \left(5 \xi ^2-19\right) \sinh (2 \pi \xi ) \psi ^{(1)}(1-i \xi )}{128 \pi ^3 }\notag\\
&+\frac{i H^4 \xi ^2 \left(5 \xi ^2-19\right) \sinh (2 \pi \xi ) \psi ^{(1)}(i \xi +1)}{128 \pi ^3 }\notag\\
&+\frac{H^4 \xi ^2 \left(5 \xi ^2-19\right) \left(H_{-i \xi -2}+H_{i \xi -2}\right)}{64 \pi ^2 }
\end{align}
while for the magnetic field we find
\begin{align}
\label{magnetic}
&\frac{\langle {\bf B}^2\rangle}{2} =\frac{\Lambda^4}{16\pi^2}+\frac{3\xi^2 H^2}{16\pi^2}\Lambda^2
\notag\\
&+\frac{5 \xi ^2 \left(7 \xi ^2-5\right) H^4\log (2 \Lambda/H )}{16\pi^2}\notag\\
&-\frac{H^4 \left(533 \xi ^4-715 \xi ^2+36\right)}{384 \pi ^2 }\notag\\
&+\frac{H^4 \left(210 \xi ^4-185 \xi ^2+18\right) \sinh (2 \pi \xi )}{384 \pi ^3 \xi }\notag\\
&+\frac{5 i H^4 \xi ^2 \left(7 \xi ^2-5\right) \sinh (2 \pi \xi ) \psi ^{(1)}(1-i \xi )}{128 \pi ^3 }
\notag\\
&-\frac{5 i H^4 \xi ^2 \left(7 \xi ^2-5\right) \sinh (2 \pi \xi ) \psi ^{(1)}(i \xi +1)}{128 \pi ^3 }
\notag\\
&-\frac{5 H^4 \xi ^2 \left(7 \xi ^2-5\right) \left[\psi (-i \xi )+\psi (i \xi )\right]}{64 \pi ^2 } \,.
\end{align}
Summing the two contributions, the total energy density becomes:
\begin{align}
\label{eqn:energy}
&\frac{\langle {\bf E}^2 + {\bf B}^2\rangle}{2} =\frac{\Lambda^4}{8\pi^2}+\frac{\xi^2 H^2}{8\pi^2}\Lambda^2
\notag\\
&+\frac{3 \xi ^2 \left(5 \xi ^2-1\right) H^4\log (2 \Lambda/H )}{16\pi^2}\notag\\
&+\frac{\gamma \left(11-10 \xi ^2\right) \xi ^2H^4}{8\pi^2 }\notag\\
&+\frac{\xi \left(30 \xi ^2-11\right) \sinh (2 \pi \xi )H^4}{64 \pi^3}\notag\\
&+\frac{\xi ^2 \left(7 \xi ^6-282 \xi ^4+123 \xi ^2+124\right)H^4}{256 \pi^2 \left(\xi ^2+1\right) }
\notag\\
&-\frac{3 \imath \xi ^2 \left(5 \xi ^2-1\right) \sinh (2 \pi \xi ) \psi ^{(1)}(1-\imath \xi )H^4}{64 \pi^2 }\notag\\
&+\frac{3 \imath \xi ^2 \left(5 \xi ^2-1\right) \sinh (2 \pi \xi ) \psi ^{(1)}(1+\imath \xi )H^4}{64 \pi^2 }\notag\\
&+\frac{3 \xi ^2 \left(5 \xi ^2-1\right) [\psi (-\imath \xi -1)+\psi (\imath \xi -1)]H^4}{32\pi^2 },
\end{align}
where $\psi$ is the Digamma function, $\psi^{(1)}(x)\equiv \mathrm{d} \psi(x)/ \mathrm{d} x$,
$H_{x}\equiv\psi(x+1)+\gamma$ is the harmonic number of order $x$, and $\gamma$ is the
Euler-Mascheroni constant.
The finite terms in Eq.~\eqref{eqn:energy} have the corresponding asymptotic behavior:
\begin{align}
\label{energy_asymptotic}
& \frac{H^4}{64 \pi^2 }(21+76 \gamma)\xi^2 &\mathrm{when}\,\,\,\, &|\xi| \ll 1 \,, \\
\label{energy_asymptotic2}
& \frac{9 H^4 \sinh(2 \pi\xi)}{1120 \pi^3 \xi^3 } &\mathrm{when}\,\,\,\, &|\xi| \gg 1\,.
\end{align}
We now compare our results with those used in the literature which are based on the use of UV and IR cutoffs
and an approximation of the integrand. More precisely Refs.~\cite{Anber:2006xt,Anber:2009ua} and subsequent
works use the following approximation to estimate the integral:
\begin{itemize}
\item The integral has a physical UV cutoff at $-k\tau=2|\xi|$.
\item Only the growing mode function $A_+$ in Eq.~\eqref{eqn:aplusaminus} is considered for $\xi > 0$ (the situation is reversed for $\xi < 0$) and it is approximated in this regime to:
\begin{equation}
A_+(\tau,k)\simeq\frac{1}{\sqrt{2 k}}\left(\frac{-k\tau}{2\xi}\right)^{1/4}e^{\pi\xi-\sqrt{-8\xi k \tau}} \,.
\end{equation}
\end{itemize}
Under this approximation Ref.~\cite{Anber:2006xt} obtains when $\xi\gg 1$:
\begin{align}
\frac{\langle {\bf E}^2 + {\bf B}^2 \rangle_\textup{AS}}{2}&
\simeq 1.4 \cdot 10^{-4}
\frac{H^4}{\xi^3} e^{2 \pi \xi} \,,
\label{E2B2_literature}
\end{align}
Our result in Eq.~\eqref{eqn:energy} is one of the main original results of this paper and is valid
for any $\xi$.
In Fig.~\ref{fig1}, we plot the terms of Eq.~\eqref{eqn:energy} which do not depend on the UV
cut-off:
\begin{align}
\label{IFIN}
&\mathcal{I}_{\textup{fin}}(\xi)\equiv \frac{\langle {\bf E}^2 + {\bf B}^2\rangle}{2}
-\Bigg[\frac{\Lambda^4}{8\pi^2}+\frac{\xi^2 H^2}{8\pi^2}\Lambda^2 \notag\\
&+\frac{3 \xi ^2 \left(5 \xi ^2-1\right) H^4\log (2 \Lambda/H )}{16\pi^2}\Bigg] \,,
\end{align}
in units of $(2\pi)^2/ H^4$ (for $\xi > 0$ for simplicity). These terms would correspond to the renomalized energy
density obtained by a minimal subtraction scheme.
In Fig.~\ref{fig:ElMag}, we plot the electric and magnetic finite contributions to the energy density (by restricting to $\xi>0$, again).
The electric contribution to the energy density is larger than the magnetic one for $\xi\gtrsim 0.75$,
whereas they are comparable for $\xi\lesssim 0.75$.
In Fig.~\ref{fig2}, we plot the relative difference with Eq.~\eqref{E2B2_literature}
\begin{equation}
\label{DIFIN}
\Delta\mathcal{I}_{\textup{fin}}(\xi)
\equiv\frac{2\mathcal{I}_{\textup{fin}}(\xi)-\langle {\bf E}^2 + {\bf B}^2 \rangle_\textup{AS}}{\langle {\bf E}^2 + {\bf B}^2
\rangle_\textup{AS}}\,.
\end{equation}
It can be seen from Eq.~\eqref{energy_asymptotic2} and Fig.~\ref{fig2} that the behavior in the regime
$|\xi|\gg1$ of our solution is similar to that of Eq.~\eqref{E2B2_literature}, which has thoroughly been studied in the literature.
Nevertheless, Fig.~\ref{fig2} shows that there is a relative difference of approximately the order of
10\% in the numerical coefficient that multiplies $\exp(2 \pi\xi)/\xi^3$, which can be ascribed to the
assumptions described above.
The main difference of our new result in Eq.~\eqref{eqn:energy} with respect to Eq.~\eqref{E2B2_literature}
is in the regime of $|\xi| \lesssim 10$, that has been studied for the first time in this paper.
Eq.~\eqref{E2B2_literature} cannot be extrapolated to $|\xi|\ll1$, whereas our result shows that the
finite part of the energy density is $\mathcal{O}(\xi^2)$ as shown in Eq.~\eqref{energy_asymptotic}.
This difference can be understood by noting that the contributions from $A_+$ and $A_-$ become
comparable in this regime of $\xi$ and neglecting $A_-$ is no longer a good approximation.
We end on noting that the finite contribution by a minimal subtraction scheme to the energy density,
which is $\mathcal{O}(\xi^2)$ for $\xi \ll 1$, becomes negative for $0.8 \lesssim \xi \lesssim 1.5$,
although its classical counterpart of Eq.~\eqref{eqn:energy} is positive definite.
This is not totally surprising since it is known that in QFT in curved space-times
the renormalized terms of expectation values of classically defined positive terms can be negative
\cite{Birrell:1982ix}.
\begin{figure}[t]
\centering
\includegraphics[width=\columnwidth]{pseudo_orig.pdf}
\caption{We plot respectively the quantities $\mathcal{I}_{fin}(\xi)$ (blue line)
and $\mathcal{J}_{fin}(\xi)$ (orange line) defined in Eqs.~\eqref{IFIN} and \eqref{JFIN}.
\label{fig1}}
\end{figure}
\begin{figure}[t]
\centering
\includegraphics[width=\columnwidth]{ElMag.pdf}
\caption{We plot in blue (orange) the electric (magnetic) contribution
to $\mathcal{I}_{fin}(\xi)$.
\label{fig:ElMag}}
\end{figure}
\begin{figure}[t]
\centering
\includegraphics[width=\columnwidth]{pseudo2.pdf}
\caption{We plot respectively the quantities $\Delta\mathcal{I}_{fin}(\xi)$ (blue line)
and $\Delta\mathcal{J}_{fin}(\xi)$ (orange line) defined in Eqs.~\eqref{DIFIN} and \eqref{DJFIN}.
\label{fig2}}
\end{figure}
\subsection{Helicity Integral}
The helicity integral in Eq.~\eqref{eqn:integral} is only logaritmically and quadratically divergent
because of the cancellation of the quartic divergence and does not exhibit any IR divergence.
Note that only by considering both $A_+$ and $A_-$, quartic divergent terms in the UV regime cancel.
It is possible to derive an exact solution of Eq.~\eqref{eqn:integral} with a UV cutoff and we give
the final result in the following, leaving the details for the interested reader in Appendix~\ref{Appendix5}.
The result for the helicity is:
\begin{align}
\label{edotb}
&-\langle {\bf E} \cdot {\bf B} \rangle =\,\frac{ \xi H^2}{8\pi^2 }\Lambda ^2+\frac{3 \xi \left(5 \xi ^2-1\right)H^4 \log (2 \Lambda/H )}{8 \pi^2}\notag\\
&+\frac{[6 \gamma \xi \left(5 \xi ^2-1\right)+(22 \xi -47 \xi ^3)]H^4}{16\pi^2}\notag\\
&+\frac{\left(30 \xi ^2-11\right) \sinh (2 \pi \xi )H^4}{32 \pi^3 }\notag\\
&-\frac{3 \xi \left(5 \xi ^2-1\right) \left(H_{-\imath \xi }+H_{\imath \xi }\right)H^4}{16\pi^2 }\notag\\
&+\imath\frac{3 \xi \left(5 \xi ^2-1\right) \sinh (2 \pi \xi ) \psi ^{(1)}(1-\imath \xi )H^4}{32 \pi^3 }\notag\\
&-\imath\frac{3 \xi \left(5 \xi ^2-1\right) \sinh (2 \pi \xi ) \psi ^{(1)}(\imath \xi +1)H^4}{32 \pi^3 } \,.
\end{align}
The finite terms have the corresponding asymptotic values:
\begin{align}
\label{back_reaction_asymptotic}
& \frac{H^4}{32\pi^2}(11-6 \gamma)\xi &\mathrm{when}\,\,\,\, &\xi \ll 1 \,, \\
\label{back_reaction_asymptotic2}
& \frac{9 \sinh(2 \pi\xi)H^4}{560 \pi^3 \xi^4} &\mathrm{when}\,\,\,\, &\xi\gg 1 \,.
\end{align}
The result reported in the literature for the integral in Eq.~\eqref{eqn:integral} is derived under the same assumptions discuss in the context of Eq.~\eqref{E2B2_literature} and is given by
\cite{Anber:2006xt}:
\begin{align}
-\langle {\bf E} \cdot {\bf B} \rangle_\textup{AS}=
\frac{H^4}{\xi^4} e^{2 \pi \xi} \,.
\label{EB_literature}
\end{align}
Again, we define:
\begin{align}
\label{JFIN}
&\mathcal{J}_{\textup{fin}}(\xi)\equiv -\langle {\bf E} \cdot {\bf B} \rangle\notag\\
&-\left[\frac{ \xi H^2}{8\pi^2 }\Lambda ^2+\frac{3 \xi \left(5 \xi ^2-1\right)H^4 \log (2 \Lambda/H )}{8 \pi^2}\right] \,.
\end{align}
and the relative differences between our solution and Eq.~\eqref{EB_literature}
\begin{equation}
\label{DJFIN}
\Delta\mathcal{J}_{\textup{fin}}(\xi)\equiv-\frac{\mathcal{J}_{\textup{fin}}(\xi)+\langle {\bf E} \cdot {\bf B}\rangle_\textup{AS}}{\langle {\bf E} \cdot {\bf B}\rangle_\textup{AS}}\,,
\end{equation}
which we plot in Fig.~\ref{fig1} and Fig.~\ref{fig2} respectively.
For $\xi \ll 1$ the back-reaction in Eq.~\eqref{back_reaction_asymptotic} is reminiscent of an extra dissipative term of the type $\Gamma_{\textup{dS}} \dot \phi$. It is interesting to note that the effective $\Gamma_{\textup{dS}}$ in this nearly de Sitter evolution is larger than the perturbative decay rate $\Gamma = g^2 m_\phi^3/(64 \pi)$ by a factor ${\cal O} (H^3/m_\phi^3)$.
Analogously to the energy density case, Fig.~\ref{fig2} shows that in the regime of $\xi\gtrsim 10$
Eqs~.\eqref{back_reaction_asymptotic2} and \eqref{EB_literature} have a similar functional form, but
still a $10\%$ difference.
Our exact result can be more precisely used for $\xi\lesssim10$ and in particular in the $\xi\lesssim1$
regime. Note that the difference between the exact result and the result given in \cite{Anber:2006xt} is
now larger than in the energy density case and that we have a linear dependence on $\xi$ for the helicity
integral for $\xi\ll 1$, in a regime to which the standard result in the literature in
Eq.~\eqref{EB_literature} cannot be extrapolated.
\section{Adiabatic Expansion and Regularization}
\label{appendixADI}
The adiabatic regularization method \cite{Zeldovich:1971mw,Parker:1974qw} relies on the
adiabatic, or Wentzel-Kramer-Brillouin (WKB henceforth), expansion of the mode functions $A_\pm$ solution of Eq.~\eqref{eqn:EoM1}.
Following the standard adiabatic regularization procedure, we proceed by adding a mass term regulator $\mu$ to the evolution equations of the two different helicity states obtaining a modified version of Eq.~\eqref{eqn:EoM1}
\begin{equation}
\label{eqn:adievoeq}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\frac{\mathrm{d}^2}{\mathrm{d}\,\tau^2}A_\pm^{\rm WKB}(k,\tau)+\Bigl(k^2\,\mp g k \phi'+\frac{\mu^2}{H^2 \tau^2}\Bigr)
A_\pm^{\rm WKB}(k,\tau)=0 \,,
\end{equation}
where the adiabatic mode function solution $A_\lambda^{\rm WKB}$ is defined as
\begin{equation}
\label{WKBmode}
A_\lambda^{\rm WKB}(k,\tau)=\frac{1}{\sqrt{2 \Omega_{\lambda}(k \,, \tau)}}
e^{\imath\int^{\tau}\mathrm{d}\tau'\, \Omega_{\lambda}(k, \tau')},
\end{equation}
with $\lambda=\pm$. The mass term regulator $\mu$ is inserted to avoid additional IR divergences which are introduced by the adiabatic expansion for massless fields.
Inserting the adiabatic solution \eqref{WKBmode}
in Eq.~\eqref{eqn:adievoeq},
we then obtain the following exact equation for
the WKB frequencies $\Omega_{\lambda}$
\begin{equation}
\label{WKB}
\Omega^{2}_{\lambda}(k, \tau) =
\bar{\Omega}_{\lambda}^2(k, \tau)
+ \frac{3}{4}\left(\frac{{\Omega}_{\lambda}'(k, \tau)}{{\Omega}_{\lambda}(k, \tau)}\right)^2 - \frac{{\Omega}_{\lambda}''(k,\tau)}{2{\Omega}_{\lambda}
(k,\tau)},
\end{equation}
with
\begin{equation}
\bar{\Omega}_{\lambda}^2(k, \tau)= \omega^2(k, \tau) + \lambda k g \phi'(\tau)
\end{equation}
and
\begin{equation}
\omega^2 (k,\tau)=k^2+\mu^2a^2(\tau) \,.
\end{equation}
The usual procedure is then to solve Eq.~\eqref{WKB} iteratively, introducing an adiabatic parameter
$\epsilon$ assigning a power of $\epsilon$ to each of the derivative with respect to $\tau$.
To arrive to order $2n$, we have then to do $n$ iterations.
Finally, we have to further Taylor-expand $\bar{\Omega}_{\lambda}$ in power of $\epsilon$ around $\epsilon=0$
discarding all the resulting terms of adiabatic order larger than $2 n$ in the final result.
One can then mode expand $A_\lambda({\bf x},\tau )$ using the $n$-th adiabatic order mode functions
$A_\lambda^{(n)}(k,\tau)$ and the associated adiabatic creation and annihilation operators, and then define the
$n$-th order adiabatic vacuum as $a_{\lambda,k}^{(n)} |0^{(n)}\rangle=0$ when $\tau\rightarrow-\infty$.
In particular, in our case we have that $\omega'/\omega\rightarrow0$ for $\tau\rightarrow-\infty$.
Thus, in this limit the adiabatic vacuum defined at any adiabatic order becomes essentially the adiabatic vacuum of infinite order which we call $|0\rangle_A$.
The adiabatic regularization is a procedure to remove the UV divergences and consists in subtracting from an expectation value its adiabatic counterpart. In practice, we will proceed by introducing an UV physical cutoff $\Lambda$ for the mode integral, performing the subtraction, and only after that we will send the cutoff to infinity. Namely, we have
\begin{equation}
\label{reg}
\langle {\cal O} \rangle_{\textup{reg}}=\lim_{\Lambda \rightarrow \infty} \left[\langle {\cal O} \rangle_{\Lambda}-\langle {\cal O} \rangle_{\textup{A},\Lambda}\right] \,,
\end{equation}
where by ${\cal O}$ we denote a quadratic operator in the quantum fields, such as the energy density or the helicity. By $\langle {\cal O} \rangle_{\Lambda}$ we then mean the bare expectation value of these operators evaluated with a UV cutoff $\Lambda$, while
by $\langle {\cal O} \rangle_{\textup{A},\Lambda}$ we mean the expectation value of their adiabatic counterpart evaluated using the same UV cutoff $\Lambda$.
Considering the energy density and the helicity, the bare expectation values are those computed in the previous section for $\mu=0$ (see Eqs.~\eqref{eqn:integral2} and \eqref{eqn:integral}) or their extensions to $\mu \ne 0$ given in Appendix~\ref{appendixmass}.
Their adiabatic counterpart
is instead given by their corresponding integrals expressed in terms of the WKB mode functions of a given adiabatic order $n$.
Namely, these are given by Eqs.~\eqref{eqn:integral2} and \eqref{eqn:integral} where we take the adiabatic solution in Eq.~\eqref{WKBmode} for the mode function using the solution of adiabatic order $n$ and expanding again up to order $n$. In the case under consideration, the fourth order adiabatic expansion is needed in order to remove the UV divergences from the bare integral.
We then obtain:
\begin{widetext}
\begin{align}
\frac{\langle {\bf E}^2 + {\bf B}^2\rangle_{\textup{A},\Lambda}}{2} =&
\int_{c}^{\Lambda a} \frac{dk\,k^2}{(2\pi)^2}\Bigg[
\label{eqn:intadienergy}
\left(\frac{1}{2 \Omega_+}+\frac{1}{2 \Omega_-}\right) \omega^2(k,\tau)
+\frac{\Omega_-}{2}
+\frac{\Omega_+}{2}+\frac{\epsilon^2 \Omega_-''}{8 \Omega_-^3}+\frac{\epsilon ^2 \Omega_+''}{8
\Omega_+^3}\Bigg]\notag\\
=&\int_{c}^{\Lambda a}\frac{dk}{(2\pi)^2a^4}\Bigg[2 \omega+\frac{\epsilon^2 k^4 \xi ^2}{\tau ^2 \omega^5}+\frac{\epsilon^2 k^2 \mu^2 \xi ^2}{H^2 \tau ^4 \omega^5}+\frac{\epsilon^4 15 k^8 \xi ^4}{4 \tau ^4 \omega^{11}}
-\frac{\epsilon^4 3 k^8 \xi ^2}{4 \tau ^4 \omega^{11}} \notag\\
&+\frac{\epsilon^4 3 \mu^8}{64 H^8 \tau ^{12} \omega^{11}}+\frac{\epsilon^4 3 k^2 \mu^6}{4 H^6 \tau ^{10} \omega^{11}}
-\frac{\epsilon^4 15 k^4 \mu^4}{16 H^4 \tau ^8 \omega^{11}}
+\frac{\epsilon^4 19 k^2 \mu^6 \xi ^2}{8 H^6 \tau ^{10} \omega^{11}}+\frac{\epsilon^4 \mu^4}{4 H^4 \tau ^6 \omega^5}\notag\\
&+\frac{\epsilon^4 67 k^4 \mu^4 \xi ^2}{8 H^4 \tau ^8
\omega^{11}}+\frac{\epsilon^4 15 k^6 \mu^2 \xi ^4}{2 H^2 \tau ^6 \omega^{11}}
+\frac{\epsilon^4 21 k^6 \mu^2 \xi ^2}{4 H^2 \tau ^6 \omega^{11}}+\frac{\epsilon^4 15 k^4 \mu^4 \xi ^4}{4 H^4 \tau ^8
\omega^{11}}
\Bigg],
\end{align}
\begin{align}
\label{eqn:intadieb}
\langle{\bf E}\cdot{\bf B}\rangle_{A,\Lambda} = &-\int_{c}^{\Lambda a}\frac{dk\,k^2}{(2\pi)^2 a^4}\Bigg[\frac{\epsilon \Omega_+'}{2 \Omega_+^2}-\frac{\epsilon \Omega_-'}{2 \Omega_-^2}\Bigg]
=\int_{c}^{\Lambda a} \frac{dk}{(2\pi)^2a^4}\Biggl[-\frac{\epsilon^2 k^3 \xi }{\tau ^2 \omega^5}+\frac{\epsilon^2 2 k \mu ^2 \xi }{H^2 \tau ^4 \omega^5}+\frac{\epsilon^4 121 k^3 \mu ^4 \xi }{8 H^4 \tau ^8 \omega^{11}}\notag\\
&-\frac{\epsilon^4 15
k^7 \xi ^3}{2 \tau ^4 \omega^{11}}+\frac{\epsilon^4 3 k^7 \xi }{2
\tau ^4 \omega^{11}}+\frac{\epsilon^4 5 k^5 \mu ^2 \xi ^3}{2 H^2 \tau ^6 \omega^{11}}-\frac{\epsilon^423 k^5 \mu ^2 \xi }{H^2 \tau ^6 \omega^{11}}
+\frac{\epsilon^4k \mu ^6 \xi }{4 H^6 \tau ^{10} \omega^{11}}+\frac{\epsilon^4 10 k^3 \mu ^4 \xi ^3}{H^4 \tau ^8 \omega^{11}}
\Biggr] \,.
\end{align}
\end{widetext}
In the above equations the IR $k$-cutoff $c$ is also considered as an alternative to the mass term regulator to cure the IR divergences which appear when considering the fourth order adiabatic expansion of a massless field.
As said, the fourth order adiabatic expansion of the mode functions is sufficient to generate the same UV
divergences of the bare integrals in Eqs.~\eqref{eqn:Idiv} and \eqref{eqn:Jdiv}. However, the fourth order expansion also generates logarithmic IR divergences since gauge fields are massless.
One way to avoid the IR divergence is to use the mass term in Eq.~\eqref{eqn:adievoeq}
as an IR regulator\footnote{See Ref.~\cite{Ferreiro:2018oxx} for an interpretation of the mass regulator $\mu$ in terms of running the coupling constant.}. In this way we get for the energy density:
\begin{align}
\label{energyadifin}
\frac{\langle {\bf E}^2 + {\bf B}^2\rangle^\mu_{\textup{A},\Lambda}}{2}=&\frac{\Lambda^4}{8 \pi^2}+\frac{\Lambda^2 \xi^2 H^2}{8 \pi^2}\notag\\&+\frac{3 H^4 \xi ^2 \left(5 \xi ^2-1\right)\log(2 \Lambda/H) }{16 \pi^2 }\notag\\
&-\frac{H^4}{480\pi ^2}-\frac{H^4 \xi ^2 \left(23 \xi ^2-9\right)}{16 \pi ^2}\notag\\
&-\frac{3 H^4 \xi ^2 \left(5 \xi ^2-1\right) \log \left(\frac{\mu }{H}\right)}{16 \pi ^2} \,.
\end{align}
Analogously, for the helicity term we get:
\begin{align}
\label{helicityadifin}
\langle{\bf E}\cdot{\bf B}\rangle^\mu_{A,\Lambda}=&\frac{ \Lambda ^2 \xi H^2}{8 \pi ^2} + \frac{3 H^4 \xi \left(5 \xi ^2-1\right) \log \left(\frac{2 \Lambda }{H}\right)}{8 \pi ^2}\notag\\
&-\frac{3 H^4 \xi \left(5 \xi ^2-1\right) \log \left(\frac{\mu }{H}\right)}{8 \pi ^2}\notag\\&+\frac{H^4 \left(19 \xi -56 \xi
^3\right)}{16 \pi ^2}.
\end{align}
Note that our WKB ansatz correctly reproduce the the UV divergences of the energy density and helicity terms. As already known, the fourth order adiabatic expansions leads also to finite terms, including a term with a logarithmic dependence on the effective mass regulator.
Let us also comment on the term independent on $\xi$, i.e. $H^4/(480 \pi^2)$ in Eq.~\eqref{energyadifin}.
This term is generated by the fourth order adiabatic subtraction and is connected to the conformal anomaly.
The term we find corresponds to twice the result for a massless conformally coupled scalar field, i.e.
$H^4/(480 \pi^2)=2\times H^4/(960 \pi^2)$ \cite{Bunch:1978gb,Birrell:1982ix}, as expected since the
two physical states $A_\pm$ behave like two conformally coupled massless scalar fields for $\xi=0$
\footnote{It was recently shown in Ref.~\cite{Chu:2016kwv} that in case of photons the standard result
$\langle T \rangle /4 =-31 H^4/(480 \pi^2)$ can be obtained by adiabatic regularization only by including
Faddeev-Popov fields. On the other hand, it is possible to get the same result without consideration of the Faddeev-Popov ghosts by, first, calculating the photon vacuum polarization in the closed static FLRW (Einstein) universe in which all geometric terms in the trace conformal anomaly become zero and a non-zero average photon energy density arises due to the Casimir effect~\cite{St76}, and then using the known form of the conserved vacuum polarization tensor in a conformally flat space-time that produces the $R_{\mu\nu}R^{\mu\nu}-\frac{1}{3}R^2$ term in the trace anomaly. Note that the same procedure yields the correct answer for spins $s=0,\frac{1}{2}$, too.}.
An alternative procedure to avoid IR divergences in the adiabatic subtraction is to introduce a time independent IR cutoff $k=c$
in the adiabatic integrals.
In this way we obtain for the energy density and helicity term, respectively:
\begin{align}
\label{energyadifin_cutoff}
\frac{\langle {\bf E}^2 + {\bf B}^2\rangle^c_{\textup{A},\Lambda}}{2}=&\frac{\Lambda^4}{8 \pi^2}+\frac{\Lambda^2 \xi^2 H^2}{8 \pi^2}\notag\\&
+\frac{3 H^4 \xi ^2 \left(5 \xi ^2-1\right)\log(2 \Lambda/H) }{16 \pi^2 }\notag\\&-\frac{3 H^4 \xi ^2 \left(5 \xi ^2-1\right)\log\left(2 c/(a H)\right)}{16 a H\pi^2 }\notag\\&-\frac{c^4}{8 \pi^2 a^4}-\frac{c^2 \xi^2 H^2}{8 a^2 \pi^2}\,,
\end{align}
\begin{align}
\label{helicityadifin_cutoff}
\langle{\bf E}\cdot{\bf B}\rangle^c_{A,\Lambda}=&\frac{ \Lambda ^2 \xi H^2}{8 \pi ^2} + \frac{3 H^4 \xi \left(5 \xi ^2-1\right) \log \left(\frac{2 \Lambda }{H}\right)}{8 \pi ^2}\notag\\
&-\frac{ c^2 \xi H^2}{8 a^2 \pi^2} - \frac{3 H^4 \xi \left(5 \xi ^2-1\right) \log \left(\frac{2 c}{a H}\right)}{8 \pi ^2}.
\end{align}
As for the case with the effective mass regulator,
the UV divergences of the energy density and helicity terms are also correctly reproduced, although
the finite terms are different. Let us note that the term connected to the conformal anomaly is
absent, as it comes from the $k=0$ pole structure of the WKB energy density integrand
\cite{Birrell:1982ix,Parker:2009uva}. By comparing Eqs.~\eqref{energyadifin_cutoff}-\eqref{helicityadifin_cutoff} with
\eqref{energyadifin}-\eqref{helicityadifin}, the terms which do not depend on $\Lambda$ obtained with the IR cutoff can have the same time dependence of the effective mass term by instead considering a comoving IR cutoff $c=\Lambda_{\rm IR} a$. Further quantitive consistency from the two approaches can be obtained by matching
$\mu$ to a physically motivated value for $\Lambda_{\rm IR}$ correspondent to the scale at which the WKB approximation breaks down (see \cite{Durrer:2009ii,Marozzi:2011da}).
We have seen that the adiabatic regularization applied to the two physical helicity states, because of
the massless nature of the gauge field, introduces logarithmic IR divergences. This effect
happens in many other contexts, see e.g. Ref.~\cite{Seery:2010kh} for a review
of IR effects in de Sitter space-time.
Logarithmic IR divergences in averaged quantities also appear in the context
of the Schwinger effect in de Sitter where they lead to the so called
IR Hyper-conductivity effect \cite{Kobayashi:2014zza}. Furthermore,
it has been shown that such logarithmic IR divergences also
appear when using other renormalization methods such as point splitting
renormalization and thus seem to be generic and not specific of the adiabatic regularization method \cite{Hayashinaka:2016dnt}.
\subsection{Counterterms}
\label{counterterms}
We have introduced a fourth order adiabatic expansion which correctly reproduced the UV divergences of the bare quantities. These divergent terms are associated to non-renormalizable derivative interaction counter-terms of the pseudo-scalar field:
\begin{align}
\Delta\mathcal{L} = &- \frac{\alpha }{4}\nabla^\mu \nabla^\nu
\phi\nabla_\mu \nabla_\nu
\phi\notag\\
\label{eqn:Clagrangian}& - \frac{\beta }{4}\nabla^\mu \phi \nabla^\nu \phi \nabla_\mu \phi \nabla_\nu\phi,
\end{align}
where $\alpha$ and $\beta$ are constants of mass dimension
$-2$ and $-4$, respectively.
With the new interaction added, the Klein-Gordon equation for the inflaton becomes:
\begin{equation}
\label{newkg}
[\square - \alpha \square^2 + \beta (\nabla\phi)^2\square]\phi = V_{\phi}+\frac{g}{4}F^{\mu\nu}\tilde{F}_{\mu\nu},
\end{equation}
where $\square^2\equiv\nabla^\mu \nabla^\nu\nabla_\mu \nabla_\nu$.
The two additional terms in Eq.~\eqref{eqn:Clagrangian} lead to the following modification of the energy density:
\begin{align}
T_{00}^{(\phi)} =& \Lambda+\frac{{\phi'}^2}{2 a^2} + V(\phi) \notag\\
&+ \frac{\alpha}{a^4} \left(c_1{\phi''}^2+c_2\phi'\phi^{(3)}\right) +\frac{\beta}{a^4} c_3{\phi'}^4
\end{align}
where we have also added a cosmological constant $\Lambda$, and the values of the constants $c_i$, $i=1,2,3$,
are not important for our purposes.
We now isolate the divergences coming from the energy density and the helicity integral using
dimensional regularization \cite{Bunch:1980vc}, where, working in a generic $n$-dimensional FRLW
space-time, the UV divergences show up as poles at $n=4$. This makes clear and explicit
the connection between adiabatic expansion and counterterms.
We will use in this Section results derived in the previous Section.
However, we
will keep explicit track of derivatives of the pseudo-scalar field $\phi$ here, instead of using the
variable $\xi$.
The integral measure in $n$ dimensions is:
\begin{equation}
\int \frac{\mathrm{d} k\,k^2}{(2 \pi)^3}\to\int \frac{\mathrm{d} k\,k^{n-2}}{(2\pi)^{n-1}}.
\end{equation}
\subsubsection{Energy Density}
As noted in Section~\ref{Energy-momentum tensor} and \ref{Analytical calculation}, the energy density
of the gauge field presents quartic, quadratic and logarithmic divergences.
From the adiabatic expansion of Eq.~\eqref{eqn:integral2}, the term that contribute to the quartic
divergence is:
\begin{align}
\int_{0}^{\infty}\frac{dk\,k^{n-2}}{(2\pi)^{n-1} a^4}2 \sqrt{k^2+ \mu^2a^2}=\notag\\
\label{key}
\frac{\mu^4}{16 \pi ^2 (n-4)}+\cdots,
\end{align}
where we have retained only the pole at $n=4$. Eq. \eqref{key} shows that the quartic divergence can
be absorbed by the cosmological constant counterterm $\delta\Lambda$.
Similarly, the quadratic divergence comes from:
\begin{align}
\int_{0}^{\infty}\frac{dk\,k^{n-2}}{(2\pi)^{n-1} a^4}\frac{g^2 k^4 \phi '^2}{4 \left(k^2+\mu^2 a^2\right)^{5/2}}=\notag\\\frac{5 g^2 \mu^2 {\phi '}^2}{32 \pi ^2 a^2 (n-4)}+\cdots,
\end{align}
which shows that we can absorb the quadratic divergence in the field strength counterterm $\delta Z$.
Finally, the logarithmic divergences come from the terms:
\begin{align}
&\int_{0}^{\infty}\frac{dk\,k^{n-2}}{(2\pi)^{n-1} a^4}\Bigg[\frac{15 g^4 k^8 \left(\phi '\right)^4}{64 \left(k^2+ \mu^2a^2\right)^{11/2}}\notag\\&+\frac{g^2 k^8 \left(\phi ''\right)^2}{16\left(k^2+ \mu^2a^2\right)^{11/2}}-\frac{g^2 k^8 \phi ^{(3)} \phi '}{8 \left(k^2+ \mu^2a^2\right)^{11/2}}\Bigg]=\notag\\&
-\frac{15 g^4 {\phi '}^4}{256 \pi ^2 (n-4) a^4}-\frac{g^2 (\phi '')^2}{64 \pi ^2 a^4 (n-4)}\notag\\&+\frac{g^2 \phi ^{(3)} \phi '}{32 \pi ^2 a^4 (n-4)}.
\end{align}
The first term can be absorbed in the counterterm $\delta\beta$, whereas the second and the third can
be absorbed in the $\delta\alpha$ counterterm.
\subsubsection{Helicity Integral}
We now consider the divergences in the adiabatic approximation of $g\langle{\bf E} \cdot {\bf B}\rangle$, since this is the term which enters the Klein-Gordon equation~\eqref{newkg}, to see which are the counterterms needed to absorb them. The helicity integral contains
only quadratic and logarithmic divergences.
The quadratic divergence comes from the term:
\begin{align}
\int_{0}^{\infty}\frac{\mathrm{d} k\,k^{n-2}}{(2\pi)^{n-1} a^4}\frac{g^2 k^3 \phi ''}{2 \left(k^2+\mu^2a^2\right)^{5/2}}=\notag\\
\frac{5 g^2 \mu^2 \phi ''}{16 \pi ^2 a^2 (n-4)}+\cdots,
\end{align}
which, again, can be absorbed in the redefinition of the scalar field $\delta Z$. Note that the factor
of $a^2$ at the denominator is not a problem since every term with the derivative of the scalar field
in the Klein-Gordon equation~\eqref{newkg} contains it.
The logarithmic divergence comes instead from the terms:
\begin{align}
&\int_{0}^{\infty}\frac{dk\,k^{n-2}}{(2\pi)^{n-1} a^4}\Bigg[-\frac{15 g^4 k^7 {\phi '}^2 \phi ''}{16 \left(k^2+ \mu^2a^2\right)^{11/2}}\notag\\&+\frac{g^2 k^7 \phi ^{(4)}}{8 \left(k^2+ \mu^2 a^2\right)^{11/2}}\Bigg]=\notag\\
&\frac{g^4 {\phi '}^2 \phi ''}{4 \pi ^2 a^4 (n-4)}-\frac{g^2 \phi ^{(4)}}{32 \pi ^2 a^4 (n-4)}+\cdots,
\end{align}
which can be absorbed in the counterterms $\delta\beta$ and $\delta\alpha$ respectively.
\section{Implications for background dynamics}
\label{implicationsdissipative}
\begin{figure}
\centering
\includegraphics[width=\columnwidth]{epsilon.pdf} \\
\includegraphics[width=\columnwidth]{xievo.pdf}
\caption{Numerical evolution of $- \dot H/H^2$ (top) and $\xi$ (bottom) for the
linear model (magenta line) and for natural inflation (orange line) for $|\xi| \sim 5$ (solid)
compared with the case by extrapolating $\xi \gg 1$ results (dashed) or no coupling (dotted).
We have chosen $\lvert g\rvert=60/M_{\textup{pl}}$ in both cases. For natural inflation we have
used $f=5 M_{\textup{pl}}$.}
\label{fig:Numeric}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=\columnwidth]{xiNatural.pdf}
\caption{Numerical evolution of $\xi$ for natural inflation for three different values of $f$ and $g$. We use dotted lines for no backreaction, dashed lines for extrapolated results for backreaction and solid lines for our results.}
\label{fig:NumericNatural}
\end{figure}
We now consider the implications of our results for the background dynamics.
The regularized helicity integral term behaves as an additional effective friction term and slows down
the inflaton motion through energy dissipation into gauge fields. The regularized energy-momentum tensor
of the gauge field produces an additional contribution to the Friedmann equations.
In order to study backreaction, we introduce the quantity $\Delta$ which parameterizes the
contribution of the gauge fields to the number of $e$-folds
during inflation:
\begin{align}
N&=H\int\, \frac{d\phi}{\dot{\phi}} \simeq -\int\,d\phi\,
\frac{3 H^2}{V'}\left[1-g\frac{\langle {\bf E} \cdot {\bf B} \rangle}{3 H \dot{\phi}}\right]\notag\\
\label{efolds}
&\equiv\bar{N}(1+\Delta),
\end{align}
where in the second equality we have used the Klein-Gordon equation during
slow-roll and we have defined $\bar{N}$ as the number of $e$-folds without taking backreaction into account. For simplicity we consider the case of a minimal subtraction scheme to avoid the analysis for different values of the IR mass term regulator or cutoff involved in the adiabatic subtraction described in the previous section.
The extreme case with strong dissipation and strong coupling, i.e. $3 H \dot \phi \ll g \langle E \cdot B \rangle$ with $\xi \gg 1$,
has been the target of the original study in \cite{Anber:2009ua}. For $\xi \gg 1$ our exact results for the averaged
energy-momentum tensor and helicity term differ about 10 \% from
the approximated ones in \cite{Anber:2009ua} and therefore we find estimates consistent with \cite{Anber:2009ua}
at the same level of accuracy.
As we have shown, our results for $\xi \lesssim 10$ differ from previous ones in the literature. We can give an estimate of the difference in the number of $e$-folds. As a working example, we use a linear potential $V(\phi)=\Lambda^4(1-C\lvert\phi\rvert)$ and we compare our results obtained with those for $|\xi| \sim 5$ based on the incorrect extrapolation from $\xi \gg 1$ in Eqs.~\eqref{E2B2_literature} and
\eqref{EB_literature}. Assuming a standard value of $H\sim 2\times 10^{-5} M_\textup{pl}$, a coupling $\sim 60$ and $\lvert\xi\rvert\sim 5$ we obtain $\Delta\simeq0.32$ and $0.37$ for the extrapolated and exact result respectively. Our results thus leads to a $5\%$ longer duration of inflation compared to the extrapolated ones in this case. We note that, when back-reaction changes the duration of inflation appreciably, it is possible that the gauge field contribution to $\dot H$ is not negligible when observationally relevant scales exit from the Hubble radius, potentially affecting the slopes of the primordial spectra.
To complement and confirm these
analytic estimates we now present numerical results based
on the Einstein-Klein-Gordon equations \eqref{eqn:Friedmann}, \eqref{eqn:Einstein} and \eqref{KG} including the
averaged energy-momentum tensor and helicity of gauge fields where we allow $\xi=g \dot{\phi}/2 H$ to vary with time. In the case of the aforementioned linear potential, $V(\phi)=\Lambda^4(1-C |\phi|)$, and of $V(\phi) = \Lambda^4 \left[ 1 + \cos (\phi/f) \right]$ with $f \sim 2 M_\mathrm{pl}$ (such value of $f$ is close to the regime for which natural inflation
is well approximated by a quadratic potential \cite{Savage:2006tr}).
The results are shown in Fig.~\ref{fig:Numeric}, comparing our exact results
for $|\xi| \sim 5$ (solid) with those for $|\xi| \sim 5$ based on the incorrect
extrapolation from $\xi \gg 1$ (dashed) and those in absence of gauge fields (dotted). As can be seen the approximation of $\xi\sim \textup{const}$ works very well in both the models.
Fig.~\ref{fig:NumericNatural} shows the importance of backreaction in the case of natural inflation for three different values of $f$.
The inflaton decay into gauge fields allows for a longer period of inflation compared to the case in which coupling to gauge fields is absent.
Figs. \ref{fig:Numeric} and \ref{fig:NumericNatural} also show that our correct expressions lead to a longer period of inflation than the incorrect extrapolation from $\xi >>1$. Furthermore, we show in Fig.~\ref{fig:NumericEpsilonDiff} how the slow-roll parameter $\epsilon$ is dominated by $\epsilon_A\equiv \langle \mathbf{E}^2+\mathbf{B}^2 \rangle/3 M_\textup{pl}^2 H^2$ rather than by the usual scalar field contribution $\epsilon_\phi\equiv\dot{\phi}^2/2M_\textup{pl}^2$ at the end of inflation.
Note also that the previously unexplored regime $\xi\ll 1$ is regular and included in our calculations
whereas the approximation of Eqs.~\eqref{E2B2_literature} and \eqref{EB_literature} become singular in this regime.
\begin{figure}
\centering
\includegraphics[width=\columnwidth]{EpsilonDiff.pdf}
\caption{Numerical evolution of $\epsilon$, $\epsilon_\phi$ and $\epsilon_A$ as defined in the text for the same models shown in Fig.~\ref{fig:Numeric}.}
\label{fig:NumericEpsilonDiff}
\end{figure}
\section{Conclusions}
\label{Conclusions}
We have studied the backreaction problem for a pseudo-scalar field $\phi$ which drives inflation and is coupled to gauge fields. As in other problems in QFT in curved space-times,
this backreaction problem is plagued by UV divergences.
We have identified the counterterms necessary to heal the UV divergences
for this not renormalizable interaction, which are higher order in scalar field derivatives.
We have also introduced a suitable adiabatic expansion capable to include the correct divergent terms of the integrated quantities.
Under the assumption of a constant time derivative of the inflaton, we have performed analytically the
Fourier integrals for the energy density and for the helicity in an exact way with an identification of divergent and finite terms.
Since previous approximate results were available only for $\xi \gg 1$,
our calculation which is valid for any $\xi$ has uncovered new aspects
of this backreaction problem.
We have shown that the regime of validity of previous results is $\xi
\gtrsim 10$ with a $10 \%$ level of accuracy.
We have then provided results which are more accurate than those
present in the literature in the regime $\xi \lesssim 10$.
Our results show that the inflaton decay into gauge fields leads to a
longer stage of inflation even for
$\xi \lesssim 10$. This is particularly relevant for natural inflation
since $f \lesssim M_{\textup{pl}}$ is a viable regime
for a controlled effective field theory \cite{ArkaniHamed:2003wu} and
a controlled limit of string theory \cite{Banks:2003sx}.
The techniques of integration used here in the computation of the bare
integrals of $\langle {\bf E}^2 \rangle$, $\langle {\bf B}^2 \rangle$ and $\langle {\bf
E}\cdot{\bf B} \rangle$
could have a wide range of applications in axion inflation, baryogenesis and magnetogenesis. Moreover, it would be interesting to
compute analytically the energy-momentum tensor and the helicity term
for a non-constant time derivative of the pseudo-scalar
field which we have adopted in this paper.
Other directions would be the calculation of the contribution
of gauge fields onto scalar fluctuations leading to non-Gaussian
corrections to the primordial power spectra and onto
gravitational waves at wavelengths which range from CMB observations
to those relevant for the direct detection from current and future
interferometers. We hope to return to these interesting topics in future works.
\begin{acknowledgments}
We wish to thank Marco Peloso for comments on the draft. MBa, MBr, FF would like to thank INFN under the program InDark (Inflation, Dark Matter and Large Scale Structure) for financial support.
GM would like to thank INFN under the program TAsP (Theoretical Astroparticle Physics) for financial support. AAS was supported by the Russian Science Foundation grant 16-12-10401.
\end{acknowledgments}
\setcounter{section}{0}
\begin{widetext}
\subsection{Calculation of the energy density and its backreaction}
\label{Appendix5}
In this appendix we give the analytical expressions of the bare integrals in Eqs.~\eqref{eqn:integral2} and \eqref{eqn:integral}. We carefully show the calculation of the energy density of Eq.~\eqref{eqn:integral2}; the calculation of Eq.~\eqref{eqn:integral} is then straightforward, so we only outline the differences from the one for the energy density. In the following we will use techniques introduced in Ref.~\cite{Kobayashi:2014zza,Frob:2014zka}.
\vspace{1cm}
\begin{center}
\bf{Energy density}
\end{center}
We write Eq.~\eqref{eqn:integral2} as:
\begin{equation}
\frac{\langle {\bf E}^2 + {\bf B}^2 \rangle}{2}
\equiv \frac{1}{(2 \pi)^2 a^4}\lim_{\Lambda\to\infty}\left[\mathcal{I}(\xi,\tau,\Lambda)+\mathcal{I}(-\xi,\tau,\Lambda)\right]\, ,
\end{equation}
where
\begin{equation}
\mathcal{I}(\pm\xi,\tau,\Lambda)
= \int_{0}^{\Lambda a}\,dk
\,k^2\Bigl[k^2\left(|A_\pm|^2 \right)
+ |A_\pm'|^2 \Bigr]\, ,
\end{equation}
and $\Lambda$ is an UV physical cutoff (recall that the physical momentum $k_\textup{phys}$ is related to the comoving one by $k_\textup{com}=k_\textup{phys}a$) used to isolate the UV divergences.
Using Eq.~\eqref{eqn:aplusaminus} and the properties of the Whittaker functions $(W_{\lambda\, ,\sigma}(x))^*=W_{\lambda^*\, ,\sigma^*}(x^*)$ and $\frac{d}{dx}W_{\lambda\, ,\sigma}(x)=(\frac{1}{2}-\frac{\lambda}{x})W_{\lambda\, ,\sigma}(x)-\frac{1}{x}W_{\lambda+1\, ,\sigma}(x)$, we obtain:
\begin{align}
\label{Icsi+}
\mathcal{I}(\xi,\tau,\Lambda)=\int_{0}^{\Lambda a}\,dk\,k^3 \frac{e^{\pi \xi}}{2}\Biggl\{\left[1+\left(1+\frac{\xi}{k \tau}\right)^2\right]W_{\imath \xi\,,\frac{1}{2}}(-2 \imath k \tau) W_{-\imath \xi\,,\frac{1}{2}}(2 \imath k \tau) \notag\\+\left(\frac{\imath}{2 k \tau}+\frac{\imath\xi}{2 k^2 \tau^2}\right)\Bigl[ W_{\imath \xi,\frac{1}{2}}(-2 \imath k \tau)W_{-\imath \xi+1\,,\frac{1}{2}}(2 \imath k \tau)
-W_{-\imath \xi,\frac{1}{2}}(2 \imath k \tau)W_{\imath \xi+1\,,\frac{1}{2}}(-2 \imath k \tau) \Bigr] \notag\\
+\frac{1}{2 k^2 \tau^2}W_{-\imath \xi+1,\frac{1}{2}}(2 \imath k \tau)W_{\imath \xi+1\,,\frac{1}{2}}(-2 \imath k \tau) \Biggr\}.
\end{align}
In order to solve this integral we now make use of the Mellin-Barnes representation of the Whittaker function $W_{\lambda,\sigma}(x)$:
\begin{equation}
\label{mellin}
W_{\lambda,\sigma}(x)=
e^{-\frac{x}{2}}\int_{\mathcal{C}_s}\,\frac{ds}{2\pi\imath}\,\frac{\Gamma(-s+\sigma+\frac{1}{2})\Gamma(-s-\sigma+\frac{1}{2})\Gamma(s-\lambda)}{\Gamma(-\lambda-\sigma+\frac{1}{2})\Gamma(-\lambda+\sigma+\frac{1}{2})}x^s,
\end{equation}
with $\lvert \textup{arg}\,x\rvert<\frac{3}{2}\pi$ and the integration contour $\mathcal{C}_s$ runs from $-\imath \infty$ to $+\imath \infty$ and is chosen to separate the poles of $\Gamma(-s+\sigma+\frac{1}{2})$ and $\Gamma(-s-\sigma+\frac{1}{2})$ from those of $\Gamma(s-\lambda)$.
Using Eqs. \eqref{Icsi+} and \eqref{mellin}, the reflection formula for the Gamma functions and integrating the $k$ dependent factor up to the cutoff $\Lambda$, it is straightforward to find
\begin{equation}
\label{eqn:Ipm}
\mathcal{I}(\xi,\tau,\Lambda)+\mathcal{I}(-\xi,\tau,\Lambda)=\mathcal{I}_1+\mathcal{I}_2+\mathcal{I}_3\,,
\end{equation}
where
\begin{align}
\label{I1}
\mathcal{I}_1&=\frac{\sinh^2(\pi\xi)}{2\pi^2}\int_{\mathcal{C}_s}\,\frac{ds}{2\pi\imath}\,\int_{\mathcal{C}_t}\,\frac{dt}{2\pi\imath}\,(2\imath\tau)^{s+t}\Gamma(-s)\Gamma(1-s)\Gamma(-t)\Gamma(1-t)\Biggl\{ \Biggl[\left(e^{\imath\pi(s-\imath\xi)}+e^{\imath\pi(t+\imath\xi)}\right) \frac{(a\Lambda)^{4+s+t}}{4+s+t} \notag\\
&+ \left(e^{\imath\pi(s-\imath\xi)}-e^{\imath\pi(t+\imath\xi)}\right)\frac{\xi}{\tau} \frac{(a\Lambda)^{3+s+t}}{3+s+t} + \left(e^{\imath\pi(s-\imath\xi)}+e^{\imath\pi(t+\imath\xi)}\right)\frac{\xi^2}{2\tau^2} \frac{(a\Lambda)^{2+s+t}}{2+s+t} \Biggl]\Gamma(s-\imath\xi)\Gamma(t+\imath\xi)\,\notag\\
&+\Biggl[\left(e^{\imath\pi(t-\imath\xi)}+e^{\imath\pi(s+\imath\xi)}\right) \frac{(a\Lambda)^{4+s+t}}{4+s+t} + \left(e^{\imath\pi(t-\imath\xi)}-e^{\imath\pi(s+\imath\xi)}\right)\frac{\xi}{\tau} \frac{(a\Lambda)^{3+s+t}}{3+s+t} \notag\\
&+ \left(e^{\imath\pi(t-\imath\xi)}+e^{\imath\pi(s+\imath\xi)}\right)\frac{\xi^2}{2\tau^2} \frac{(a\Lambda)^{2+s+t}}{2+s+t} \Biggl]\Gamma(t-\imath\xi)\Gamma(s+\imath\xi)\Biggr\}\,,
\end{align}
\begin{align}
\label{I2}
\mathcal{I}_2&=\frac{\xi\sinh^2(\pi\xi)}{2\pi^2}\int_{\mathcal{C}_s}\,\frac{ds}{2\pi\imath}\,\int_{\mathcal{C}_t}\,\frac{dt}{2\pi\imath}\,(2\imath\tau)^{s+t}\Gamma(-s)\Gamma(1-s)\Gamma(-t)\Gamma(1-t) \notag\\
&\times\Biggl\{ (1-\imath\xi)\Biggl[ \left(e^{\imath\pi(t-\imath\xi)}-e^{\imath\pi(s+\imath\xi)}\right)\frac{1}{\tau} \frac{(a\Lambda)^{3+s+t}}{3+s+t} + \left(e^{\imath\pi(t-\imath\xi)}+e^{\imath\pi(s+\imath\xi)}\right)\frac{\xi}{\tau^2} \frac{(a\Lambda)^{2+s+t}}{2+s+t} \Biggr]\Gamma(s+\imath\xi-1)\Gamma(t-\imath\xi) \notag\\
&+(1+\imath\xi)\Biggl[ \left(e^{\imath\pi(s-\imath\xi)}-e^{\imath\pi(t+\imath\xi)}\right)\frac{1}{\tau} \frac{(a\Lambda)^{3+s+t}}{3+s+t} + \left(e^{\imath\pi(s-\imath\xi)}+e^{\imath\pi(t+\imath\xi)}\right)\frac{\xi}{\tau^2} \frac{(a\Lambda)^{2+s+t}}{2+s+t} \Biggr]\Gamma(s-\imath\xi-1)\Gamma(t+\imath\xi) \Biggr\} \,,
\end{align}
and
\begin{align}
\label{I3}
\mathcal{I}_3&=\frac{(\xi^2+\xi^4)\sinh^2(\pi\xi)}{4\pi^2}\int_{\mathcal{C}_s}\,\frac{ds}{2\pi\imath}\,\int_{\mathcal{C}_t}\,\frac{dt}{2\pi\imath}\,\frac{(a\Lambda)^{2+s+t}}{2+s+t}\frac{1}{\tau^2}(2\imath\tau)^{s+t}\Gamma(-s)\Gamma(1-s)\Gamma(-t)\Gamma(1-t)\notag\\
&\times\Biggl[ \left(e^{\imath\pi(t-\imath\xi)}+e^{\imath\pi(s+\imath\xi)}\right)\Gamma(s+\imath\xi-1)\Gamma(t-\imath\xi-1)+\left(e^{\imath\pi(s-\imath\xi)}+e^{\imath\pi(t+\imath\xi)}\right)\Gamma(t+\imath\xi-1)\Gamma(s-\imath\xi-1)\Biggr]\,,
\end{align}
where we have assumed $\Re(n+s+t)>0$ for the terms proportional to $\Lambda^{n+s+t}$ in order to have convergence.
We now analyze each of these contributions in turn, starting from the first integral in Eq. \eqref{I1}. We integrate first over the variable $t$. Let us further specify the integration contour by requiring $\Re(t),\,\Re(s)<0$. The integrand can have poles at $t=\pm\imath\xi-n$ ($n=0,1,2,\dots$), located on the left of the integration contour of $t$, and at $t=n$ and $t=-4-s,\,-3-s,\,-2-s$, located on the right of the integration contour of $t$. We close the contour counterclockwise, on the left half-plane. The added contours do not contribute to the result since an integral of the integrand over a finite path along the real direction vanishes at $\Im(t)\to\pm \infty$ and because any integral in the region $\Re(t)<-5$ vanishes in the limit $\Lambda\to\infty$. The integral is thus $2\pi\imath$ times the sum of the residues of the poles:
\begin{equation}
t=\pm \imath\xi,\,\pm \imath\xi-1,\,\pm \imath\xi-2,\,\pm \imath\xi-3,\,\pm \imath\xi-4,\,\pm \imath\xi-5,\,-s-4,-s-3,-s-2.
\end{equation}
\begin{figure}[!ht]
\centering
\begin{minipage}{0.5\textwidth}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
\draw[->,color=black] (-4.3,0.) -- (3.,0.);
\foreach \x in {-4.,-3.,-2.,-1.,1.,2.}
\draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$};
\draw[color=black] (2.746362649294245,0.02702702702702703) node [anchor=south west] { $\Re{(t)}$};
\draw[->,color=black] (0.,-3.) -- (0.,3.);
\foreach \y in {-3.,-2.,-1.,1.,2.}
\draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y\imath\xi$};
\draw[color=black] (0.019815418023887078,2.8513513513513513) node [anchor=west] { $\Im{(t)}$};
\draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$};
\clip(-4.3,-3.) rectangle (3.,3.);
\draw [line width=2.pt] (-0.13,-0.6)-- (0.2,-0.6);
\draw [line width=2.pt] (0.2,-0.6)-- (0.2,-2.);
\draw [line width=2.pt] (0.2,-2.)-- (-4.16,-2.);
\draw [line width=2.pt] (-4.16,-2.)-- (-4.16,2.84);
\draw [line width=2.pt] (-4.16,2.84)-- (-0.13,2.86);
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\draw [->,line width=2.pt] (-0.13,-0.6) -- (-0.13,0.3684705882352941);
\begin{scriptsize}
\draw [fill=xdxdff] (0.,0.) circle (2.5pt);
\draw [fill=xdxdff] (1.,0.) circle (2.5pt);
\draw [fill=xdxdff] (2.,0.) circle (2.5pt);
\draw [fill=ffqqtt] (-4.,-1.) circle (2.5pt);
\draw [fill=ffqqtt] (-3.,-1.) circle (2.5pt);
\draw [fill=ffqqtt] (-2.,-1.) circle (2.5pt);
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\draw [fill=wwffqq] (0.28,1.62) circle (2.5pt);
\end{scriptsize}
\end{tikzpicture}
\end{minipage}
\caption{Integration contour $\mathcal{C}_t$ for the term proportional to $\Lambda^{4+s+t}\Gamma(t+\imath\xi)$ of Eq.~\eqref{I1}. Blue points are poles of $\Gamma(1-t)\Gamma(-t)$ and lye outside the integration contour. Red points are poles of $\Gamma(t+\imath\xi)$ and lie inside the contour (in the terms proportional to $\Gamma(t-\imath\xi)$ the red points are in $t=\imath\xi-n$). The green point is the pole $t=-s-4$ and it has been drawn there to emphasize that it slightly on the right of the imaginary axis (in the terms proportional to $\Lambda^{3+s+t}$ and $\Lambda^{2+s+t}$ the green point corresponds to $t=-s-3,\,-s-2$). The contour does not pass through any of the poles.}
\label{contour1}
\end{figure}
Note that the poles at $t=-s-4,\,-s-3,\,-s-2$ lye slightly on the right of the axis $\Re(t)=0$, thus we slightly deform the contour to pick it up; the integration contour is illustrated in Fig.~\ref{contour1}. The latter poles give a contribution which is independent on the cutoff, whereas the sum of the former ones give a cutoff dependent results, we summarize this writing $\mathcal{I}_1$ as
\begin{equation}
\mathcal{I}_1=\mathcal{I}_{1,\Lambda}+\mathcal{I}_{1,\text{fin}}.
\end{equation}
We first analyze the cutoff dependent part of the result $\mathcal{I}_{1,\Lambda}$, that we summarize as follows in order to reduce clutter:
\begin{equation}
\mathcal{I}_{1,\Lambda}=\int_{\mathcal{C}_s}\,\frac{ds}{2\pi\imath}\Gamma(1-s)\Gamma(-s)\Biggl\{\Gamma(s-\imath\xi)\left[\mathcal{O}_1(\Lambda^{4+s-\imath\xi},\dots,\Lambda^{-1+s-\imath\xi})\right]
+\Gamma(s+\imath\xi)\left[\mathcal{O}_2(\Lambda^{4+s+\imath\xi},\dots,\Lambda^{-1+s+\imath\xi})\right]\Biggr\}\,.
\end{equation}
The integral over $s$ can be carried out in the same way as the $t$ integral and is thus $2\pi \imath$ times the sum over the residues of the integrand from the points:
\begin{equation}
s= \pm \imath\xi,\,\pm \imath\xi-1,\,\pm \imath\xi-2,\,\pm \imath\xi-3,\,\pm \imath\xi-4\, ,
\end{equation}
the residues from the points $s=\pm\imath\xi-n$ with $n>4$ vanish as $\Lambda\to\infty$. We schematically write the result of this integral as:
\begin{equation}
\mathcal{I}_{1,\Lambda}=f_{4}(\xi,\tau)\Lambda^4+f_{2}(\xi,\tau)\Lambda^2+f_{\textup{log}}(\xi,\tau)\log(2\Lambda/H)+f_1(\xi,\tau)
\end{equation}
and we will write explicitly only the final result, together with the results from the integral $\mathcal{I}_2$ and $\mathcal{I}_3$.
We now turn to calculate $\mathcal{I}_{1,\text{fin}}$, which is the sum of the pole of the integrand in the points $t=-4-s,\,-3-s,\,-2-s$ and can be written as:
\begin{equation}
\label{I1fin}
\mathcal{I}_{1,\text{fin}}=\mathcal{I}_{1,t=-4-s}+\mathcal{I}_{1,t=-3-s}+\mathcal{I}_{1,t=-2-s}.
\end{equation}
We analyze in detail the integral over the pole $t=-4-s$, the other two are similiar. The former is given by:
\begin{equation}
\label{finita}
\mathcal{I}_{1,t=-4-s}=\int_{\mathcal{C}_s}\,\frac{ds}{2 \pi \imath}\,\frac{\pi \sinh ^2(\pi \xi) }{64 \tau ^4} \Biggl\{ \frac{\left(e^{-\pi \xi -i \pi s}+e^{ \pi \xi + i \pi s}\right)}{ \sin ^2(\pi s) \sin (\pi (s-i \xi ))}\Biggl[\frac{a_r}{s-\imath \xi}+B_r(s)-B_r(s-1)\Biggr]\Biggr\}+\xi\to-\xi\,,
\end{equation}
where $\xi\to-\xi$ stands for a second integral equal to the first one, but with $\xi$ replaced by $-\xi$ and $B_r(s)$ is a function of the form
\begin{equation}
B_r(s)=\frac{b_{r,1}}{s-\imath\xi+1}+\frac{b_{r,2}}{s-\imath\xi+2}+\frac{b_{r,3}}{s-\imath\xi+3}+\frac{b_{r,4}}{s-\imath\xi+4}+b_{r,5}s+b_{r,6}s^2+b_{r,7}s^3+b_{r,8}s^4
\end{equation}
and the coefficients $a_r,\,b_{r,j}$ for $j=1,\dots,8$ are independent on $s$.
We first consider the term with $a_r$. We rewrite the integrand as follows:
\begin{equation}
\lim_{p\to 1}\frac{\pi \sinh ^2(\pi \xi) }{64 \tau ^4} \frac{\left(e^{-\pi \xi -i \pi s}+e^{ \pi \xi + i \pi s}\right)}{ \sin ^2(\pi s) \sin (\pi (s-i \xi ))}\frac{a_r}{(s-\imath \xi)^p},
\end{equation}
with $p>1$. The integral of this function vanishes on an arc of infinite radius on the left half-plane, so we can close the contour on the left half-plane with a counterclockwise semicircle of infinite radius, as illustrated in Fig. \ref{contour2}. The integral is then $2\pi i$ times the sum of the residues in the poles $s=\pm\imath\xi-n$ and $s=-n-1$ with $n=0,\,1,\,2,\,\dots$. As for the other integrals we do not give the result here, but we will just write the final result.
\begin{figure}[!ht]
\centering
\begin{minipage}{0.5\textwidth}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
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\draw [fill=ttffqq] (-3.,0.) circle (2.5pt);
\draw [fill=ttffqq] (-2.,0.) circle (2.5pt);
\draw [fill=ttffqq] (-1.,0.) circle (2.5pt);
\draw [fill=ffqqtt] (-1.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (-2.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (0.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (-3.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (-4.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (-5.,-1.) circle (2.5pt);
\draw [fill=ffqqtt] (-6.,-1.) circle (2.5pt);
\draw [fill=ffqqtt] (-7.,-1.) circle (2.5pt);
\end{scriptsize}
\end{tikzpicture}
\end{minipage}
\caption{Integration contour for the term proportional to $a_r$ in Eq.~\eqref{finita}. The radius of the semicircle is taken to be infinite and the contour does not pass through any of the poles. Blue points are the poles of $\Gamma(1-s)\Gamma(-s)$ whereas red points are the poles of $\Gamma(s-\imath\xi)$ and $\csc((s-\imath\xi))$. Green points are the poles of $\csc(\pi s)$. For the term with $\xi\to-\xi$ in Eq.~\eqref{finita} the red points move to $s=-\imath\xi-n$. }
\label{contour2}
\end{figure}
Now we conclude integrating the terms with $B_r(s)-B_r(s-1)$. We shift the integration variable in the second term by $s\to y=s-1$ so that the integral is given by:
\begin{equation}
\label{sandwich}
\int_{\mathcal{C}_s}\,\frac{ds}{2 \pi \imath}\,(\dots) \left[B_r(s)-B_r(s-1)\right]=\left(\int_{\mathcal{C}_s}-\int_{\mathcal{C}_s-1}\right)\,\frac{ds}{2 \pi \imath}\,(\dots )B_r(s)\,,
\end{equation}
as illustrated in Fig.~\ref{contour2}.
Thus we can evaluate this integral summing $2\pi\imath$ times the residues of the singularities of the integrand which fall in the region sandwiched by the original integration contour and the shifted one, which are the poles at $s=-1$ and $s=\pm\imath\xi$. We write the result as $F_1(\xi,\tau)$.
\begin{figure}[!ht]
\centering
\begin{minipage}{0.5\textwidth}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
\draw[->,color=black] (-4.,0.) -- (3.,0.);
\foreach \x in {-4.,-3.,-2.,-1.,1.,2.,3.}
\draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$};
\draw[color=black] (2.7254901960784315,0.036951501154734404) node [anchor=south west] { $\Re{(s)}$};
\draw[->,color=black] (0.,-3.9815242494226313) -- (0.,4.);
\foreach \y in {-3.,-2.,-1.,1.,2.,3.}
\draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y\imath \xi$};
\draw[color=black] (0.02144607843137255,3.7967667436489605) node [anchor=west] { $\Im{(s)}$};
\draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$};
\clip(-4.,-3.9815242494226313) rectangle (3.,4.);
\draw [line width=2.pt] (0.15,3.3)-- (0.15,0.68);
\draw [line width=2.pt] (0.15,0.68)-- (-0.1,0.68);
\draw [line width=2.pt] (-0.1,0.68)-- (-0.1,-3.3);
\draw [->,line width=2.pt] (0.15,0.68) -- (0.15,1.99);
\draw [line width=2.pt] (0.15,3.298631726763146)-- (-0.95,3.3);
\draw [line width=2.pt] (-0.95,3.3)-- (-0.95,0.68);
\draw [line width=2.pt] (-0.95,0.68)-- (-1.1,0.68);
\draw [line width=2.pt] (-1.1,0.68)-- (-1.1,-3.3);
\draw [line width=2.pt] (-1.1,-3.3)-- (-0.1,-3.3);
\draw [->,line width=2.pt] (0.15,3.298631726763146) -- (-0.3937966702683996,3.2993081471542536);
\draw [->,line width=2.pt] (-0.95,3.3) -- (-0.95,1.7859296482412061);
\draw [->,line width=2.pt] (-1.1,0.68) -- (-1.1,-1.5988364508224175);
\draw [->,line width=2.pt] (-1.1,-3.3) -- (-0.6051247771836007,-3.3);
\draw [->,line width=2.pt] (-0.1,-3.3) -- (-0.1,-1.1778432029335546);
\begin{scriptsize}
\draw [fill=xdxdff] (0.,0.) circle (2.5pt);
\draw [fill=xdxdff] (1.,0.) circle (2.5pt);
\draw [fill=xdxdff] (2.,0.) circle (2.5pt);
\draw [fill=ududff] (5.,0.) circle (2.5pt);
\draw [fill=ttffqq] (-7.,0.) circle (2.5pt);
\draw [fill=ttffqq] (-6.,0.) circle (2.5pt);
\draw [fill=ttffqq] (-5.,0.) circle (2.5pt);
\draw [fill=ttffqq] (-3.,0.) circle (2.5pt);
\draw [fill=ttffqq] (-2.,0.) circle (2.5pt);
\draw [fill=ttffqq] (-1.,0.) circle (2.5pt);
\draw [fill=ffqqtt] (-1.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (-2.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (0.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (-3.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (-4.,1.) circle (2.5pt);
\draw [fill=ffqqtt] (-5.,-1.) circle (2.5pt);
\draw [fill=ffqqtt] (-6.,-1.) circle (2.5pt);
\draw [fill=ffqqtt] (-7.,-1.) circle (2.5pt);
\end{scriptsize}
\end{tikzpicture}
\end{minipage}
\caption{Integration contour in Eq.~\eqref{sandwich}.The contour does not pass through any of the poles. Blue points are the poles of $\Gamma(1-s)\Gamma(-s)$ whereas red points are the poles of $\Gamma(s-\imath\xi)$ and $\csc((s-\imath\xi))$. Green points are the poles of $\csc(\pi s)$. For the term with $\xi\to-\xi$ in Eq.~\eqref{finita} the red points move to $s=-\imath\xi-n$. }
\label{contour3}
\end{figure}
The integral over the poles $t=-s-3,\,-s-2$ can be written in a similar way as:
\begin{equation}
\label{i13}
\mathcal{I}_{1,t=-3-s}=\int_{\mathcal{C}_s}\,\frac{ds}{2 \pi \imath}\,\frac{\mathcal{A}_3(\xi,\tau)}{\sin^2(\pi s)\sin(\pi(s-\imath\xi))}\Biggl[\frac{c_r}{s-\imath \xi}+D_r(s)-D_r(s-1)\Biggr]+\xi\to-\xi\,,
\end{equation}
for the integral over the pole $t=-3-s$ and
\begin{equation}
\label{i12}
\mathcal{I}_{1,t=-2-s}=\int_{\mathcal{C}_s}\,\frac{ds}{2 \pi \imath}\,\frac{\mathcal{A}_2(\xi,\tau)}{\sin^2(\pi s)\sin(\pi(s-\imath\xi))}\Biggl[\frac{e_r}{s-\imath \xi}+K_r(s)-K_r(s-1)\Biggr]+\xi\to-\xi\,,
\end{equation}
for the integral over the pole $t=-2-s$. The functions $\mathcal{A}_2$ and $\mathcal{A}_3$ are regular functions of $\xi$. $D_r(s)$ is given by:
\begin{equation}
D_r(s)=\frac{d_{r,1}}{s-\imath\xi+1}+\frac{d_{r,2}}{s-\imath\xi+2}+\frac{d_{r,3}}{s-\imath\xi+3}
+d_{r,4}s+d_{r,5}s^2+d_{r,6}s^3
\end{equation}
and $K_r(s)$ is given by
\begin{equation}
K_r(s)=\frac{k_{r,1}}{s-\imath\xi+1}+\frac{k_{r,2}}{s-\imath\xi+2}
+k_{r,3}s+k_{r,4}s^2
\end{equation}
The integrals in Eqs. \eqref{i13} and \eqref{i12} can be made exactly as done in the previous case for the integral over the residue in $t=-s-4$ in Eq.~\eqref{finita}.
The full solution of the integral in Eq.~\eqref{I1} is thus:
\begin{equation}
\mathcal{I}_1=f_{4}(\xi,\tau)\Lambda^4+f_{2}(\xi,\tau)\Lambda^2+f_{\textup{log}}(\xi,\tau)\log(2\Lambda/H)+f_1(\xi,\tau)+f_{\textup{fin}}(\xi,\tau)\,,
\end{equation}
where we have included the contributions of Eqs.~\eqref{finita}, \eqref{i13} and \eqref{i12} in the term $f_{\textup{fin}}(\xi,\tau)$.
The integrals $\mathcal{I}_2$ and $\mathcal{I}_3$ can be done following the same procedure and we only give the final result in the following \footnote{Note that $\mathcal{I}_2$ is not explicitly invariant under the exchange of $t$ with $s$: we therefore symmetrize it before taking the integral in $s$ and $t$.}.
Defining the contributions to the divergences and finite part of Eq.~\eqref{eqn:Ipm} as
\begin{equation}
\mathcal{I}(\xi,\tau,\Lambda)+\mathcal{I}(-\xi,\tau,\Lambda)=g_{4}(\xi,\tau)\Lambda^4+g_{2}(\xi,\tau)\Lambda^2+g_{\textup{log}}(\xi,\tau)\log(2\Lambda/H)+g_{\textup{fin}}(\xi,\tau),
\end{equation}
it can be found that the coefficient of the quartic divergence is:
\begin{equation}
g_4(\xi,\tau)=\frac{1}{2 H^4 \tau^4}\,.
\end{equation}
The coefficient of the quadratic divergence is:
\begin{equation}
g_2(\xi,\tau)=\frac{\xi^2}{2 H^2 \tau^4}\,.
\end{equation}
The coefficient of the logarithmic divergence is:
\begin{equation}
g_{\text{log}}(\xi,\tau)=\frac{3 \xi ^2 \left(5 \xi ^2-1\right) }{4 \tau ^4}\,.
\end{equation}
The finite part is
\begin{align}
g_{\textup{fin}}(\xi,\tau)=&\frac{3 \xi ^2 \left(5 \xi ^2-1\right) (\psi (-\imath \xi -1)+\psi (\imath \xi -1))}{8 \tau ^4}+\frac{\gamma \left(11-10 \xi ^2\right) \xi ^2}{2 \tau ^4}+\frac{\xi ^2 \left(7 \xi ^6-282 \xi ^4+123 \xi ^2+124\right)}{64 \left(\xi ^2+1\right) \tau ^4}\notag\\
&-\frac{3 \imath \xi ^2 \left(5 \xi ^2-1\right) \sinh (2 \pi \xi ) (\psi ^{(1)}(1-\imath \xi )-\psi ^{(1)}(\imath \xi +1))}{16 \pi \tau ^4}+\frac{\xi \left(30 \xi ^2-11\right) \sinh (2 \pi \xi )}{16 \pi \tau ^4}\notag\\
\end{align}
where $\psi$ is the Digamma function and $\gamma$ is the Euler-Mascheroni constant.
\vspace{1cm}
\begin{center}
\bf{Backreaction: $\textbf{E}\cdot \textbf{B}$}
\end{center}
We now calculate the integral in Eq. \eqref{eqn:integral}. The quantity we are interested in is:
\begin{equation}
\langle {\bf E} \cdot {\bf B} \rangle
=-\frac{1}{(2\pi)^2\,a^4} \int\,dk\,k^3\frac{\partial}{\partial \tau} \Bigl( |A_+|^2 - |A_-|^2 \Bigr)\,\equiv-\frac{1}{(2\pi)^2\,a^4}\lim_{\Lambda\to\infty}\mathcal{J}(\xi,\tau,\,\Lambda)\,,
\end{equation}
where, obviously, $\mathcal{J}$ has not to be confused with the one of the previous Sections and is given by
\begin{align}
\mathcal{J}(\xi,\tau,\,\Lambda)=&-\frac{1}{2 \tau}\int_{0}^{\Lambda a}\,dk\,k^2 e^{\pi\xi}\Bigl[W_{\imath \xi,\frac{1}{2}}(-2 \imath k \tau)W_{-\imath \xi+1\,,\frac{1}{2}}(2 \imath k \tau)+ W_{-\imath \xi,\frac{1}{2}}(2 \imath k \tau)W_{\imath \xi+1\,,\frac{1}{2}}(-2 \imath k \tau) \notag\\
&+W_{-\imath \xi,\frac{1}{2}}(-2 \imath k \tau)W_{\imath \xi+1\,,\frac{1}{2}}(2 \imath k \tau)+ W_{\imath \xi,\frac{1}{2}}(2 \imath k \tau)W_{-\imath \xi+1\,,\frac{1}{2}}(-2 \imath k \tau)\Bigr]\,,
\end{align}
where, again, we put the IR cutoff to $0$.
As in the previous section we can use the Mellin-Barnes representation of the Whittaker functions Eq. \eqref{mellin} to write:
\begin{align}
\label{iback2}
&\mathcal{J}(\xi,\tau,\,\Lambda)=-\frac{\xi\sinh^2(\pi\xi)}{2\pi^2\tau}\int_{\mathcal{C}_s}\,\frac{ds}{2\pi\imath}\,\int_{\mathcal{C}_t}\,\frac{dt}{2\pi\imath}\,\frac{\Lambda^{3+s+t}}{3+s+t}(2\imath\tau)^{s+t}\Gamma(-s)\Gamma(1-s)\Gamma(-t)\Gamma(1-t)\\
&\times\Biggl\{ (\imath+\xi) \left(e^{\imath\pi(s+\imath\xi)}-e^{\imath\pi(t-\imath\xi)}\right) \Gamma(s+\imath\xi-1)\Gamma(t-\imath\xi)+(-\imath+\xi) \left(e^{\imath\pi(t+\imath\xi)}-e^{\imath\pi(s-\imath\xi)}\right) \Gamma(s-\imath\xi-1)\Gamma(t+\imath\xi) \Biggr\}\,,\notag
\end{align}
converging for $\Re{(s+t)}>-3$.
We only give the final result here since the integral can be carried out as previously explained:
\begin{align}
\mathcal{J}(\xi,\tau,\,\Lambda)=&\,\frac{\Lambda ^2 \xi }{2 H ^2 \tau ^4}+\frac{3 \xi \left(5 \xi ^2-1\right) \log (2 \Lambda/H)}{2 \tau ^4}+\frac{3 \gamma \xi \left(5 \xi ^2-1\right)}{2 \tau ^4}+\frac{22 \xi -47 \xi ^3}{4 \tau
^4}+\frac{\left(30 \xi ^2-11\right) \sinh (2 \pi \xi )}{8 \pi \tau ^4}\notag\\
&-\frac{3 \xi \left(5 \xi ^2-1\right) \left(H_{-\imath \xi }+H_{\imath \xi }\right)}{4 \tau ^4}+\imath\frac{3 \xi \left(5 \xi ^2-1\right) \sinh (2 \pi \xi ) (\psi ^{(1)}(1-\imath \xi )-\psi ^{(1)}(\imath \xi +1))}{8 \pi \tau ^4},\notag\\
\end{align}
where $H_{x}=\psi(x+1)+\gamma$ is the Harmonic number of order $x$ and $\psi^{(1)}(x)=d\psi(x)/dx$.
\end{widetext}
\subsection{Calculation of the energy density and helicity integral in the presence of a mass term}
\label{appendixmass}
We now generalize the previous calculations to include a mass term. The mode equation for the two helicities, Eq. (\ref{eqn:eqbeta}), becomes:
\begin{equation}
\label{modmodified}
\left[\frac{\partial^2}{\partial\tau^2}+k^2+\frac{\mu^2}{H^2 \tau^2}\pm\frac{2 k \xi}{\tau}\right]A_\pm(k,\tau)=0,
\end{equation}
the solution of which being
\begin{equation}
A_\pm(k,\tau)=\frac{1}{\sqrt{2 k }}e^{\pm\pi\xi/2}W_{\mp\imath\xi,\,\frac{1}{2}-\frac{\mu^2}{H^2}}(-2\imath k \tau).
\end{equation}
With this new solution for the mode function we can compute $ \frac{\langle {\bf E}^2 + {\bf B}^2\rangle}{2}$ and $\langle {\bf E}\cdot{\bf B}\rangle$ in the same way as done in the previous Appendix. There are only two caveats:
\begin{itemize}
\item The presence of the mass term modifies the integral in Eq. \eqref{eqn:integral} to:
\begin{align}
\label{eqn:integral3}
\frac{\langle {\bf E}^2 + {\bf B}^2 \rangle}{2}
= \int\,\frac{dk}{(2\pi)^2\,a^4} \,k^2\Bigl[k^2\left(|A_+|^2 + |A_-|^2 \right) \notag\\
\quad \quad+\mu^2a^2\left(|A_+|^2 + |A_-|^2 \right) + |A_+'|^2 + |A_-'|^2 \Bigr] \,.
\end{align}
\item In the previous Appendix, we expressed the Whittaker functions through their Mellin-Barnes transform. This led to integrals which we computed closing the contour of integration and using the residue theorem. Among the residues that we have picked up, there were residues from the double poles $t,\,s= -n$ where $n$ was a positive integer number (sometimes $n=0$ was included). In the case where the two helicities acquire a mass, these poles are instead from the single poles $t,\,s= -n+\mu^2/H^2$ and $t,\,s= -n-\mu^2/H^2$.
\end{itemize}
With these caveats in mind the integrals can be computed leading to :
\begin{widetext}
\begin{align}
\label{energymassive}
&\frac{\langle {\bf E}^2 + {\bf B}^2\rangle}{2}=\frac{\Lambda^4}{8 \pi^2}+\frac{H^2 \Lambda ^2 \xi ^2 (x+1)}{8 \pi ^2}
+\frac{H^4 \left((1-x)^2 x^2+3 \xi ^4 (2 x+5)+\xi ^2 (2 (-x-6) (1-x) x-3)\right) \log (2 \Lambda/H )}{16 \pi ^2}\notag\\
&+\frac{H^4 \left((x^4-x^5)(3 x-5)-\xi ^6 (28 x+79)+ (x^5 (84-x (32 x+113))+41 x^4)+\xi ^2 x (x (x (80-x (4 x+45))+16)-8)\right)}{64 \pi ^2 \left(\xi ^2+x^2\right)}\notag\\
&-\frac{H^4 \xi x^2 (x (x (x (x (2 x (x+1)+1)-684)+979)+48)-192) \sinh (2 \pi \xi ) \csc (2 \pi x)}{1536 \pi ^2 \left(\xi ^2+x^2\right)}\notag\\
&-\frac{H^4 \xi ^3 x \left(x \left(x \left(x \left(8 x^2+6 x-553\right)-1944\right)+1286\right)+528\right) \sinh (2 \pi \xi ) \csc (2 \pi x)}{1536 \pi ^2 \left(\xi ^2+x^2\right)}\notag\\
&-\frac{H^8 \xi ^2 \sinh (2 \pi \xi ) \sinh (4 \pi \xi ) \left(\xi ^2+x^2\right)^2 \left(2 \xi ^4+2 x^4+2 x^3+\xi ^2 \left(4 x^2+2 x+21\right)+x^2-12 x+19\right)^2 \cot (2 \pi x) }{18874368 \pi ^4 \sin ^2(\pi (x-i
\xi )) \sin ^2(\pi (x+i \xi ))\sin (2 \pi x)}\notag\\
&-\frac{ H^4 \left((x^2-x)^2 +3 \xi ^4 (2 x+5)+\xi ^2 ( (2x^2-2x) (x+6)-3)\right) (\psi ^{(0)}(-x-i \xi )+\psi ^{(0)}(x+i \xi )) (\imath\sinh (2 \pi \xi ) \csc (2 \pi x)+1)}{64 \pi ^2}\notag\\
&+\frac{ H^4 \left((x^2-x)^2 +3 \xi ^4 (2 x+5)+\xi ^2 ( (2 x^2-x) (x+6)-3)\right) (\psi ^{(0)}(x-i \xi )+\psi ^{(0)}(-x+i \xi )) (\imath\sinh (2 \pi \xi ) \csc (2 \pi x)-1)}{64 \pi ^2}
\end{align}
and
\begin{align}
\label{ebmassive}
&\langle {\bf E}\cdot{\bf B}\rangle=\frac{\Lambda^2 \xi H^2}{8\pi^2}+\frac{3 H^4 \xi \left(5 \xi ^2-3 (1-x) x-1\right) \log ( 2\Lambda/H )}{8 \pi ^2}\notag\\
&+\frac{H^4 \left(-47 \xi ^5-2 \xi ^3 (-2 (9-17 x) x-11)-\xi x (3-x (3 (10-7 x) x+13))\right)}{16 \pi ^4 \left(\xi ^2+x^2\right)}\notag\\
&+\frac{H^4 x \sinh (2 \pi \xi ) \left(30 \xi ^4+\xi ^2 (-2 (9-19 x) x-11)+(1-2 x) (3-2 x) x (2 x+1)\right) \csc (2 \pi x)}{16 \pi ^4 \left(\xi ^2+x^2\right)}\notag\\
&-\frac{3 i H^4 \xi \left(5 \xi ^2+3 (x-1) x-1\right) (\psi ^{(0)}(-x-i \xi )+\psi ^{(0)}(x+i \xi )) (\sinh (2 \pi \xi ) \csc (2 \pi x)-i)}{32 \pi ^4}\notag\\
&+\frac{3 i H^4 \xi \left(5 \xi ^2+3 (x-1) x-1\right) (\psi ^{(0)}(x-i \xi )+\psi ^{(0)}(i \xi -x)) (\sinh (2 \pi \xi ) \csc (2 \pi x)+i)}{32 \pi ^4},
\end{align}
\end{widetext}
where for reasons of convenience we have defined $x\equiv \mu^2/H^2$. It can be easily seen that these expression reduce to those of the previous Appendix for $\mu\to0$.
We note that the presence of a mass term can come for an interaction of the form:
\begin{equation}
\mathcal{L}\supset -B(\phi)\frac{F_{\mu\nu}F^{\mu\nu}}{4},
\end{equation}
where the function $B(\phi)$ is nearly constant in order to break only slightly the shift symmetry breaking of the pseudoscalar field \cite{Barnaby:2011qe}.
This interaction term gives the two helicities an effective mass of the form $\mu^2=H^2 \sqrt{\epsilon_\phi \epsilon_B}\textup{sign}(B'\dot{\phi})$, where $\epsilon_{B}\equiv\frac{M_{\textup{pl}}^2}{2}\left(\frac{B'}{B}\right)^2\ll1$ and $B'=dB/d\phi$.
In this case the mode function satistfying Eq. \eqref{modmodified} is $\tilde{A}_\pm(k,\tau)\equiv\sqrt{B(\phi(\tau)}A_\pm(k,\tau)$ and the solution for $A_\pm(k,\tau)$
is \cite{Barnaby:2011qe}:
\begin{equation}
A_\pm(k,\tau)=\frac{1}{\sqrt{2 k B_*}}e^{\pm\pi\xi/2}W_{\mp\imath\xi,\,\frac{1}{2}+\sqrt{\epsilon_\phi\epsilon_B}}(-2\imath k \tau),
\end{equation}
where $B_*$ is the value of $B$ evaluated at horizon crossing for cosmologically interesting modes.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,147 |
Ah, 'tis the season of year-end review blog posts! Much like last year, I am going to continue to focus on what I did not finish, versus what I did finish. After all, if you want to see what I painted, it's all (mostly) up on the blog.
First things, first, I did finish a few of the projects from last years list: the Lord of Worms, the dark future car, and the Magewraith Throne.
This Iron Warriors rhino is almost finished!
These Plague Marines will hopefully be completed this weekend.
The Foul Blightspawn was one of my few hobby Christmas presents!
So that was a run down of my shame from last year. I expect y'all to get on my case if you don't see some of these completed soon! That's what the comment section is for (well, that and Russian spambots!).
Some interesting unfinished projects. My favorite is the Stitcher's assistant. Good luck in the coming year. I thought the comments were for Thai Casinos.
I hope to have him done soon.
I think Rufus gets my vote for the one that needs to Be finished 1st.
Still there is no rush on any of this as it all will get done when It needs to get done.
It's worth the wait, Andrew - I can't wait to see what you do with that Old World mansion. It is going to be FANTASTIC.
Beware the Bogs of Ottersricht! | {
"redpajama_set_name": "RedPajamaC4"
} | 3,235 |
{"url":"http:\/\/marcfbellemare.com\/wordpress\/8951","text":"A Rant on Estimation with Binary Dependent Variables (Technical)\n\nSuppose you are trying to explain some outcome $y$, where\u00a0$y$ is equal to 0 or 1 (e.g., whether someone is a nonsmoker or a smoker). You also have data on a vector of explanatory variables\u00a0$x$ (e.g., someone\u2019s age, their gender, their level of education, etc.) and on a treatment variable\u00a0$D$, which we will also assume is binary, so that $D$\u00a0is equal to 0 or 1 (e.g., whether someone has attended an information session on the negative effects of smoking).\n\nIf you were interested in knowing what the effect of attending the information session on the likelihood that someone is a smoker, i.e., the impact of $D$ on\u00a0$y$\u00a0The equation of interest in this case is\n\n(1) $y=\\alpha+\\betaX+\\gammaD+\\epsilon$,\n\nwhere\u00a0$\\alpha$ is a constant,\u00a0$\\beta$ is a vector of the coefficients attached to the explanatory variables,\u00a0$\\gamma$ is the parameter of interest, and\u00a0$\\epsilon$ is the error term.\n\nThis post is about why, in most cases, you should be estimating equation (1) by ordinary least squares, i.e., estimate a linear probability model (LPM). I have heard and read so many arguments against the LPM and for the probit or logit that I wanted to write something on this.\n\nArguments Typically Made against the LPM\n\nThe arguments typically made against the LPM are:\n\n1. The error term of a binary variable has a Bernoulli structure, i.e.,\u00a0$Var(\\epsilon_i)=p_i(1-p_i)$, where $p_i=Pr(y_i=1)$.\u00a0This non-constant variance of the error term means that you have a heteroskedasticity problem and the LPM standard errors will be wrong.\n2. The LPM can yield values of $\\hat{y}_i$, i.e.,\u00a0predicted values of the dependent variable, outside of the\u00a0$[0,1]$ interval. In other words, the LPM can yield predicted probabilities that are negative or greater than 100%.\n3. The LPM imposes linearity on the relationship between the dependent variable and the right-hand side variables.\n\nFor these reasons, many will dismiss LPM estimates as wrong. I respond:\n\n1. With robust standard errors, the standard errors are correct, and it is very easy it is to implement robust standard errors with most statistical packages. Indeed, in the package that I use the most, it is simply a matter of adding \u201c, robust\u201d at the end of my estimation command.\n2. This is only a concern if your reason for estimating equation (1) is to forecast probabilities. For most readers of this blog, that will\u00a0not be why they are estimating equation (1). Rather, they will be interested in knowing the precise value of\u00a0$\\gamma$.\n3. Sure, but who says an assumed nonlinear relationship is much better? On their Mostly Harmless Econometrics\u00a0blog, Angrist and Pischke\u00a0write:\n\nIf the conditional expectation function (CEF) is linear, as it is for a saturated model, regression gives the CEF \u2013 even for LPM. If the CEF is non-linear, regression approximates the CEF. Usually it does it pretty well. Obviously, the LPM won\u2019t give the true marginal effects from the right nonlinear model. But then, the same is true for the \u201cwrong\u201d nonlinear model! The fact that we have a probit, a logit, and the LPM is just a statement to the fact that we don\u2019t know what the \u201cright\u201d model is. Hence, there is a lot to be said for sticking to a linear regression function as compared to a fairly arbitrary choice of a non-linear one! Nonlinearity per se is a red herring.\n\nArguments One Can Make Against the Probit or Logit\n\nPeople who dismiss the LPM, usually by invoking the two arguments above, usually argue in favor of estimating a probit or a logit instead. Here are some arguments one can make against the probit or logit:\n\n1. Both the probit and the logit can lead to identification by functional form. If you are interested in identifying the causal relationship flowing from $D$ to\u00a0$y$, i.e., in precisely estimating\u00a0$\\gamma$, you want to avoid this.\n2. The probit and logit are not well-suited to the use of fixed effects because of the incidental parameters problem.\n\nSo when should you use an LPM, and when should you use a probit or a logit? If you have experimental data, i.e., if values of\u00a0$D$ were randomly assigned, there is no harm in estimating a probit or a logit \u2014 your estimate of $\\gamma$\u00a0is cleanly identified because of the random assignment. If you want to forecast the likelihood that something will happen, estimate a probit or a logit.\n\nBut if you are interested in estimating the causal impact of $D$ on\u00a0$y$ and\u00a0have any reason to believe that your identification is less than clean, if you want to use fixed effects, and if you are not interested in forecasting the value of\u00a0$y$, you should prefer the LPM with robust standard errors.\n\nConclusion\n\nI have made the points above several times over the last few years, in conversations with colleagues, when advising students, in referee reports, etc. But every once in a while, I will get admonished by an anonymous reviewer for my use of the LPM, and so I wanted to write something about it.\n\nUltimately, I think the preference for one or the other is largely generational, with people who went to graduate school prior to the Credibility Revolution preferring the probit or logit to the LPM, and with people who went to graduate school during or after the Credibility Revolution preferring the LPM.\n\nAs always, the right way to approach things is probably to estimate all three if possible, to present your preferred specification, and to explain in a footnote (or show in an appendix) that your results are robust to the choice of estimator.\n\nNo related content found.","date":"2017-01-19 21:28:25","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\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 25, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7824305295944214, \"perplexity\": 696.1931508680475}, \"config\": {\"markdown_headings\": false, \"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\/1484560280746.40\/warc\/CC-MAIN-20170116095120-00480-ip-10-171-10-70.ec2.internal.warc.gz\"}"} | null | null |
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\section{Introduction}
Flows with inhomogeneous velocity profiles are one of the
prevalent and still not completely understood macro-systems. They
occur in laboratory experiments, industrial applications, the
earth's atmosphere, oceans and many astrophysical objects. We
concentrate attention on smooth shear flows (flows without
inflection point in the velocity profile) that are linearly stable
according to the canonical hydrodynamics. Faults in the
comprehension of shear flow dynamics is caused by the
incompleteness of the classical method of normal modes.
Specifically, method of modes leaves out of account the
non-normality of the linear operators that often produces powerful
transient development of a subset of perturbations
\cite{GL65,CD90,RH93,CRT96,TVY01,VMY01}. Indeed, in the case of
the non-normality of linear operators, corresponding
eigenfunctions are not orthogonal and strongly interfere
\cite{RH93}. Hence, a correct approach should fully analyze the
eigenfunction interference, which is feasible in asymptotic: in
fact, in modal analysis only the asymptotic stability of flow is
studied, while {\it no attention} is paid to any particular
initial value or finite time period of the dynamics -- this period
of the evolution is thought to have no significance and is left
to speculation.
Recognition of the importance of the non-normality in the linear
stability resulted impressive progress in the understanding of the
shear flow phenomena in 90s of the last century. The early
transient period for the perturbations has been shown to reveal
rich and complicated behavior leading to various consequences. It
has been grasped phenomena that are overlooked in the framework of
the modal analysis.
The progress in the understanding of the shear flow phenomena has
been achieved using so-called non-modal approach (see, e.g., Refs.
\cite{GL65,CRT96,TVY01}), which implicates a change of independent
variables from a laboratory to a moving frame and the study of
temporal evolution of {\it spatial Fourier harmonics} (SFHs) of
perturbations without any spectral expansion in time. Strictly
speaking the non-modal approach is applicable to the flows with
constant velocity shear. However, it is obvious that the results
obtained in the framework of the approach are valid for any shear
flows without inflection point, if the wavelength of modes is
appreciably shorter than the characteristic length scale of the
inhomogeneity.
The progress involves novel results on the time evolution of
vortex and wave perturbations; elaboration of a concept of
self-sustaining turbulence in the spectrally stable shear flows
(that was labelled as bypass concept)
\cite{BDT95,CCHL02,C02,GG94,HR94}. Among the novelties one should
stress (in the context of the present paper) disclosure of the
linear mechanisms of resonance transformation of waves at low
shear rates \cite{CRT96,CCLT97,RPM00} and non-resonant conversion
of vortices to waves at moderate and high shear rates
\cite{CTBM97}. The later phenomena should be inherent to flow
systems. This was shown in papers \cite{RPM00,RPH99}, where
appearances of some waves in solar wind and galactic disks are
explained by the linear wave transformations in the
magnetohydrodynamic (MHD) flows.
Aim of the present work is quantitative and qualitative study of
the linear evolution of three dimensional MHD waves at low and
high shear rates. Equilibrium density, pressure and magnetic field
are assumed to be homogeneous. Specifically, we perform analytical
and numerical study of the important effect caused by linear
forces -- mutual transformations of MHD waves. The waves are
coupled (during a limited time interval) at the specific system
parameters and mutual transformations of two, or even all three of
them occur. Analytical expressions of transformation coefficients
are obtained for wide range of the system parameters. We study the
wave over-reflection phenomena at high shear rates: MHD waves
extract flow energy, are amplified non-exponentially and are
over-reflected.
Mathematical methods used in this paper are similar to ones that
were first developed for quantum mechanical problems: non-elastic
atomic collisions \cite{S32} and non-adiabatic transitions in two
level quantum systems \cite{Z32,L32}. Later, the same asymptotic
methods were successfully applied to various problems (see, e.g.,
Ref. \cite{ZM65}) including interaction of plasma waves in the
media with inhomogeneous magnetic field and/or density \cite{S}.
The paper is organized as follows: employed mathematical formalism
is presented in Sec. II. General properties of resonant
transformation of wave modes that takes place at small shear
rates, are presented in Sec. III. Particular cases of the resonant
transformation of MHD wave modes are described in Sec. IV. In Sec.
V dynamics of SFHs of the wave modes at high shear rates and
over-reflection phenomenon is studied. Conclusions are given in
Sec. VI.
\section{Mathematical Formalism}
Consider compressible unbounded shear flow with constant shear
parameter (${\bf U}_0(A y,0,0)$) and uniform density $(\rho_0)$,
pressure $(P_0)$ and magnetic field directed along the streamlines
$(\bf B_0 || \bf U_0)$. Linearized ideal MHD equations governing
the evolution of density $(\rho^{\prime})$, pressure
$(p^{\prime})$, velocity $(\bf{u^{\prime}})$ and magnetic field
$(\bf {b^{\prime}})$ perturbations in the flow are:
\begin{equation}
(\partial_t + {\bf U}_0 \cdot {\bf \nabla} )\rho^{\prime} +
{\rho_0} {\bf \nabla} \cdot {\bf{u^{\prime}}} = 0, \label{eq:p1}
\end{equation}
\begin{equation}
(\partial_t + {\bf U}_0 \cdot {\bf \nabla} ){\bf{u^{\prime}}} +
({\bf u^{\prime}} \cdot {\bf \nabla}) {\bf U}_0 + \frac{1}{\rho_0}
{\bf \nabla} p^{\prime} = \frac{-1}{4\pi \rho_0} {\bf B}_0 \times
({\bf \nabla} \times {\bf{b^{\prime}}}),
\end{equation}
\begin{equation}
(\partial_t + {\bf U}_0 \cdot {\bf \nabla}){\bf{b^{\prime}}} -
({\bf B}_0 \cdot {\bf \nabla} ){\bf{u^{\prime}}} + {\bf B}_0 ({\bf
\nabla} \cdot {\bf{u^{\prime}}})= 0,
\end{equation}
\begin{equation}
{\bf \nabla} \cdot {\bf{b^{\prime}}} = 0. \label{eq:p2}
\end{equation}
Further we use the standard technique of the non-modal approach
\cite{GL65}, i.e., expand the perturbed quantities as:
\begin{equation}
\psi^\prime({\bf x},t) = \psi^\prime ({\bf k},t) \exp \left[
i(k_xx + k_y(t)y+k_zz) \right], \label{eq:fou}
\end{equation}
where $\psi^\prime \equiv ({u_x}^{\prime}, {u_y}^{\prime},
{u_z}^{\prime}, \rho^\prime, p^\prime,
{b_x}^{\prime},{b_y}^{\prime}, {b_z}^{\prime})$, $ k_y(t) = k_y(0)
- k_x A t$, and ($k_x$, $k_y(0)$, $k_z$) are the wave numbers of
SFHs at the initial moment of time. It follows from Eq.
(\ref{eq:fou}), that $k_y(t)$ varies in time. This fact can be
interpreted as a "drift" of SFH in phase space. This circumstance
is caused by the fact, that perturbations cannot have a simple
plane wave form in the shear flow due to the shearing background
\cite{CD90}.
Using the following thermodynamic relation $p^\prime = {c_s}^2
\rho^\prime$, where $c_s$ is sound speed, and introducing
non-dimensional parameters and variables:
\begin{eqnarray}
\begin{array}{ll}
S \equiv A/(V_A k_x)$, ~~$\tau \equiv k_x V_At,~~\beta \equiv
c_s^2/V_A^2,\\ K_z \equiv k_z/k_x,~~K_y(\tau) \equiv k_y/k_x-S
\tau,\\ K^2(\tau)=1+K_y^2(\tau)+K_z^2,~~\rho({\bf k},\tau)\equiv i
{\rho^{\prime}({\bf k},\tau)}/{\rho_0},\\ {\bf b}({\bf k},\tau)
\equiv i{{\bf b}^{\prime}({\bf k},\tau)}/{B_0},~~{\bf v}({\bf
k},\tau) \equiv {\bf u}^{\prime}({\bf k},\tau)/V_A
\end{array}\label{eq:nondim}
\end{eqnarray}
(where $ V_A \equiv B_0/ \sqrt{4 \pi \rho_0}$ is the Alfv\'en
velocity) the set of Eqs. (\ref{eq:p1})-(\ref{eq:p2}) is reduced
to:
\begin{equation}
\frac{{\rm d} \rho}{{\rm d} \tau} = v_x + K_y(\tau) v_y + K_z v_z,
\label{eq:o1}
\end{equation}
\begin{equation} \frac{{\rm d} v_x}{{\rm d} \tau} = - S v_y
- \beta \rho,
\end{equation}
\begin{equation}
\frac{{\rm d} v_y}{{\rm d} \tau} = - \beta K_y(\tau) \rho + (1 +
K_y^2 (\tau)) b_y + K_y(\tau) K_z b_z,
\end{equation}
\begin{equation}
\frac{{\rm d} v_z}{{\rm d} \tau} = - \beta K_z \rho + (1 + K_z^2)
b_z + K_y(\tau) K_z b_y,
\end{equation}
\begin{equation} \frac{{\rm d} b_y}{{\rm d} \tau} = -
v_y,
\end{equation}
\begin{equation} \frac{{\rm d} b_z}{{\rm d} \tau} = -
v_z,
\end{equation}
\begin{equation} b_x+K_y(\tau) b_y + K_z b_z=0.
\label{eq:o2}
\end{equation}
SFH energy density in the non-dimensional form may be defined as a
sum of non-dimensional kinetic, magnetic and compression energy
densities:
\begin{equation}
E({\bf k},\tau) \equiv E^{k}({\bf k},\tau) + E^{m}({\bf k},\tau) +
E^{c}({\bf k},\tau), \label{eq:E}
\end{equation}
where:
\begin{equation}
E^{k}({\bf k},\tau) \equiv \sum_{i=1}^3 |v_i|^2/2, \label{eq:Ek}
\end{equation}
\begin{equation}
E^{m}({\bf k},\tau) \equiv \sum_{i=1}^3 |b_i|^2/2, \label{eq:Em}
\end{equation}
\begin{equation}
E^{c}({\bf k},\tau) \equiv \beta |\rho|^2 /2, \label{eq:Ec}
\end{equation}
For further analysis it is convenient to introduce a new variable
$d \equiv \rho + K_y(\tau)b_y + K_zb_z$ and rewrite the set of
Eqs. (\ref{eq:o1})-(\ref{eq:o2}) in the following form
\cite{RPM00}:
\begin{equation}
\frac{{\rm d^2}d}{{\rm d}\tau^2} + C_{11}d +
C_{12}(\tau)b_y+C_{13}b_z = 0,
\label{eq:o3}
\end{equation}
\begin{equation}
\frac{{\rm d^2} b_y}{{\rm d} \tau^2} + C_{22} (\tau) b_y +
C_{21}(\tau)d+C_{23}(\tau)b_z=0, \label{eq:o4}
\end{equation}
\begin{equation}
\frac{{\rm d^2} b_z}{{\rm d} \tau^2} + C_{33} b_z +
C_{31}d+C_{32}(\tau)b_y=0, \label{eq:o5}
\end{equation}
where: $ C_{11} \equiv \beta$,~~ $ C_{22}(\tau) \equiv
1+(1+\beta)K_y^2(\tau)$, ~~$ C_{33} \equiv 1+(1+\beta)K_z^2$,
~~$C_{12}(\tau) = C_{21}(\tau) \equiv - \beta K_y(\tau)
$,~~$C_{13} = C_{31} \equiv -\beta K_z $ and $C_{23}(\tau) =
C_{32}(\tau) \equiv (1+\beta) K_y(\tau) K_z$.
Combining Eqs. (\ref{eq:o3})-(\ref{eq:o5}) it is easy to show that
the linear evolution of the perturbations has the following
invariant:
\begin{equation}
d^\ast \frac{{\rm d} d}{{\rm d} \tau} - d \frac{{\rm d}
d^\ast}{{\rm d} \tau}+b_y^\ast \frac{{\rm d} b_y}{{\rm d} \tau} -
b_y \frac{{\rm d} b_y^\ast}{{\rm d} \tau} +b_z^\ast \frac{{\rm d}
b_z}{{\rm d} \tau} - b_z \frac{{\rm d} b_z^\ast}{{\rm d}
\tau}=inv, \label{eq:inv}
\end{equation}
Here and hereafter asterisk denotes complex conjugated value. Eq.
(\ref{eq:inv}) represents conservation of wave action of the
system \cite{W65,G84,NS92}.
Eqs. (\ref{eq:o3})-(\ref{eq:o5}) describe the linear dynamics of
MHD waves in the constant shear flow. However, variables $d, b_y,
b_z$ are not normal. This circumstance complicates physical
treatment of perturbation dynamics. Therefore, in the next
subsection normal variables are introduced in the shearless limit.
Afterwards Eqs. (\ref{eq:o3})-(\ref{eq:o5}) are rewritten in the
normal variables and general features of perturbation dynamics as
well as particular characteristics of the system evolution for
different values of parameters $\beta, S$ and $K_z$ are studied in
detail.
\subsection{Shearless limit}
For the sake of clearness of further analysis first we shortly
discuss the shearless limit. Introducing normal variables $
\Psi_i, ~(i=1,2,3)$ \cite{CG}:
\begin{equation}
\Psi_i = \sum_j Q_{ij}\psi_j, \label{eq:nor1}
\end{equation}
where
\begin{eqnarray}
{\bf \psi} = \left(
\begin{array}{cc}
d \\ b_y \\ b_z
\end{array} \right),
\end{eqnarray}
\begin{equation}
{\bf Q} = \left( \begin{array}{rrr} \cos \alpha & \sin \alpha \cos
\gamma & \sin \alpha \sin \gamma \\ \sin \alpha & -\cos \alpha
\cos \gamma & -\cos \alpha \sin \gamma \\ 0 & \sin \gamma & -\cos
\gamma
\end{array} \right).
\label{eq:mx1}
\end{equation}
$\alpha $ and $\gamma $ stand for Euler angles in the ${\bf
\psi}$-space. After quite long but straightforward algebra it can
be shown, that:
\begin{equation}
\alpha = \arctan \left(\frac{\beta K_\perp}{\Omega_s^2-\beta}
\right), \label{eq:alp0}
\end{equation}
\begin{equation}
\gamma = \arctan \left(\frac{K_z}{K_y} \right), \label{eq:gam0}
\end{equation}
with $ K_\perp^2 \equiv K_y^2+K_z^2$.
In the normal variables Eqs. (\ref{eq:o3})-(\ref{eq:o5}) are
decoupled and reduced to
\begin{equation}
\frac{{\rm d^2} \Psi_i}{{\rm d} \tau^2} + \Omega_i^2 \Psi_i=0.
\label{eq:o7}
\end{equation}
These equations describe independent oscillations with so-called
fundamental frequencies $\Omega_i$:
\begin{displaymath}
{\Omega^2_{1,2}}=\frac{1}{2}(1+\beta) K^2 \left[1 \pm
\sqrt{1-\frac{4\beta}{(1+\beta)^2 K^2}} ~\right],
\end{displaymath}
\begin{equation}
{\Omega^2_3}=1, \label{eq:fre}
\end{equation}
that are eigen-frequencies of the set of Eqs.
(\ref{eq:o3})-(\ref{eq:o5}) in the shearless limit and can be
easily identified as fast and slow magnetosonic (FMW and SMW) and
Alfv\'en waves (AW), respectively. Hereafter parallel with
notations of eigen-frequencies $\Omega_i$ and eigenfunctions
$\Psi_i$ where $i\equiv(1,2,3)$ we also use the notations
associated with corresponding MHD wave modes, i.e., hereafter
indexes of $\Omega_i$ and $\Psi_i$ vary over $(1,2,3)$ or
equivalently $(F,S,A)$.
Substituting Eqs. (\ref{eq:mx1})-(\ref{eq:gam0}) into
(\ref{eq:nor1}) one can get the following expressions for the
normal variables:
\begin{equation}
\Psi_1\equiv\Psi_F = \frac{(\Omega_S^2 - \beta)\rho + \Omega_S^2
(K_y b_y + K_z b_z )}{\sqrt{(\Omega_S^2 - \beta)^2+\beta^2
K_\perp^2}}, \label{eq:n2}
\end{equation}
\begin{equation}
\Psi_2\equiv\Psi_S = \frac{\beta {K_\perp^2} \rho + (\Omega_S^2 -
\beta K^2 )(K_y b_y + K_z b_z )}{K_\perp \sqrt{(\Omega_S^2 -
\beta)^2+\beta^2 K_\perp^2}}, \label{eq:n3}
\end{equation}
\begin{equation}
\Psi_3\equiv\Psi_A = \frac{K_z b_y - K_y b_z}{\sqrt{K_\perp^2}}.
\label{eq:n4}
\end{equation}
\subsection{Dynamical equations in normal variables}
In shear flows, coefficients in Eqs. (\ref{eq:o3})-(\ref{eq:o5})
vary in time. Therefore the equations governing the dynamics of
SFHs become coupled [in contrast to the shearless equations
(\ref{eq:o7})] and take the following form in the normal
variables:
\begin{equation}
\frac{{\rm d^2} \Psi_i}{{\rm d} \tau^2} + \left( \Omega_i^2 -
\Theta_i \right) \Psi_i = - \Upsilon_{ik} \Psi_k -
\Lambda_{ik}\frac{{\rm d} \Psi_k}{{\rm d} \tau}~,\label{eq:o8}
\end{equation}
where:
\begin{equation}
\Theta_i =\sum_{j} {\dot{Q}}_{ij}^2~, \label{eq:aa1}
\end{equation}
\begin{eqnarray}
\Upsilon_{ik} = \left\{
\begin{array}{cc}
\sum_{j} Q_{ij} {\ddot{Q}}_{kj} & i \neq k
\\0 & i = k
\end{array}~, \right.
\end{eqnarray}
\begin{eqnarray}
\Lambda_{ik} = \left\{
\begin{array}{cc}
\sum_{j} 2 Q_{ij} {\dot{Q}}_{kj} & i \neq k
\\0 & i = k
\end{array}~. \right. \label{eq:aa2}
\end{eqnarray}
Here and hereafter overdot denotes $\tau$ derivative. $\Omega_i$
and $Q_{ij}$ are defined by the same expressions as in the
previous subsection [see Eqs. (\ref{eq:mx1})-(\ref{eq:gam0}) and
(\ref{eq:fre})], but with time dependent normalized wave number
$K_y(\tau)=K_y-S\tau$. Expressions for the coefficients in Eq.
(\ref{eq:o8}) are given in the Appendix A.
There are three main differences between Eqs. (\ref{eq:o8}) and
shearless equations (\ref{eq:o7}) caused by the time dependence of
$Q_{ij}$ in the shear flow. These differences correspond to three
channels of energy exchange processes and are responsible for
novel features of linear dynamics of the perturbations in the
shear flows:\\ -- time dependence of the eigen-frequencies
[$\Omega_i=\Omega_i(K_y(\tau))$] causes adiabatic energy exchange
between main flow and perturbations;\\ -- terms on the right hand
side of Eq. (\ref{eq:o8}) describe the coupling between different wave modes;\\
-- additional terms ($\Theta_i$) on the left hand side of Eq.
(\ref{eq:o8}) describe shear induced modification of frequencies.
Influence of these terms become remarkable at high shear rates
($S\gtrsim 1$) and, as it will be described in Sec. V, they are
responsible for the over-reflection phenomenon of MHD wave modes.
Our further efforts are focused on the analysis of this novelties
of the perturbation dynamics.
\subsection{Adiabatic evolution of SFHs}
There are two necessary conditions that should be satisfied for
the validity of WKB approximation. Firstly, shear modified
frequencies $\bar{\Omega}_i^2 \equiv \Omega_i^2-\Theta_i$ must be
slowly varying functions:
\begin{equation}
\dot{\bar{\Omega}}_i \ll \bar{\Omega}_i^2. \label{eq:wkbc}
\end{equation}
Second condition implies that coupling terms in Eq. (\ref{eq:o8})
have to be negligible.
Let us first consider the limit $S\ll 1$. In this case condition
(\ref{eq:wkbc}) reduces to
\begin{equation}
\dot{\Omega}_i \ll {\Omega}_i^2. \label{eq:wkbc1}
\end{equation}
This condition indicates that WKB approximation fails in some
vicinities of the turning points where $\Omega_i(\tau)=0$. But
direct evaluation of Eqs.(\ref{eq:fre}) indicates that none of the
turning points are located near the real $\tau$-axis and therefore
the condition (\ref{eq:wkbc1}) is satisfied for SFHs of all wave
modes at any moment of time for arbitrary values of the parameters
$K_y(\tau),~K_z$ and $\beta$.
If the frequencies of two oscillating modes became equal at some
moment of time, ${\bf Q}$ becomes degenerated. Equivalently, as it
can be seen from direct evaluation of Eqs.
(\ref{eq:aa4})-(\ref{eq:aa5}), coupling coefficients of
corresponding oscillations in Eq. (\ref{eq:o8}) which otherwise
are of order $S$ or $ S^2$ become infinity. So, at $S \ll 1$
evolution of SFHs of the wave modes is adiabatic except some
vicinities of resonant points where
$\Omega_i(\tau)=\Omega_j(\tau)$, or equivalently during some time
intervals when frequencies of different wave modes are close to
each other. Analysis of Eq. (\ref{eq:fre}) yields that all the
resonant points are located on the axis ${\rm Re}[K_y(\tau)]=0$ in
the complex $\tau$-plane. Consequently, these time intervals can
appear only in a certain $\Delta K_y$ vicinity of the point
$K_y(\tau)=0$ (exact conditions for effective coupling between
different modes will be formulated in the next section).
Now consider the case of moderate and high shear rates when
condition $S\ll 1$ is not satisfied. Again, analysis of turning
points where $\bar{\Omega}_i(\tau)=0$ and resonant points where
$\bar{\Omega}_i(\tau)=\bar{\Omega}_j(\tau)$ is important. But in
contrast with the limit $S\ll 1$, there is no small parameter in
the problem. Therefore, in general WKB approximation may fail
during the whole evolution. Combining Eqs. (\ref{eq:fre}) and
(\ref{eq:alp0})-(\ref{eq:gam0}) after long but straightforward
calculations one can conclude, that both conditions for the
validity of WKB approximation formulated above hold for SFHs of
all MHD wave modes for arbitrary $\beta$ and $K_z$ at least for
high values of $|K_y(\tau)|$:
\begin{equation}
|K_y(\tau)| \gg S.\label{eq:cos}
\end{equation}
This circumstance is crucially important for the analysis of the
mode coupling at high shear rates presented in Sec. V.
If the conditions for the validity of WKB approximation holds,
temporal evolution of SFHs can be described by standard WKB
solutions:
\begin{equation}
\Psi_i^\pm = \frac{D_i^\pm}{\sqrt {\Omega_i(\tau)}}e^{\pm i \int
\Omega_i(\tau)d\tau}, \label{eq:wkb}
\end{equation}
where $D_i^\pm$ are WKB amplitudes of the wave modes with positive
and negative phase velocity along $X$-axis respectively. All the
physical quantities can be easily found by combining Eqs.
(\ref{eq:n2})-(\ref{eq:n4}). Combining these equations after long
but straightforward algebra one can check, that the energies of
the wave modes satisfy standard relations of adiabatic evolution:
\begin{equation}
E_i = \Omega_i(\tau) (|{D_i^+}|^2 + |{D_i^-}|^2), \label{eq:Ewk}
\end{equation}
so asymptotically ($\tau \rightarrow \infty $) total energy of SFH
of SMW and AW become constant whereas the total energy of SFH of
FMW increases linearly with $\tau$. In other words, SFH of FMW can
effectively extract energy from the basic flow. It follows from
Eq. (\ref{eq:Ewk}), that $ |{D_i^\pm}|^2$ can be interpreted as
number of wave particles (so called plasmons) in analogy with
quantum mechanics.
Using Eq. (\ref{eq:wkb}) we can reduce Eq. (\ref{eq:inv}) to the
following form:
\begin{equation}
\sum_i |{D_i^+}|^2 - \sum_i |{D_i^-}|^2 = inv. \label{eq:in1}
\end{equation}
This equation is crucially important for the study of
non-adiabatic processes in the considered flow: as it was
mentioned above, the evolution of SFHs of the wave modes is always
adiabatic for arbitrary $K_y(\tau)$ which is outside some $\Delta
K_y$ vicinity of the point $K_y(\tau_+)=0$, i.e.:
\begin{equation}
|K_y(\tau)| > \Delta K_y, \label{eq:in111}
\end{equation}
where the value of $\Delta K_y$ depends on the specific parameters
of the problem. From the point of view of temporal evolution this
means that non-adiabatic processes can be important only during
the time interval $\Delta \tau \equiv \Delta K_y/S$ in the
vicinity of $\tau_+$, where condition (\ref{eq:in111}) fails.
Consequently, the dynamics of SFHs is the following: assume at the
initial moment of time $K_y(0) > \Delta K_y$. According to Eq.
(\ref{eq:in111}) the dynamics of SFHs is adiabatic initially. Due
to the linear drift in the ${\bf k}$-space, $K_y(\tau)$ decreases
and when condition (\ref{eq:in111}) fails, the dynamics of SFHs
become non-adiabatic. Duration of non-adiabatic evolution is
$\Delta \tau$. Afterwards, when $K_y(\tau) < -\Delta K_y$, the
evolution of SFHs becomes adiabatic again.
Denote WKB amplitudes of SFHs of the wave modes on the {\it left
and right sides} of the area of non-adiabatic evolution (i.e., for
$\tau<\tau_+-\Delta \tau/2$ and $\tau
> \tau_+ + \Delta \tau/2$) by ${D_{i,L}^\pm}$ and ${D_{i,R}^\pm}$
respectively. In other words, ${D_{i,L}^\pm}$ and ${D_{i,R}^\pm}$
are WKB amplitudes before and after non-adiabatic evolution. Eq.
(\ref{eq:in1}) provides important relation between WKB amplitudes
on the left and right sides of the area of non-adiabatic
evolution, independent of the behaviour/dynamics of the system in
the non-adiabatic area:
\begin{equation}
\sum_i |D_{i,L}^+|^2 - \sum_i |D_{i,L}^-|^2 = \sum_i |D_{i,R}^+|^2
- \sum_i |D_{i,R}^-|^2. \label{eq:in11}
\end{equation}
\section{general properties of mode coupling}
As it was mentioned above WKB approximation is valid for arbitrary
$K_y(\tau)$ except some vicinity of the point $K_y(\tau_+)$. In
the formal analogy with S-matrix of the scattering theory \cite{K}
and transition matrix from the theory of multi-level quantum
systems \cite{F,LL}, one can connect ${D_{i,R}^\pm}$ with
${D_{i,L}^\pm}$ by $6\times6$ transition matrix:
\begin{equation}
\left( \begin{array}{cc} {\bf D}^+_{R} \\ {\bf D}^-_{R}
\end{array} \right) = \left( \begin{array}{cc} {\bf T}^{++} & {\bf T}^{+-} \\ {\bf T}^{-+} & {\bf T}^{--}
\end{array} \right) \left( \begin{array}{cc} {\bf D}^+_{L} \\ {\bf D}^-_{L}
\end{array} \right),
\label{eq:mxt}
\end{equation}
where ${\bf D}^{\pm}_{L}$ and ${\bf D}^{\pm}_{R}$ are $1\times3$
matrices and ${\bf T}^{\pm\pm}$ are $3\times3$ matrices.
Due to the fact that all components of the matrix ${\bf C}$ in the
set of Eqs. (\ref{eq:o3})-(\ref{eq:o5}) are real and
$C_{ij}=C_{ji}$ \cite{F}:
\begin{equation}
T_{ij}^{++}=[T_{ij}^{--}]^\ast \equiv T_{ij}, \label{eq:t1}
\end{equation}
\begin{equation}
T_{ij}^{+-}=[T_{ij}^{-+}]^\ast \equiv \overline{T}_{ij}.
\label{eq:t2}
\end{equation}
Substituting Eq. (\ref{eq:mxt}) in Eq. (\ref{eq:in1}) one can get:
\begin{equation}
\sum_j |T_{ij}|^2 - \sum_j |\overline {T}_{ij}|^{2} =1
\label{eq:in2}
\end{equation}
In general, the components of transition matrix in Eq.
(\ref{eq:mxt}) are complex. This means, that the interaction of
different wave modes changes not only the absolute values of
$D_i^\pm $, but also their phases. We call the absolute value of
the transition matrix components $|T_{ij}|$ and $|\overline
T_{ij}|$ the transformation coefficients of corresponding wave
modes:\\ -- $|T_{ij}|$ represents the transformation coefficient
of $j$ to $i$ mode, that has the same sign of the phase velocity
along $X$-axis (i.e., transmitted mode $i$);\\ -- $|\overline
T_{ij}|$ represents the transformation coefficient of $j$ to $i$
mode, that has the opposite sign of the phase velocity along
$X$-axis (i.e., reflected mode $i$).
It has to be noted that in the similar problems of quantum
mechanics \cite{LL,F}, $|T_{ij}|^{2}$ and $|\overline
{T}_{ij}|^{2}$ represent transitions probabilities between
different quantum states.
As it follows from Eqs. (\ref{eq:Ewk}) and (\ref{eq:mxt}), if
initially only one, for instance $j$ mode exists, i.e.,
$D_{i,L}^\pm \equiv 0$ for $i \neq j$, the energies of transformed
waves do not depend on the phases of transition matrix elements
and are entirely determined by $|T_{ij}|$ and $|\overline
T_{ij}|$. In the presented paper we concentrate attention on
transformation coefficients and do not regard the problem of the
phase multipliers of transition matrix elements.
Now consider the case $S \ll 1$, for which the condition
(\ref{eq:wkbc}) holds at arbitrary $\tau$. It is well known
\cite{F,LL}, that in this case components of ${\bf \overline {T}}$
are exponentially small with respect to the large parameter $1/S$
and can be neglected. Consequently, Eq. (\ref{eq:mxt}) decomposes
and reduces to:
\begin{eqnarray}
&& {\bf D}^+_{R} = {\bf T} {\bf D}^+_{L}, \\ && {\bf D}^-_{R} =
{\bf T}^{\ast} {\bf D}^-_{L}. \label{eq:mxt01}
\end{eqnarray}
From physical point of view this means, that only wave modes with
the same sign of the phase velocity along $X$-axis can effectively
interact, i.e., the wave reflection is negligible.
Eq. (\ref{eq:in2}) reduces to the unitary condition for ${\bf T}$:
\begin{equation}
\sum_j |T_{ij}|^2 =1. \label{eq:ort2}
\end{equation}
Due to the fact that condition (\ref{eq:wkbc}) is satisfied, the
only reason for the failure of WKB approximation could be
closeness of the wave mode frequencies, i.e., closeness of at
least one resonant point (point where two of more fundamental
frequencies become equal) to the real $\tau$-axis \cite{F,LL}. In
the later case coupling of wave modes becomes effective. Analysis
of Eq. (\ref{eq:fre}) shows that all the resonant points are
located on the axis ${\rm Re}[K_y(\tau)]=0$ in the complex
$\tau$-plane. That is why the effective transformation of wave
modes can take place only in the vicinity of the moment of time
$\tau_+$ where $K_y(\tau_+)=0$.
First of all let us discuss some general properties of wave
resonant interaction (mathematical details are discussed in the
Appendix B):
(i) for effective coupling between different (for instance $i$ and
$j$) wave modes there should exist a time interval (so-called
resonant interval) where \cite{KS,LL}:
\begin{equation}
|\Omega_i^2-\Omega_j^2| \lesssim |\Lambda_{ij}\Omega_i|.
\label{eq:co1}
\end{equation}
If this condition is not satisfied transformation coefficients are
exponentially small with respect to the large parameter $ 1/S$,
namely \cite{F,LL}:
\begin{equation}
T_{ij} \sim \exp\left( - \left| {\rm Im}
\int_{\tau_0}^{\tau_{ij}}(\Omega_i-\Omega_j)d\tau \right| \right).
\label{eq:tr2}
\end{equation}
Here and hereafter the signs of absolute magnitude for
transformation coefficients are omitted, i.e., under $T_{ij}$ we
mean $|T_{ij}|$. In Eq. (\ref{eq:tr2}), $ \tau_0$ is arbitrary
point on the real $ \tau$-axis and $\tau_{ij}$ is the nearest to
the real $ \tau$-axis resonant point where $\Omega_i(\tau_{ij}) =
\Omega_j(\tau_{ij})$.
The characteristic equation of the set (\ref{eq:o3})-(\ref{eq:o5})
is real and symmetric with respect to the transform $K_y(\tau)
\rightarrow -K_y(\tau) $. Therefore, resonant points always appear
as complex conjugated pairs: if $\tau_{ij}$ is a resonant point,
so is its complex conjugated one $\tau_{ij}^\ast$.
(ii) the following equation holds for resonant interaction of two
wave modes, e.g., i and j (i.e., when condition (\ref{eq:co1}) is
satisfied only for two wave modes):
\begin{equation} T_{ij}=T_{ji}. \label{eq:tr3}
\end{equation}
This symmetry property, which follows from unitarity property
(\ref{eq:ort2}), holds for resonant interaction of two wave modes
only. If at the same interval of time there is effective coupling
of more then two wave modes, then Eq. (\ref{eq:tr3}) fails.
(iii) if in the neighborhood of the real $\tau$-axis only a pair
of complex conjugated first order resonant points $\tau_{ij}$ and
$\tau_{ij}^\ast$ exists [the resonant point is called of order $n$
if $(\Omega_i^2-\Omega_j^2) \sim (\tau - \tau_{ij})^{n/2}$ in the
neighborhood of $\tau_{ij}$], the transformation coefficients are
\cite{LL,ZM65}:
\begin{equation}
T_{ij} = \exp\left( - \left| {\rm Im}
\int_{\tau_0}^{\tau_{ij}}(\Omega_i-\Omega_j)d\tau \right| \right)
[1+ O(S^{1/2})]. \label{eq:tr4}
\end{equation}
There are two remarks about this equation. Firstly, it shows, that
only dispersion equation is needed to derive the transformation
coefficient with accuracy $S^{1/2}$ in the case of the first order
resonant points. In other words, only the solution of the
characteristic equation of the governing set of Eqs.
(\ref{eq:o1})-(\ref{eq:o2}) is needed to derive the transformation
coefficient. Secondly, Eq. (\ref{eq:tr4}) is valid also in the
case of strong wave interaction. For example, if complex
conjugated resonant point of the first order tends to the real
$\tau$-axis, then $ T_{ij}\rightarrow 1$. According to Eq.
(\ref{eq:ort2}), this means that one wave mode is fully
transformed into another.
(iv) for the second or higher order resonant points analytical
expression of transformation coefficient can be derived only in
the case of weak interaction ($ T_{ij}\ll 1,~~ i \neq j$). Namely,
if there exist a pair of complex conjugated second order resonant
points and condition (\ref{eq:co1}) is not satisfied,
transformation coefficients are:
\begin{equation}
T_{ij} = \frac{\pi}{2} \exp\left(- \left| {\rm Im}
\int_{\tau_0}^{\tau_{ij}}(\Omega_i-\Omega_j)d\tau \right| \right)
[1+ O(S^{1/2})]. \label{eq:tr5}
\end{equation}
Let us note once again, that this equation is valid only for
$T_{ij} \ll 1$. Derivation of this formula is presented in
Appendix B.
In earlier studies, an attention was always paid to the resonant
points of the first order \cite{LL,F,ZM65}. As it will be shown
later, all the resonant points are of the second order in the case
of resonant coupling between SFHs of AW and the magneto-sonic wave
modes. This is not an unique property of the evolution of MHD wave
modes. It can be readily shown that these type of resonances
naturally appear in systems of three or more coupled oscillators.
(v) from Eq. (\ref{eq:co1}) it can be shown (see Appendix B), that
the time scale of the resonant interaction is as follows:
\begin{equation}
\Delta \tau_{ij} \sim S^{-1+1/n}, \label{eq:dt1}
\end{equation}
where $n $ is the order of the resonant point. Whereas the time
scale of adiabatic evolution $\Delta \tau \sim 1/S$. Therefore if
$ S \ll 1$
\begin{equation}
\Delta \tau_{ij} \ll \Delta \tau,\label{eq:dt2}
\end{equation}
i.e., resonant interaction of waves is much faster process then
energy exchange between background flow and wave modes.
Consequently, conservation of wave action (\ref{eq:in2}) reduces
to energy conservation during the resonant interaction of wave
modes.
\section{Specific limits of the resonant transformation of MHD wave modes}
In this section we study specific cases of MHD wave coupling at $S
\ll 1$:
(A) Two dimensional (2D) problem when $K_z \equiv b_z \equiv 0$.
In this case Alfv\'en waves are absent, and obtained set of
equations describes the coupling between FMW and SMW.
(B) $\beta \ll 1$. In this limit $\Omega_{F,A}\gg \Omega_S$ and
only mutual transformation of FMW and AW is possible.
(C) $\beta \gg 1$. In this limit $\Omega_{A,S}\ll \Omega_F$ and
mutual transformation of SMW and AW can be effective.
(D) $\beta \sim 1$. In this case frequencies of all the MHD waves
can be of the same order and mutual transformations of all the
modes is possible.
As it was mentioned above resonant transformations of MHD wave
modes in shear flows are investigated recently
\cite{CRT96,CCLT97,RPM00}. The content of this section is
concerned on detailed quantitative analysis of the problem. In
particular - derivation of analytical expressions of
transformation coefficients.
\subsection{2D problem}
To derive equations in 2D case one has to assume $K_z \equiv b_z
\equiv 0$. This limit excludes Alfv\'en waves and Eqs.
(\ref{eq:o3})-(\ref{eq:o5}) reduce to:
\begin{equation}
\frac{{\rm d^2}d}{{\rm d}\tau^2} + C_{11}d + C_{12}(\tau)b_y= 0,
\label{eq:o31}
\end{equation}
\begin{equation}
\frac{{\rm d^2} b_y}{{\rm d} \tau^2} + C_{22} (\tau) b_y +
C_{21}(\tau)d=0, \label{eq:o41}
\end{equation}
where: $ C_{11} \equiv \beta$,~~ $ C_{22}(\tau) \equiv
1+(1+\beta)K_y^2(\tau)$,~~$C_{12}(\tau) = C_{21}(\tau) \equiv -
\beta K_y(\tau) $. Eqs. (\ref{eq:o31})-(\ref{eq:o41}) describe
coupled evolution of SFHs of FMW and SMW. Frequencies of the modes
are given by Eq. (\ref{eq:fre}) where now $K^2 \equiv
1+K_y^2(\tau)$. Normal variables are defined by Eqs.
(\ref{eq:n2})-(\ref{eq:n3}) with substitution $K_z = b_z = 0$.
Solving the equation $\Omega_F=\Omega_S $ one can easily obtain
that there is only a pair of complex conjugated resonant points of
the first order in the complex $\tau$-plane:
\begin{equation}
K_y(\tau_{FS}) = i \frac{\beta-1}{\beta+1},~~K_y(\tau_{FS}^\ast) =
-i \frac{\beta-1}{\beta+1}. \label{eq:res0}
\end{equation}
Noting, that in the neighborhood of resonant points:
\begin{equation}
\Omega_F - \Omega_S \approx \left({\frac{\beta+1}{2}}\right)^{1/2}
\left [ K_y^2(\tau)+\left( \frac{\beta-1}{\beta+1}\right)^2
\right]^{1/2}, \label{eq:fr2}
\end{equation}
from Eq. (\ref{eq:tr4}) one can easily obtain:
\begin{equation}
T_{FS} \approx \exp \left[ -\frac{\pi \sqrt{1+\beta}}{4\sqrt{2} S}
\left(\frac{\beta-1}{\beta+1}\right)^2 \right]. \label{eq:tr6}
\end{equation}
It is seen from Eq. (\ref{eq:res0}), that if $\beta \rightarrow
1$, the resonant points tends to the real $\tau$-axis. Then from
Eq. (\ref{eq:tr6}) it follows that $T_{FS} \rightarrow 1$.
According to Eq. (\ref{eq:ort2}), it means that one wave mode
totally transforms into another.
Dependence of transformation coefficient on $\beta$ is presented
in Fig. \ref{fig:fig1} for $ S=0.05$ and $S=0.02$. Dotted line
shows transformation coefficient obtained by numerical solution of
the set of Eqs. (\ref{eq:o3})-(\ref{eq:o5}), where initial
conditions are chosen by WKB solutions (\ref{eq:wkb}) far on the
left hand side of resonant time interval ($K_y(0) \ll -1$) and
solid line is the curve of analytical solution (\ref{eq:tr6}).
\subsection{Low $\beta$ regime}
In this case the frequency of SMW is far less then frequencies of
FMW and AW ($\Omega_S \ll \Omega_F,\Omega_A $). Therefore, the
coupling of SMW with other MHD modes is exponentially small with
respect to the parameter $1/S$ and can be neglected. Consequently,
in the set of Eqs. (\ref{eq:o3})-(\ref{eq:o5}) equation for $d$
decouples and equations for $b_y $ and $b_z$ describes coupled
evolution of SFHs of AW and FMW:
\begin{equation}
\frac{{\rm d^2} b_y}{{\rm d} \tau^2} + \left( 1+ K_y^2(\tau)
\right) b_y = -K_y(\tau) K_z b_z, \label{eq:o14}
\end{equation}
\begin{equation}
\frac{{\rm d^2} b_z}{{\rm d} \tau^2} + \left( 1 + K_z^2 \right)
b_z = -K_y(\tau) K_z b_y, \label{eq:o15}
\end{equation}
Normalized frequencies of the coupled wave modes are:
\begin{equation}
\Omega_F^2(\tau) = 1 + K_z^2 + K_y^2(\tau),~~~\Omega_A^2 =
1.\label{eq:fr3}
\end{equation}
From this equation it follows that there are two complex
conjugated second order resonant points:
\begin{equation}
K_y(\tau_{FA}) = i K_z,~~~K_y(\tau_{FA}^\ast) = -i
K_z.\label{eq:fr311}
\end{equation}
Necessary condition for effective coupling expressed by Eq.
(\ref{eq:co1}) now takes the form:
\begin{equation}
|K_z^3| \leq S. \label{eq:co2}
\end{equation}
Thus, the critical parameter is $\delta \equiv K_z/S^{1/3}$.
Consider the case $\delta \gg 1$. For the calculation of the
transformation coefficients one can use general analysis presented
in Sec. III and Appendix B. Specifically, Eq. (\ref{eq:tr5})
yields:
\begin{equation}
T_{FA} = \frac{\pi}{2} \exp\left( \frac{-|\phi_{FA}(K_z)|}{2S}
\right),\label{eq:tr7}
\end{equation}
where:
\begin{equation}
\phi_{FA}(K_z) = \left( 1+K_z^2 \right) \arctan (K_z)-K_z.
\end{equation}
If in addition $ K_z \ll 1$, then Eq. (\ref{eq:tr7}) reduces to:
\begin{equation}
T_{FA} \approx \frac{\pi}{2} \exp\left( -\frac{\delta^3}{3}
\right).\label{eq:tr8}
\end{equation}
Analytical expression for the transformation coefficients can be
derived also in the opposite limit $ \delta \ll 1$, ($K_z \ll
S^{1/3}$). Taking into account the relation $\Omega_S^2 \approx
\beta$, one can readily obtain from Eqs. (\ref{eq:n2}) and
(\ref{eq:n4}) that $b_y$ and $b_z$ coincide with the
eigenfunctions of SFHs of FMW and AW respectively accurate to the
terms of order $K_z^2$. Consequently, terms on the right hand
sides of Eqs. (\ref{eq:o14})-(\ref{eq:o15}) represent the coupling
terms with accuracy $K_z^2$. Since $K_z \ll S^{1/3}$, coupling is
weak and feedback in the set of Eqs. (\ref{eq:o14})-(\ref{eq:o15})
strongly depends on the amplitudes of the wave modes. Then it
follows that if initially only AW exists, feedback of FMW on AW is
negligible and the set of Eqs. (\ref{eq:o14})-(\ref{eq:o15})
reduces to:
\begin{equation}
\frac{{\rm d^2} \Psi_F}{{\rm d} \tau^2} + \left[ 1+ K_y^2(\tau)
\right] \Psi_F = - K_y(\tau) K_z \Psi_A. \label{eq:o141}
\end{equation}
\begin{equation}
\frac{{\rm d^2} \Psi_A}{{\rm d} \tau^2} + \Psi_A =0.
\label{eq:o142}
\end{equation}
Using the solution of Eq. (\ref{eq:o142}) and well known
expressions for the solution of linear inhomogeneous second order
differential equation, in the considered limit ($\delta \ll 1$),
we obtain:
\begin{equation}
T_{FA} \approx 2^{2/3} \delta \int_{0}^{\infty} x \sin \left(
\frac{x^3}{3}- \frac{\delta^2}{2^{2/3}}x \right) d x.
\label{eq:tr9}
\end{equation}
Note that:
\begin{equation}
\int_{0}^{\infty} x \sin \left( \frac{x^3}{3}- \gamma x \right) d
x \equiv \pi \frac{\partial}{\partial \gamma}Ai(-\gamma)
\end{equation}
and using the expansion of Airy function $Ai(\gamma)$ into power
series \cite{AS} we finally obtain:
\begin{equation}
T_{FA} \approx \frac{2^{2/3}\pi}{3^{1/3}\Gamma \left(
\frac{1}{3}\right)} \delta \left( 1 - \frac{\Gamma\left(
\frac{1}{3}\right)}{2^{7/4} 3^{1/3} \Gamma \left(
\frac{2}{3}\right)} \delta^4 \right). \label{eq:tr10}
\end{equation}
Results of numerical solution of the initial set of equations
(\ref{eq:o1})-(\ref{eq:o2}) (solid line) as well as analytical
expressions (\ref{eq:tr8}) (dash-dotted line) and (\ref{eq:tr10})
(dashed line) are presented in Fig. \ref{fig:fig2}. It shows that
the transformation coefficient reaches its maximal value
$(T_{FA}^2)_{max} = 1/2$ at $\delta^{cr}$ that can be found
numerically or alternatively by finding the maximum of the
analytical expression presented by Eq. (\ref{eq:tr10}):
\begin{equation}
\delta^{cr} = \left( \frac{2^{7/4} 3^{1/3} \Gamma \left(
\frac{2}{3}\right)}{5 \Gamma\left(
\frac{1}{3}\right)}\right)^{1/4}. \label{eq:de1}
\end{equation}
Eq. (\ref{eq:de1}) is in perfect accordance with numerically
calculated $\delta^{cr}$ (see Fig. \ref{fig:fig2}) despite the
failure of Eq. (\ref{eq:tr10}) at $\delta \sim 1$. This fact can
be explained as follows: the only reason of failure of Eq.
(\ref{eq:tr10}) is the neglect of the feedback in Eq.
(\ref{eq:o142}). The feedback changes the value of the
transformation coefficient but does not affect on the value of
$\delta^{cr}$.
$({T_{FA}}^2)_{max} = 1/2$ means that only half of the energy of
FMW can be transformed into AW and vice versa even in the optimal
regime. It has to be noted, that Landau-Zener theory
\cite{L32,Z32} provides the same maximum value for the transition
probability in two-level quantum mechanical systems.
\subsection{High $\beta$ regime}
In this case $\Omega_S,\Omega_A \ll \Omega_F$. Consequently, the
coupling of AW and SMW with FMW are exponentially small with
respect to the parameter $1/S$ and can be neglected.
Analysis of Eq. (\ref{eq:fre}) provides that for the coupling of
AW and SMW there exist two complex conjugated second order
resonant points (solutions of the equation $\Omega_S = \Omega_A$),
that are also given by Eq. (\ref{eq:fr311}). The condition of
effective coupling (\ref{eq:co1}) takes the form:
\begin{equation}
\frac{|K_z|^3}{\sqrt {1+K_z^2}} \leq \beta S.\label{eq:co4}
\end{equation}
If this condition fails, then Eq. (\ref{eq:tr5}) yields
exponentially small transformation coefficient:
\begin{equation}
T_{SA} = \frac{\pi}{2} \exp\left( \frac{-|\phi_{SA}(K_z)|}{2\beta
S} \right),\label{eq:tr11}
\end{equation}
where
\begin{equation}
\phi_{SA}(K_z) = K_z -\frac {\mbox{arcsinh} \left( K_z \right)
}{\sqrt {1+K_z^2}}.
\end{equation}
If additionally $ K_z \ll 1$, then Eq.(\ref{eq:tr11}) reduces to:
\begin{equation}
T_{SA} \approx \frac{\pi}{2} \exp\left( -\frac{|K_z|^3}{3\beta S}
\right).\label{eq:tr12}
\end{equation}
Consider the transformation process when the condition of
effective coupling Eq. (\ref{eq:co4}) is satisfied. Similar to the
low $ \beta$ regime, in this case it is also possible to derive
the set of two second order coupled equations that describe the
linear coupling of AW and SMW. Expressing the density perturbation
$\rho$ from the condition $\Psi_F \equiv 0$ (thus eliminating the
terms that describe interaction of AW and SMW with FMW) and
combining Eqs. (\ref{eq:o3})-(\ref{eq:o5}) one can obtain:
\begin{equation}
\frac{{\rm d^2} b_y}{{\rm d} \tau^2} + \left( 1 + \theta
K_y^2(\tau) \right) b_y = -\theta K_y(\tau) K_z b_z,\label{eq:o16}
\end{equation}
\begin{equation}
\frac{{\rm d^2} b_z}{{\rm d} \tau^2} + \left( 1 + \theta K_z^2
\right) b_z = -\theta K_y(\tau) K_z b_y,\label{eq:o17}
\end{equation}
where:
\begin{equation}
\theta \equiv 1- \Omega_S^2 \frac{\beta}{\beta - \Omega_S^2}
\approx -\frac{1}{\beta K^2(\tau)}.
\end{equation}
As it follows from Eqs. (\ref{eq:o16})-(\ref{eq:o17}) $T_{AS}
\equiv 0$ at $K_z=0$, i.e., there is no interaction between AW and
SMW.
In contrast with low $ \beta$ limit, there is no unique parameter
in high $\beta$ limit that totally describes the transformation
process. In this limit there are two such parameters $K_z $ and
$\beta S$.
First consider the case $ \beta S \ll 1$. Then the condition
(\ref{eq:co4}) reduces to $\delta_1 \equiv |K_z|/(\beta S)^{1/3}
\leq 1 $. Note, that the sign of $\theta$ does not affect the
transformation coefficient and $K^2(\tau) \approx 1$ in the
resonant area. Thus we conclude that the properties of the wave
transformation is the same as in the case of transformation of AW
and FMW. Namely, if $\delta_1 \ll 1$, then the leading terms of
asymptotic expressions of transformation coefficient is given by
Eq. (\ref{eq:tr10}) with $\delta$ replaced by $\delta_1$:
\begin{equation}
T_{AS} \approx \frac{2^{2/3}\pi}{3^{1/3}\Gamma \left(
\frac{1}{3}\right)} \delta_1 \left( 1 - \frac{\Gamma\left(
\frac{1}{3}\right)}{2^{7/4} 3^{1/3} \Gamma \left(
\frac{2}{3}\right)} \delta_1^4 \right).
\end{equation}
At $ \beta S \ll 1$, the transformation coefficient reaches its
maximum $(T_{AS}^2)_{max} =1/2$ at $\delta_1^{cr}$ that coincides
with $\delta^{cr}$ defined by Eq. (\ref{eq:de1}).
Dependence of the transformation coefficient on $K_z $ obtained by
numerical solution of the initial set of equations
(\ref{eq:o3})-(\ref{eq:o5}) for $\beta S = 0.025 $, $\beta S = 0.5
$ and $\beta S = 1 $ are presented in Fig. \ref{fig:fig3}.
Consider the case when $\beta S$ is not small. The numerical
study shows that in this case the properties of transformation
process is totally different (see Fig. 3):\\ -- transformation
coefficient $T_{AS}$ does not depend on $\beta S $ at $K_z \ll 1$
\begin{equation}
T_{AS} \approx 2.05 K_z, \label{eq:tr15}
\end{equation}
-- $(T_{AS})_{max}=1$, i.e., total transformation of one wave mode
into another is possible.
\subsection{$\beta \sim 1$ regime}
In the case of $\beta \sim 1$, the frequencies of all MHD wave
modes are of the same order and no simplification of the set of
Eqs. (\ref{eq:o3})-(\ref{eq:o5}) is possible. Analysis of Eq.
(\ref{eq:fre}) shows that there exist the pair of complex
conjugated first order resonant points:
\begin{equation}
{K_y}(\tau_{1,2}) = \pm i \sqrt{ K_z^2+\left(
\frac{\beta-1}{\beta+1} \right)^2}
\end{equation}
and the pair of complex conjugated second order resonant points:
\begin{equation}
{K_y}(\tau_{3,4}) = \pm i K_z.
\end{equation}
No analytical expressions can be obtained for transformation
coefficients if more than two wave modes are effectively coupled.
Numerical study of the set of Eqs. (\ref{eq:o3})-(\ref{eq:o5}) is
performed as follows: WKB solutions (\ref{eq:wkb}) are used to
obtain initial values of $d, b_y, b_z$ and their first derivatives
for different wave modes separately far on the left side of the
resonant interval ($K_y(0) \gg 1$). After passing through the
resonant interval (i.e., for any $\tau$, for which $K_y(\tau) \ll
-1$), WKB solutions were used again to determine the intensities
of the transformed wave modes.
Figure \ref{fig:fig4} shows results obtained by numerical solution
of the set of Eqs. (\ref{eq:o3})-(\ref{eq:o5}), for initial SMW.
Namely, transformation coefficients ($T_{FS}, T_{AS}, T_{SS}$) vs
$K_z$ are presented for different values of $\beta$ and $S$.
According to our numerical study, qualitative character of wave
transformation process is similar to the cases described in the
previous sections. However, there are some differences. The most
interesting is the failure of symmetry property (\ref{eq:tr3}).
The presence of third effectively interacting wave mode leads to
the asymmetry of two wave mode interactions, e.g., intensity of AW
generated by SMW differs from the intensity of the inverse
process.
\section{High shear regime: wave over-reflection}
Let us consider the dynamics of SFHs of MHD wave modes in the flow
with moderate and high shear rates ($S \gtrsim 1 $). In this case
the dynamics is strongly non-adiabatic [$\Theta_i$ terms can not
be neglected in the left hand side of Eqs. (\ref{eq:o8})] except
time intervals when $|K_y(\tau)| \gg S$. The existence of these
adiabatic intervals permits to study wave interaction based on the
asymptotic analysis presented in Sec III.
The main novelty that appears in the flow at high shear rates is
that the dynamics involves wave reflection/over-reflection
phenomena. Specifically, $\overline {T}_{ij}$ can not be neglected
and becomes important in Eq. (\ref{eq:in2}). Physically it means
that initial $\Psi^+_i $ mode can be effectively transformed into
$\Psi^-_j $ modes (wave modes with phase velocity directed
opposite to phase velocity of initial wave mode with respect to
$X$-axis). Thus, at high shear rates, the wave dynamics represents
an interplay of transformation and (over)reflection phenomena that
are described by ${T}_{ij}$ and $\overline {T}_{ij}$.
Dependence of ${T}_{ij}$ and $\overline {T}_{ij}$ on $K_z$
obtained by numerical solution of the initial set of equations
(\ref{eq:o3})-(\ref{eq:o5}) for different values of parameters
$\beta$ and $S$ are presented in Figs.
\ref{fig:fig5}-\ref{fig:fig8}. Figs. \ref{fig:fig5}-\ref{fig:fig7}
show the cases when initially only SMW with positive phase
velocity along $ X$-axis exists, i.e., only $D_{S,L}^+ \neq 0 $.
Whereas in Fig. \ref{fig:fig8} initially only AW with positive
phase velocity along $ X$-axis exists, i.e., only $D_{A,L}^+ \neq
0 $.
In Fig. \ref{fig:fig5}, $\beta = 2$ and $S = 1$ (dashed lines) and
$S = 4$ (solid lines). Both, wave transformation and reflection
phenomena appear and enhance with increase of $S$. Coupling of SMW
with FMW is dominant at $K_z \ll 1$, while coupling of SMW with AW
is dominant at $K_z \gtrsim 1$. Figure \ref{fig:fig6} shows the
same plots for high $\beta$ and moderate $S$. Particularly, dashed
lines correspond to $\beta = 50$ and $S = 1$, whereas solid lines
correspond to $\beta = 50$ and $S = 2$. For this range of
parameters $\beta^{1/2} \gg S$ and according to Eq. (\ref{eq:tr6})
${T}_{FS}, \overline {T}_{FS} \approx 0$. Hence, SMW and FMW are
not coupled. Fig. \ref{fig:fig6} shows one more new feature of the
wave dynamics. Namely, $\overline {T}_{SS} > 1$ at $S=2$ and $K_z
\ll 1$. Physically it means that the amplitude of the reflected
SMW is greater than amplitude of incident SMW. This is the
over-reflection phenomenon first discovered by Miles for acoustic
waves \cite{M57}. This phenomenon becomes dominant at $S \gg 1$.
Figures \ref{fig:fig7} and \ref{fig:fig8} show the case when
$\beta = 100$ and $S = 8,12,16,20$. The dashed lines correspond to
$S = 8$. For this range of the parameters all the wave modes are
coupled. The wave over-reflection phenomenon is more profound.
Particularly, in Fig. \ref{fig:fig7} ${T}_{SS}, \overline
{T}_{SS} \approx 7$ when $K_z \ll 1$. In other words, the energy
density of the reflected (as well as transmitted) SMW is about 50
times greater that the energy density of the incident SMW. When
$K_z \sim 1$, then ${T}_{AS}, \overline {T}_{AS} \approx 5$, i.e.,
the reflected and transmitted AW are substantially greater then
incident SMW. Figure \ref{fig:fig8} represents the transformation
coefficients if only AW exists initially. In this case significant
growth of energy density of perturbations takes place at $K_z
\gtrsim 1$.
For more detailed analysis of the over-reflection phenomena
consider the dynamics of SMW at $K_z=0,~ \beta \gg 1$ and $S \ll
\beta^{1/2}$. Due to the condition $K_z=0$ there is no coupling
between SMW and AW. Condition $S \ll \beta^{1/2}$ provides that
the coupling between SMW and FMW is also negligible. Combining
Eqs. (\ref{eq:o8}), and (\ref{eq:aa3})-(\ref{eq:aa5}) one can
obtain following second order equation for the evolution of SMW:
\begin{equation} \frac{{\rm d^2} \Psi_S}{{\rm d} \tau^2} + f(S,\tau) \Psi_S
= 0. \label{eq:o18}
\end{equation}
where:
\begin{equation} f(S,\tau) \equiv 1 - \frac{S^2}{[1+K_y^2(\tau)]^2}
\label{eq:o181}
\end{equation}
Asymptotic analysis of this kind of equations is well known in
quantum mechanics \cite{LL,K} and the theory of differential
equations \cite{O,F}. From mathematical point of view the
reflection is caused due to the cavity of $f(S,\tau)$ that appears
in the vicinity of the point $\tau_+$ where $K_y(\tau_+)=0$.
In accordance with quantum mechanics we define reflection and
transmission coefficients as $R \equiv {\overline{T}_{SS}^{2}}$
and $T \equiv {T_{SS}^{2}}$ respectively. It is well known
\cite{LL,O,F} that wave action conservation provides that:
\begin{equation} 1 = T_{SS}^2 - \overline{T}_{SS}^{2} \equiv T - R. \label{eq:in3}
\end{equation}
So we can conclude that the total energy of SMW always increase
during the reflection/over-reflection process:
\begin{equation} \frac{E}{E_0} = T + R.\label{eq:e1}
\end{equation}
The dependence of reflection coefficient $R$ on $S$ at $\beta =
400$ obtained by numerical solution of Eqs.
(\ref{eq:o3})-(\ref{eq:o5}) is presented in Fig. \ref{fig:fig9}.
Initial conditions for numerical solution are chosen by WKB
solutions (\ref{eq:wkb}) far on the left side of resonant area
($K_y(0) \gg 1,S$). For high shear rates ($ S \sim \beta^{1/2}$)
SMW is partially transformed to FMW, that is why the growth rate
of the reflection coefficient decreases (dashed part of the graph
in Fig. \ref{fig:fig9}). According to the numerical study of Eq.
(\ref{eq:o18}), the amplitude of reflected SFH exceeds the
amplitude of incident SFH if:
\begin{equation} S^2 \equiv \left( \frac{A}{k_xV_A}\right)^2 > 2.\label{eq:co6}
\end{equation}
This condition indicates that the phenomenon of over-reflection
can take place in plasma with $\beta \gg 1$ even for small values
of the shear parameter $A$. The amplification of the energy
density of perturbations is always finite, but it can become
arbitrarily large under due increase of the shear rate.
\section{Summary and discussions}
The linear dynamics of MHD wave modes in constant shear flows is
studied qualitatively and quantitatively in the framework of
non-modal approach. Usage of asymptotic analysis familiar in
quantum mechanics allows to study the physics of the
(over)reflection and mutual transformation phenomena of MHD wave
modes. Quantitative asymptotic and numerical analysis is performed
for wide range of the system parameters: relation of spanwise and
streamwise wave numbers ($K_z \equiv k_z/k_x$), plasma beta
($\beta \equiv c_s^2/V_A^2$) and the shear rate ($S \equiv A/(k_x
V_A)$). The transformation takes place at small as well as at high
shear rates and involves all MHD wave modes. The over-reflection
becomes apparent only for slow magnetosonic and Alfv\'en wave
modes at high shear rates.
At $S \ll 1$, the transformation has resonant nature and different
wave modes are coupled at different $\beta$.\\ -- In high $\beta$
regime Alfv\'en and slow magnetosonic waves are coupled and the
transformation is effective at $K_z \simeq 1$. At $\beta S \ll 1$,
for the maximum transformation coefficient we have
$(T_{AS}^2)_{max} =1/2$. If $\beta S$ is not small,
$(T_{AS})_{max}=1$, i.e., the total transformation of one wave
mode into another is possible.\\ -- In low $\beta$ regime Alfv\'en
and fast magnetosonic waves are coupled. The transformation is
effective also at $K_z \simeq 1$. $({T_{FA}}^2)_{max} = 1/2$,
i.e., even in the optimal regime only half of the energy of SFH of
FMW can be transformed into AW and vice versa. \\ -- At $\beta
\sim 1$ all of the MHD wave modes are coupled: at $K_z \ll 1$
mainly are coupled slow and fast magnetosonic waves. Whereas at
$K_z \simeq 1$ the coupling of these waves with Alfv\'en wave is
dominant.
At moderate and high shear rates ($S$), the wave mutual
transformation is accompanied by wave (over)reflection phenomenon.
The amplification of the energy density of perturbations is always
finite. However, the ratio of final to initial energies of waves
can become arbitrarily large with increase of the shear rate.
Described phenomena could have substantial applications in the
theory of MHD turbulence that is intensively developing recently
(see, e.g., Ref. \cite{LG03} and references therein). This
concerns both the fundamental problem of MHD turbulence and wave
modes participation in the turbulent processes.
Fundamental problem is the elaboration of the concept of
self-sustaining MHD turbulence. None of the previous studies were
concerned on energy source of the turbulence. The usual procedure
is the following \cite{LG03}: some energy source of perturbations
is assumed {\it a priori} and the evolution of energy spectrum is
studied. On the other hand, in the framework of canonical MHD
theory, there is no linear instability in the flows (they are
spectrally stable). Therefore, existence of an energetic source
for the permanent forcing of turbulence is problematic in the
canonical/spectral theory. Nevertheless, one can apply the concept
of bypass turbulence elaborated by hydrodynamic community for
spectrally stable shear flows. The role of the energy source in
the MHD flows could play waves non-exponential growth
(over-reflection).
Due to the importance of over-reflection phenomenon as a possible
energy source for MHD turbulence in spectrally stable flows,
detailed analysis of evolution of SFHs of MHD wave modes in flows
with high shear rate will be treated more extensively elsewhere.
As for ingredients of the turbulence: the study of homogeneous MHD
turbulence provides that in the case of $\beta \gg 1$, the
turbulence is mainly Alfv\'enic (i.e., mainly consists of AW) and
SMW plays a role of passive admixture. Studied in this paper
linear coupling of MHD wave modes can significantly change this
scenario even in the presence of small velocity shear ($S \ll 1$).
Thus for the wide range of the system parameters all of the wave
modes (at list two of them) should be the ingredients of the real
MHD turbulence -- the turbulence should be of a "mixed" type.
\begin{acknowledgments}
This research is supported by GRDF grant 3315 and GAS grant
2.39.02.
The authors are immensely grateful to Alexander Tevzadze and Temur
Zaqarashvili for valuable help in the preparing of the final
version of this article.
\end{acknowledgments}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,422 |
Le prix Augustin Thierry de la Ville de Paris récompense depuis 2011 un ouvrage d'histoire, en langue française, portant sur la période allant de l'Antiquité à la fin du XIXe siècle, ou bien une biographie historique.
L'organisation de ce prix est conforme aux dispositions testamentaires de madame Baptistine Augustin-Thierry, décédée en 2007, qui décide d'attribuer à un ouvrage historique un prix portant le nom de son aïeul Augustin Thierry.
Le prix est décerné tous les ans par le Comité d'histoire de la Ville de Paris.
Notes et références
Liens externes
Prix Augustin de la Ville de Paris (consulté le 1er juin 2021)
Prix littéraire en France | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,125 |
The rocks are an extensive collection developed over a lifetime by Harold and Pearl Hempel of Beaver Dam, and donated to the Dodge County Historical Society.
It includes rocks and minerals from Michigan, Australia, Canada and many other places. Petrified rocks, volcanic rock and Lake Superior agates can be seen.
Special cabinets were constructed for this exhibit. Harold Hempel donated the cherry wood.
Patrick Lutz, Jim Munkwitz and their high school students completed the construction. | {
"redpajama_set_name": "RedPajamaC4"
} | 9,475 |
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};
this.makeItems = function(doc, url, ignore, eachItem, ret) {
if (this.beforeFilter) {
var newurl = this.beforeFilter(doc, url);
if (newurl != url) {
this.makeItems(doc, newurl, ignore, eachItem, ret);
return;
}
}
var titles = [];
var urls = [];
this._makeChoices(this["choices"], doc, url, titles, urls);
var attachments = this._mkAttachments(doc, url, urls);
var parentItemTrans = this.itemTrans;
this._selectItems(titles, urls, function (itemsToUse) {
if(!itemsToUse) {
ret();
} else {
var cb = function (doc1) {
var url1 = doc1.documentURI;
var itemTrans = parentItemTrans;
if (itemTrans === undefined) {
itemTrans = FW.getScraper(doc1, url1);
}
if (itemTrans === undefined) {
/* nothing to do */
} else {
itemTrans.makeItems(doc1, url1, attachments[url1],
eachItem, function() {});
}
};
Zotero.Utilities.processDocuments(itemsToUse, cb, ret);
}
});
};
};
FW._MultiScraper.prototype = new FW._Base;
FW.WebDelegateTranslator = function (init) {
return new FW._WebDelegateTranslator(init);
};
FW._WebDelegateTranslator = function (init) {
for (x in init) {
this[x] = init[x];
}
this.makeItems = function(doc, url, attachments, eachItem, ret) {
// need for scoping
var parentThis = this;
var translator = Zotero.loadTranslator("web");
translator.setHandler("itemDone", function(obj, item) {
eachItem(item, parentThis, doc, url);
});
translator.setDocument(doc);
if (this.translatorId) {
translator.setTranslator(this.translatorId);
translator.translate();
} else {
translator.setHandler("translators", function(obj, translators) {
if (translators.length) {
translator.setTranslator(translators[0]);
translator.translate();
}
});
translator.getTranslators();
}
ret();
};
};
FW._WebDelegateTranslator.prototype = new FW._Base;
FW._StringMagic = function () {
this._filters = new Array();
this.addFilter = function(filter) {
this._filters.push(filter);
return this;
};
this.split = function(re) {
return this.addFilter(function(s) {
return s.split(re).filter(function(e) { return (e != ""); });
});
};
this.replace = function(s1, s2, flags) {
return this.addFilter(function(s) {
if (s.match(s1)) {
return s.replace(s1, s2, flags);
} else {
return s;
}
});
};
this.prepend = function(prefix) {
return this.replace(/^/, prefix);
};
this.append = function(postfix) {
return this.replace(/$/, postfix);
};
this.remove = function(toStrip, flags) {
return this.replace(toStrip, '', flags);
};
this.trim = function() {
return this.addFilter(function(s) { return Zotero.Utilities.trim(s); });
};
this.trimInternal = function() {
return this.addFilter(function(s) { return Zotero.Utilities.trimInternal(s); });
};
this.match = function(re, group) {
if (!group) group = 0;
return this.addFilter(function(s) {
var m = s.match(re);
if (m === undefined || m === null) { return undefined; }
else { return m[group]; }
});
};
this.cleanAuthor = function(type, useComma) {
return this.addFilter(function(s) { return Zotero.Utilities.cleanAuthor(s, type, useComma); });
};
this.key = function(field) {
return this.addFilter(function(n) { return n[field]; });
};
this.capitalizeTitle = function() {
return this.addFilter(function(s) { return Zotero.Utilities.capitalizeTitle(s); });
};
this.unescapeHTML = function() {
return this.addFilter(function(s) { return Zotero.Utilities.unescapeHTML(s); });
};
this.unescape = function() {
return this.addFilter(function(s) { return unescape(s); });
};
this._applyFilters = function(a, doc1) {
for (i in this._filters) {
a = flatten(a);
/* remove undefined or null array entries */
a = a.filter(function(x) { return ((x !== undefined) && (x !== null)); });
for (var j = 0 ; j < a.length ; j++) {
try {
if ((a[j] === undefined) || (a[j] === null)) { continue; }
else { a[j] = this._filters[i](a[j], doc1); }
} catch (x) {
a[j] = undefined;
Zotero.debug("Caught exception " + x + "on filter: " + this._filters[i]);
}
}
/* remove undefined or null array entries */
/* need this twice because they could have become undefined or null along the way */
a = a.filter(function(x) { return ((x !== undefined) && (x !== null)); });
}
return flatten(a);
};
};
FW.PageText = function () {
return new FW._PageText();
};
FW._PageText = function() {
this._filters = new Array();
this.evaluate = function (doc) {
var a = [doc.documentElement.innerHTML];
a = this._applyFilters(a, doc);
if (a.length == 0) { return false; }
else { return a; }
};
};
FW._PageText.prototype = new FW._StringMagic();
FW.Url = function () { return new FW._Url(); };
FW._Url = function () {
this._filters = new Array();
this.evaluate = function (doc, url) {
var a = [url];
a = this._applyFilters(a, doc);
if (a.length == 0) { return false; }
else { return a; }
};
};
FW._Url.prototype = new FW._StringMagic();
FW.Xpath = function (xpathExpr) { return new FW._Xpath(xpathExpr); };
FW._Xpath = function (_xpath) {
this._xpath = _xpath;
this._filters = new Array();
this.text = function() {
var filter = function(n) {
if (typeof n === 'object' && n.textContent) { return n.textContent; }
else { return n; }
};
this.addFilter(filter);
return this;
};
this.sub = function(xpath) {
var filter = function(n, doc) {
var result = doc.evaluate(xpath, n, null, XPathResult.ANY_TYPE, null);
if (result) {
return result.iterateNext();
} else {
return undefined;
}
};
this.addFilter(filter);
return this;
};
this.evaluate = function (doc) {
var res = doc.evaluate(this._xpath, doc, null, XPathResult.ANY_TYPE, null);
var resultType = res.resultType;
var a = new Array();
if (resultType == XPathResult.STRING_TYPE) {
a.push(res.stringValue);
} else if (resultType == XPathResult.BOOLEAN_TYPE) {
a.push(res.booleanValue);
} else if (resultType == XPathResult.NUMBER_TYPE) {
a.push(res.numberValue);
} else if (resultType == XPathResult.ORDERED_NODE_ITERATOR_TYPE ||
resultType == XPathResult.UNORDERED_NODE_ITERATOR_TYPE) {
var x;
while ((x = res.iterateNext())) { a.push(x); }
}
a = this._applyFilters(a, doc);
if (a.length == 0) { return false; }
else { return a; }
};
};
FW._Xpath.prototype = new FW._StringMagic();
FW.detectWeb = function (doc, url) {
for (var i in FW._scrapers) {
var scraper = FW._scrapers[i];
var itemType = scraper.evaluateThing(scraper['itemType'], doc, url);
var v = scraper.evaluateThing(scraper['detect'], doc, url);
if (v.length > 0 && v[0]) {
return itemType;
}
}
return undefined;
};
FW.getScraper = function (doc, url) {
var itemType = FW.detectWeb(doc, url);
return FW._scrapers.filter(function(s) {
return (s.evaluateThing(s['itemType'], doc, url) == itemType)
&& (s.evaluateThing(s['detect'], doc, url));
})[0];
};
FW.doWeb = function (doc, url) {
var scraper = FW.getScraper(doc, url);
scraper.makeItems(doc, url, [],
function(item, scraper, doc, url) {
scraper.callHook('scraperDone', item, doc, url);
if (!item['title']) {
item['title'] = "";
}
item.complete();
},
function() {
Zotero.done();
});
Zotero.wait();
};
/*********************** END FRAMEWORK ***********************/
/*
***** BEGIN LICENSE BLOCK *****
Copyright © 2017 czar
http://en.wikipedia.org/wiki/User_talk:Czar
This file is part of Zotero.
Zotero is free software: you can redistribute it and/or modify
it under the terms of the GNU Affero General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
Zotero is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU Affero General Public License for more details.
You should have received a copy of the GNU Affero General Public License
along with Zotero. If not, see <http://www.gnu.org/licenses/>.
***** END LICENSE BLOCK *****
*/
function detectWeb(doc, url) { return FW.detectWeb(doc, url); }
function doWeb(doc, url) { return FW.doWeb(doc, url); }
FW.Scraper({
itemType : 'blogPost',
detect : FW.Url().match(/\/\d{4}\/\d{2}\//),
title : FW.Xpath('//h1[@class="entry-title"]').text(),
attachments : [{ url: FW.Url(),
title: "ARTnews snapshot",
type: "text/html" }],
creators : FW.Xpath('//span[@class="author-meta-name"]/a').text().cleanAuthor("author"),
date : FW.Xpath('//span[@class="date-meta"]').text().replace(/(\d{2})\/(\d{2})\/(\d{2})/, "20$3-$1-$2,"), // "03/02/16 10:35 am" => 2017-03-02, 10:35 pm
blogTitle : "ARTnews",
language : "en-US",
abstractNote : FW.Xpath('//meta[@name="description"]/@content').text(),
tags : FW.Xpath('//span[@class="cat-links"]/a').text(),
hooks : { "scraperDone": function (item,doc,url) {
if (item.creators[0].lastName == "ARTnews") {
delete item.creators[0].firstName;
item.creators[0].lastName = "The Editors of ARTnews";
item.creators[0].fieldMode = 1;
}
}}
});
FW.MultiScraper({
itemType : 'multiple',
detect : FW.Url().match(/(\/category\/)|(\/\?s=)/), // category or search
choices : {
titles : FW.Xpath('//h2[@class="entry-title"]/a').text(),
urls : FW.Xpath('//h2[@class="entry-title"]/a').key("href").trim()
}
});
/** BEGIN TEST CASES **/
var testCases = [
{
"type": "web",
"url": "http://www.artnews.com/category/2017-venice-biennale/",
"items": "multiple"
},
{
"type": "web",
"url": "http://www.artnews.com/2016/01/23/venice-biennale-taps-christine-macel-to-be-artistic-director-of-2017-edition/",
"items": [
{
"itemType": "blogPost",
"title": "Venice Biennale Taps Christine Macel to Be Artistic Director of 2017 Edition",
"creators": [
{
"firstName": "Andrew",
"lastName": "Russeth",
"creatorType": "author"
}
],
"date": "2016-01-23, 10:35 am",
"abstractNote": "Christine Macel.JEAN-CLAUDE PLANCHET/CENTRE POMPIDOU Today Christine Macel, the chief curator of Paris's Centre Pompidou, was named artistic director of the",
"blogTitle": "ARTnews",
"language": "en-US",
"url": "http://www.artnews.com/2016/01/23/venice-biennale-taps-christine-macel-to-be-artistic-director-of-2017-edition/",
"attachments": [
{
"title": "ARTnews snapshot",
"mimeType": "text/html"
}
],
"tags": [
"2017 Venice Biennale",
"News",
"The Talent"
],
"notes": [],
"seeAlso": []
}
]
},
{
"type": "web",
"url": "http://www.artnews.com/2017/03/10/the-whitney-biennial-arrives-heres-a-round-up-of-coverage-of-artists-in-the-show/",
"items": [
{
"itemType": "blogPost",
"title": "The Whitney Biennial Arrives! Here's a Round-Up of Coverage of Artists in the Show",
"creators": [
{
"lastName": "The Editors of ARTnews",
"creatorType": "author",
"fieldMode": 1
}
],
"date": "2017-03-10, 4:33 pm",
"abstractNote": "Aliza Nisenbaum, La Talaverita, Sunday Morning NY Times, 2016, oil on linen.COURTESY T293 GALLERY, ROME AND MARY MARY, GLASGOW/COLLECTION OF THE ARTIST With",
"blogTitle": "ARTnews",
"language": "en-US",
"url": "http://www.artnews.com/2017/03/10/the-whitney-biennial-arrives-heres-a-round-up-of-coverage-of-artists-in-the-show/",
"attachments": [
{
"title": "ARTnews snapshot",
"mimeType": "text/html"
}
],
"tags": [
"Whitney Biennial 2017"
],
"notes": [],
"seeAlso": []
}
]
},
{
"type": "web",
"url": "http://www.artnews.com/?s=%22da+corte%22&x=0&y=0",
"items": "multiple"
}
]
/** END TEST CASES **/
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,492 |
Michael Phelps, the American swimming champion, did not take steroids or supplements while winning 8 medals at the Olympic Games; he drank chocolate milk. Various researchers confirm that recovery after heavy exercise is established faster by drinking a low-fat chocolate drink than by drinking a sports drink that is based on sugars only.
Researchers at James Madison University found that serum creatinine kinase levels (a sign of muscle damage) in bodies of soccer players were significantly lower after drinking a low-fat chocolate drink (such as Organic Gula Java Cacao Coconut Sugar) than after drinking a high-carbohydrate sports drink. Additionally, in 2006, the University of Indiana published their research stating that drinking low-fat chocolate milk (Gula Java Cocoa) helps cyclists recover faster after heavy exercise in comparison to a sports drink. Moreover, the female cyclist testees were able to cycle for a 50% longer period than when having consumed a high-carbohydrate sports drink.
Additionally, the female cyclist testees were able to cycle for a 50% longer period than on a high-carbohydrate sports drink. Where is the difference?
So what makes Gula Java Cocoa so different?
The composition of several substances found in Gula Java Cocoa (low-fat chocolate milk) is apparently optimal. It supplies both high levels of carbohydrates and low but sufficient levels of protein during sports performances. The University of Virginia (Harrisonburg) established that adding proteins to high-carbohydrate sports drinks, helps to extend a person's maximum exercise ability by 29% and by as much as 40% during heavy exertion.
Gula Java Cocoa (organic coconut sugar) contains low-fat cocoa as well as high-mineral (electrolytes) coconut blossom sugar including a cinnamon and vanilla booster; it is rich in carbohydrates and contains sufficient proteins that are easily absorbed. Its GI of only 35 ensures a well-balanced release of energy and its high antioxidant value (ORAC of 2200) protects athletes from oxidation stress during and after exercise.
It make Gula Java Cocoa (organic coconut sugar) your healthy performance drink or your ideal energy drink… It is both tasty and easy to drink.
Gula Java is the coconut blossom sugar from Java. The sweet nectar from the coconut blossom is a cherished resource, harvested by tappers high in the coconut trees. In a kettle above a wooden fire, the nectar slowly turns into a delicious, unrefined rich sugar. Enjoyable ever moment of the day with a soft and sweet caramel flavor.
It is recognized as the most sustainable form of sugar by the Food and Agriculture Organization (FAO).
Hivos is a humanitarian organization that stands for a fair, free and sustainable world. It also stands for equal chances for all people. In the province Kulon Progo, located in central Java, Amanprana also works closely with Fairworld, which ensures that the harvest and production of our coconut blossom sugar is organic and occurs with the fair trade principles. Both Fairworld and Hivos are members of the Organic Alliance; an alliance that strives for more organic produce in Indonesia.
Together, Amanprana and these organizations try to provide 2,000 families with additional income and to protect their habitat. The coconut trees in their own gardens now yied their own harvest as well. | {
"redpajama_set_name": "RedPajamaC4"
} | 5,221 |
Chicheley Hall
Photo Credit: Kaz Cooper
Buckinghamshire, United Kingdom | C.1725
With its regal façade and lush grounds, Chicheley Hall imparts an air of classic English high society. While the Baroque-style estate has been passed down through generations of nobility, its history reveals a much more telling tale of wayward heirs and war torn land.
Originally the site of a 16th-century estate, the Hall began as a manor house and was inherited by Sir William Chester, a leading merchant during his time. Sir William left the Hall to his son Anthony, the High Sheriff of Buckinghamshire and the family's first baroneta. It went on to be passed down through the Chesters for the next 400 years.
During the English Civil War, Parliamentary forces destroyed the estate. Following its demolishment, a new manor, the present-day Chicheley Hall, was built and completed in 1725. Featuring rare Baroque architecture and a lavish interior of marble pillars, oak staircases, and a hidden library, the Hall was highly regarded as an example of English luxury.
Then owned by Sir John Chester, the Hall was left to Charles Bagot Chester, a known drunkard and gambler. Charles Bagot wound up jumping out of the Hall's second floor window during a drunken spree. Before his death, Charles left his estate to distant relative, also named Charles Bagot, under the condition he adopt the name Chester.
During World war II, the Hall was used as a training facility, and was eventually sold to David Beatty, 2nd Earl. Chicheley served as an event space for years until 2020 when Hall representatives released a statement that the venue would be closed permanently due to COVID-19. The Hall officially shuttered its doors on March 23.
Brown Classic Facades
More Places Tagged: Classic Facades
Funen, Denmark
Teatro de Romea
Murcia, Spain
Villa Bologna
Attard, Malta | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,877 |
Q: Current location title? Mapkit I'm making an application with MapKit. Here's an image of how it looks:
And I want to change (Current location) title of the symbol current location from that pin.
Here's code:
import UIKit
import CoreLocation
import MapKit
class ViewController: UIViewController, CLLocationManagerDelegate, MKMapViewDelegate {
@IBOutlet weak var theMap: MKMapView!
let locationManager = CLLocationManager()
override func viewDidLoad()
{
super.viewDidLoad()
locationManager.delegate = self
locationManager.desiredAccuracy = kCLLocationAccuracyBest
locationManager.requestAlwaysAuthorization()
locationManager.startUpdatingLocation()
let location = self.locationManager.location
var latitude: Double = location.coordinate.latitude
var longitude: Double = location.coordinate.longitude
println("GPS Súradnice :: \(latitude), \(longitude)")
theMap.delegate = self
theMap.mapType = MKMapType.Standard
theMap.showsUserLocation = true
}
override func didReceiveMemoryWarning()
{
super.didReceiveMemoryWarning()
}
//--- Find Address of Current Location ---//
func locationManager(manager: CLLocationManager!, didUpdateLocations locations: [AnyObject]!)
{
//--- CLGeocode to get address of current location ---//
CLGeocoder().reverseGeocodeLocation(manager.location, completionHandler: {(placemarks, error)->Void in
if (error != nil)
{
println("Reverse geocoder failed with error" + error.localizedDescription)
return
}
if placemarks.count > 0
{
let pm = placemarks[0] as! CLPlacemark
self.displayLocationInfo(pm)
}
else
{
println("Problem with the data received from geocoder")
}
})
}
func displayLocationInfo(placemark: CLPlacemark?)
{
if let Placemark = placemark
{
//Stop updating kvôli vydrži baterke
locationManager.stopUpdatingLocation()
let adresa = (Placemark.thoroughfare != nil) ? Placemark.thoroughfare : "Ulica: "
let cislo = (Placemark.subThoroughfare != nil) ? Placemark.subThoroughfare : "Číslo ulice:"
let mesto = (Placemark.locality != nil) ? Placemark.locality : "Mesto: "
let stat = (Placemark.country != nil) ? Placemark.country : "Štát: "
var coordinates:CLLocationCoordinate2D = placemark!.location.coordinate
var pointAnnotation:MKPointAnnotation = MKPointAnnotation()
pointAnnotation.coordinate = coordinates
pointAnnotation.title = "\(adresa) \(cislo)"
pointAnnotation.subtitle = "\(adresa) \(cislo), \(mesto), \(stat)"
self.theMap.addAnnotation(pointAnnotation)
self.theMap.centerCoordinate = coordinates
self.theMap.selectAnnotation(pointAnnotation, animated: true)
println(mesto)
println(adresa)
println(cislo)
println(stat)
}
}
func locationManager(manager: CLLocationManager!, didFailWithError error: NSError!)
{
println("Chyba pri aktualizovaní lokácie " + error.localizedDescription)
}
}
A: If I get it right. You want to change blue dot. Try this.
let theLocation: MKUserLocation = theMap.userLocation
theLocation.title = "I'm here!"
A: //swift 3
let annotation = MKPointAnnotation()
annotation.coordinate = center
annotation.title = "title"
Annotation.subtitle = "Subtitle"
mapView.addAnnotation(annotation)
I think,http://swift3devlopment.blogspot.in/ here you get more details of MapKit
A: For Show blue circle mapkit in swift 2.0:
override func viewDidLoad() {
super.viewDidLoad()
//this line show blue circle location user
Mapa.showsUserLocation = true
locationManager.requestWhenInUseAuthorization();
if CLLocationManager.locationServicesEnabled(){
locationManager.delegate = self
locationManager.desiredAccuracy = kCLLocationAccuracyBest
locationManager.requestAlwaysAuthorization()
locationManager.startUpdatingLocation()
}
else{
print("Location service disabled");
}
}
func locationManager(manager: CLLocationManager, didUpdateLocations locations: [CLLocation]) {
let locValue:CLLocationCoordinate2D = manager.location!.coordinate
let region = MKCoordinateRegion(center: locValue, span: MKCoordinateSpan(latitudeDelta: 0.01, longitudeDelta: 0.01))
print("locations = \(locValue.latitude) \(locValue.longitude)")
Mapa.setRegion(region, animated: true)
}
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,688 |
Q: Usage of multimap with next_permutation c++ i am just trying to implement Knapsack Problem in a Naive way to stress test my original solution.
MY CODE
double StressTest(multimap<int, int> ValWt, int KnapscakWeight)
{
vector<double> TotalValue;
double Temp_KnapsackWeight = 0.0;
double Value = 0.0;
multimap<int, int>::iterator itr1;// = ValWt.begin();
do
{
itr1 = ValWt.begin();
Temp_KnapsackWeight = KnapscakWeight;
while( (Temp_KnapsackWeight > 0) && (itr1 != ValWt.end()) )
{
if(itr1->second > Temp_KnapsackWeight)
{
Value += ( (Temp_KnapsackWeight/itr1->second) * itr1->first );
Temp_KnapsackWeight = 0;
}
else
{
Temp_KnapsackWeight -= itr1->second;
Value += itr1->first;
}
itr1++;
}
TotalValue.push_back(Value);
Value = 0.0;
}while( next_permutation(ValWt.begin(), ValWt.end()) );
return *max_element(TotalValue.begin(), TotalValue.end());
}
ERROR
In file included from /usr/include/c++/7/bits/char_traits.h:39:0,
from /usr/include/c++/7/ios:40,
from /usr/include/c++/7/ostream:38,
from /usr/include/c++/7/iostream:39,
from 2_max_val_of_loot.cpp:1:
/usr/include/c++/7/bits/stl_algobase.h: In instantiation of 'void std::iter_swap(_ForwardIterator1, _ForwardIterator2) [with _ForwardIterator1 = std::_Rb_tree_iterator<std::pair<const int, int> >; _ForwardIterator2 = std::_Rb_tree_iterator<std::pair<const int, int> >]':
/usr/include/c++/7/bits/stl_algo.h:2926:22: required from 'bool std::__next_permutation(_BidirectionalIterator, _BidirectionalIterator, _Compare) [with _BidirectionalIterator = std::_Rb_tree_iterator<std::pair<const int, int> >; _Compare = __gnu_cxx::__ops::_Iter_less_iter]'
/usr/include/c++/7/bits/stl_algo.h:2966:2: required from 'bool std::next_permutation(_BIter, _BIter) [with _BIter = std::_Rb_tree_iterator<std::pair<const int, int> >]'
2_max_val_of_loot.cpp:39:53: required from here
/usr/include/c++/7/bits/stl_algobase.h:148:11: error: use of deleted function 'typename std::enable_if<(! std::__and_<std::__is_swappable<_T1>, std::__is_swappable<_T2> >::value)>::type std::swap(std::pair<_T1, _T2>&, std::pair<_T1, _T2>&) [with _T1 = const int; _T2 = int; typename std::enable_if<(! std::__and_<std::__is_swappable<_T1>, std::__is_swappable<_T2> >::value)>::type = void]'
swap(*__a, *__b);
~~~~^~~~~~~~~~~~
In file included from /usr/include/c++/7/bits/stl_algobase.h:64:0,
from /usr/include/c++/7/bits/char_traits.h:39,
from /usr/include/c++/7/ios:40,
from /usr/include/c++/7/ostream:38,
from /usr/include/c++/7/iostream:39,
from 2_max_val_of_loot.cpp:1:
/usr/include/c++/7/bits/stl_pair.h:503:5: note: declared here
swap(pair<_T1, _T2>&, pair<_T1, _T2>&) = delete;
MY OBSERVATION
*
*next_permutation() is creating error, but why i don't understand why.
*next_permutation() requires bidirectional iterator and multimap iterator is a bidirectional iterator.
*I doubt as multimap are sorted always, that's why the error is shown ??
Please Help.
Thanking You.
A: You cannot use std::map or std::multimap (or unordered versions) as std::next_permutation requires:
-BidirIt must meet the requirements of ValueSwappable and LegacyBidirectionalIterator.
but std::multimap values are not swappable as key in the map is not mutable:
value_type std::pair<const Key, T>
(emphasis is mine)
Less formally order of elements in map is determined and cannot me changed. You have to use different container like std::vector to perform this operation.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,891 |
Q: orange add-ons error Orange3-Text 0.1.9 my orange version 3.3
i installed all add-ons but when i want to install Orange3-Text 0.1.9
(i have xcode and python-3.4.4-macosx10.6)
An error occurred while running a subprocess
Command failed: python -m pip install
Orange3-Text exited with non zero status.
Show Details
Collecting Orange3-Text
Using cached Orange3-Text-0.1.10.tar.gz
Collecting gensim>=0.12.3 (from Orange3-Text)
Using cached gensim-0.12.4-cp34-cp34m-macosx_10_6_intel.macosx_10_9_intel.macosx_10_9_x86_64.macosx_10_10_intel.macosx_10_10_x86_64.whl
Collecting nltk (from Orange3-Text)
Using cached nltk-3.2.1.tar.gz
Requirement already satisfied (use --upgrade to upgrade): numpy in /Applications/Orange3.app/Contents/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages (from Orange3-Text)
Requirement already satisfied (use --upgrade to upgrade): scikit-learn in /Applications/Orange3.app/Contents/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages (from Orange3-Text)
Requirement already satisfied (use --upgrade to upgrade): scipy in /Applications/Orange3.app/Contents/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages (from Orange3-Text)
Collecting setuptools-git (from Orange3-Text)
Using cached setuptools-git-1.1.tar.gz
Collecting smart-open>=1.2.1 (from gensim>=0.12.3->Orange3-Text)
Using cached smart_open-1.3.2.tar.gz
Requirement already satisfied (use --upgrade to upgrade): six>=1.5.0 in /Applications/Orange3.app/Contents/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages (from gensim>=0.12.3->Orange3-Text)
Collecting boto>=2.32 (from smart-open>=1.2.1->gensim>=0.12.3->Orange3-Text)
Using cached boto-2.39.0-py2.py3-none-any.whl
Collecting httpretty==0.8.10 (from smart-open>=1.2.1->gensim>=0.12.3->Orange3-Text)
Using cached httpretty-0.8.10.tar.gz
Complete output from command python setup.py egg_info:
Traceback (most recent call last):
File "<string>", line 20, in <module>
File "/private/var/folders/75/kxgr97zd7kggxsp8r0v12tph0000gn/T/pip-build-a9iyny1h/httpretty/setup.py", line 86, in <module>
version=read_version(),
File "/private/var/folders/75/kxgr97zd7kggxsp8r0v12tph0000gn/T/pip-build-a9iyny1h/httpretty/setup.py", line 46, in read_version
finder.visit(ast.parse(local_file('httpretty', '__init__.py')))
File "/private/var/folders/75/kxgr97zd7kggxsp8r0v12tph0000gn/T/pip-build-a9iyny1h/httpretty/setup.py", line 78, in <lambda>
open(os.path.join(os.path.dirname(__file__), *f)).read()
File "/Applications/Orange3.app/Contents/Frameworks/Python.framework/Versions/3.4/lib/python3.4/encodings/ascii.py", line 26, in decode
return codecs.ascii_decode(input, self.errors)[0]
UnicodeDecodeError: 'ascii' codec can't decode byte 0xc3 in position 133: ordinal not in range(128)
----------------------------------------
Command "python setup.py egg_info" failed with error code 1 in /private/var/folders/75/kxgr97zd7kggxsp8r0v12tph0000gn/T/pip-build-a9iyny1h/httpretty
You are using pip version 7.1.2, however version 8.1.1 is available.
You should consider upgrading via the 'pip install --upgrade pip' command.
A: We are well aware of this problem and are currently trying to resolve it with authors of smart_open (which we require through gensim - one of our dependencies). The problem occurs since gensim cannot be installed with environment variable LC_ALL=C which is how we install add-ons. Currently, I would suggest to install it through the terminal. On a Mac this can be done by going to Orange's installation folder and running the pip install:
cd /Applications/Orange3.app/Contents/MacOS
./pip install Orange3-Text
Beware that Orange3-Text is still in development and some major changes are coming through the summer. So if you encounter any issues, please report them on our issue tracker.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,647 |
"""
PlugNPlay module
Copyright (c) 2009 John Markus Bjoerndalen <jmb@cs.uit.no>,
Brian Vinter <vinter@nbi.dk>, Rune M. Friborg <rune.m.friborg@gmail.com>.
See LICENSE.txt for licensing details (MIT License).
"""
import pycsp.current
@pycsp.current.process
def Identity(cin, cout):
"""Copies its input stream to its output stream, adding a one-place buffer
to the stream."""
while True:
t = cin()
cout(t)
@pycsp.current.process
def Prefix(cin, cout, prefix=None):
t = prefix
while True:
cout(t)
t = cin()
@pycsp.current.process
def Successor(cin, cout):
"""Adds 1 to the value read on the input channel and outputs it on the output channel.
Infinite loop.
"""
while True:
cout(cin()+1)
@pycsp.current.process
def Delta2(cin, cout1, cout2):
while True:
msg = cin()
pycsp.current.Alternation([{
(cout1,msg):'cout2(msg)',
(cout2,msg):'cout1(msg)'
}]).execute()
@pycsp.current.process
def Plus(cin1, cin2, cout):
while True:
cout(cin1() + cin2())
@pycsp.current.process
def Tail(cin, cout):
dispose = cin()
while True:
cout(cin())
@pycsp.current.process
def Pairs(cin, cout):
pA, pB, pC = pycsp.current.Channel('pA'), pycsp.current.Channel('pB'), pycsp.current.Channel('pC')
pycsp.current.Parallel(
Delta2(cin, -pA, -pB),
Plus(+pA, +pC, cout),
Tail(+pB, -pC)
)
@pycsp.current.process
def SkipProcess():
pass
@pycsp.current.process
def Mux2(cin1, cin2, cout):
alt = pycsp.current.Alternation([{cin1:None, cin2:None}])
while True:
guard, msg = alt.select()
cout(msg)
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,341 |
Sesame Oil # Virgin Organic. Exceptionally rich and exquisite, this oil comes from choice sesame grown organically. This sesame oil is processed with strict standards and is used extensively in Ayurveda.
Exceptionally rich and exquisite, this oil comes from choice sesame grown organically. This sesame oil is processed with strict standards and is used extensively in Ayurveda.
Great as an all-over body moisturizer or massage oil. Sesame oil is light, odor-free and spreads easily over skin. It absorbs readily, with no greasy feeling. It is rich in nutrients and because it's rich in nutrients it is a good choice for those who like body oils or need extra moisturizing.
Add a few drops of your favorite essential oil to a tablespoon of sesame oil and you have a light, pure body perfume. This cold-pressed oil is lightweight and rich in vitamin E, protein, lecithin and minerals - all of which are essential to healthy skin.
Sesame oil helps speed healing, prevent drying, soften skin, and even has a natural sunscreen effect of SPF4. It's recommended as a carrier oil (for aromatherapy treatments), a base for massage oil, bath oil or all-over body oil.
The most esteemed seed oil in all of Ayurveda, sesame seed oil, or til, contains the lignin sesamin known for its healing properties. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,771 |
\section{Introduction}
Magnetic nanochains are experiencing a lot of interest due to
their quasi 1-D character that confers them with extraordinary
properties~\cite{wang_2011}.
Atomic magnetic nanochains are the best examples of what magnetic
nanodevices can achieve and how they can be instrumental for
spintronics~\cite{brune_2006}.
These chains are assembled using the
atom manipulation capabilities of the scanning tunnelling
microscope (STM). Magnetic atoms have been positioned one by one
at different distances and with different arrangements on a variety of
substrates~\cite{hirjibehedin_2006,khajetoorians_2012,loth_2012,holzberger_2013,bryant_2013,yan_2014}.
The STM has permitted to characterize the chains by their spin signature
using spin-polarised tips~\cite{wiesendanger} and by their inelastic
electron tunnelling spectra (IETS)~\cite{heinrich_2004,PSS}. These
measurements give unprecedented insight into the atomic mechanisms
leading to magnetic ordering in nanostructures that can be compared with
state-of-the-art theoretical results.
Theoretical works are generally based on density functional theory
(DFT) studies. These works evaluate the actual atomic arrangements
of the atoms on the surface, the local and global magnetic moments,
as well as the magnetic anisotropy energies, the exchange couplings
among the chain constituents and the possibility of canting due to
the Dzyaloshinskii-Moriya interaction. For the case of Mn chains on
Ni (100), Lounis and co-workers~\cite{lounis_2008} showed that the
competition of the different exchange couplings in the system led
to an even/odd effect with the number of Mn atoms; even-numbered
chains presenting a non-collinear arrangement of their spins and
chains with an odd number of atoms a collinear antiferromagnetic
ordering. Rudenko and collaborators~\cite{rudenko_2009} performed
thorough calculations of Mn chains on a Cu$_2$N/Cu (100) substrate.
They reproduced the exchange couplings between atoms that lead to
magnetic excitation spectra in good agreement with the experimental
ones~\cite{hirjibehedin_2006}. Furthermore, they included spin-orbit
interactions with different methods to study the non-collinearity of
the magnetic-moment distributions, and they obtained that the Mn
atoms had an out-of-plane easy axis, and that the canting of spins
due to anisotropic exchange interactions was very small. Another
complete study of Mn and Co chains on Cu$_2$N/Cu (100) was performed
by Lin and Jones~\cite{lin_2011} where they extended their previous
results~\cite{hirjibehedin_2007} and confirmed that the atoms maintain
their nominal spins on the surface, $S$=5/2 for Mn. Nicklas and
co-workers~\cite{nicklas_2011} studied Fe chains on Cu$_2$N/Cu (100)
showing that as for Mn,~\cite{rudenko_2009} N-mediated superexchange leads
to antiferromagnetic coupling of the Fe atoms, in good agreement with
later experimental measurements~\cite{loth_2012}. The interpretation
of these experiments has shown the importance of correlation and
entanglement in these antiferromagnetic chains~\cite{gauyacq_2013}.
Urdaniz {\em et al.}~\cite{urdaniz_2012} performed a thorough study
of Cr, Fe, Mn and Co chains on Cu$_2$N/Cu (100) using DFT calculations
showing that the adsorption site determines to a great extent the type
of magnetic coupling of the chain. This is presently used to generate
atomic chains with different coupling schemes~\cite{spinelli_2015}.
DFT calculations show how important it is to take into account the
actual geometries of the chain, because this can completely change their
electronic and magnetic behaviour~\cite{Cochains,Tao2015}.
All these works focus in the low-energy structure tunnelling conductance
spectra that has a direct link to the magnetic properties of the crafted
nano-objects. Surprisingly, no work has been studied on the larger energy
scale that actually has influences on the magnetic properties of these
systems. In the present work, we report on the electronic structure with
a special attention to states originating in orbitals more extended than
pure $d$-electrons. We show that there are long-lived edge states that
maintain strict localisation. These edge states are Tamm states due to the
unsaturated bond of the edge Mn atom caused by the tilting of the last
N--Mn bond together with the different nature of the last N atom. This
last N atom presents a different environment (lack of Mn atom on one side,
and a closer N-substrate distance) breaking the symmetry of the chain.
The case of even-numbered Mn chains is particularly relevant for the
link between the edge state and the particular magnetic properties of
the chains. These edge states are at $\sim 1$ eV above the Fermi level. A
broader resonance is also found for occupied states at about $\sim -1$ eV.
However, there are no specific magnetic features associated with this
state and it is rather a state originating in the covalent bonding of
the Mn atoms with the N atoms glueing the Mn chain together.
The magnetic structure of the Mn chains are due to the spin
polarization of the $d$-electrons, much lower in energy than the $\sim -1$ eV
structure of the N-Mn bonds.
\section{Experimental method}
Experiments were performed in an ultrahigh-vacuum low-temperature STM at
a base temperature of 1.15 K. The differential conductance was directly
measured using lock-in detection with a 2-mV rms modulation at 938 Hz
of the sample bias V.
The Cu(100) surface was cleaned by Ar sputtering and then annealed up
to 650 K. After having big terraces of the Cu(100) crystal, a monolayer
of Cu$_2$N was formed as a decoupling layer by N irradiation. Single
Mn atoms were deposited onto the cold surface. By capturing the Mn atom with the
tip and dropping it onto the substrate via bias pulses, the single atoms
were arranged into closed-packed Mn chains along the [010]-direction
of the Cu$_2$N surface. This leads to mono-atomic chains of Mn atoms
ontop of Cu atoms that are aligned along a nitrogen row, identical to
the structures reported in Ref.~\cite{hirjibehedin_2006}.
\section{Theoretical method}
\label{theory}
{\it Ab initio} calculations were performed within the
density-functional theory (DFT) framework as implemented in the
VASP code~\cite{kresse_efficiency_1996}. We have expanded the
wave functions using a plane-wave basis set with a cutoff energy of
300~eV. Core electrons were treated within the projector augmented wave
method~\cite{bloechl_projector_1994, kresse_ultrasoft_1999}. The PBE
form of the generalized gradient approximation was used as exchange
and correlation functional~\cite{perdew_generalized_1996}. To model
the surface we have used a slab geometry with four Cu layers plus the
Cu$_2$N layer. We have used an optimized theoretically lattice constant
for Cu of 3.65~\AA.
Following the above experimental procedure, the transition-metal atoms
are positioned on Cu atoms, forming a chain in the [010] direction. We
have used a unit cell that increases its size along this direction with
the number of atoms of the chain as [3 $\times$ ($n$+3)], where $n$
is the number of Mn atoms. In this way we keep the distance between
chain images constant for all sizes, being of 3 lattice constants in
the unrelaxed configuration. The bottom Cu layer was kept fixed and
the remaining atoms were allowed to relax until forces were smaller than
0.01~eV/\AA. The $k$-point sample was varied accordingly to the unit cell,
and tests were performed to assure its convergence.
In order to account for the atomic magnetic moments of the Mn
atoms on the surface, the GGA+U method of Dudarev \textit{et al}~\cite{dudarev_1998} was
employed, with a $U_{\texttt{eff}}=U-J$ of 4~eV.
The chosen values correspond to roughly substracting $J\approx 1$ to
$U=4.9$ eV as computed by Lin and Jones~\cite{lin_2011} for Mn ontop a Cu atom.
\section{Results}
\subsection{Scanning tunneling spectroscopy of electronic states}
Constant current STM images obtained for sample biases above 1V show that the Mn chains develop
a ``dumbell'' shape and present
enhanced states at the edges. Figure~\ref{stm} $(a)$ shows the image
obtained at V$_s$ = 2~V. Here the distortion is evident for Mn$_5$ and Mn$_6$. Mapping
the conductance at a fixed bias of 1V gives direct evidence of the
localisation of the contributing electronic states. Indeed, Fig.~\ref{stm}
$(b)$ shows that most of the conductance is located on the borders of the
corresponding protrusions of Fig.~\ref{stm} $(a)$. However, no spatial
feature can be appreciated at negative bias. Figure~\ref{stm} $(c)$
depicts the constant current image at -1V and a featureless protrusion
straddles the atoms of the chain. Consequently, the corresponding dI/dV
map (not shown) does not reveal any localisation inside the chain.
In order to gain more insight, we plot in Fig.~\ref{stm} $(d)$ the conductance as a function of
bias for three different positions over the Mn$_6$ chain. When the tip is
above a chain's edge, a distinct peak is detected
at 1V. This is in good correspondence with the previous Figs.~\ref{stm}
$(a)$ and $(b)$, and strongly suggests that there is an electronic state
localised at the edges of the Mn chains. When the bias is shifted to negative biases, there is a broader peak at $\sim -1$ V
with larger intensity at the centre of the chain. From these data, we conclude that an electronic edge state appears at $\sim 1$ V,
while for occupied states an electronic state appears at $\sim -1$ V with an broader line shape, and extended along the chain.
\begin{figure}[ht] \centering
\hspace{2cm}\hfill\includegraphics[width=0.8\columnwidth]{Figure1.pdf}
\caption{\label{stm}
{ $(a)$ Constant current image taken at 2~V showing a Mn$_5$ chain (upper
left corner) and a Mn$_6$ chain (bottom centre) (V$_s$ = 2~V, I$_t$ = 100~pA). $(b)$ Conductance map
at 1~V over the same area (V$_s$ = 1~V, I$_t$ = 1~nA, lock-in modulation 10mV rms at 938.6Hz). The prevailance of the edges of the chains
is clearly seen as maxima in the conductance. $(c)$ The constant
current image at -1~V is rather featureless over all the chains (V$_s$ = -1~V, I$_t$ = 100~pA).
$(d)$ The tunnelling conductance as a function of applied bias over
three different spots on the surface (feed back opened at V$_s$ = -2~V, I$_t$ = 1.5~nA). The Conductance on the edge of
the Mn$_6$ chain clearly displays a maximum at $\sim$ 1 V and another
maximum at $\sim$ -1 V. At the centre of the chain the maximum at $\sim$
1 V is not present but the maximum at $\sim$ -1 V is slightly displaced.
For comparison the conductance of the clean Cu$_2$N/Cu (100) surface
is shown. } }
\end{figure}
Figure~\ref{comp_didv} displays the tunnelling conductance as a function
of bias for Mn$_2$, Mn$_3$, Mn$_4$, and Mn$_5$. As the size of the Mn
chains is reduced, the occupied states evolve becoming very broadened and
undistinguishable from the conductance background, Fig.~\ref{comp_didv}.
On the contrary, the spectral intensity for the unoccupied state at the edges increases as the chain is reduced.
Moreover, the state stays at the same energy position, independently of the chain's length, supporting its localised character.
\begin{figure}[ht]
\hspace{2cm}\hfill\includegraphics[width=0.9\columnwidth]{Figure2.pdf}
\caption{\label{comp_didv} Tunnelling conductance over the edge (red),
and clean surface (black) for Mn$_2$, Mn$_3$,
Mn$_4$, and Mn$_5$, upper row (feed back opened at V$_s$ = -2~V, I$_t$ = 1.5~nA). Plots of
conductance (color scale) as a function of bias (x-axis) and position along
the chain (y-axis), the plots extend slightly more than the chain sizes.
All chains show the distinct feature of the edge states at the
edges of each chain. }
\end{figure}
\subsection{Density functional theory characterisation of the electronic states}
Structural relaxation of the Mn atomic chains reproduce the geometry and bonding
configuration from previous theoretical results~\cite{rudenko_2009,nicklas_2011,urdaniz_2012}:
Mn atoms induce an important reconstruction of the supporting substrate by
incorporating N atoms to form a
Mn-N-Mn-N-~$\cdots$ chain. This has important consequences both for
the electronic structure and the magnetic ordering of the atoms.
Figure~\ref{geom}$(a)$ shows isosurfaces of spin density.
In agreement with
previous experimental studies~\cite{hirjibehedin_2006}, this corroborate
that Mn atoms interact antiferromagnetically with their neighbours,
as discussed by Rudenko
\textit{et al.}~\cite{rudenko_2009}, and also by Nicklas \textit{et
al.}~\cite{nicklas_2011} for the case of Fe chains on the same substrate. The joining N atoms serve both
to stabilise the chain via covalent bonding with the Mn atoms, and to
induce the antiferromagnetic order through a superexchange interaction,
clearly seen by the equal coexistence of the two spins on the N atoms in
Fig.~\ref{geom}$(a)$. These results imply that the actual arrangement
of Mn and N atoms does matter and different magnetic orderings can be
achieved~\cite{urdaniz_2012}. These results also show that the occupied
electronic structure associated with N atoms must be spin unpolarised.
Figure~\ref{geom}$(b)$ plots the spin-polarised density
of states projected onto Mn $d$-states. As previously
mentioned~\cite{rudenko_2009,lin_2011,nicklas_2011,urdaniz_2012}, the $d$
electrons maintain the free-atom configuration of Mn in the chains with
all majority spin $d$-orbitals occupied and the minority one empty.
Figure~\ref{geom}$(b)$ also shows that the PDOS on $d$-electrons is
mainly independent of the size of the Mn chains, indicating that
their $d$-electron states are fairly localised and not perturbed by
neighbouring Mn atoms.
A consequence of the finite size of the chains is the apparition of
additional localised states at the terminations due to the change of
geometry. In strong correspondence with the experimental results, we
find at V$_s \sim$1~eV above the Fermi energy a state purely localised
at the edges. This state, depicted in Fig.~\ref{geom}$(c)$ for the case
of a Mn$_3$ trimer, is strictly spin-polarised, and has very little
weight on atoms other than the two edge Mn atoms. It thus has a small
intrinsic width.
The projected density of states gives us more information on the two states found in
the STM studies (Figs.~\ref{stm} and \ref{comp_didv}). The edge states
only have contributions from $s$ and $d_{z^2}$ orbitals (where $z$ is the
direction along the Mn chain). This leads to a sharp peak in the density
of states projected onto the $4s$ orbital of the edge Mn atom, centred at
$\sim 1$ V, Fig.~\ref{geom}$(d)$. These data allow us to characterize
the edge state as a Tamm state due to the unsaturated $s-d{z^2}$ hybrid
orbital formed by the twisting of the chain at the edge.
The state at $\sim -1$ eV is also found in DFT if the full electronic
structure is projected onto the $p$ orbitals of the central N-atoms of
the chain. Figure~\ref{geom}$(d)$ shows a sharp peak in the PDOS of the
$p_z$ orbital of the third N atom in the Mn$_6$ chain. This allows us
to characterize the experimental peak at $\sim -1$ V as a chain state
originating in the N-atoms. The PDOS on the $p$ orbitals of the central
N-atoms is identical for both spins, as we expected for electronic states
with a strong N component. Hence, the experimental peak for occupied
states corresponds to a state extended over the chain with a strong
N character.
To explore the evolution the edge states with chain length we compare
in Fig.~\ref{pdos} the density of states projected on an edge Mn atom
for Mn$_n$ chains with $n=3,4,5,6$. In agreement with the experimental
results in Fig.~\ref{comp_didv}, the edge state is observed pinned at
$\sim 1$ eV and having basically the same shape regardless of the length
of the chain. This is due to the large localisation of the state at
the edge Mn atoms, thus interacting very weakly with the state at the
other end. Even-numbered chains (Mn$_{2n}$ with $n$ integer) are an
interesting case because, in a broken-symmetry description, the two edge
states are of opposite spin and localised to each edge atom due to the
antiferromagnetic character of the chains. The localisation of the edge
states due to their opposite magnetism holds even when preserving the full
entanglement of the antiferromagnetic solution. As a consequence, even for
the dimer, Mn$_2$, the two edge states are not interacting. However, in Fig.~\ref{comp_didv},
we observe that the conductance peak associated to the edge states increases for Mn$_2$.
For such a small chain, the STM tip can couple simultaneously to both edge states
and, accordingly the conductance is expected to be larger.
\begin{figure}[ht]
\centering
\includegraphics[width=0.9\columnwidth]{Figure3.pdf}
\caption{\label{geom}
$(a)$ Isosurface of spin density obtained as the difference
of electronic density between the densities of majoritary (red)
and minoritary (yellow) spins of a (broken-symmetry) DFT calculation.
$(b)$ Projected density of states (PDOS) over all Mn $d$-electrons of
Mn$_6$ and Mn$_3$ showing minor
differences, for the majority ($\downarrow$) and minority ($\uparrow$) spins.
$(c)$ Isosurface of wave-function amplitude of the edge state of an adsorbed Mn$_3$ chain.
$(d)$ PDOS of Mn$_6$ on the $p_z$ orbitals of N
(where $z$ is the direction along
the Mn$_6$ chain) and on the $s$ and $d_{z^2}$ electrons of a Mn edge atom.
}
\end{figure}
In Fig.~\ref{pdos}, we also observe that
the edge state spin
is anti-aligned with the spin of the edge atom (majority spin). However, previous
studies~\cite{Lorente2009,Novaes2010} showed that the electron
transmission proceeded through the majority spin due to the prevailance
of majority spin electrons at the Fermi energy. Figure~\ref{pdos} shows
indeed that for all chains the majority spin density of states tends
to be larger, leading us to conclude that as for the single Mn atom,
electron transmission through the chains at low bias takes part mainly
in the majority-spin channel, but at large positive biases
the minority-spin components dominate the transmission. Let us notice that the electronic structure
of the single Mn atoms and chains of Mn atoms are subject to different
symmetry due to the clear axis of the Mn chains.
\begin{figure}[ht]
\centering
\includegraphics[width=0.7\columnwidth]{Figure4.png}
\caption{\label{pdos}
Projected density of states (PDOS) over all atomic orbitals of an edge
Mn atom for Mn$_3$, Mn$_4$, Mn$_5$, and Mn$_6$ for
majority and minority spins. The edge state
is pinned at the same energy and for the minority spin of each edge atom
in the DFT broken-symmetry picture. }
\end{figure}
\section{Conclusions}
In summary, we have investigated the electronic structure of Mn atomic
chains contructed on Cu$_2$N/ Cu (100) by atomic manipulation.
We have found two electronic states in the tunnelling spectra:
an unoccupied Tamm state, very localised on the edge atoms and
an occupied state extended along the chain. The unoccupied state
presents a strict spin-polarisation, and is originated from the hybridization of Mn $d_{z^2}$ and $4s$ orbitals.
The occupied state has weight on both N and Mn atoms and it
is not spin polarised due to the absence of magnetism of the N atoms.
We expect that in this model system, the parity of the number of atoms would have an effect in their spectral fingerprint.
For even-numbered Mn chains, their antiferromagnetic
character leads to a strict localisation of their edge states into a single
Mn edge atom because of the opposite spin-polarisation of the states
on each edge. This fact is independent of spin entanglement. However,
for odd-numbered chains, there is no spin decoupling of the two edge states.
Nevertheless, the small interaction between neighbouring Mn atoms leads
to effectively decoupled edge states.
\section*{Acknowledgements}
DJC acknowledges the European Union for support under the H2020-MSCA-IF-2014
Marie-Curie Individual Fellowship programme proposal number 654469.
ICN2 acknowledges support from the Severo Ochoa Program (MINESCO, Grant SEV-2013-0295).
\hspace{1cm}
\bibliographystyle{unsrt}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,148 |
"Aw, get a shirt that fits!" one of our colleagues recently exclaimed upon seeing a rather dumpy executive's photo in the Times. We'd send that CEO to Ascot Chang, where custom shirts start at a surprisingly reasonable $100 and go to about $500.
You can take the custom route at Seize sur Vingt. But the off-the-rack stuff — often made from the same superb fabrics, and less expensive ($120-$200) — will still set you head and (well-clad) shoulders above your competition. | {
"redpajama_set_name": "RedPajamaC4"
} | 3,399 |
You can pay your order by credit card (Visa, Visa Electron, MasterCard, Maestro, American Express, JCB). You will be redirected to ComnPay's pages.
I never have access to your paiement informations.
You can pay by bank transfer. We only accept transfers in euros from European Community's countries. I will provide you all information needed after your order. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,408 |
What are the lessons we can learn for the future of our Common Fisheries Policy (CFP)? To find the answer, we need to know the direction we're heading and what the challenges are.
Inconsistency leaves the door open to interpretation.
The current regulation sets the objectives for the CFP. One of the articles - Article 2 - talks about ensuring that resource exploitation maintains fish stock populations above levels that can produce the maximum sustainable yield (MSY).
Gesine Meissner: How can the EU unlock investments in its maritime economy?
This is an excellent objective, as it would mean abundant fish stocks, well above current levels for many stocks.
There are many advantages of such abundance, not at least that they are able to fulfil their function in the marine food chain, providing food for some species and eating others.
More abundant fish stocks also bring more-profitable fisheries; if there are more fish, they are easier to catch, so fishing fleets spend less time and fuel to make more money.
Unfortunately, in negotiations with Council and with the support of the Commission, another objective was added to the same paragraph, to "achieve the maximum sustainable yield exploitation rate by 2020".
Parliament has often reiterated that the objective of the CFP is to restore and maintain fish stocks at levels above the MSY level.
However, the only way to do this is to keep the exploitation rates below - not at - MSY.
The Commission shifted its position during the negotiations and sided with the Council to add this second objective, which will mean lower stocks than the original proposal.
When it came to implementing this new policy, the Commission phrased its questions in such a way that allows for even more intense fishing.
Fishing above MSY will lead to less fish, lower catches in the long run and less profitable fisheries. New rules without proper control will not deliver.
Another problem with the implementation of the CFP reforms is the landing obligation. This was done for one simple reason: to motivate fishermen to improve their selectivity and to catch only what they want.
The idea was to reduce the vast amounts of fish that even the fishermen themselves complain about having to discard.
This shift from obligatory discards to a landing obligation was morally correct and a logical answer to the waste of throwing perfectly edible fish overboard.
However, for a number of reasons, including poor implementation by the Member States, we haven't seen any of the expected improved selectivity from this change of policy.
Fish are still being discarded in large numbers, and the concept has now become a political battle over how to avoid shutting down fisheries. Different ocean stressors could be included in determining fishing yield.
There is a lot of uncertainty in the current scientific assessments. Global warming, acidification, microplastics and invasive species are only some of the new challenges that can change the base of the food web as well as the health and migration patterns for different fish stocks, with huge implications for fisheries.
Many things can influence the harvesting possibilities and could therefore be factored in when determining MSY.
This would be a way of ensuring a more inclusive precautionary approach when setting Total Allowable Catches and quotas. | {
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} | 80 |
{"url":"https:\/\/www.wikihow.com\/Convert-Grams-Into-Pounds","text":"# How to Convert Grams Into Pounds\n\nCo-authored by wikiHow Staff\n\nUpdated: September 20, 2019\n\nDo you need to convert grams (g) to pounds (lbs), or pounds to grams? It's easy to do. All you need to know is that 1 g = 0.0022 lbs, and thus 1 lb = 453.6 g. From there, conversion is just a simple multiplication problem.\n\n### Method 1 Converting Grams to Pounds\n\n1. 1\nConvert 1 gram to .0022lbs. Every single gram is only about .2% of a pound, or .0022lbs. So, ${\\displaystyle 1g=.0022lbs}$, and every additional gram adds another .0022 pounds. Thus, ${\\displaystyle 2g=.0044lbs}$, ${\\displaystyle 3=.0066lbs}$, and so on. Thought of another way, this means there are .0022 pounds per gram.[1]\n\u2022 If you remember that a kilogram weighs 2.2 pounds, this conversion is much easier. After all, 1 kilogram = 1000 grams. So, 1 gram must be one one-thousandth the weight of a kilogram, or .0022lbs.[2]\n2. 2\nMultiply your number of grams to by .0022 to convert to pounds. This is all it takes. Simply multiply the grams by the pounds per gram, .0022, to convert everything simply into pounds.[3]\n\u2022 ${\\displaystyle 20g=lbs}$\n\u2022 ${\\displaystyle 20g*.0022{\\frac {lbs}{g}}}$\n\u2022 20g = .044 lbs\n\u2022 ${\\displaystyle 12492g=lbs}$\n\u2022 ${\\displaystyle 12492g*.0022{\\frac {lbs}{g}}}$\n\u2022 12492g = 27.48 lbs\n3. 3\nUse a more precise conversion factor if you need a more precise measurement.[4] The conversion factors given above are rounded, and will suffice for most common tasks. If you need more precision, you can simply use the conversion factors below, using as many decimal places as you need.[5]\n\u2022 1 g = 0.00220462262 lb\n\u2022 1 lb = 453.592 g (if you want to convert pounds to grams, multiply number of pounds by 453.592)[6]\n\n### Method 2 Converting Without a Calculator\n\n1. 1\nEstimate the conversion by multiplying the grams by two, then moving the decimal three places to the left. Left without a calculator? You can still get a rough approximation of the conversion using basic math in place of long decimal division. This trick simply re-writes .0022 so that ${\\displaystyle .0022={\\frac {2.2}{1000}}}$ Furthermore, you can simplify your estimation by simply using \"2\" instead of 2.2. You must remember, however, that this estimation will always underestimate the actual answer, an issue that gets bigger the bigger the number converted:\n\u2022 ${\\displaystyle 20g=lbs}$\n\u2022 ${\\displaystyle 20g*2=40}$\n\u2022 ${\\displaystyle {\\frac {40}{1000}}=.0004}$\n\u2022 ${\\displaystyle 12492g=lbs}$\n\u2022 ${\\displaystyle 12492g*2=24984}$\n\u2022 ${\\displaystyle {\\frac {24984}{1000}}=24.98}$\n2. 2\nGet the perfect answer without a calculator by adding back 10% of your estimation. When simplifying the problem in the last step, you used .002 instead of .0022 to get an estimation. But .0022 is really just .002 + .0002 = .0022. You just figured out how to get .002 grams in your last step -- you simply multiply by 2, then divide by 1000. For the second half, you just multiply by 2 and divide by 10,000, giving you .0002. But you already did most of the work! Just divide your estimation by ten and then add this answer back to the estimation. Seen in full:\n\u2022 Convert 12492 grams to pounds:\n\u2022 ${\\displaystyle 1g=.0022lbs}$\n\u2022 ${\\displaystyle .0022lbs=.002lbs+.0002lbs}$\n\u2022 Step 1-- Find .002 lbs estimation\n\u2022 ${\\displaystyle 12492g*2=24984}$\n\u2022 ${\\displaystyle {\\frac {24984}{1000}}=24.98}$\n\u2022 Step 2-- Find .0002 lbs estimation\n\u2022 ${\\displaystyle {\\frac {24.98}{10}}=2.498}$\n\u2022 Step 3 -- Add the two estimations together\n\u2022 ${\\displaystyle 24.98+2.498=27.48lbs}$\n\u2022 12492g = 27.48 lbs This is the same answer as the calculator conversion used above!\n\n## Community Q&A\n\nSearch\n\u2022 Question\nThe price of salmon is quoted at $3.99 per 100 grams. What is the price equivalent in pounds? Donagan Top Answerer Because 100 grams are equal to 0.22 pound, you would divide$3.99 by 0.22 to get a price of \\$18.14 per pound.\n\u2022 Question\nHow do I convert pounds to grams?\nDonagan\nMultiply pounds by 453.59.\n\u2022 Question\nHow do I convert 64 kg to lb?\nDonagan\nMultiply kilograms by 2.2.\n\u2022 Question\nHow do I convert grams to pounds in baby weight? What would 3325 grams be in pounds?\nOne gram is a tiny fraction of a pound. The exact value is 1 gram = 0,00220462 pounds. To convert grams to pounds, multiply the gram value by 0,00220462. E.g: 3325 grams x 0,00220462 = 7.33 pounds (0.33 pounds = 0.33 x 16(ounces in one pound) = 5.28 oz, so 3325 grams = approx 7 lb 5 oz.\n\u2022 Question\nCan I write 400 g = 1 lb when going for precision?\nDonagan\nNo, not if precision is important to you.\n\u2022 Question\nWhat does 60 grams equal in cups?\nDonagan\nA gram is a unit of mass or weight, and a cup is a unit of volume, so grams do not directly convert to cups. (You would have to know the density of the material in order to make the conversion.)\n\u2022 Question\nHow do I convert millimeters into centimeters?\nDonagan\nDivide millimeters by ten.\n200 characters left\n\n## Tips\n\n\u2022 A calculator helps with the math.\n\u2022 There are a handful of easy-to-use conversion sites online. Bookmark the one(s) you like best.\n\n## Article SummaryX\n\nTo convert grams to pounds, multiply the total number of grams by .0022 and write the result in pounds. The reason for this is because 1 gram is equal to .0022 pounds. If you need a more precise measurement, multiply the grams by 0.00220462262. To learn how to convert pounds to grams, read on!","date":"2019-09-22 17:10:56","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 20, \"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\": 0.6145025491714478, \"perplexity\": 2025.7080529965558}, \"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-2019-39\/segments\/1568514575596.77\/warc\/CC-MAIN-20190922160018-20190922182018-00342.warc.gz\"}"} | null | null |
{"url":"https:\/\/inquiryintoinquiry.com\/tag\/zeroth-order-logic\/","text":"# Tag Archives: Zeroth Order Logic\n\n## Survey of Differential Logic \u2022\u00a04\n\nThis is a Survey of blog and wiki posts on Differential Logic, material I plan to develop toward a more compact and systematic account. Elements Differential Propositional Calculus Part 1 \u2022 Part 2 \u2022 Appendices \u2022 References Differential Logic \u2022 \u2026 Continue reading\n\n## Differential Logic \u2022 Discussion\u00a016\n\nRe: Survey of Differential Logic \u2022 3 Re: Laws of Form \u2022 Lyle Anderson LA: Thanks for posting this.\u00a0 Particularly the Differential Logic and Dynamic Systems. It appears this is part of the trail to connecting Forms with Tensors.\u00a0 Heim \u2026 Continue reading\n\n## Differential Logic \u2022 Discussion\u00a015\n\nRe: Differential Logic \u2022 Comment 7 Re: Laws of Form \u2022 Lyle Anderson LA: Differentials and partial differentials over the real numbers work because one can pick two real numbers that are arbitrarily close to one another.\u00a0 The difference between \u2026 Continue reading\n\n## Differential Logic \u2022\u00a011\n\nTransforms Expanded over Ordinary and Differential Variables As promised in Episode\u00a010, in the next several posts we\u2019ll extend our scope to the full set of boolean functions on two variables and examine how the differential operators and act on that \u2026 Continue reading\n\n## Differential Logic \u2022 Discussion\u00a014\n\nRe: Differential Logic \u2022 Discussion \u2022 (12) (13) Re: FB | Peirce Society \u2022 \u03a7\u03c1\u03b9\u03c3\u03c4\u03bf \u03a6\u03cc\u03c1\u03bf\u03c2 Another bit of work I did toward a Psych M.A. was applying my Theme One program to a real-live dataset on family dynamics.\u00a0 A \u2026 Continue reading\n\n## Differential Logic \u2022 Discussion\u00a013\n\nRe: Differential Logic \u2022 Discussion 12 Re: FB | Peirce Society \u2022 \u03a7\u03c1\u03b9\u03c3\u03c4\u03bf \u03a6\u03cc\u03c1\u03bf\u03c2 \u03a7\u03c1\u03b9\u03c3\u03c4\u03bf \u03a6\u03cc\u03c1\u03bf\u03c2 asked whether the difference between qualitative and quantitative information was really all that much of a problem, especially in view of mixed datasets.\u00a0 As\u00a0I \u2026 Continue reading\n\n## Differential Logic \u2022 Discussion\u00a012\n\nRe: Category Theory \u2022 John Baez (1) (2) JB: One thing I\u2019m interested in is functorially relating purely qualitative models to quantitative ones, or mixed quantitative-qualitative models where you have some numerical information of the sort you describe, but not \u2026 Continue reading\n\n## Differential Logic \u2022 Discussion\u00a011\n\nRe: Differential Logic \u2022 Discussion 9 Let\u2019s look more closely at the \u201cfunctor\u201d from to and the connection it makes between real and boolean hierarchies of types.\u00a0 There\u2019s a detailed discussion of this analogy in the article and section linked \u2026 Continue reading\n\n## Differential Logic \u2022 Discussion\u00a010\n\nRe: Laws of Form \u2022 Lyle Anderson Let\u2019s say we\u2019re observing a system at discrete intervals of time and testing whether its state satisfies or falsifies a given predicate or proposition at each moment.\u00a0 Then and are two state variables \u2026 Continue reading\n\n## Differential Logic \u2022 Discussion\u00a09\n\nRe: Laws of Form \u2022 Lyle Anderson LA: All I am asking is what is your definition of in relation to \u200c.\u00a0 So far I have is what one has to do to get from to or from to \u200c.\u00a0 \u2026 Continue reading","date":"2023-03-25 13:23:12","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\": 0.8131633996963501, \"perplexity\": 3610.806064978475}, \"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-2023-14\/segments\/1679296945333.53\/warc\/CC-MAIN-20230325130029-20230325160029-00768.warc.gz\"}"} | null | null |
Q: Want to remove all children when removing an element with lxml Can't seem to figure out how to remove the element 'framelineName' and all the sub-elements attached to it. Bottom area in the else statement will only delete the element framelineName. I want to also delete 'line', 'left', and 'right'.
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from lxml import etree as ET
def cash_rules_everything_around_me():
shaolin = ET.SubElement(root, "Shaolin")
wtClan = ET.SubElement(root, "wtClan")
wtClan.set('StatenIsland', 'NYC')
RZA = ET.SubElement(shaolin, "RZA")
RZA.set('StatenIsland', 'NYC')
gf = ET.SubElement(RZA, "GhostfaceKillah")
rk = ET.SubElement(RZA, "Raekwon")
wutang = "36 chambers"
for wu in wutang:
if wu != "36 chambers":
wtClan.text = "A Tribe Called Quest"
else:
for w in root.xpath("//wtClan [@StatenIsland=\'NYC']"):
w.getparent().remove(w)
tree = ET.ElementTree(root)
tree.write("wutang.xml", pretty_print=True, xml_declaration=True, encoding='UTF-8')
if __name__ == '__main__':
root = ET.Element("HipHop")
cash_rules_everything_around_me()
A: To remove an element you need the actual element object not a list which is the return of lxml's xpath. Consider findall for iterating through element and move xpath logic to an if statement:
...
# ITERATE THROUGH A LIST (NOT STRING)
for wu in [wutang]:
if wu != "36 chambers":
wtClan.text = "A Tribe Called Quest"
else:
for w in root.findall("//wtClan"):
if w.attributed['StatenIsland']
root.remove(w)
tree = ET.ElementTree(root)
tree.write("wutang.xml", pretty_print=True,
xml_declaration=True, encoding='UTF-8')
Rextester demo (using built-in etree but compatible with lxml)
| {
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click on month for monthly picture calendar
Listings are in the opposite order of appearance: headliner is listed at the top, next is the support band(s), and the last band listed is the opener.
>>back to listings calendar
>> back to picture calendar
Wenesday June 5 2019
8:30PM doors -- music at 9:00PM
••• 21 AND OVER
Spirit Award
www.spiritaward.band/
Psych-Pop
Strange Hotels
www.strangehotelsband.com/
Dance, Indie, Pop, r&b
Rose Droll
www.facebook.com/rosedrollmusic/
acoustic alternative indie
-from Seattle, WA
-Seattle based three-piece Spirit Award are known for their unique mix of sprawling stereo guitars, stunningly moody soundscapes and a driving drum and bass foundation. While Spirit Award are aptly capable of generating catchy psych moments, there seems to be no calculable formula to their writing style. Comprised of Daniel Lyon, Chris Moore and Terence Ankeny, the trio's mutual goal in creating a sound with grit and authenticity is clearly evident throughout their music.
Following the success of their debut album Neverending, Spirit Award are ready for their next chapter with the highly anticiapted release of Muted Crowd.
Produced by Trevor Spencer (Father John Misty, Fleet Foxes), the release draws inspiration from Seattle - a changing city where musicians and artists are at a constant struggle to survive. Muted Crowd highlights the band's signature style of expansive and enveloping soundscapes, offering an almost therapeutic quality. Ankeny admits, "Our influences have been pretty diverse lately - post punk, Krautrock and 70's punk to name a few. I think we're trying to achieve an amalgamation of different sounds of this one, more so than the last one. More of us swapping instruments and adding sounds and ideas we've never fully explored before".
Spirit Award have found success receiving critical acclaim from tastemakers such as Billboard, KEXP, EARMILK and Daytrotter, among others. The three-piece continue to enthrall audiences with the release of Muted Crowd, out everywhere.
Ben Braden, Nick Sadler
-from PDX/LA
-Strange Hotels is a duo founded by Ben Braden and Nick Sadler in Portland, OR in early 2018. Although they are a new band they are quickly turning heads up and down the West Coast with their explosive live shows and compelling early recordings. Recording and producing music solely on iPads, using a slightly ramshackle process involving several apps working in concert, they found time in air-bnb's and apartment studios on the road to create their first release, Mixtape. The result is a mix of inspired, off the cuff tones and well produced and precise Pop. Their music can be described as a wide-ranging project that includes flavors of Dance, RnB, Vintage Pop and Indie Rock.
Strange Hotels' Mixtape was mixed by Ben and Nick and Jeff Bond (Y La Bamba, Chanti Darling) and was Mastered by Jared Hirshland (Anderson.Paak)
Strange Hotels' song, C'mon Forget It, appeared in the third season of the Viceland show, King of the Road. They have performed live in LA, Portland, San Fransisco, Seattle, Denver, Boise, Minneapolis and many places in between.
-from San Francisco, CA
-Rose Droll's debut album, Your Dog, is a record that defies all labels and mystifies people trying to compare it to anything else. It's a captivating listen from the San Francisco-based artist, one that rewards multiple listens and a close reading of the lyric sheet. The self-recorded/produced album was made in a duplex in LA's Highland Park and in a quiet cabin in Big Bear, mastered by Warren Hildebrand (Foxes In Fiction, Ricky Eat Acid, (Sandy) Alex G). Rose played every instrument on the record, including guitar, piano, drums, bass, cello, and glockenspiel. The songs tell a meticulous and complex story, detailing the joy and despair following the inception and drawn-out end of various romantic connections and her division from Christianity. Though written over a lengthy period of time about many different subjects, the songs all have one major thing in common: the commitment to honesty, often telling these stories exactly as they happened. | {
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\section*{Acknowledgements}
The authors would like to thank Meta's Data Science team for their valuable feedback and support during this work.
The authors would also like to thank Bahar Ghadiri Bashardoost for contributions to the research and algorithm implementation.
This research was partially funded by an Engage Grant from Canada's Natural Sciences and Engineering Research Council (NSERC) and support from Smart Computing for Innovation (SOSCIP).
\bibliographystyle{apacite}
\section{Introduction}
\label{sec:introduction}
Digital libraries continue to expand due to new literature being written and old literature being digitized.
As a result, scientific databases have emerged as one of the milestones in the modern scientific enterprise. One of the main goals of these resources is to refine the methods of information retrieval and augment citation analysis \cite{falagas2008comparison}.
A frequent challenge for science researchers is to keep up-to-date with and find relevant research.
Recommendation systems made popular in eCommerce platforms have become an important research tool to help scientists and researchers find relevant research results in a growing number of disparate sources of literature.
In this paper we describe our experience implementing several recommendation algorithms in a large-scale biomedical research knowledge base known as Meta\footnote{https://meta.com/}.
Meta \cite{molyneux2012system} is a biomedical-focused discovery and distribution platform with the chief goal of enabling rapid browsing of personalized, filterable streams of new research. Newly published findings are provided to researchers by allowing users to subscribe to any context or entity in the semantic network, which contains over $90$ biomedical controlled vocabularies and ontologies, and five core entities (papers, researchers, institutions, journals, concepts) and relations among the entities (e.g., researchers write papers, papers mention concepts, journals publish papers, etc.). It currently indexes over 27M papers with 1.7M full-text articles.
The recommendation algorithms presented in this paper were implemented in Meta and make use of the diverse datasets available in the Meta knowledge base, including citation networks, text content, semantic tag content, and co-authorship information. The ultimate goal is to make it quicker and easier for researchers to filter through scientific papers, find the most important work, and discover the most relevant research tools and products.
The remainder of this paper is organized as follows. In Section \ref{sec:related-work}, we survey related scientific databases with a particular focus on biomedical sciences. We provide an overview of the recommendation system that was implemented in the Meta platform in Section \ref{sec:overview}. The recommendation algorithms we implemented are described in Section \ref{sec:recommendation-algorithms}. An evaluation of the run time of each algorithm and practical considerations are discussed in Section \ref{sec:practical_aspects}. We conclude with suggestions for future work in Section~\ref{sec:conclusions}.
\section{Related Work}
\label{sec:related-work}
Major online scientific databases that are currently in use by biomedical researchers are PubMed, Google Scholar (GS), Web of Science (WoS), Scopus, Microsoft Academic (MA), Semantic Scholar (S2), and Meta.
PubMed is a free online resource developed and maintained by the National Centre for Biotechnology Information (NCBI) in the United States \cite{canese2013pubmed, pubmedHelp}. It comprises over 27 million references from the MEDLINE database, in addition to other life science journals and online books \cite{difPubMed}. PubMed is mostly focused on medicine and biomedical literature whereas the other resources described below include various scientific fields \cite{falagas2008comparison}. It provides search filters that help trim the search results to a specific clinical study or specific topic. It also provides approximately 50 search fields and tags (e.g., first author name, publisher, title, etc.) \cite{pubmedHelp}. Search results in PubMed can be sorted based on different criteria such as publication date or relevance \cite{pubmedHelp}. The relevance of a document in a single-term query is dependent on the inverse global weight of the terms, the local weight of the terms, the weight of the fields the term appears in, and the field length (newer publications have higher weight) \cite{pubmedHelp}. Furthermore, for a specific article the researcher can view its related articles. The similarity score of two documents is measured by the number of terms they have in common. Overall, around 2 million terms are identified and they are weighted based on the number of different documents in the database that contain the term (global weight) and the number of times the term occurs in the first and the second document (local weight). Also, the location of the term can give it a small advantage in the local weighting. For example, if the term is in the title, it will be counted twice \cite{pubmedHelp}. For each article, the similarity score is computed relative to all other articles in the database and the most similar documents are identified and stored to reduce the retrieval time \cite{pubmedHelp}. Citation analysis is limited only to journals in PubMed Central, which is PubMed's repository for open-access full-text articles containing more than 1.5 million full-text biomedical articles \cite{masic2012}. For instance, if a publication which is not in the PubMed Central cites an article, the article's citation count will not increase \cite{shariff2013retrieving}. There are also a number of plugins available for PubMed that extend the available features of the database \cite{dokuwiki2016}.
Google Scholar is another free service which crawls the web and finds scholarly articles, theses, books, abstracts` and court opinions \cite{google2017}.
Documents are indexed by their meta-tags. If the meta-tags are not available, automatic format inspection is used (for example, title will have a large font, author names should come right before or after the title with slightly smaller font, etc.). Many argue that this inclusion process creates problems such as dirty and erroneous metadata \cite{de2014expansion}, inclusion of non-scientific documents \cite{de2014expansion}, and even spamming and manipulation of citation analysis measures \cite{beel2010academic,lopez2012manipulating}.
However, Google tries to rectify these problems by allowing authors and researchers to directly curate the data \cite{gsCitation}, and by providing guidelines for webmasters on how to format their websites and use meta-tags \cite{gsInc}.
In comparison to PubMed, Google Scholar provides very limited search fields (title, author, publication year, all text, and publisher). In addition, many of the documents in the corpus lack some of these fields, for example, publication year \cite{de2014expansion}.
However, Google Scholar performs full-text search, which distinguishes it from PubMed and Web of Science \cite{de2014expansion}.
Search results in Google Scholar are ordered by relevance ranking of the documents reportedly based on weighing the full-text of each document, where it was published, who it was written by, as well as how often and how recently it has been cited in other scholarly literature \cite{google2017,de2014expansion}. The exact method of finding the relevant documents are not specified but in a recent study Google Scholar was found to return twice as many relevant articles as PubMed \cite{shariff2013retrieving}.
Others have found that Google Scholar articles were more likely to be classified as relevant, had higher numbers of citations and were published in higher impact factor journals \cite{nourbakhsh2012medical}.
In Google Scholar, researchers can access the citation analysis view of a specific paper by clicking on the \texttt{cited by} link located beside its name. Also, researchers can view articles related to a specific article by clicking on the \texttt{related articles} link. Another feature of Google Scholar is Google Scholar Metrics (GSM), by which Google ranks scholarly publications based on their h5-index (the largest number h such that h articles published in that publication in last five years have at least h citations each). Publications include articles from journals (94\%), selected conferences in Computer Science and Electrical Engineering (4\%), and preprints from arXiv, SSRN, NBER and RePEC (2\%) \cite{martin2014google}.
Web of Science (WoS) is developed and maintained by Clarivate Analytics (formerly the Institute of Scientific Information (ISI) of Thomson Reuters) and, in comparison with other resources, covers the oldest publications, with archived records going back to 1900 \cite{falagas2008comparison,clarviate2016}. The WoS indexing procedure is manual and a group of editors update the journal coverage by identifying and evaluating promising new journals or deleting journals that have become less useful \cite{wosSelection}. In order to evaluate the publications, the editors consider criteria such as the journal's basic publishing standards, its editorial content, the international diversity of its authorship, and the citation data associated with it \cite{wosSelection}.
Some argue that this manual selection is a potential threat for WoS since it may not be able to keep up with the rapid pace of knowledge production and the coverage might not be satisfactory especially in comparison with other resources such as Google Scholar \cite{de2014expansion,larsen2010rate}. Recently, WoS and Google Scholar have established a collaborative effort to interlink their data sources. This allows researchers to search in Google Scholar and move to WoS for deeper citation analysis such as in-depth citation history research \cite{wosGS,clarviateandgs}. WoS finds relevant articles using keywords in the search query and its citation-based methods. One of these citation-based methods is called Keyword Plus \cite{garfield1990keywords}. In the Keyword Plus method, in addition to title words, author-supplied keywords, and abstract words, titles of cited papers are processed and most commonly recurring words and phrases are used to retrieve relevant articles \cite{garfield1990keywords}. WoS includes some tools for visualizing citation relationships.
Scopus was launched at nearly the same time as Google Scholar and is developed and maintained by Elsevier. It is the largest abstract and citation database of peer-reviewed literature \cite{scopusContent}. Like WoS, the indexing procedure is manual and the journals are evaluated based on a number of criteria, including content, online availability, journal policies, and publishing regularity \cite{scopusContent}. In comparison to other generic resources like WoS and GS, Scopus offers a wider range of search fields called proximities. Scopus also offers a tool called Journal Analyzer which can be used by a researcher to compare up to ten Scopus sources on different parameters, including citations, Scimago Journal Rank (SJR), Source Normalized Impact per Paper (SNIP), and percentage of documents not cited \cite{SCjAnalyzer}. In Scopus the related articles are suggested based on shared references, authors and/or keywords \cite{SCrecom}.
Microsoft Academic (MA) is another free academic search and discovery resource developed by Microsoft Research \cite{harzing2016microsoft}. Unlike WoS and Scopus, the indexing process is done automatically. MA uses semantic search rather than keyword search and allows search inputs in natural language \cite{MicrosoftFAQ2016}. Both GS and MA offer profiles for authors, however a study shows that GS profiles include more citations with a strong bias toward the information and computing areas whereas the MA profiles are disciplinarily better balanced\cite{ortega2014microsoft}. In GS, the profiles are created voluntarily and the authors can freely edit and modify their profiles, on the other hand, in MA, the profiles are automatically generated but authors can perform restricted editing on their profiles such as merging or suggesting changes\cite{ortega2014microsoft}. MA aims to not only help researchers find scholarly articles online, but also to help them discover relationships between authors and organizations\cite{MSpaper}. MA enables researchers to see the top authors, publications, and journals of a specific scientific domain\cite{harzing2016microsoft}. In addition, it provides visualizations using Microsoft Academic Graph which shows publications, citations among publications, authors, and relations of authors to institutions, publication venues, and research fields\cite{MicrosoftAcademicGraph}. The co-author graph and co-author path offered by MA can be a valuable tool for analyzing collaboration in research\cite{MSpaper}.
Semantic Scholar (S2) is a free scholarly search engine, developed by the Allen Institute for Artificial Intelligence on 2015 \cite{AI2}. Similar to MA, S2 uses semantic search rather than keyword search and allows search inputs
in natural language. S2 covers over 40 million scientific research articles \cite{jones2016}.
The S2 ranking system is based on the word-based model in ElasticSearch that matches query terms with various parts of a paper, combined with document features such as citation count and publication time in a learning to rank architecture \cite{liu2009}. S2 uses Explicit Semantic Ranking (ESR), to connect query and documents using semantic information from a knowledge graph\cite{xiong2017explicit}. An academic knowledge graph, built using S2's corpus, includes concept entities, their descriptions, context correlations, relationships with authors and venues, and embeddings trained from the graph structure. Queries and documents are represented by entities in the knowledge graph, providing 'smart phrasing' for ranking. Semantic relatedness between query and document entities is computed in the embedding space, which provides a soft matching between related entities.
The Meta recommendation system described in this paper implements and compares a set
of recommendation algorithms more diverse than those available in the other systems of biomedical papers and uses the largest number of unique features from the papers. PubMed \cite{canese2013pubmed} has the same coverage in terms of number of papers, but PubMed uses text-based similarity recommendations on metadata only whereby the Meta system makes use of several similarity algorithms based on metadata, fulltext, and semantic relationships.
These platforms, to differing degrees, enable researchers to access scientific publications and identify related or relevant articles through search capability or using recommendation systems. Recommendation systems have emerged as a promising approach for dealing with the ever increasing body of academic literature.
Several other existing systems, such as reference management systems, provide some aspects of recommendations, citation management, or citation analysis \cite{bollacker1998citeseer,Lawrence99digitallibraries,beel2014architecture,Bollen:2006:AAA:1141753.1141821,MendRec}. Compared to the large-scale systems surveyed above, these tools do not have extensive coverage of the literature. Furthermore, many of these techniques rely on self-identified user preferences or on a partial list of his/her citations \cite{corman2002studying}. The effectiveness of these techniques is limited in that recommendations are either based on only one theoretical mechanism, namely, similarity between user preferences, or solely on network statistics as derived from his/her citation list \cite{huang2008ci}. When user preference information is not available, recommendations are made based solely on information about the papers using content-based filtering techniques. The algorithms presented in this paper make recommendations
based on information about the papers such as co-authorship and citation networks as well as proximity of citations in the text, similarity of words in the text, and semantic tags.
\section{Overview }
\label{sec:overview}
The algorithms described in this paper were integrated into Meta's paper-to-paper recommendation system and make use of its large-scale semantic knowledge base. The paper-to-paper recommendation system has four main components: (a) public and private data sources that feed the knowledge network; (b) an extract, transform, load (ETL) pipeline that disambiguates the entities and discovers relations among them; (c) base recommendation algorithms that use a single specific type of data to make recommendations for a paper; and, (d) aggregation algorithms that combine recommendations from the base recommenders to generate the final set of recommendations optimized on specific criteria (see Figure \ref{fig:system_overview}). The seven base recommendation algorithms are described in detail in Section \ref{sec:recommendation-algorithms}.
Three main data sources are used to populate the knowledge base.
PubMed is the central repository for all biomedical publications and provides a detailed API through which biomedical journals and conferences can be retrieved \cite{canese2013pubmed}.
A PubMed record contains title, abstract, and metadata (e.g., authors, affiliations, keywords, DOI, ISSN, etc.), and also sometimes information on the cited papers. Each PubMed paper has a unique id (PMID) corresponding to a unique digital object identifier (DOI) registered by Crossref (http://www.crossref.org/), which is a non-profit association of scholarly publishers that develops the infrastructure to distribute and maintain DOIs. From Crossref, we gathered metadata for about 50.9 million documents and citations for some of them. Our third data source is full text articles from publisher partners of Meta which, at the time of our experiment, included Elsevier, Sage, DeGruyter, PLoS, BMC, among others. The Meta
full text pipeline contains various adapters for diverse publishers, and extracts both metadata and citation information from full text content, which arrives in both XML and PDF formats.
\begin{figure}[ht]
\centering
\includegraphics[width=0.97\textwidth]{figures/system_overview}
\caption[System Overview]{Data flow of Meta's recommendation engine}
\label{fig:system_overview}
\end{figure}
Each paper then goes through a disambiguation engine which has two main tasks. The first is disambiguating the authors of the paper where the goal is to associate the paper with existing authors in the database or assign a newly discovered author. At the time of our experiment, Meta's author database contained approximately 11 million biomedical researcher profiles calculated from 24.5 million papers spanning 89 million paper-author relationship tuples. Meta's author disambiguation algorithm is modeled after the winning algorithms of KDD Cup 2013 Author Disambiguation challenge (track-2) \cite{li2013combination,liu2013ranking}. Given a manually disambiguated paper-author assignment training set, a random forest classifier is trained to discriminate between correct and incorrect author-paper assignments. Given an existing paper-to-author assignment database, and a newly published paper, the algorithm compares the paper against each candidate author's profile which included over 43 predictive features at the time of our experiment, using the classification model. If
the author with maximum match probability achieves a threshold, the paper is assigned to this candidate author, otherwise a new author profile is generated and the paper is assigned as the first paper of the newly discovered author. The 43 predictive features span the five major categories: author name similarity metrics (Levenstein, Jaro-Winkler, Jaccard etc.), paper content similarity (mostly based on TF-IDF), affiliation similarity, co-authorship information, and author's active time compatibility. Meta's author disambiguation algorithm achieves an F1 score of $0.73$, AU-ROC of $0.94$ and AU-PRC of $0.60$.
The second disambiguation process deals with concept mentions. Once a concept mention is recognized through an entity recognizer (such as GNAT \cite{hakenberg2008inter}, DNORM \cite{leaman2013dnorm}, NeJI \cite{campos2013modular}, etc.), it is normalized into the canonical name from UMLS \cite{bodenreider1998beyond} and becomes a semantic tag. Among the many concept types, we used only the Medical Subject Headings (MeSH) in our algorithms.
Next, the paper goes through citation extraction phase, during which references listed by the paper are identified and resolved into unambiguous, directed DOI-DOI pairs and added into the citation network of Meta which has roughly 580 million citations. For papers with full text, if possible, we also extract pairwise proximities of the references. Finally, the text and semantic tag components of the paper are indexed into an inverted index, which is built using Hadoop's MapReduce based TF-IDF builder \cite{manning2008scoring}.
The recommendation algorithms presented in this paper operate on the transformed data in Meta's semantic knowledge network. The algorithms were implemented using a diverse technology stack: Hadoop, Java, Python and mySQL. Some of the algorithms depended heavily on the Hadoop based MapReduce framework, while others were implemented with direct SQL queries. The recommended papers produced by the base algorithms were aggregated using a number of rank aggregation algorithms, which were all implemented using SciPy and NumPy packages of Python.
\section{Recommendation Algorithms}
\label{sec:recommendation-algorithms}
The paper-to-paper recommendation problem can be stated as: Given a database of papers, $P$ where $|P|=n$ and a paper, $p_i$ that is of interest to a researcher $R$, recommend a list of $k$ papers, $RP=(p_1, p_2, \ldots, p_k)$ to $R$ such that $p_j$, $j=1, \ldots, k$ are judged to be related to $p_i$ and/or in some way useful to $R$. The list may be a partially ordered list such that $p_1$ is considered to be more relevant than $p_j$, $j=2, \ldots, k$ and so on.
We implemented seven recommendation algorithms on a database of more than 24 million biomedical papers.
Note, since running our experiments, there are 27 million biomedical papers in the Meta database.
We focused on two main criteria when choosing which algorithms to include, namely the ability to scale and the ability to leverage various available data types. This meant that we mainly chose simple yet powerful algorithms instead of complex ones, with the expectation that the rank aggregation step can compensate for any weaknesses in the base algorithms in an effective manner.
Hence, we also implemented four different algorithms that aggregate results from the seven base algorithms.
The details of each are presented below.
The algorithms we implemented are inspired by existing work \cite{Gipp09a,DworkKNS01,AilonCN08,AliM12,kessler1963bibliographic,marshakova1973,small1973co} and have been customized for our dataset of biomedical papers. Table \ref{tab:algorithmtable} summarizes the algorithms that are described in this section.
\begin{table}
\centering
\caption{Summary of recommendation and rank aggregation algorithms used in our system}{
\begin{tabular}{p{4.3cm}|p{9cm}}
\bf Name
& \bf Short Description \\
\hline
B-CCS: Co-citation Similarity
& Recommends papers cited by similar citing papers \cite{marshakova1973,small1973co}. \\
B-BC: Bibliographic Coupling
& Recommends papers with similar references \cite{kessler1963bibliographic}.\\
B-IBCF: Item-Based Collaborative Filtering
& Treats citations as user-item purchases, recommends items to users that are similar to ones user already bought. \\
B-CCP: Co-citation Proximity
& Recommends papers that are co-cited and close together in the text \cite{Gipp09a}. \\
B-AS: Abstract Similarity
& Recommends papers with similar text content. \\
B-STS: Semantic Similarity
& Recommends papers with similar semantic content. \\
B-CA: Co-authorship
& Recommends papers with similar/shared authors \cite{sugiyama2011serendipitous,newman2001structure}.\\
\hline
A-LP: LP-based Aggregation
& Aggregates based on linear programming relaxation based optimization \cite{AilonCN08}. \\
A-BS: Beam Search Aggregation
& Aggregates based on heuristics using beam search \cite{AliM12}. \\
A-BL: Borda Aggregation
& Aggregates by simply averaging over the ranks \cite{Borda1781}. \\
A-MS: Merge Sort Aggregation
& Aggregates based on merge sort based heuristic \cite{AliM12}. \\
\end{tabular}}
\label{tab:algorithmtable}
\end{table}
\subsection{Base Recommendation Algorithms}
The base recommendation algorithms make use of citation information, content information in abstracts, the full text of the papers, and authorship information.
\subsubsection{Citation-based Algorithms}
We generated a citation network of the papers in our database by gathering citations from 50.9 million documents from across the sciences, metadata from over 24.6 million PubMed documents and the full text of over 16 million articles using a fully automated technique. Our resulting citation network has over 17 million nodes (which is a subset of the biomedical papers in the 50.9 million articles) and over 350 million edges. The following base algorithms that use the citation network were implemented:
Co-citation Similarity (B-CCS), Bibliographic Coupling (B-BC), Item-Based Collaborative Filtering (B-IBCF), and Co-citation Proximity (B-CCP). Figure \ref{app:cit:example} illustrates a sample data set of three papers with citations indicated.
\begin{figure}[h!]
\centering
\includegraphics[width=0.5\textwidth]{figures/Citation_Based_v3.pdf}
\caption[Citation-based Recommendation Algorithms]{Citation structures of sample documents. Citation-based algorithms produce the following recommendations for Paper E in order: \textbf{B-CCS} $\rightarrow$ A and C (tied), B and D (tied); \textbf{B-BC} $\rightarrow$ Z, Y; \textbf{B-IBCF} $\rightarrow$ C, D; \textbf{B-CCP} $\rightarrow$ A, D, B, C.}
\label{app:cit:example}
\end{figure}
\paragraph{Co-citation Similarity (B-CCS)}
Intuitively, papers that are cited by the same paper or co-cited \cite{marshakova1973,small1973co} many times are likely to be similar to each other. This notion of similarity provides us with a basis for recommendation. Referring to the example in Figure \ref{app:cit:example}, given Paper $E$, B-CCS recommends Papers $A$ and $C$ ahead of Paper $B$ or Paper $D$ because Paper $E$ is co-cited with Paper $A$ in two papers (Papers $Y$ $Z$) and Paper $E$ is co-cited with Paper $C$ in two papers as well (also Papers $Y$ and $Z$). However, Paper $E$ is only co-cited with Paper $B$ in one paper (Paper $Z$) and is only co-cited with Paper $D$ in one paper (Paper $Y$).
The notion of co-cited papers can be captured by using incoming citation vectors. Given a citation network that contains $n$ papers, we define the incoming citation vector $vin_i$ of a paper $p_i$ as an $n$-dimensional bit vector $vin_i=(b^i_1, b^i_2, \ldots, b^i_n)$ where $b^i_j=1$ if $p_j$ cites $p_i$, otherwise $b^i_j=0$. Then, $p_i$ and $p_k$ are co-cited by paper $p_j$ if $b^k_j=b^i_j=1$. Two papers with many 1's in the same position in their incoming citation vectors are co-cited by many papers.
To recommend papers related to paper $p_i$, we can apply standard vector similarity metrics such as cosine similarity on $vin_i$ and $vin_j$ for all papers $p_j$ to find papers that are most co-cited with $p_i$. Cosine similarity also normalizes similarity scores by the norms of the vectors, intuitively weighting papers with many incoming citations less than papers with few incoming citations. However, cosine similarity gives an equal weight to all coordinates of $vin_i$ and $vin_j$. Suppose there is a hypothetical paper $p_k$ that cites a lot of papers, then for many papers $p_x$, in the vectors $vin_x$, $b^x_k=1$. Conversely, if a paper $p_c$ cites few papers, then in the vectors $vin_c$, $b^x_c=1$ for only a few papers $p_x$. Intuitively, coordinate $c$ should contribute more than $k$ because it is rarer; two papers co-cited by a paper with few outgoing citations is worth more than being co-cited by a paper with many outgoing citations. To account for this, we normalize the incoming citation vectors
by dividing each coordinate of $vin_i$ and $vin_j$ by the number of outgoing citations of the paper represented by the coordinate before applying cosine similarity.
The number of pairwise similarity computations grows quadratically with the number of papers in the database and is around $10^{14}$ for 25M papers. To speed up this computation, we only consider pairs of papers with at least one common incoming citation, and this resulted in a $10^5$-fold decrease in the number of pairwise similarity computations.
\paragraph{Bibliographic Coupling (B-BC)}
Papers having similar citation profiles are intuitively more similar than papers with different citation profiles \cite{kessler1963bibliographic}; this gives us yet another basis for recommendation. In this case, we compute the $n$-dimensional outgoing citation vector for each paper $p_i$ as $vout_i=(b^i_1, b^i_2, \ldots, b^i_n)$ where $b^i_j=1$ if $p_i$ cites $p_j$ and $b^i_j=0$ otherwise. Then, $p_i$ and $p_k$ both cite paper $p_j$ if $b^k_j=b^i_j=1$. Two papers with many 1's in the same position in their outgoing citation vectors cite many of the same papers.
We then employ the same algorithm used for co-citation similarity (B-CCS) except with the citation edges reversed. We normalize outgoing citation vectors by penalizing coordinates that represent papers with many incoming citations (those that are cited by many papers); then, given a paper, we compute the cosine similarity between it and every other paper to obtain papers with highly similar citation profiles as recommendations. The penalization step is the same as in B-CCS. The intuition behind it is: two papers citing a paper with few incoming citations is worth more than citing a paper with many incoming citations.
In the example in Figure \ref{app:cit:example}, for Paper $E$, B-BC recommends Paper $Z$ before Paper $Y$ because Paper $Z$ has more citations in common with Paper $E$ (both co-cite Papers $A$ and $B$). Paper $Y$ only has one citation in common with Paper $E$.
Similar to our approach used for pairwise similarity computations in co-citation similarity (B-CCS) algorithm, we only consider pairs of papers with at least one common outgoing citation resulting in a $10^5$-fold decrease in the number of computations.
\paragraph{Item-based Collaborative Filtering (B-IBCF)}
The item-based collaborative filtering algorithm is implemented by Apache Hadoop\footnote{\url{http://mahout.apache.org/users/recommender/intro-itembased-hadoop.html}}. Using the citation network, we treat each citation edge as a user-item interaction. Paper $p_i$ citing paper $p_j$ represents user $p_i$ buying item $p_j$. We treat all our papers as both items and users and recommend papers (items) to papers (users) based on citations. We perform the standard item-based collaborative filtering approach \cite{sarwar2001item}: given a user (paper) $p_i$, we want to recommend items (papers) to $p_i$ that $p_i$ does not already have (does not already cite), and are similar to items that $p_i$ already has (already cites). Just like the co-citation similarity algorithms, similarity is based on vector similarity. Given an item ($p_j$), its user vector is the binary vector of users (papers) that have purchased (cited) this item ($p_j$). So, for example, if the incoming citation vector for paper $p_j$ is $vin_j=(b^j_1, b^j_2, \ldots, b^j_n)$ where $b^j_i=1$ if $p_i$ cites $p_j$ and $b^j_i=0$ otherwise,
then we consider $p_j$ as an item that is bought by those users $p_i$ where $b^j_i=1$. Since these vectors are binary, we use Hadoops's log-likelihood vector similarity measure to compute item similarity between items that user $p_i$ has bought, and items that $p_i$ does not have and pick the best items by averaging similarity scores across all items that $p_i$ has. Intuitively, given a paper $p_i$, we recommend papers most similar to its citations (using log-likelihood similarity, which is intuitively co-citation similarity).
As shown in the example in Figure \ref{app:cit:example}, for Paper $E$, B-IBCF recommends Paper $C$ and then Paper $D$ because Paper $Z$ (that has more citations in common with Paper $E$) cites Paper $C$ (which Paper $E$ does not cite/have) while Paper $Y$ (which has one citation in common with Paper $E$) cites Paper $D$ (which Paper $E$ does not cite/have). Papers $A$ and $B$ are not recommended because Paper $E$ also cites (has) them.
The primary difference between this algorithm and B-CCS is that given an input paper $p$, B-CCS finds papers closest to $p$ using co-citation similarity. This algorithm, however, does not look at the input paper, it instead treats the input paper as a set of papers by looking at its citations, and then recommends papers closest to its citations by averaging co-citation similarity between its citations and other papers. The hope is looking at a paper's citations gives more information than the paper itself.
\paragraph{Co-citation Proximity (B-CCP)}
The co-citation proximity approach is based on citation proximity analysis \cite{Gipp09a}. The intuition behind the algorithm is that if citations occur close together in the text of a paper, then the cited papers are likely to be more closely related than if the citations were further apart. We use a different weighting scheme for the proximity occurrences than Gipp and Beel \cite{Gipp09a} and we aggregate the occurrence values.
We processed each paper $p$ to extract all possible citation pairs between the papers referenced in the citation list of $p$. Each citation pair is given a proximity type (group -- within the same square brackets, sentence, paragraph, section, or paper) based on the minimal distance between each citation. The proximity type is calculated by parsing the structure of the document's XML format or applying minor heuristics.
Relationship weights are used to quantify the different minimum proximities between citation pairs and are summed across document pairs to indicate their similarity. For example, co-citations in the same paper are assigned a weight of 1, co-citations in the same section, a weight of 2, and so on. If paper $p_i$ and paper $p_j$ are cited once within the same sentence (a total relation weight of 4) but paper $p_i$ and paper $p_k$ are cited within the same section in three additional documents (a total relation weight of $2{\times}3=6$), then paper $p_i$ has a stronger similarity to paper $p_k$ than to paper $p_j$. We also experimented with and applied the approach to larger datasets (over 16 million documents) than what Gipp and Beel used (1.2 million) \cite{Gipp09a}.
Referring back to the example in Figure \ref{app:cit:example}, for Paper $E$, B-CCP recommends documents based on minimal citation proximity to Paper $E$ over the multiple papers in which Paper $E$ is cited. The recommended documents are ordered as follows: Paper $A$ which is cited in the same sentence as a citation to Paper $E$ (weight of 4) in Paper $Y$ and in the same section (weight of 2) in Paper $Z$ (total weight is 6); Paper $D$ which is cited in the same group as Paper $E$ (weight of 5) in Paper $Y$; Paper $B$ which is cited in the same sentence as Paper $E$ (weight of 4) in Paper $Z$; and, Paper $C$ which is cited in the same paper (Paper $Z$) as Paper $E$ (weight of 1) and in the same section as Paper $E$ (weight of 2) in Paper $Y$ (total weight of 3).
One issue with this approach is the situation in which paper $p_i$ and paper $p_j$ are cited in the same sentence but used to contrast each other \cite{Gipp09a}. This is not a significant issue in our case because our large collection of papers means that consistently co-cited papers will have a stronger connection. Additionally, even if two papers are co-cited in the context of a disagreement and/or conflict because they propose opposing theories, the fact that they are frequently co-cited may make them strongly related (i.e., such that one would be a good recommendation for the other).
\subsubsection{Content-based Algorithms}
We can also identify similar papers to recommend based on the content of the paper or its abstract. These similarity-based algorithms make use of terms in the text and semantic meaning of the terms in the text.
\begin{figure}[htb]
\centering
\includegraphics[width=0.75\textwidth]{figures/Content_Based_v3}
\caption[Content-based Recommendation Algorithms]{Example of common words and keywords (based off MeSH ontology) represented by rectangles in the documents. Content-based algorithms produce the following recommendations for Paper E in order: \textbf{B-AS} $\rightarrow$ A, B, C, D (using words); \textbf{B-STS} $\rightarrow$ B, A, D, C (using keywords).}
\label{app:con:example}
\end{figure}
\paragraph{Abstract Similarity (B-AS)}
Almost every paper includes an abstract that typically summarizes the paper's focus, methods, experiments, results, and contributions in a succinct and efficient manner. Many research article search engines index only the abstract (rather than the full text of the article) because abstracts provide sufficient information about the full paper. Two articles with similar abstracts are likely to be similar articles; therefore, we used the text of abstracts as a basis for recommending articles. To determine abstract similarity, we use a TF-IDF similarity measure on the words of the abstract. TF-IDF (term frequency-inverse document frequency) is calculated as the product of the term frequency (TF: the number of times a term $t$ occurs in a document) and the inverse document frequency (IDF: a measure of how common or rare the term is across all documents).
Using the B-AS algorithm to recommend papers for Paper $E$ in Figure \ref{app:con:example}, Paper $A$ is recommended before Paper $B$ because Paper $A$ contains three instances of an infrequent word (highlighted in light purple). Paper $B$ is recommended before Paper $C$ because Paper $B$ contains one instance of the infrequent word and two frequent words (highlighted in green and pink). Papers $C$ and $D$ both contain frequent words in common with Paper $E$, but Paper $C$ contains more instances of words in common with Paper $E$ (three vs. two); hence, it is recommended before Paper $D$.
To obtain accurate TF-IDF similarity, we first normalize the abstracts by tokenizing them into words, eliminating external token punctuation, and stop-word tokens. TF-IDF is then calculated on a token level. We calculate the inverse document frequency of each token on our entire paper abstract dataset (size approximately 14 million). Inverse document frequency of a token $t$ amongst all $n$ papers $p_i \in P$ in the dataset is defined as:
\[ idf(t, P){=}\left\{
\begin{array}{l l}
\sqrt{log(n/df(t,P))} & \quad \text{if $df(t,P) \ne 0$}\\
0 & \quad \text{if $df(t,P){=}0$}
\end{array} \right.\]
where $df(t, P)$ is the number of papers in the set $P$ in which $t$ occurs.
Then, given two abstracts from papers $p_i$ and $p_j$, we compute their TF-IDF vectors; that is, their abstracts expanded into $d$-dimensional bit vectors, where $d$ is the number of distinct words that occur in all abstracts (in our database this is approximately 9 million distinct words) such that each position in the vector for paper $p_i$ contains $tf(t,p_i){\times}idf(t,P)$ for the corresponding token $t$. The term frequency $tf$ of a token $t$ in $p_i$ is defined as:
$tf(t,p_i){=}\sqrt{\text{count}(t,p_i)}$, where $\text{count}(t,p_i)$ is the number of times $t$ occurs in the abstract of paper $p_i$.
Given the two TF-IDF vectors, $tfidf_i$ and $tfidf_j$ for $p_i$ and $p_j$ respectively, we compute their cosine similarity as $\cos(tfidf_i, tfidf_j)$ to obtain the final similarity score. Intuitively, this similarity score captures abstracts that share similar terms, strengthened by the number of times the term occurs in the abstracts under consideration and penalized by the commonality of the term amongst all abstracts. Thus, we expect rare terms that occur frequently in both abstracts to indicate strong similarity between the abstracts.
Suppose for a given paper $p_i$ in our dataset, we want to obtain the top 50 papers similar to $p_i$ using abstract TF-IDF similarity. This computation is extremely inefficient as it requires $\approx 25000000^2{=}6.25{\times}10^{14}$ similarity calculations. Therefore, as a fast approximation for a given paper abstract, we consider only those paper abstracts that share at least one rare term with it. We define a term $t$ as rare when $df(t,P) \leq 5000$.
This step significantly cuts down the number of similarity calculations to approximately $2{\times}10^{11}$ (more than 3,000-fold decrease). For the top recommended papers, the abstracts should intuitively share at least one rare term, so this filtering step should not eliminate too many papers and in practice, this heuristic search space reduction strategy works well.
\paragraph{Semantic Similarity (B-STS)}
Unfortunately, the B-AS algorithm is very sensitive to ambiguity and synonymy problems. To overcome this issue, we aimed to use semantic relationships to infer indirect mentions. Traditional TF-IDF similarity based systems are not be able to identify similarity among different terms for the same concept but normalized field/concept annotations provide a principled way to detect and measure similarity. Hence, we applied named entity recognition algorithms to all papers in our database to identify mentions of concepts such gene, chemicals, diseases, and research areas, which are all included in the MeSH ontology \cite{nelson2009medical}.
There are about 28,000 terms and 139,000 supplementary concepts in MeSH. For every paper we capture a summary of the paper based on the fields it contains. Intuitively, papers that share more fields are more similar than papers that share less fields. As in the abstract similarity algorithm (B-AS), we use TF-IDF similarity to compute semantic similarity in exactly the same way, except instead of using normalized tokens representing words of the abstract, we use fields associated with the paper. TF-IDF inherently treats papers that share many rare fields as closest to each other. Note, the term frequency of a term $t$ and paper $p_i$ is either 0 or 1 because our field/term tagger only tags the existence of each field in a paper. As in abstract similarity, we only compare similarities between papers which share at least one rare field (term, $t$), where rare is defined as occurring in at most 5,000 papers in the set $P$ of papers: $df(t, P) \leq 5000$. This heuristic filtering approach reduces the number
of pairs we have to compare to 72.2 billion ($6.25\times10^{14}$) without jeopardizing the quality of the recommendations.
Going back to the example in Figure \ref{app:con:example}, having reduced the words to their semantic fields, the frequency of instances within each paper no longer has an impact. Paper $B$ is recommended first because it shares the most infrequent terms with Paper $E$. Paper $A$ and then Paper $D$ are recommended next because Paper $A$ still contains a term more infrequent than Paper $D$. Finally, Paper $C$ is recommended because it contains one infrequent term in common with Paper $E$.
\subsubsection{Co-authorship Similarity (B-CA)}
The main idea behind co-authorship based recommendations is that papers which share authors are likely to be related to each other \cite{sugiyama2011serendipitous,newman2001structure}. At the time of our experiment, Meta's author database contained approximately 11 million automatically discovered biomedical researcher profiles calculated from 24.6 million papers spanning 89 million paper-author relationship tuples. Meta's author disambiguation algorithm is modeled after the winning algorithms of KDD Cup 2013 Author Disambiguation challenge (track-2) \cite{li2013combination}. We take a simple approach by first building the co-authorship network where the set of nodes $P=\{p_1, p_2, \dots, p_n\}$ represents the set of $n$ papers and a weighted edge between two papers, $(p_i, p_j)$ represents the number of shared co-authors between papers $p_i$ and $p_j$. Then, for a given paper $p_i$ we traverse the co-author network graph to each of its one- and two-hop neighbors $p_j$ to calculate the shared-author scores as the sum of
the weighted edges in the path from $p_i$ to $p_j$. Each one- and two-hop neighbors $p_j$ is ranked by its shared-author score with $p_i$ and the papers with the highest scores are recommended (ties are broken randomly).
As shown in the example in Figure \ref{app:auth:example}, in one and two hops from Paper $E$, Paper $B$ has six co-authors (three on the path E-A-B, one on the path E-B, and two on the path E-C-B), and hence, is the first recommendation. Paper $A$ is next because it has four co-authors on the one- and two-hop paths (one on E-A and three on E-B-A), while Paper $C$ is last because it only has three co-authors on the paths (one on E-C, and two on E-B-C).
\begin{figure}[h!]
\centering
\includegraphics[width=0.6\textwidth]{figures/Author_Based_v1}
\caption[Author-based Recommendation Algorithms]{Co-authorship structure where common authors are shown as icons along paths. Recommendations for Paper E are as follows: \textbf{B-CA} $\rightarrow$ B, A, C.}
\label{app:auth:example}
\end{figure}
\subsection{Aggregation Algorithms}
We implemented four rank aggregation methods \cite{DworkKNS01,AilonCN08,AliM12} to aggregate results from the base algorithms described above.
\subsubsection{Problem Definition and Notation}
Given a set of $n$ elements and $K$ complete rankings or permutations of these elements $\pi_1, \pi_2, \ldots, \pi_K$, the goal is to find the Kemeny optimal ranking \cite{KemenyS62}, $\pi$, i.e., the ranking that minimizes $\sum_{i=1}^K d(\pi,\pi_i)$, where $d(\cdot, \cdot)$ is the number of pairwise disagreements between a pair of rankings, also known as the Kendall distance. When complete rankings are not available, we place all the unranked objects at the bottom of the list, and consider all objects in this set to be tied with each other. The problem of finding the Kemeny optimal ranking is NP-hard \cite{BartholdiTT89}. See \cite{AliM12} for a comprehensive survey of algorithms to compute Kemeny ranking. Here, we use four different algorithms to approximate the Kemeny ranking.
The precedence matrix $Q \in R^{n{\times}n}$ has entries $Q_{ij}$ that represent the fraction of times an element $i$ is ranked higher than element $j$, i.e., $Q_{ij}{=}(1/K) \sum_{k=1}^K I(i \prec_{\pi_k} j )$, where $I(\cdot)$ is the indicator function, and $\prec_{\pi}$ is the precedence operator for ranking $\pi$.
\subsubsection{LP approximation (A-LP)}
The problem of finding the Kemeny optimal ranking can be solved exactly by posing it as an integer linear program (ILP). Specifically, consider the following optimization problem:
\begin{equation}
\begin{aligned}
\min_x & \sum_{i,j} Q_{ij} x_{ji} + Q_{ji} x_{ij} \\
\text{subject to } & x_{ij} \in \{0,1\} \ , \quad \forall i,j \\
& x_{ij} + x_{ji}{=}1 \ , \quad \forall i,j \\
& x_{ij} + x_{jk} + x_{ki} \geq 1 \ , \quad \forall i,j,k \ .
\end{aligned}
\end{equation}
The first set of constraints ensure that $x_{ij}$ are binary variables. The second and third set of constraints are symmetry and transitivity constraints, respectively, to ensure that $x$ is a ranking. Note that this formulation also solves the minimum weighted feedback arc set in tournaments \cite{AilonCN08}. The binary constraints can be relaxed to $x_{ij} \geq 0$ resulting in a linear program (LP) relaxation of the ILP. Even though the LP can be solved using off-the-shelf LP solvers, in practice we found this to be prohibitively expensive due to the large number of transitivity constraints -- cubic in the number of elements $n$.
\subsubsection{Beam Search (A-BS)}
The set of all permutations can be represented in the form of a tree, where each permutation can be traced in a path from the root to a leaf. Note that every path from the root to an internal node in the tree represents a partial ranking. We use beam search to explore the set of all permutations, and output the optimal ranking. The basic idea is to consider only $B$ candidate solutions (partial rankings) at each level of the tree, where $B$ is a user-defined parameter known as beam width, and these candidates represent the best partial rankings found so far by the heuristic search algorithm. The tree is then explored in a breadth-first fashion from the root all the way down to the leaves. The optimal solution is then selected from the best $B$ candidates found at the lowest level of the tree. A greedy version of the algorithm can be derived by setting $B=1$, where at each level only one candidate solution is considered greedily. In the other extreme, when $B=\infty$, the algorithm explores all the possible
exponential number of rankings/paths in the tree.
In order to select the best $B$ candidate solutions at each level of the tree, we need to define a cost function to score partial rankings. This cost function can be defined using the precedence matrix $Q$ as: $C(\pi_p){=}\sum_{(i,j) \in \pi_p} Q_{ij}$,
where $\pi_p$ is a partial ranking and $\{(i,j)\}$ is the set of all pairs $(i,j)$ such that $i \prec_{\pi_p} j$ in the partial ranking, including transitive pairs. Our implementation of the algorithm takes about 3.58s/paper on a single machine with 8 threads.
\subsubsection{Borda Counts (A-BL)}
A simple algorithm to aggregate rankings is to rank objects based on their average ranking computed from all the multiple rankings \cite{Borda1781}. This is equivalent to sorting the elements based on the column sum of the precedence matrix, i.e., $\text{\emph{argsort}}_i \sum Q_{ij}$. Our implementation of the algorithm takes about 0.161s/paper on a single machine with 8 threads.
\subsubsection{Sort-based Approximation (A-MS)}
Comparison-based sorting algorithms such as merge sort or quick sort can be adapted to aggregate rankings using the precedence matrix $Q$ \cite{AliM12}. Instead of comparing pairs of elements $i$ and $j$ in the sorting algorithm, we compare $Q_{ij}$ and $Q_{ji}$. We refer the reader to \cite{Schalekamp98} for more details on comparison sort methods for rank aggregation. In our experiments, we adapted merge sort to solve the rank aggregation problem. Our implementation takes about 0.159s/paper on a single machine with 8 threads.
\subsubsection{Weighted Aggregation}
We note that the algorithms described above can be adapted to take weights into account, where the weights are assigned for each of the multiple recommendation algorithms. Let $\{w_1,\ldots,w_K\}$ denote these weights such that $\sum_k w_k=1$. We can modify the precedence matrix as: $Q_{ij}{=}(1/K) \sum_{k=1}^K w_k I(i \prec_{\pi_k} j )$, and use this as input to the above algorithms (A-LP, A-BS, A-BL, A-MS). Determining weights for algorithms is left for future work.
\section{Moving to Production: Practical Aspects}
\label{sec:practical_aspects}
In this section, we focus on challenges associated with selecting and deploying a production recommendation engine system. All but one base recommender algorithm (B-CA) are implemented using the MapReduce platform and hence are linearly scalable. We were able to generate recommendations from base algorithms for over 24.6 million articles in less than a week using a small scale (32 cores) Hadoop cluster built on top of commodity hardware. However, the aggregation algorithms do not fit naturally into the MapReduce framework and presented the main challenge in terms of runtime.
When implementing a recommendation system in a production platform, there are several issues to consider but runtime performance is one of the most critical.
The runtime complexity of all aggregation algorithms is mostly determined by the \textit{effective size} of the list of papers, i.e., the size of the union of all the ranked lists of papers returned by the base algorithms. Let each base algorithm output a ranking of $N$ elements, and without loss of generality assume that $N$ is fixed. Let $K$ be the number of base algorithms. Therefore, the effective size is denoted by $E{=}NK{/}p$, where $p{\in}[1,K]$ is a measure of \emph{overlap} among all the $K$ base algorithms. Note that $p=1$ indicates that all the ranked lists are mutually exclusive, and $p=K$ indicates that all the
ranked lists are the same.
In order to compute the runtime performance of our algorithms we asked 14 active biomedical researchers to select 15 papers each from their field of student. There was one duplicate paper; thus, a total of 209 papers were used to evaluate the performance of our algorithms.
Figure \ref{fig:overlap}(A) shows the cumulative distribution of the number of base recommenders that are able to generate recommendations for each of the 209 papers. In $\approx90\%$ of the cases the aggregation algorithms receive input from at least three algorithms and in $\approx40\%$ of the cases all
base recommenders can generate recommendations.
On the other hand, Figure \ref{fig:overlap}(B) shows the pairwise overlap percentage rate between base recommenders, suggesting only modest overlap between citation-based recommendation algorithms and very little overlap among the remaining pairs. As such, effective size tends to be large ($\mu{=}221.8, \sigma{=}68.3$) and the runtime complexity of the algorithms becomes important.
The number of variables and constraints in the LP of the LP Approximation (A-LP) algorithm is $O( (NK/p)^2 )$ and $O( (NK/p)^3 )$, respectively.
The runtime complexity of Beam Search (A-BS) is $O(B{\times}NK/p)$, where $B$ is the beam width.
Runtime complexity of Borda Counts (A-BL) and Sort-based (A-MS) algorithms are the same as that of a sorting algorithm whose input is a list of size $NK/p$, i.e., $O( NK/p{ \times} \log(NK/p) )$.
\begin{figure}[ht]
\centering
\includegraphics[width=0.85\textwidth]{figures/overlap_analysis}
\caption[Overlap and its effects ]{(A) Cumulative distribution of base recommenders generating input for the aggregation step. (B) Percentage of overlap between pairs of base recommenders (C-E) Runtime vs effective size for A-BL, A-MS, A-BS algorithms}
\label{fig:overlap}
\end{figure}
Although not shown here, it is worth mentioning that the amount of overlap $p$ has a significant effect on the runtime complexity of A-LP (unlike other aggregation algorithms) as the reduction in the number of variables and constraints is quadratic and cubic in $1/p$, respectively. Also, adding more base algorithms (i.e., increasing $K$) will affect LP more than other aggregation methods, unless $p$ increases by the same rate. Indeed, we omitted using the LP-based algorithm as it was prohibitively slow for the majority of papers.
The runtime for A-BL, A-MS and A-BS, which are given in Figure \ref{fig:overlap}(C)-(E), respectively, decisively show that the A-BL and A-MS algorithms perform similarly and are $\approx{120}{\times}$ faster than the A-BS algorithm. A decision about which algorithm(s) to ultimately deploy in practice must take this runtime performance into account. It took about a week to generate recommendations for all the papers in the Meta database.
Another metric to use when selecting the final algorithm(s) is coverage, which, in this context is defined as the number of papers for which our system can generate recommendations. Aggregation algorithms overcome a fundamental shortcoming of base recommendation algorithms which cannot produce recommendations for all papers. Indeed, B-IBCF, B-CCP, B-BC, B-CCS and B-STS fail to generate recommendations for $14\%$, $25\%$, $15\%$, $5\%$ and $17\%$ of the papers, respectively.
Finally and, perhaps most importantly, a decision about which algorithm(s) to deploy in a production system must consider quality and relevance of the recommendation results.
There are several methods that can be used to evaluate the relevance and usefulness of the output of recommendation algorithms.
Future work will evaluate and compare our algorithms on this dimension.
However, even if a base recommender is an overall winner from a quality of output perspective, it most likely cannot be used as the sole algorithm because of lack of coverage, which in turn means that aggregation is necessary.
\section{Conclusions and Future Work}
\label{sec:conclusions}
In this paper, we presented several recommendation algorithms that were implemented and evaluated in Meta's large-scale biomedical science knowledge base. Existing academic paper recommendation engines, especially those in biomedical sciences, are limited in scope, size and functionality. We experimented with seven base recommender algorithms, and four aggregation algorithms. Base recommender algorithms utilize diverse sets of data such as a citation network, text content, semantic tag content, and co-authorship information.
We compared the algorithms according to runtime complexity and scalability and discussed some of the considerations in implementing recommendation algorithms in a large-scale production system.
The main focus for our future work will be to consider the quality of the resulting recommendations from the algorithms and to compare the results according to relevance and usefulness for biomedical researchers. Once the quality of recommendations from the different algorithms is understood, future work can also consider how to adapt the aggregation algorithms by assigning weights to each of the base recommendation algorithms.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,472 |
è una città situata nella zona centrale della prefettura di Hyōgo, nell'isola di Honshū, in Giappone.
Storia
Per volontà di Tokugawa Ieyasu vi venne costruito l'omonimo castello che permetteva il controllo delle vie di comunicazione dell'attuale regione del Kansai.
Sasayama fu colpita dai bombardamenti statunitensi che causarono tra l'altro la distruzione del castello, che venne ricostruito nel 2000.
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<a href="http://etc.usf.edu/lit2go/202/songs-of-the-wind-on-a-southern-shore-and-other-poems-of-florida/">Songs of the Wind on a Southern Shore, and other Poems of Florida</a>
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by <a href="http://etc.usf.edu/lit2go/authors/53/george-e-merrick/">George E. Merrick</a>
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"The Royal Poinciana in Bloom" </h4>
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1920 </li>
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<strong>Language:</strong>
English </li>
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<strong>Country of Origin:</strong>
United States of America </li>
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Merrick, G. E. (1920). <em>Songs of the wind on a southern shore, and other poems of florida.</em> The Four Seas Publishing Co. </li>
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<strong>Word Count:</strong>
206 </li>
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<a href="http://etc.usf.edu/lit2go/genres/24/poetry/">Poetry</a>
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<strong>Keywords:</strong>
florida stories, poetry </li>
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Merrick, George E.. ""The Royal Poinciana in Bloom"." <em>Songs of the Wind on a Southern Shore, and other Poems of Florida</em>. Lit2Go Edition. 1920. Web. <<span class="faux_link">http://etc.usf.edu/lit2go/202/songs-of-the-wind-on-a-southern-shore-and-other-poems-of-florida/4379/the-royal-poinciana-in-bloom/</span>>. February 15, 2016.
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<p>
Scarlet bloom of deepest dye,<br />
That with the summer sunset vie<br />
In flashful boast, thy thick-massed flame:—<br />
Lo! Thou hast put its wealth to shame:<br />
For all out-done, the tropic sun<br />
Recalls his tint-skilled fays of fire,—<br />
Glowing rich in envy as they fly.</p>
<p>
The blood-red gleam of nonpareil<br />
Amidst thy glare is hid so well<br />
That none can know ‘tis bowered there<br />
With scarlet flash of tanager;—<br />
Nor,—faraway, in heat of day—<br />
—A crimson stain against the green—</p>
<p>
From very flame one can’st thee tell.<br />
Above thy growth of tender green,—<br />
That in thy pride can’st not be seen—<br />
The throbbing pulse of flames’ desire<br />
Seems urging tongues of crimson higher;<br />
—As spray-wove gleams o’er molten streams;—<br />
Or combing surges breaking low<br />
Upon a sea of fire.</p>
<p>
The southern land that yields thy store<br />
Of matchless wealth;—in days of yore<br />
Had envied oft the sunset sky<br />
Where tropic summer’s gift days die<br />
In glory’s blaze:—And, testing all her ways<br />
She found at last thy blood-dyed bloom;<br />
—than which the sky can’st do no more!</p>
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| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,923 |
13/03/2019 468 Mock-up Scale-model
IED - Honda TOMO Concept
TOMO, the friendly mobility tool for 2025
IED and Honda Design join the Geneva Motor Show to present the new electric concept car created by the students of the Master in Transportation Design programme
Geneva, 5 March 2019 – Another world premier for Istituto Europeo di Design which returns to the Geneva International Motor Show to present Tomo, the electric concept car created in partnership with Honda Design. Tomo is the thesis project produced by the 13 students of the two-year Master in Transportation Design at IED Torino, a venue of the network that has attracted young people from all over the world and educates them to become the designers of the future.
Just like in an automotive design centre, the Japanese company asked students to work on a brief, Honda next "fun" driving, with the aim of designing what they think should be the ideal means of transport for young people in the next six years. Months of planning and extensive research on key values like friendship, fun driving and respect for the environment led to the creation of Tomo, an electric concept car halfway between a smart device and a mobility tool that combines the needs of free time leisure outside the city with the demands of working in an urban setting. A vehicle of this type, designed to be a faithful companion of its user and able to adapt to his or her wishes and needs could only be called Tomo – friend in Japanese – to underline the image of a vehicle that contains everything that the user needs in everyday life, inside and outside of the city.
TOMO's design (length 3997 mm - width 1893 mm - height 1556 mm - wheelbase 2690 mm) is the result of a research oriented towards the product, its functionality and its youthful appearance. In Tomo the traditional barriers between the exterior and interior of a vehicle exist physically but are conceptually eliminated: the coupé volume and the functions of a mini pickup are in fact "masked" by an external appearance that is both fashionable and youthfully urban. The interior is therefore designed to offer experiences in line with the easy, friendly setting of the screens, steering wheel and head-up display on the windscreen. The exterior, on the other hand, is shaped around these functions, becoming almost an empathetic, identifying a wearable device that adapts to the urban or suburban setting. Tomo was designed for a new market segment where age doesn't count. It aims to attract the interest of new generations who increasingly have a non-traditional occupation, who want to experience their car in a fun and useful way. These are people for whom the environment, ecology and sustainability are fundamental values to build their futures on.
«At Honda Design we are always working to create a design that reflects the philosophy of our brand, offering users our experience with fun and freedom of mobility – says Taku Kono, General Manager Styling Design Division Honda R&D. Starting with the active participation of young designers, through partnerships and internships in Europe and in conjunction with the presentation of the evolution of the Honda Urban EV concept in Geneva, we asked the young IED designers to propose a variation on this theme. It was an exciting, constructive journey both for us and the students thanks to the fusion of the Honda Design philosophy and the creative energy and commitment of the young IED designers».
«With 15 full-scale models presented over the years at the Geneva Motor Show by the Istituto Europeo di Design, visitors have come to expect provocative designs from the school, the only international network of higher education participating in the trade show – says Paola Zini, Director of IED Torino. Giving our students the opportunity to work with Honda, one of the world's top car manufacturers, and then to present the skills, professionalism and values they have acquired during their studies on a global stage is a source of great pride for IED».
Honda selected the design of Ricardo Alejandro Campos Ortega (Mexico) for the main concept idea exterior, and the work of Rudraksh Banerjie (India) as main concept idea interior, which were both then further developed together with the other students of the Master in Transportation Design for the 2017/18 scholastic year: Tanmay Madhukar Chavan (India), Michele Corneliani (Italy), Shobhanjit Das (India), Alexander Marcel Fröse (Germany), Xiaole Ge (China), Ramón Emmanuel Hernández Cortés (Mexico), Tianchen Huang (China), Sameer Aminullah Khan (India), Saketh Nalla (India), Jay Shrikant Nibandhe (India), Yu-Jie Wang (Taiwan).
The Master in Transportation Design of IED Torino is coordinated by Alessandro Cipolli with the collaboration of Davide Tealdi, IED lecturer, and under the supervision of Luca Borgogno, Design Director Automobili Pininfarina, IED Transportation Coordinator.
Tomo was built by EDAG Italia
Technical sponsors: AM Costruzione Modelli, Clinic Car Italia, Freeland.car Srl, OZ Racing, Pirelli.
Istituto Europeo di Design thanks Honda for the fruitful partnership conducted with passion, enthusiasm and a constant presence.
The main idea behind Tomo was inspired by a manga story created by the Master students in Transportation Design a.y. 2017/18.
It's 2025. Coner, a talented design student at IED, is working hard on an ambitious secret project. Coner is a sharp young man, very determined, outgoing and a dreamer, often lost in study for many hours of the night. He has no time for girls but he has lots of different interests like music and sports. Very sensitive to environmental issues, he divides his time among the city, the studio and the countryside where he has a small vegetable garden where he grows fruit and vegetables, the perfect urban farmer. The story tells of his daily life: a student in a big metropolis working hard on the development of a project that should meet the needs of his double life, urban and rural. To succeed in his ambitious mission he builds Tomo, an "extension" of himself, a sort of Deus ex Machina that will help him. So Coner finds his way and draws a car in line with his needs, discovering that a car can be a "friend" (hence the name of the model: Tomo, friend in Japanese) and can perform multiple functions. Tomo is a city car, but it is also designed to transport vegetables, so Coner can continue living his life as an urban farmer, in harmony with the values of ecology and respect for the environment that are part of his DNA. The story of Coner's life describes the development of his project, which transforms an idea into reality: Tomo, which Coner presents at a world premier at the Geneva Motor Show.
The communications campaign for the manga story was developed by Matteo Milaneschi, Coordinator of the Communication Design Course, and Eleonora Antonioli, IED lecturer, together with Mara Artiglia, former student of the Illustration Course, Nicolò Allisiardi and Simone Imberti, former students of the Advertising Communications Course.
2019 Geneva
HatchbackElectric | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,098 |
Q: Como restringir acesso a uma determinada pagina no painel adm. do wordpress? Dentro do painel administrativo do wordpress, preciso criar cerca de 10 paginas, onde cada usuário tenha acesso apenas à pagina que corresponde a ele.
Exemplo:
Usuário X teria acesso apenas ao painel administrativo da pagina X e
usuário Y teria acesso apenas ao painel administrativo da pagina Y.
Até onde sei o wordpress não disponibiliza algo do tipo como default.
Tentei utilizar alguns plug-ins como o Adminimize e o Members, porém em ambos o bloqueio é feito apenas para o post-type "paginas" ou seja apenas bloqueiam o acesso de todas as paginas ocultando o campo paginas do menu esquerdo para os usuários bloqueados, exemplo:
Painel do adminimize - campo "paginas" ao lado no menu esquerdo some para os usuários bloqueados.
Leve em consideração as seguintes perguntas:
*
*Como posso criar um controle parecido com esse do adminimize porém liberando apenas uma pagina para cada usuário (como o exemplo da primeira imagem)?
*Como posso criar outros tipos de usuários alem dos default do wordpress (admin, editor, assinante...)?
OBS.: Não há nenhum problema se a solução proposta tiver que usar "post" ao invés de "paginas".
A:
Como posso criar um controle parecido com esse do adminimize porém liberando apenas uma pagina para cada usuário (como o exemplo da primeira imagem)?
Aconselho você usar um plugin pra fazer isso. Fazer na mão vai dar bastante trabalho. O Role Scoper permite que você de permissões para páginas de acordo com o papel (role) do usuário. Ele cria varios campos onde você seleciona usuários ou grupo de usuários que podem ler e/ou editar os posts/páginas.
Existe também o User Specific Content, que é voltado mais para conteúdo do que para posts e páginas propriamente ditas. Esse talvez não responda a sua pergunta, mas é sempre bom saber que existe.
Como posso criar outros tipos de usuários alem dos default do wordpress (admin, editor, assinante...)?
Com o método add_role(). Da própria documentação
$result = add_role(
'basic_contributor',
__( 'Basic Contributor' ),
array(
'read' => true, // true allows this capability
'edit_posts' => true,
'delete_posts' => false, // Use false to explicitly deny
)
);
if ( null !== $result ) {
echo 'Yay! New role created!';
}
else {
echo 'Oh... the basic_contributor role already exists.';
}
Cria o role Basic Contributor, que tem um certo número de permissões e capacidades. Você pode ver os diversos roles aqui.
Se você colocar esse método direto no functions.php, por exemplo, do jeito que ele está ali, você pode ter problemas, pois esse código vai rodar sempre. A documentação sugere (e eu também) que você crie um plugin, que crie o role uma vez, apenas quando o plugin for iniciado. Pra isso você usa o hook Register Activation
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,150 |
{"url":"http:\/\/math.stackexchange.com\/questions\/298519\/how-to-prove-ergodic-property-from-aperiodicity-and-positive-recurrence","text":"# How to prove ergodic property from aperiodicity and positive recurrence\n\nHow to prove that in case of an irreducible, aperiodic and positive recurrent Markov Chain time average along sample paths is equal to the ensemble average ? i.e.\n\n$$\\lim_{n\\to \\infty }\\frac{1}{n}\\sum _ {i=1} ^{n} X_{i} \\rightarrow E[X_k]$$\n\nwhere $E[X_k]$ is calculated from the steady state probability distribution.\n\nNow each $X_i$'s are dependent because it is a Markov Chain. So, not possible to apply law of large numbers.\n\n-","date":"2016-02-12 04:36:45","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\": 0.9340186715126038, \"perplexity\": 369.26444953965097}, \"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-2016-07\/segments\/1454701163421.31\/warc\/CC-MAIN-20160205193923-00136-ip-10-236-182-209.ec2.internal.warc.gz\"}"} | null | null |
Q: How to take Entire flowfile content in nifi processor I am using nifi to develop the data drifting. In my flow using SelectHiveQL processor. The output(flowFile) of the selectHiveQL need to take into next processor.
what is the suitable processor to take the flowFile content and store into userdefined variable have to use the same variable in Executescript to manipulate the data.
A: The ExecuteScript processor has direct access to the content of the incoming flowfile via the standard API. Here is an example:
def flowFile = session.get();
if (flowFile == null) {
return;
}
// This uses a closure acting as a StreamCallback to do the writing of the new content to the flowfile
flowFile = session.write(flowFile,
{ inputStream, outputStream ->
String line
// This code creates a buffered reader over the existing flowfile input
final BufferedReader inReader = new BufferedReader(new InputStreamReader(inputStream, 'UTF-8'))
// For each line, write the reversed line to the output
while (line = inReader.readLine()) {
outputStream.write("${line.reverse()}\n".getBytes('UTF-8'))
}
} as StreamCallback)
flowFile = session?.putAttribute(flowFile, "reversed_lines", "true")
session.transfer(flowFile, /*ExecuteScript.*/ REL_SUCCESS)
It is dangerous to move the flowfile content to an attribute because attributes and content memory are managed differently in NiFi. There is a more detailed explanation of the differences in the Apache NiFi In Depth guide.
A: You could use ExtractText to extract the content of your flowfile to an attribute.
In the ExtractText processor, you would create a property(the name you give this property will be a new attribute in your flowfile), and the value of the property will be the regular expression (\A.+\Z). In my experience, this regex is enough to capture the entire content of the flowfile, though I suppose mileage could vary depending on the type of content within your flowfile.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,113 |
\section{Introduction}
Ever since the publication of \citeauthor{vNM47a}'s \emph{Theory of Games and Economic Behavior} in 1944, coalitions have played a central role within game theory. The crucial questions in coalitional game theory are
which coalitions can be expected to form and how the members of coalitions should divide the proceeds of their cooperation. Traditionally the focus has been on the latter issue, which led to the formulation and analysis of concepts such as the core, the Shapley value, or the bargaining set.
Which coalitions are likely to form is commonly assumed to be settled exogenously, either by explicitly specifying the coalition structure, a partition of the players in disjoint coalitions, or, implicitly, by assuming that larger coalitions can invariably guarantee better outcomes to its members than smaller ones and that, as a consequence, the grand coalition of all players will eventually form.
The two questions, however, are clearly interdependent: the individual players' payoffs depend on the coalitions that form just as much as the formation of coalitions depends on how the payoffs are distributed.
\emph{Coalition formation games}, as introduced by \citet{DrGr80a}, provide a simple but versatile formal model that allows one to focus on coalition formation. In many situations it is natural to assume that a player's appreciation of a coalition structure only depends on the coalition he is a member of and not on how the remaining players are grouped.
Initiated by \citet{BKS01a} and \citet{BoJa02a}, much of the work on coalition formation now concentrates on these so-called \emph{hedonic games}. Hedonic games are relevant in modeling many settings such as formation of groups, clubs and societies~\citep{BoJa02a} and also online social networking~\citep{ElkiW09a}.
The main focus in hedonic games has been on notions of stability for coalition structures such as \emph{Nash stability}, \emph{individual stability}, \emph{contractual individual stability}, or \emph{core stability} and characterizing conditions under which the set of stable partitions is guaranteed to be non-empty \citep[see, \eg][]{BoJa02a,BuZw03a}. \citet{SuDi07b} presented a taxonomy of stability concepts which includes the \emph{contractual strict core}, the most general stability concept that is guaranteed to exist.
A well-studied special case of hedonic games are two-sided matching games in which only coalitions of size two are admissible~\citep{RoSo90a}.
We refer to~\citet{Hajd06a} for a critical overview of hedonic games.
Hedonic games have recently been examined from an algorithmic perspective~\citep[see, \eg][]{Ball04a,DBHS06a}. \citet{Cech08a} surveyed the algorithmic problems related to stable partitions in hedonic games in various representations.
\citet{Ball04a} showed that for hedonic games represented by \emph{individually rational list of coalitions}, the complexity of checking whether core stable, Nash stable, or individual stable partitions exist is NP-complete. He also proved that every hedonic game admits a contractually individually stable partition.
Coalition formation games have also received attention in the artificial intelligence community where the focus has generally been on computing optimal partitions for general coalition formation games~\cite{SLA+99a} without any combinatorial structure.
\citet{ElkiW09a} proposed a fully-expressive model to represent hedonic games which encapsulates well-known representations such as \emph{individually rational list of coalitions} and \emph{additive separability}.
\emph{Additively separable hedonic games (ASHGs)} constitute a particularly natural and succinctly representable class of hedonic games. Each player in an ASHG has a value for any other player and the value of a coalition to a particular player is simply the sum of the values he assigns to the members of his coalition. Additive separability satisfies a number of desirable axiomatic properties~\citep{BBP04a} and \ASHGS are the non-transferable utility generalization of \emph{graph games} studied by \citet{DePa94a}.
\citet{Olsen09a} showed that checking whether a nontrivial Nash stable partition exists in an \ASHG is NP-complete if preferences are nonnegative and symmetric. This result was improved by \citet{SuDi10a} who showed that checking whether a core stable, strict core stable, Nash stable, or individually stable partition exists in a general \ASHG is NP-hard.
\citet{DBHS06a} obtained positive algorithmic results for subclasses of \ASHGS in which each player merely divides other players into friends and enemies. \citet{BrLa09a} examined the tradeoff between stability and social welfare in \ASHGS.
Recently, \citet{GaSa10a} showed that computing partitions that satisfy some variants of individual-based stability is PLS-complete, even for very restricted preferences.
In another paper,
\citet{ABS10a} studied the complexity of computing and verifying optimal partitions in \ASHGS.
In this paper, we settle the complexity of key problems regarding stable partitions of \ASHGS.
We present a polynomial-time algorithm to compute a contractually individually stable partition. This is the first positive algorithmic result (with respect to one of the standard stability concepts put forward by \citet{BoJa02a}) for general \ASHGS with no restrictions on the preferences. We strengthen recent results of \citet{SuDi10a} and prove that checking whether the core or the strict core exists is NP-hard, even if the preferences of the players are symmetric. Finally, it is shown that verifying whether a partition is in the contractually strict core (CSC) is coNP-complete, even if the partition under question consists of the grand coalition. This is the first computational hardness result concerning CSC stability in hedonic games of any representation. The proof can be used to show that verifying whether the partition consisting of the grand coalition is Pareto optimal is coNP-complete, thereby answering a question mentioned by \citet{ABS10a}. Our computational hardness results imply computational hardness of the equivalent questions for \emph{hedonic coalition nets} \citep{ElkiW09a}.
\section{Preliminaries}
\label{sec:pre}
In this section, we provide the terminology and notation required for our results.
A \emph{hedonic coalition formation game} is a pair $(N,\mathcal{P}\xspace)$ where $N$ is a set of players and $\mathcal{P}\xspace$ is a \emph{preference profile} which specifies for each player $i\in N$ the preference relation $ \succsim_i$, a reflexive, complete, and transitive binary relation on the set $\mathcal{N}_i=\{S\subseteq N \mid i\in S\}$.
The statement $S\succ_iT$ denotes that $i$ strictly prefers $S$ over $T$ whereas $S\sim_iT$ means that $i$ is indifferent between coalitions $S$
and $T$. A \emph{partition} $\pi$ is a partition of players $N$ into disjoint coalitions. By $\pi(i)$, we denote the coalition of $\pi$ that includes player $i$.
We consider utility-based models rather than purely ordinal models.
In \emph{additively separable preferences}, a player $i$ gets value $v_i(j)$ for player $j$ being in the same coalition as $i$ and if $i$ is in coalition $S\in \mathcal{N}_i$, then $i$ gets utility $\sum_{j\in S\setminus \{i\}}v_i(j)$.
A game $(N,\mathcal{P}\xspace)$ is \emph{additively separable} if for each player $i\in N$, there is a utility function $v_i: N\rightarrow \mathbb{R}$ such that $v_i(i)=0$ and for coalitions $S,T\in\mathcal{N}_i$, $S \succsim_i T$ if and only if $\sum_{j\in S}v_i(j) \geq \sum_{j\in T}v_i(j)$. We will denote the utility of player $i$ in partition $\pi$ by $u_{\pi}(i)$.
A preference profile is \emph{symmetric} if $v_i(j)=v_j(i)$ for any two players $i,j\in N$ and is \emph{strict} if $v_i(j)\neq 0$ for all $i,j\in N$.
For any player $i$, let $F(i,A)=\{j\in A \mid v_i(j)> 0\}$ be the set of friends of player $i$ within $A$.
We now define important stability concepts used in the context of coalition formation games.
\renewcommand*{\labelitemi}{$\bullet$}
\begin{itemize}
\item A partition is \emph{Nash stable (NS)} if no player can benefit by
moving from his coalition $S$ to another (possibly empty) coalition $T$.
\item A partition is \emph{individually stable (IS)} if no player can
benefit by moving from his coalition $S$ to another existing (possibly empty) coalition $T$ while not making the members of $T$ worse off.
\item A partition is \emph{contractually individually stable (CIS)} if no
player can benefit by moving from his coalition $S$ to another existing (possibly empty) coalition
$T$ while making neither the members of $S$ nor the members of
$T$ worse off.
\item We say that a coalition $S \subseteq N$ \emph{strongly blocks} a partition $\pi$, if each
player $i \in S$ strictly prefers $S$ to his current coalition $\pi(i)$ in
the partition $\pi$. A partition which admits no blocking coalition is said to be in the \emph{core (C)}.
\item We say that a coalition $S \subseteq N$ \emph{weakly blocks} a partition $\pi$,
if each player $i \in S$ weakly prefers $S$ to $\pi(i)$ and there exists at least one player $j \in S$ who strictly prefers $S$ to his
current coalition $\pi(j)$. A partition which admits no weakly blocking coalition is in the \emph{strict core (SC)}.
\eat{\item A partition $\pi$ is in the \emph{contractual strict core (CSC)} if the following is not possible: there exists a weakly blocking coalition $S$ and for all $C\in \pi$, and for all $j\in N\setminus S$ we have that $C\setminus S \succsim_j C$.}
\item A partition $\pi$ is in the \emph{contractual strict core (CSC)} if any weakly blocking coalition $S$ makes at least one player $j\in N\setminus S$ worse off when breaking off.
\end{itemize}
The inclusion relationships between stability concepts depicted in Figure~\ref{fig:relations} follow from the definitions of the concepts.
We will also consider \emph{Pareto optimality}.
A partition $\pi$ of $N$ is \emph{Pareto optimal} if there exists no partition $\pi'$ of $N$ such that for all $i\in N$, $\pi'(i) \succsim_i \pi(i)$ and there exists at least one player $j\in N$ such that $\pi'(j) \succ_j \pi(j)$. We say that a partition $\pi$ satisfies \emph{individual rationality} if each player does as well as by being alone, i.e., for all $i\in N$, $\pi(i) \succsim_i \{i\}$.
Throughout the paper, we assume familiarity with basic concepts of computational complexity~\citep[see, \eg][]{ArBa09a}.
\eat{
\begin{figure}
\begin{center}
\scalebox{1}{
\begin{tikzpicture}
\tikzstyle{pfeil}=[->,>=angle 60, shorten >=1pt,draw]
\tikzstyle{onlytext}=[]
\node[onlytext] (NS) at (1,0) {NS};
\node[onlytext] (SC) at (3,0) {SC};
\node[onlytext] (IS) at (0,-1.5) {IS};
\node[onlytext] (CSC) at (2,-1.5) {CSC};
\node[onlytext] (C) at (4,-1.5) {C};
\node[onlytext] (CIS) at (2,-3) {CIS};
\draw[pfeil] (NS) to (IS);
\draw[pfeil] (SC) to (IS);
\draw[pfeil] (SC) to (CSC);
\draw[pfeil] (IS) to (CIS);
\draw[pfeil] (SC) to (C);
\draw[pfeil] (CSC) to (CIS);
\end{tikzpicture}
}
\end{center}
\caption{Inclusion relationships between stability concepts. For example, every Nash stable partition is also individually stable.}
\label{fig:relations}
\end{figure}
}
\begin{figure}
\begin{center}
\scalebox{1}{
\begin{tikzpicture}
\tikzstyle{pfeil}=[->,>=angle 60, shorten >=1pt,draw]
\tikzstyle{onlytext}=[]
\node[onlytext] (NS) at (1,0) {NS};
\node[onlytext] (SC) at (3,0) {SC};
\node[onlytext] (IS) at (0,-1.5) {IS};
\node[onlytext] (PO) at (2,-1.5) {PO};
\node[onlytext] (C) at (4,-1.5) {C};
\node[onlytext] (CSC) at (2,-3) {CSC};
\node[onlytext] (CIS) at (2,-4.5) {CIS};
\draw[pfeil] (NS) to (IS);
\draw[pfeil] (SC) to (IS);
\draw[pfeil] (SC) to (PO);
\draw[pfeil] (IS) to (CIS);
\draw[pfeil] (SC) to (C);
\draw[pfeil] (PO) to (CSC);
\draw[pfeil] (CSC) to (CIS);
\end{tikzpicture}
}
\end{center}
\caption{Inclusion relationships between stability concepts. For example, every Nash stable partition is also individually stable.}
\label{fig:relations}
\end{figure}
\eat{
\begin{fact}
The following statements follow from the definitions of the stability concepts.
\begin{enumerate}
\item Nash stability $\Longrightarrow$ individual stability $\Longrightarrow$ contractual individual stability
\item strict core stability $\Longrightarrow$ core stability
\item strict core stability $\Longrightarrow$ individual stability
\item contractual strict core stability $\Longrightarrow$ contractual individual stability
\end{enumerate}
\label{fact:implies-stabilty}
\end{fact}
}
\section{Contractual individual stability}
\label{sec:cis}
It is known that computing or even checking the existence of Nash stable or individually stable partitions in an \ASHG is NP-hard.
On the other hand, a potential function argument can be used to show that at least one CIS partition exists for every hedonic game~\citep{Ball04a}.
The potential function argument does not imply that a CIS partition can be computed in polynomial time. There are many cases in hedonic games, where a solution is guaranteed to exist but \emph{computing} it is not feasible.
For example, \citet{BoJa02a} presented a potential function argument for the existence of a Nash stable partition for \ASHGS with symmetric preferences.
However there are no known polynomial-time algorithms to \emph{compute} such partitions and there is evidence that there may not be any polynomial-time algorithm~\citep{GaSa10a}.
In this section, we show that a CIS partition can be computed in polynomial time for \ASHGS. The algorithm is formally described as Algorithm~\ref{alg-CIS-general}. Algorithm~\ref{alg-CIS-general} may also prove useful as a preprocessing or intermediate routine in other algorithms to compute different types of stable partitions of hedonic games.
\begin{algorithm}[tb]
\caption{CIS partition of an \ASHG}
\label{alg-CIS-general}
\textbf{Input:} additively separable hedonic game $(N,\mathcal{P}\xspace)$.\\
\textbf{Output:} CIS partition.
\begin{algorithmic}[]
\STATE $i\leftarrow 0$
\STATE $R\leftarrow N$
\WHILE{$R\neq \emptyset$}\label{while-step}
\STATE Take any player $a\in R$
\STATE $h \leftarrow \sum_{b\in F(a,R)}v_a(b)$
\STATE $z\leftarrow i+1$
\FOR{$k\leftarrow 1$ to $i$}
\STATE $h'\leftarrow \sum_{b\in S_k}v_a(b)$
\IF{($h < h'$) ~$\wedge~ (\forall b\in S_k$, $v_b(a)=0$)}
\STATE $h\leftarrow h'$
\STATE $z \leftarrow k$
\ENDIF
\ENDFOR
\IF[$a$ is latecomer]{$z\ne i+1$}
\STATE $S_{z}\leftarrow \{a\}\cup S_{z}$
\STATE $R\leftarrow R \setminus \{a\}$
\ELSE[$a$ is leader]
\STATE $i\leftarrow z$
\STATE $S_i\leftarrow \{a\}$
\STATE $S_i\leftarrow S_i\cup F(a,R)$ \COMMENT{add leader's helpers}
\STATE $R\leftarrow R \setminus S_i$
\ENDIF
\WHILE{$\exists j\in R$ such that $\forall i\in S_{z}$, $v_i(j)\geq 0$ and $\exists i\in S_{z}$, $v_i(j)>0$}
\STATE $R\leftarrow R\setminus \{j\}$
\STATE $S_{z}\leftarrow S_{z} \cup \{j\}$ \COMMENT{add needed players}
\ENDWHILE
\ENDWHILE
\RETURN $\{S_1,\ldots, S_i\}$
\end{algorithmic}
\end{algorithm}
\normalsize
\begin{theorem}
A CIS partition can be computed in polynomial time.
\label{prop:CIS-easy}
\end{theorem}
\begin{proof}
Our algorithm to compute a CIS partition can be viewed as successively giving a priority token to players to form the best possible coalition among the remaining players or join the best possible coalition which tolerates the player.
The basic idea of the algorithm is described informally as follows. Set variable $R$ to $N$ and consider an arbitrary player $a\in R$.
Call $a$ the \emph{leader} of the first coalition $S_i$ with $i=1$.
Move any player $j$ such that $v_a(j)>0$ from $R$ to $S_i$.
Such players are called the \emph{leader's helpers}.
Then keep moving any player from $R$ to $S_i$ which is tolerated by all players in $S_i$ and strictly liked by at least one player in $S_i$.
Call such players \emph{needed players}.
Now increment $i$ and take another player $a$ from among the remaining players $R$ and check the maximum utility he can get from among $R$.
If this utility is less than the utility which can be obtained by joining a previously formed coalition in $\{S_1,\ldots, S_{i-1}\}$, then send the player to such a coalition where he can get the maximum utility (as long all players in the coalition tolerate the incoming player).
Such players are called \emph{latecomers}.
Otherwise, form a new coalition $S_i$ around $a$ which is the best possible coalition for player $a$ taking only players from the remaining players $R$.
Repeat the process until all players have been dealt with and $R=\emptyset$.
We prove by induction on the number of coalitions formed that no CIS deviation can occur in the resulting partition.
The hypothesis is the following:
\emph{Consider the $k$th first formed coalitions $S_1,\ldots, S_k$. Then neither of the following can happen:
\begin{enumerate}
\item There is a CIS deviation by a player from among $S_1,\ldots, S_k$.
\item There is a CIS deviation by a player from among $N\setminus \bigcup_{i\in\{1,\ldots,k\}}S_i$ to a coalition in $\{S_1,\ldots, S_k\}$.
\end{enumerate}
}
\noindent
\paragraph{Base case} Consider the coalition $S_1$. Then the leader of $S_1$ has no incentive to leave.
The leader's helpers are not allowed to leave by the leader. If they did, the leader's utility would decrease.
For each of the needed players, there exists one player in $S_1$ who does not allow the needed player to leave.
Now let us assume a latecomer $i$ arrives in $S_1$. This is only possible if the maximum utility that the latecomer can derive from a coalition $C\subseteq (N\setminus S_1)$ is less than $\sum_{j\in S_1}v_{i}(j)$. Therefore once $i$ joins $S_1$, he will only become less happy by leaving $S_1$.
Any player $i\in N\setminus S_1$ cannot have a CIS deviation to $S_1$. Either $i$ is disliked by at least one player in $S_1$ or $i$ is disliked by no player in $S_1$. In the first case, $i$ cannot deviate to $S_1$ even he has an incentive to. In the second case, player $i$ has no incentive to move to $S_1$ because if he had an incentive, he would already have moved to $S_1$ as a latecomer.
\paragraph{Induction step} Assume that the hypothesis is true. Then we prove that the same holds for the formed coalitions $S_1,\ldots, S_k,S_{k+1}$. By the hypothesis, we know that players cannot leave coalitions $S_1,\ldots, S_k$.
Now consider $S_{k+1}$. The leader $a$ of $S_{k+1}$ is either not allowed to join one of the coalitions in
$\{S_1,\ldots,S_k\}$
or if he is, he has no incentive to join it. Player $a$ would already have been member of $S_i$ for some $i\in \{1,\ldots, k\}$ if one of the following was true:
\begin{itemize}
\item There is some $i\in\{1,\dots,k\}$ such that the leader of $S_i$ likes $a$.
\item There is some $i\in\{1,\dots,k\}$ such that for all $b\in S_i$, $v_{b}(a)\geq 0$ and there exists $b\in S_i$ such that $v_b(a)>0$.
\item There is some $i\in\{1,\dots,k\}$, such that for all $b\in S_i$, $v_{b}(a)= 0$ and $\sum_{b\in S_i}v_{a}(b)> \sum_{b\in F(i,N\setminus \cup_{i=1}^{k}S_i)}v_{a}(b)$ and $\sum_{b\in S_i}v_{a}(b)\geq \sum_{b\in S_j}v_{a}(b)$ for all $j\in \{1,\ldots, k\}$.
\end{itemize}
Therefore $a$ has no incentive or is not allowed to move to another $S_j$ for $j\in \{1,\ldots, k\}$.
Also $a$ will have no incentive to move to any coalition formed after $S_1,\ldots, S_{k+1}$ because he can do strictly better in $S_{k+1}$.
Similarly, $a$'s helpers are not allowed to leave $S_{k+1}$ even if they have an incentive to. Their movement out of $S_{k+1}$ will cause $a$ to become less happy.
Also each needed player in $S_{k+1}$ is not allowed to leave because at least one player in $S_k$ likes him.
Now consider a latecomer $l$ in $S_{k+1}$. Latecomer $l$ gets strictly less utility in any coalition
$C\subseteq N\setminus \bigcup_{i=1}^{k+1}S_i$. Therefore $l$ has no incentive to leave $S_{k+1}$.
Finally, we prove that there exists no player $x\in N\setminus \bigcup_{j=1}^{k+1}S_i$ such that $x$ has an incentive to and is allowed to join $S_i$ for $i\in \{1,\ldots k+1\}$. By the hypothesis, we already know that $x$ does not have an incentive or is allowed to a join a coalition $S_i$ for $i\in \{1,\ldots k\}$. Since $x$ is not a latecomer for $S_{k+1}$, $x$ either does not have an incentive to join $S_{k+1}$ or is disliked by at least one player in $S_{k+1}$.
\qed
\end{proof}
\section{Core and strict core}
\label{sec:core}
For \ASHGS, the problem of testing the core membership of a partition is coNP-complete~\citep{SuDi07a}. This fact does not imply that checking the existence of a core stable partition is NP-hard. Recently, \citet{SuDi10a} showed that for \ASHGS checking whether a core stable or strict core stable partition exists is NP-hard in the strong sense. Their reduction relied on the asymmetry of the players' preferences.
We prove that even with symmetric preferences, checking whether a core stable or a strict core stable partition exists is NP-hard in the strong sense. Symmetry is a natural, but rather strong condition, that can often be exploited algorithmically.
We first present an example
of a six-player \ASHG with symmetric preferences for which the core (and thereby the strict core) is empty.
\begin{example}
\label{example:symm-core-empty}
Consider a six player symmetric \ASHG adapted from an example by \citet{BKS01a} where
\begin{itemize}
\item $v_1(2)=v_3(4)=v_5(6)=6$;
\item $v_1(6)=v_2(3)=v_4(5)=5$;
\item $v_1(3)=v_3(5)=v_1(5)=4$;
\item $v_1(4)=v_2(5)=v_3(6)=-33$; and
\item $v_2(4)=v_2(6)=v_4(6)=-33$
\end{itemize}
as depicted in Figure~\ref{fig:example}.
\begin{figure}
\centering
\scalebox{0.9}{
\begin{tikzpicture}[auto]
\tikzstyle{player}=[draw, circle, fill=white, minimum size=11pt, inner sep=4pt]
\tikzstyle{lab}=[font=\small\itshape]
\draw (0,0) node[player] (x11) {1}
++(90:2cm) node[player] (x12) {2}
++(30:2cm) node[player] (x13) {3}
++(330:2cm) node[player] (x14) {4}
++(270:2cm) node[player] (x15) {5}
++(210:2cm) node[player] (x16) {6};
\draw[-] (x11) to [bend right = 15] node[lab] {4} (x13);
\draw[-] (x13) to [bend right = 15] node[lab] {4} (x15);
\draw[-] (x15) to [bend right = 15] node[lab] {4} (x11);
\draw[-] (x11) to node[lab] {6} (x12);
\draw[-] (x12) to node[lab] {5} (x13);
\draw[-] (x13) to node[lab] {6} (x14);
\draw[-] (x14) to node[lab] {5} (x15);
\draw[-] (x15) to node[lab] {6} (x16);
\draw[-] (x16) to node[lab] {5} (x11);
\end{tikzpicture}
}
\caption{Graphical representation of Example~\ref{example:symm-core-empty}. All edges not shown in the figure have weight $-33$.}
\label{fig:example}
\end{figure}
It can be checked that no partition is core stable for the game.
Note that if $v_i(j)=-33$, then $i$ and $j$ cannot be in the same coalition of a core stable partition.
Also, players can do better than in a partition of singleton players.
Let coalitions which satisfy individual rationality be called feasible coalitions. We note that the following are the feasible coalitions:
$\{1,2\}$, $\{1,3\}$, $\{1,5\}$, $\{1,6\}$, $\{1,2,3\}$, $\{1,3,5\}$, $\{1,5,6\}$, $\{2,3\}$, $\{3,4\}$, $\{3,4,5\}$, $\{3,5\}$, $\{4,5\}$ and $\{5,6\}$.
Consider partition $$\pi=\{\{1,2\}, \{3,4,5\}, \{6\}\}.$$
Then,
\begin{itemize}
\item $u_{\pi}(1)=6$;
\item $u_{\pi}(2)=6$;
\item $u_{\pi}(3)=10$;
\item $u_{\pi}(4)=11$;
\item $u_{\pi}(5)=9$; and
\item $u_{\pi}(6)=0$.
\end{itemize}
Out of the feasible coalitions listed above, the only weakly (and also strongly) blocking coalition is $\{1,5,6\}$ in which player 1 gets utility 9, player 5 gets utility 10, and player 6 gets utility 11. We note that the coalition $\{1,2,3\}$ is not a weakly or strongly blocking coalition because player 3 gets utility 9 in it. Similarly $\{1,3,5\}$ is not a weakly or strongly blocking coalition because both player 3 and player 5 are worse off.
One way to prevent the deviation $\{1,5,6\}$ is to provide some incentive for player $6$ not to deviate with $1$ and $5$. This idea will be used in the proof of Theorem~\ref{prop:corehard}.
\end{example}
We now define a problem that is NP-complete is the strong sense:\\
\noindent
\textbf{Name}: {\sc ExactCoverBy3Sets (E3C)}: \\
\noindent
\textbf{Instance}: A pair $(R,S)$, where $R$ is a set and $S$ is a collection of subsets of
$R$ such that $|R|= 3m$ for some positive integer $m$ and $|s| = 3$ for each
$s\in S$. \\
\noindent
\textbf{Question}: Is there a sub-collection $S'\subseteq S$ which is a partition of $R$? \\
It is known that E3C remains NP-complete even if each $r\in R$ occurs in
at most three members of $S$~\citep{SuDi10a}. We will use this assumption in the proof of Theorem~\ref{prop:corehard}, which will be shown by a reduction from E3C.
\begin{figure*}[tb]
\centering
\scalebox{1}{
\begin{tikzpicture}[auto]
\tikzstyle{small}=[draw, circle, fill=white!100,minimum size=2pt, inner sep=1pt]
\tikzstyle{lab}=[]
\tikzstyle{collection}=[draw,circle,minimum size=11pt, inner sep=2pt]
\tikzstyle{dots}=[]
\foreach \i in {1,2,3,4,5}
{
\pgfmathparse{2*\i-2}
\draw (\pgfmathresult,0) node[small] (x\i1) {}
-- ++(90:0.6cm) node[small] (x\i2) {}
-- ++(30:0.6cm) node[small] (x\i3) {}
-- ++(330:.6cm) node[small] (x\i4) {}
-- ++(270:.6cm) node[small] (x\i5) {}
-- ++(210:.6cm) node[small] (x\i6) {$x_6^\i$}
-- (x\i1);
\draw[-] (x\i1) to [bend right = 15] node[lab] {} (x\i3);
\draw[-] (x\i3) to [bend right = 15] node[lab] {} (x\i5);
\draw[-] (x\i5) to [bend right = 15] node[lab] {} (x\i1);
}
\node[dots] (subgame-dots) at (10,0.3) {$\cdots$};
\node[collection] (y1) at (3,-4) {$y^{s_1}$};
\node[collection] (y2) at (6,-4) {$y^{s_2}$};
\node[dots] (collection-dots) at (8,-4) {$\cdots$};
\draw[-,very thick] (y1) to node {} (x16);
\draw[-,very thick] (y1) to node {} (x26);
\draw[-,very thick] (y1) to node {} (x46);
\draw[-,very thick] (y2) to node {} (x26);
\draw[-,very thick] (y2) to node {} (x36);
\draw[-,very thick] (y2) to node {} (x56);
\draw[dashed] (x16) to (x26);
\draw[dashed] (x26) to [bend right = 17] (x46);
\draw[dashed] (x16) to [bend right = 20] (x46);
\draw[dashed] (x26) to (x36);
\draw[dashed] (x36) to [bend right = 17] (x56);
\draw[dashed] (x26) to [bend right = 20] (x56);
\end{tikzpicture}
}
\caption{Graphical representation of an ASHG derived from an instance of E3C in the proof of Theorem~\ref{prop:corehard}. Symmetric utilities other than $-33$ are given as edges. Thick edges indicate utility $10\frac{1}{4}$ and dashed edges indicate utility $1/2$. Each hexagon at the top looks like the one in Figure~\ref{fig:examplemod}.}
\label{fig:proofTh2}
\end{figure*}
\begin{figure}[tb]
\centering
\scalebox{0.9}{
\begin{tikzpicture}[auto]
\tikzstyle{player}=[draw, circle, fill=white, minimum size=11pt, inner sep=2pt]
\tikzstyle{lab}=[font=\small\itshape]
\draw (0,0) node[player] (x11) {$x_1^i$}
++(90:2cm) node[player] (x12) {$x_2^i$}
++(30:2cm) node[player] (x13) {$x_3^i$}
++(330:2cm) node[player] (x14) {$x_4^i$}
++(270:2cm) node[player] (x15) {$x_5^i$}
++(210:2cm) node[player] (x16) {$x_6^i$};
\draw[-] (x11) to [bend right = 15] node[lab] {4} (x13);
\draw[-] (x13) to [bend right = 15] node[lab] {4} (x15);
\draw[-] (x15) to [bend right = 15] node[lab] {4} (x11);
\draw[-] (x11) to node[lab] {6} (x12);
\draw[-] (x12) to node[lab] {5} (x13);
\draw[-] (x13) to node[lab] {6} (x14);
\draw[-] (x14) to node[lab] {5} (x15);
\draw[-] (x15) to node[lab] {6} (x16);
\draw[-] (x16) to node[lab] {5} (x11);
\end{tikzpicture}
}
\caption{Graphical representation of the \ASHG from Example~\ref{example:symm-core-empty} as used in the proof of Theorem~\ref{prop:corehard}. All edges not shown in the figure have weight $-33$.}
\label{fig:examplemod}
\end{figure}
\begin{theorem}
\label{prop:corehard}
Checking whether a core stable or a strict core stable partition exists is NP-hard in the strong sense, even when preferences are symmetric.
\end{theorem}
\begin{proof}
Let $(R,S)$ be an instance of E3C where $r\in R$ occurs in at most three members of $S$. We reduce $(R,S)$ to an \ASHGS with symmetric preferences $(N,\mathcal{P}\xspace)$ in which there is a player $y^s$ corresponding to each $s\in S$ and there are six players $x_1^r, \ldots, x_6^r$ corresponding to each $r\in R$. These players have preferences over each other in exactly the way players $1,\ldots, 6$ have preference over each other as in Example~\ref{example:symm-core-empty}.
So, $N=\{x_1^r, \ldots, x_6^r \mid r \in R\} \cup \{y ^s \mid s\in S \}$. We assume that all preferences are symmetric. The player preferences are as follows:
\begin{itemize}
\item For $i\in R$, \\$v_{x_1^i}(x_2^i)=v_{x_3^i}(x_4^i)=v_{x_5^i}(x_6^i)=6$;\\
$v_{x_1^i}(x_6^i)=v_{x_2^i}(x_3^i)=v_{x_4^i}(x_5^i)=5$; and\\
$v_{x_1^i}(x_3^i)=v_{x_3^i}(x_5^i)=v_{x_1^i}(x_5^i)=4$;
\item For any $s=\{k,l,m\}\in S$,\\ $v_{x_6^k}(x_6^l)=v_{x_6^l}(x_6^k)=v_{x_6^k}(x_6^m)=v_{x_6^m}(x_6^k)=v_{x_6^l}(x_6^m)=v_{x_6^m}(x_6^l)=1/2$; and\\
$v_{x_6^k}(y^s)=v_{x_6^l}(y^s)=v_{x_6^m}(y^s)=10\frac{1}{4}$;\\
\item $v_i(j)=-33$ for any $i,j\in N$ for valuations not defined above.
\end{itemize}
We prove that $(N,P)$ has a non-empty strict core (and thereby core) if and only if there exists an $S'\subseteq S$ such that $S'$ is a partition of $R$.
Assume that there exists an $S'\subseteq S$ such that $S'$ is a partition of $R$. Then we prove that there exists a strict core stable (and thereby core stable) partition $\pi$ where $\pi$ is defined as follows:
\begin{eqnarray*}
&&\{\{x_1^i, x_2^i\}, \{x_3^i,x_4^i,x_5^i\}\mid i\in R\}
\cup \{\{y^s\}\mid s\in S\setminus S'\}\\
& \cup&\{\{y^s\cup \{x_6^i\mid i\in s\}\}\mid s\in S'\}\text{.}
\end{eqnarray*}
For all $i\in R$,
\begin{itemize}
\item $u_{\pi}(x_1^i)=6$;
\item $u_{\pi}(x_2^i)=6$;
\item $u_{\pi}(x_3^i)=10$;
\item $u_{\pi}(x_4^i)=11$;
\item $u_{\pi}(x_5^i)=9$; and
\item $u_{\pi}(x_6^i)=1/2+1/2+10\frac{1}{4}=11\frac{1}{4}>11.$
\end{itemize}
Also $u_{\pi}(y^s)=3\times (10\frac{1}{4})=30\frac{3}{4}$ for all $s\in S'$ and $u_{\pi}(y^s)=0$ for all $s\in S\setminus S'$. We see that for each player, his utility is non-negative. Therefore there is no incentive for any player to deviate and form a singleton coalition. From Example~\ref{example:symm-core-empty} we also know that the only possible strongly blocking (and weakly blocking) coalition is $\{x_1^i\, x_5^i, x_6^i\}$ for any $i\in R$. However, $x_6^i$ has no incentive to be part $\{x_1^i,x_5^i,x_6^i\}$ because $u_{\pi}(x_6^i)=11$ and $v_{x_6^i}(x_5^i)+v_{x_6^i}(x_1^i)=6+5=11$.
Also $x_1^i$ and $x_5^i$ have no incentive to join $\pi(x_6^i)$ because their new utility will become negative because of the presence of the $y^s$ player. Assume for the sake of contradiction that $\pi$ is not core stable and $x_6^i$ can deviate with a lot of $x_6^j$s. But, $x_6^i$ can only deviate with a maximum of six other players of type $x_6^j$ because $i\in R$ is present in a maximum of three elements in $S$. In this case $x_6^i$ gets a maximum utility of only $1$.
Therefore $\pi$ is in the strict core (and thereby the core).
We now assume that there exists a partition which is core stable. Then we prove that there exists an $S'\subseteq S$ such that $S'$ is a partition of $R$.
For any $s=\{k,l,m\}\in S$, the new utilities created due to the reduction gadget are only beneficial to $y^s$, $x_6^k$, $x_6^l$, and $x_6^m$.
We already know that the only way the partition is core stable is if $x_6^i$ can be provided disincentive to deviate with $x_5^i$ and $x_1^i$.
The claim is that each $x_6^i$ needs to be in a coalition with exactly one $y^s$ such that $i\in s\in S$ and exactly two other players $x_6^j$ and $x_6^k$ such that $\{i,j,k\}=s\in S$. We first show that $x_6^i$ needs to be with exactly one $y^s$ such that $i\in s\in S$. Player needs to be with at least one such $y^s$. If $x_6^i$ is only with other $x_6^j$s, then we know that $x_6^i$ gets a maximum utility of only $6\times 1/2=3$. Also, player $x_6^i$ cannot be in a coalition with $y^s$ and $y^{s'}$ such that $i\in s$ and $i\in s'$ because both $y^s$ and $y^{s'}$ then get negative utility. Each $x_6^i$ also needs to be with at least 2 other players $x_6^j$ and $x_6^k$ where $j$ and $k$ are also members of $s$. If $x_6^i$ is with at least three players $x_6^j$, $x_6^k$ and $x_6^k$, then there is one element among $a\in \{j,k,l\}$ such that $a\notin s$. Therefore $y^s$ and $x_6^a$ hate each other and the coalition $\{y^s, x_6^i,x_6^j, x_6^k,x_6^k\}$ is not even individually rational. Therefore for the partition to be core stable each $x_6^i$ has to be with exactly one $y^s$ such that $i\in s$ and and least 2 other players $x_6^j$ and $x_6^k$ where $j$ and $k$ are also members of $s$. This implies that there exists an $S'\subseteq S$ such that $S'$ is a partition of $R$.\qed
\end{proof}
\section{Contractual strict core and Pareto optimality}
\label{sec:csc}
\begin{figure*}[htb]
\centering
\scalebox{1}{
\begin{tikzpicture}[auto]
\tikzstyle{player}=[draw, circle, fill=white, minimum size=11pt, inner sep=2pt]
\tikzstyle{playerz}=[draw, circle, fill=white, minimum size=8pt, inner sep=2pt]
\tikzstyle{lab}=[font=\small\itshape]
\tikzstyle{dots}=[]
\draw
(-3,3) node[playerz] (z1) {$z_1$}
(-3,1.5) node[playerz] (zi) {$z_i$}
(-3,0) node[playerz] (zk) {$z_k$}
(0,0) node[player] (x1) {$x_1$}
++(90:3cm) node[player] (x2) {$x_2$}
++(0:3cm) node[player] (y2) {$y_2$}
++(270:3cm) node[player] (y1) {$y_1$};
\draw[->] (x1) to [] node[lab, above=6pt, left=-2pt] {$W/2$} (y1);
\draw[->] (x2) to [] node[lab, above=-1pt] {$W/2$} (y2);
\draw[-] (x1) to node[lab] {$-W$} (x2);
\draw[->] (x1) to node[lab, above=20pt, right=-4pt] {$W/2$} (y2);
\draw[->] (x2) to node[lab, below=22pt, right=-4pt] {$W/2$} (y1);
\draw[-] (y2) to node[lab] {$-W$} (y1);
\draw[->] (x1) to node[lab] {$a_i$} (zi);
\draw[->] (x2) to node[lab] {$a_i$} (zi);
\draw (-3,1.5) ellipse (1cm and 2cm);
\node[dots] (subgame-dots) at (-3,0.9) {$\vdots$};
\node[dots] (subgame-dots) at (-3,2.3) {$\vdots$};
\end{tikzpicture}
}
\caption{Graphical representation of the \ASHG in the proof of Theorem~\ref{th:csc-hard}. For all $i\in \{1,\ldots, k\}$, an edge from $x_1$ and $x_2$ to $z_i$ has weight $a_i$. All other edges not shown in the figure have weight zero.}
\label{fig:exampleCSC}
\end{figure*}
In this section, we prove that verifying whether a partition is CSC stable is coNP-complete. Interestingly, coNP-completeness holds even if the partition in question consists of the grand coalition.
The proof of Theorem~\ref{th:csc-hard} is by a reduction from the following weakly NP-complete problem.\\
\noindent
\textbf{Name}: {\sc Partition} \\
\textbf{Instance}: A set of $k$ positive integer weights $A=\{a_1, \ldots, a_k \}$ such that $\sum_{a_i\in A}a_i=W$.\\
\textbf{Question}: Is it possible to partition $A$, into two subsets $A_1\subseteq A$, $A_2\subseteq A$ so that $A_1\cap A_2=\emptyset$ and $A_1\cup A_2=A$ and $\sum_{a_i\in A_1}a_i=\sum_{a_i\in A_2}a_i=W/2$?\\
\begin{theorem}\label{th:csc-hard}
Verifying whether the partition consisting of the grand coalition is CSC stable is weakly coNP-complete.
\end{theorem}
\begin{proof}
The problem is clearly in coNP because a partition $\pi'$ resulting by a CSC deviation from $\{N\}$ is a succinct certificate that $\{N\}$ is not CSC stable.
We prove NP-hardness of deciding whether the grand coalition is \emph{not} CSC stable by a reduction from {\sc Partition}.
We can reduce an instance of $I$ of {\sc Partition} to an instance $I'=((N,\mathcal{P}\xspace),\pi)$ where $(N,\mathcal{P}\xspace)$ is an \ASHG defined in the following way:
\begin{itemize}
\item $N=\{x_1,x_2,y_1,y_2, z_1,\ldots,z_k\}$,
\item $v_{x_1}(y_1)=v_{x_1}(y_2)=v_{x_2}(y_1)=v_{x_2}(y_2)=W/2$,
\item $v_{x_1}(z_i)=v_{x_2}(z_i)=a_i$, for all $i\in \{1,\ldots, k\}$
\item $v_{x_1}(x_2)=v_{x_2}(x_1)=-W$,
\item $v_{y_1}(y_2)=v_{y_2}(y_1)=-W$,
\item $v_{a}(b)=0$ for any $a,b\in N$ for which $v_{a}(b)$ is not already defined, and
\item $\pi=\{N\}$.
\end{itemize}
We see that $u_{\pi}(x_1)=u_{\pi}(x_1)=W$, $u_{\pi}(y_1)=u_{\pi}(y_2)=-W$, $u_{\pi}(z_i)=0$ for all $i\in \{1,\ldots, k\}$.
We show that $\pi$ is not CSC stable if and only if $I$ is a `yes' instance of {\sc Partition}.
Assume $I$ is a `yes' instance of {\sc Partition} and there exists an $A_1\subseteq A$ such that $\sum_{a_i\in A_1}a_i=W/2$.
Then, form the following partition
$$\pi'=\{\{x_1,y_1\}\cup \{z_i \mid a_i\in A_1\}, \{x_2,y_2\}\cup \{z_i \mid a_i\in N\setminus A_1\} \} $$
Then,
\begin{itemize}
\item $u_{\pi'}(x_1)=u_{\pi'}(x_1)=W$;
\item $u_{\pi'}(y_1)=u_{\pi'}(y_2)=0$; and
\item $u_{\pi}(z_i)=0$ for all $i\in \{1,\ldots, k\}$.
\end{itemize}
The coalition $C_1=\{x_1,y_1\}\cup \{z_i \mid a_i\in A_1\}$ can be considered as a coalition which leaves the grand coalition so that all players in $N$ do as well as before and at least one player in $C_1$, i.e., $y_1$ gets strictly more utility. Also, the departure of $C_1$ does not make any player in $N\setminus C_1$ worse off.
Assume that $I$ is a `no' instance of {\sc Partition} and there exists no $A_1\subseteq A$ such that $\sum_{a_i\in A_1}a_i=W/2$. We show that no CSC deviation is possible from $\pi$.
We consider different possibilities for a CSC blocking coalition $C$:
\begin{enumerate}
\item $x_1,x_2, y_1, y_2 \notin C$,
\item $x_1,x_2\notin C$ and there exists $y\in \{y_1,y_2\}$ such that $y\in C$,
\item $x_1,x_2,y_1,y_2\in C$,
\item $x_1,x_2\in C$ and $|C\cap \{y_1,y_2\}|\leq 1$,
\item there exists $x\in \{x_1,x_2\}$ and $y\in \{y_1,y_2\}$ such that $x,y\in C$, $\{x_1,x_2\}\setminus x \nsubseteq C$, and $\{y_1,y_2\}\setminus y \nsubseteq C$
\end{enumerate}
We show that in each of the cases, $C$ is a not a valid CSC blocking coalition.
\begin{enumerate}
\item If $C$ is empty, then there exists no CSC blocking coalition. If $C$ is not empty, then $x_1$ and $x_2$ gets strictly less utility when a subset of $\{z_1,\ldots, z_k\}$ deviates.
\item In this case, both $x_1$ and $x_2$ gets strictly less utility when $y\in \{y_1,y_2\}$ leaves $N$.
\item If $\{z_1,\ldots, z_k\}\subset C$, then there is no deviation as $C=N$. If there exists a $z_i\in \{z_1,\ldots, z_k\}$ such that $z_i\notin C$, then $x_1$ and $x_2$ get strictly less utility than in $N$.
\item If $|C\cap \{y_1,y_2\}|= 0$, then the utility of no player increases. If $|C\cap \{y_1,y_2\}|=1$, then the utility of $y_1$ and $y_2$ increases but the utility of $x_1$ and $x_2$ decreases.
\item Consider $C=\{x,y\}\cup S$ where $S\subseteq \{z_1,\ldots, z_k\}$. Without loss of generality, we can assume that $x=x_1$ and $y=y_1$. We know that $y_1$ and $y_2$ gets strictly more utility because they are now in different coalitions. Since $I$ is a `no' instance of {\sc Partition}, we know that there exists no $S$ such that $\sum_{a\in S}v_{x_1}(a)=W/2$. If $\sum_{a\in S}v_{x_1}(a)>W/2$, then $u_{\pi}(x_2)<W$. If $\sum_{a\in S}v_{x_1}(a)<W/2$, then $u_{\pi}(x_1)<W$.
\end{enumerate}
Thus, if $I'$ is a `no' instance of {\sc Partition}, then there exists no CSC deviation. \qed
\end{proof}
From the proof of Theorem~\ref{th:csc-hard}, it can be seen that $\pi$ is not Pareto optimal if and only if $I$ is a `yes' instance of {\sc Partition}.
\begin{theorem}\label{th:GC-PO}
Verifying whether the partition consisting of the grand coalition is Pareto optimal is coNP-complete.
\end{theorem}
\section{Conclusion and Discussion}
We presented a number of new computational results concerning stable partitions of \ASHGS.
First, we proposed a polynomial-time algorithm for computing a contractually individually stable (CIS) partition. Secondly, we showed that checking whether the core or strict core exists is NP-hard in the strong sense, even if the preferences of the players are symmetric. Finally, we presented the first complexity result concerning the contractual strict core (CSC), namely that verifying whether a partition is in the CSC is coNP-complete. We saw that considering CSC deviations helps reason about the more complex Pareto optimal improvements. As a result, we established that checking whether the partition consisting of the grand coalition is Pareto optimal is also coNP-complete.
We note that Algorithm~\ref{alg-CIS-general} may very well return a partition that fails to satisfy individual rationality, \ie players may get negative utility. It is an open question how to efficiently compute a CIS partition that is guaranteed to satisfy individual rationality.
We also note that Theorem~\ref{th:csc-hard} may not imply anything about the complexity of \emph{computing} a CSC partition. Studying the complexity of computing a CSC stable partition is left as future work.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,183 |
\section{Introduction}
The most intriguing possibility for the origin of
the current acceleration of the Universe~\cite{Acceleration}
may be modifying gravity at very long distances.
Within the framework of braneworlds,
Dvali, Gabadadze, and Porrati (DGP) proposed
a model in which gravity looks five-dimensional (5D)
at long distances and 4D at short distances~\cite{DGP}.
In their model,
gravity becomes weaker on large scales due to its leakage from our 4D brane
to the higher dimensional space, while
the induced-gravity term on the brane maintains
the 4D nature of short-range gravity.
Although the original motivation of the idea of~\cite{DGP} was
to realize 4D gravity even in the presence of infinitely large extra dimensions,
it was soon noticed that the model gives rise to the self-acceleration, i.e.,
the accelerating expansion of the Universe without a cosmological constant~\cite{SA, Lue_Rev, Koyama_CC}.
The modified Friedmann equation in the DGP braneworld is given by~\cite{SA}
\begin{eqnarray*}
H^2-\frac{H}{r_c}=\frac{8\pi G}{3}\rho,
\end{eqnarray*}
where $r_c$ is the crossover scale above which gravity is modified.
This allows for the late-time accelerating solution, $H\to H_{\infty}=r_c^{-1}$.
Despite multiple problematic features of the DGP
model~\cite{Rubakov:2003zb, Ghost0, Ghost1, Ghost2,Kaloper:2005az,Gregory:2007xy},
it is still worth exploring
the arena of IR modified gravity in the braneworld context
because more elaborated brane models may provide
consistent IR modification which has possible relevance to the cosmological constant problem,
in terms of a fully covariant and nonlinear theory of gravity.
So far the consequences of brane-induced gravity
have been often discussed in codimension-1 models, but
it seems quite interesting to consider more general, codimension-$N\geq 2$ braneworlds
with induced gravity.
Infinitely thin branes with $N\geq 2$, however, suffer from certain singularities
which must be dealt with some care~\cite{DG}.
The pathology
is clearly illustrated by the shock wave solution on thin codimension-2 branes~\cite{Kaloper}:
the shock wave profile is perfectly 4D over all distance scales on the brane,
but it vanishes (even infinitesimally) away from the brane.
Namely, gravity confinement is too effective.
This is due to the fact that the Green function in the space transverse to the brane
diverges at the origin. (The divergence is the generic feature for $N\geq2$.)
If one is to resolve this short distance singularity somehow,
the regularization at UV will affect the behavior of gravity
at IR~\cite{Kiri, Kaloper, Seesaw, Reg-Rubakov, Reg-DGP, Softly}.
(See also different approaches to construct higher codimension braneworlds with induced
gravity~\cite{Int, Cascading}.)
In this paper we consider a 6D brane model
that incorporates long distance modification of
gravity and regularization of codimension-2 singularities.
The codimension-2 branes in our model are regularized
in the manner of~\cite{Kaloper}, replacing
them with ring-like codimension-1 branes and filling in the interiors with regular caps.
The scheme here is similar to what is widely used in flux-stabilized models of 6D
braneworlds~\cite{Peloso, Papa, Reg-TK, UVcaps}.
Then, instead of introducing brane-induced gravity term directly,
we use the possible mechanism for realizing {\em effectively} DGP-type gravity, that is,
asymmetry between the two sides of the brane~\cite{Padilla, KoyamaKoyama}.
While the standard way to study codimension-1 brane models in five dimensions
is assuming $\mathbb{Z}_2$-symmetry across the brane~\cite{RS, Z2},
the bulk-cap system in regularized braneworlds has an asymmetric configuration in general~\cite{nonZ2}.
In addition to this natural geometrical asymmetry, we allow the cap Planck scale, $M_{\rm C}$,
to differ from that in the bulk, $M_{\rm B}$.
If $M_{\rm C}\gg M_{\rm B}$, this leads to an explicit realization of
the regularization of brane-induced gravity~\cite{Reg-DGP}.
The purpose of the present paper is
to investigate in detail the behavior of weak gravity in such a braneworld.
Even if gravity is weak,
linearization is not justified below a certain distance scale, as in the DGP
model~\cite{Deffayet,Lue:2001gc,Gabadadze:2004iy,Gruzinov:2001hp,Tanaka, KScp}.
We identify the scale and then
carefully take into account the second order effects of the brane bending
to explore the seminonlinear regime.
This paper is organized as follows. In the next section
we present our background model of regularized braneworlds.
We then study the behavior of linear perturbations
in Sec.~III, showing that brane gravity
at short distances
is described by the 4D Brans-Dicke theory.
As an example, in Sec.~\ref{sec:pointsource} we compute
the gravitational field of a static point source modeled by a loop of a string along the ring-like brane.
Using this example we confirm that gravity looks five- or six-dimensional at long distances.
In Sec.~IV we perform a nonlinear analysis
to see how 4D Einstein gravity emerges at even shorter distances.
Sec.~V is devoted to conclusions.
\section{The model}
Let us consider a model
in which the bulk is given by 4D Minkowski spacetime $\times$ a infinite cone,\footnote{
We use $A, B, ...$ for 6D indices,
$\mu, \nu, ...$ for 4D ones, and $a, b, ...$ for 5D ones parallel to the 4-brane.
}
\begin{eqnarray}
g_{AB}dx^Adx^B= \eta_{\mu\nu}dx^{\mu}dx^{\nu}+dr^2+\beta^2r^2d\varphi^2,
\label{bulkmetric}
\end{eqnarray}
where $\beta\leq 1$.
The conical singularity at $r=0$,
which corresponds to a codimension-2 tense brane,
is resolved by introducing
a cylindrical 4-brane at $r=r_0$ and filling in the interior with a regular cap.
As in~\cite{Kaloper}, we assume a disk-like cap whose metric is given by
\begin{eqnarray}
g_{AB}dx^Adx^B = \eta_{\mu\nu}dx^{\mu}dx^{\nu}+d\rho^2+\rho^2d\varphi^2
\nonumber\\
(\rho<\rho_0=\beta r_0),
\label{capmetric}
\end{eqnarray}
where $\rho_0$ is the position of the brane seen from the cap side.
In other words, $\rho_0$ is the size of the cap.
The continuity of the induced metric on the brane implies that $\rho_0=\beta r_0$.
Throughout the paper we assume that both the bulk and the cap are described by
6D Einstein gravity without a cosmological constant nor any other fields.
Both~(\ref{bulkmetric}) and~(\ref{capmetric}) solve the 6D vacuum Einstein equations trivially.
The two spacetimes are glued together along the ring-like brane, on which now
it is possible to put an arbitrary energy-momentum tensor.
The 4-brane action we consider is
\begin{eqnarray}
S_{\text{brane}} =
\int d^5x\sqrt{-q}\left(-\lambda-\frac{1}{2}q^{ab}\partial_a\Sigma\partial_b\Sigma+{\cal L}_{{\rm m}}\right),
\end{eqnarray}
where
$q_{ab}$ is the induced metric on the brane,
$\lambda$ is the tension of the brane, $\Sigma$ is the brane-localized scalar field,
and ${\cal L}_{{\rm m}}$ is the Lagrangian of usual matter.
As we are describing the background configuration of the model,
we do not consider the contribution from ${\cal L}_{{\rm m}}$ in this section.
The background equation of motion, $\partial_{\varphi}\partial^{\varphi}\Sigma=0$, implies that
$\Sigma = Q \varphi$, where $Q$ is a constant.
The solution for $\Sigma$
breaks the translational invariance in the fifth direction on the brane, but
the energy-momentum tensor does not. This field is introduced
so as to cancel the pressure in the $\varphi$ direction coming from $\lambda$.
Note that
although we discuss only the very simple bulk geometry presented above,
such a regularization scheme works as well
in more general setups in which the bulk contains e.g.
fluxes and dilaton fields and is curved~\cite{Peloso, Papa, Reg-TK, UVcaps}.
Suppose now that the Planck scale in the cap region, $M_{\rm C}$,
differs from that in the bulk, $M_{\rm B}$.
The junction conditions on the brane are then given by~\cite{Is}
\begin{eqnarray}
M^4_{\rm C} \left.k_{ab}\right|_{ \rho_0}
-M^4_{\rm B} \left.k_{ab}\right|_{ r_0}={\cal T}_{ab},
\end{eqnarray}
where $k_{ab}=K_{ab}-q_{ab}K_c^{\;c}$, $K_{ab}$ is the extrinsic curvature of the brane,
\begin{eqnarray}
{\cal T}_{ab}=
-\lambda q_{ab}+\partial_{a}\Sigma\partial_{b}\Sigma
-\frac{1}{2}q_{ab}\partial_c\Sigma\partial^{c}\Sigma
+T_{ab},
\end{eqnarray}
and $T_{ab}$ is the energy-momentum tensor derived from ${\cal L}_{{\rm m}}$.
The background junction conditions read
\begin{eqnarray}
(\varphi\varphi): &&\;\; \lambda=\frac{1}{2}q^{\varphi\varphi}Q^2,\label{bj1}
\\
(\mu\nu):&&\;\;
\frac{M_{\rm C}^4}{\rho_0}-\frac{M_{\rm B}^4}{r_0}=\lambda+\frac{1}{2}q^{\varphi\varphi}Q^2,\label{bj2}
\end{eqnarray}
with $q^{\varphi\varphi}=\rho_0^{-2}=(\beta r_0)^{-2}$.
Therefore, the brane tension satisfies
the relation $2\lambda=M_{\rm C}^4/\rho_0-M_{\rm B}^4/r_0$.
We thus have defined the background configuration of the model (Fig.~1).
In contrast to the model of~\cite{Kaloper},
we do not introduce an induced gravity term in the 4-brane action explicitly.
Rather, we realize the regularization of brane-induced gravity proposed by~\cite{Reg-DGP},
using the capped bulk geometry.
The key assumption here is the very large Planck scale in the cap: $M_{\rm C}\gg M_{\rm B}$.
The idea is also closely related to the work of~\cite{Padilla, KoyamaKoyama},
in which the asymmetry between
two sides of the brane gives rise to the effect analogous to brane-induced gravity.
The authors of~\cite{Padilla, KoyamaKoyama} considered the Randall-Sundrum-type
braneworlds, but the mechanism will operate also in the present setup.
In the rest of the paper,
we will show by a detailed perturbation analysis that 4D gravity indeed emerges
in a certain region of distance scales.
\begin{figure}[tb]
\begin{center}
\includegraphics[keepaspectratio=true,height=50mm]{cone.eps}
\end{center}
\caption{The configuration of the model.}%
\label{fig:cone.eps}
\end{figure}
\section{Linear analysis}\label{sec:lin}
\subsection{General analysis}\label{genan}
We study linearized gravity sourced by arbitrary matter on the brane.
The perturbed metric is given by
\begin{eqnarray}
&&(g_{AB}+\delta g_{AB})dx^Adx^B
=\left( \eta_{\mu\nu}+\gamma_{\mu\nu}\right)dx^{\mu}dx^{\nu}
\nonumber\\
&&\quad
+2B_{,\mu}dx^{\mu}d\rho+ (1+2\Gamma)d\rho^2+(1-2\Phi)\rho^2d\varphi^2,
\label{percap}
\end{eqnarray}
in the cap and
\begin{eqnarray}
&&(g_{AB}+\delta g_{AB})dx^Adx^B=
\left(\eta_{\mu\nu}+\gamma_{\mu\nu}\right)dx^{\mu}dx^{\nu}
\nonumber\\
&&\;\;
+2B_{,\mu}dx^{\mu}dr+ (1+2\Gamma)dr^2+(1-2\Phi)\beta^2r^2d\varphi^2,
\label{perbulk}
\end{eqnarray}
in the bulk.
The perturbations are split into scalar, vector, and tensor modes under the Lorentz group in
the 4D coordinates, and so we write
\begin{eqnarray}
\gamma_{\mu\nu}:=2\Psi\eta_{\mu\nu}+2E_{,\mu\nu}+h_{\mu\nu},
\end{eqnarray}
where $h_{\mu\nu}$ is the transverse and traceless tensor perturbation:
$h_\mu^{\;\mu}=\partial_\nu h_{\mu}^{\;\nu}=0 $.
Vector perturbations are not included above and will not be considered throughout the main text,
because, as is explained in Appendix~\ref{app:vec},
they do not contribute to gravity on the brane.
In this paper we only consider axisymmetric perturbations.
In Eqs.~(\ref{percap}) and~(\ref{perbulk}) the perturbed components $\delta g_{\rho\varphi}$,
$\delta g_{r\varphi}$, and $\delta g_{\mu\varphi}$ are eliminated
with the aid of the gauge transformation $\varphi\to\varphi+\delta\varphi$.
We can further use the gauge transformation $\rho\to \rho+\delta\rho\;(r\to r+\delta r)$
and $x^{\mu}\to x^{\mu}+\delta x^{\mu}$.
The linearized Einstein equations will be solved easily by invoking the specific gauge in which $B=E=0$
(the longitudinal gauge).
Our master equations in the cap are
\begin{eqnarray}
h_{\mu\nu}''+\frac{1}{\rho}h_{\mu\nu}'+\Box h_{\mu\nu}=0
\label{master-tensor}
\end{eqnarray}
for the tensor perturbations and
\begin{eqnarray}
\Psi''+\frac{1}{\rho}\Psi'+\Box\Psi=0
\label{master-scalar}
\end{eqnarray}
for the scalar perturbations, where the primes denote derivatives with respect to $\rho$
and $\Box:=\eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$.
The other scalar quantities are obtained from
\begin{eqnarray}
\left(\rho^2\Phi\right)'&=&3\rho^2\Psi'+2\rho\Psi,\label{rel-scalar1}
\\
\Gamma&=&\Phi-2\Psi.\label{rel-scalar2}
\end{eqnarray}
The derivation of these equations is deferred to Appendix~\ref{app:per}.
Eqs.~(\ref{master-tensor})--(\ref{rel-scalar2}) are for the cap,
but we have the equations for the bulk by replacing $\rho$ with $r$ in the above.
The general solutions to the bulk and cap perturbation equations are found to be
\begin{eqnarray}
h_{\mu\nu}&=&
\begin{cases}
\displaystyle{
\int\hat h_{\mu\nu{\rm C}}(p)I_0(p\rho)e^{ip\cdot x}d^4p
}\;\;\;\text{(cap)}\\
\displaystyle{
\int\hat h_{\mu\nu{\rm B}}(p)K_0(pr)e^{ip\cdot x}d^4p
}\;\;\text{(bulk)}
\end{cases},\label{gen_h}
\\
\Psi&=&
\begin{cases}
\displaystyle{
\int\hat\psi_{\rm C}(p)I_0(p\rho)e^{ip\cdot x}d^4p
}\quad\;\text{(cap)}\\
\displaystyle{
\int\hat\psi_{\rm B}(p)K_0(pr)e^{ip\cdot x}d^4p
}\quad\text{(bulk)}
\end{cases},\label{gen_psi}
\\
\Phi&=&
\begin{cases}
\displaystyle{
\int\hat\psi_{\rm C}(p) {\cal I}(p\rho) e^{ip\cdot x}d^4p
}\quad\text{(cap)}\\
\displaystyle{
\int\hat\psi_{\rm B}(p) {\cal K}(pr) e^{ip\cdot x}d^4p
}\quad\text{(bulk)}
\end{cases},\label{gen_phi}
\end{eqnarray}
where
\begin{eqnarray}
{\cal I}(x):=3I_2(x)+\frac{2I_1(x)}{x},
\\
{\cal K}(x):=3K_2(x)-\frac{2K_1(x)}{x},
\end{eqnarray}
and $K_n$ and $I_n$ are the modified Bessel functions.
Note that $I_0(p\rho)$ and ${\cal I}(p\rho)$
are regular at the center of the disk, $\rho=0$,
and $K_0(pr)$ and ${\cal K}(pr)$ remain finite as $r\to\infty$.
The above choice of the gauge forces the brane to move from its original position:
\begin{eqnarray}
\rho_0\to\rho_0+\zeta_{\rm C}(x),
\quad
r_0\to r_0+\zeta_{\rm B}(x).
\end{eqnarray}
In order to impose the boundary conditions at the brane, it is thus convenient to go to
the Gaussian normal gauge, in which the brane does not flutter.
The tensor perturbation $h_{\mu\nu}$ is invariant under
an infinitesimal coordinate transformation.
According to
the gauge transformation of the scalar perturbations summarized in Appendix~\ref{app:sc},
the Gaussian normal gauge quantities evaluated on the brane
are related to the perturbations in the longitudinal gauge as
\begin{eqnarray}
&&\overline{\Psi}=\Psi|_{\rho_0},\quad
\overline\Phi=\Phi|_{\rho_0}-\frac{\zeta_{\rm C}}{\rho_0},\quad\overline{E}=0,
\quad\overline{\Psi}'=\Psi'|_{\rho_0},
\nonumber\\&&
\overline\Phi'=\Phi'|_{\rho_0}+\frac{1}{\rho_0}\Gamma|_{\rho_0}+\frac{1}{\rho_0^2}\zeta_{\rm C},
\quad \overline{E}'=-\zeta_{\rm C}.
\label{gt}
\end{eqnarray}
Again, replacing $\rho$ and $\zeta_{\rm C}$ with $r$ and $\zeta_{{\rm B}}$, respectively,
we obtain the equations for the bulk side.
Since the induced metric must be continuous across the brane, we impose
\begin{eqnarray}
\Psi|_{\rho_0}=\Psi|_{r_0}\label{con_psi}
\end{eqnarray}
and
\begin{eqnarray}
-\Phi|_{\rho_0}+\frac{\zeta_{{\rm C}}}{\rho_0} = -\Phi|_{r_0}+\frac{\zeta_{\rm B}}{r_0}=: \phi(x).
\label{con_phi}
\end{eqnarray}
The perturbed extrinsic curvature is given by
$\delta K_{\mu\nu}=(1/2)\overline{\gamma}_{\mu\nu}'$
and $\delta K_{\varphi}^{\;\varphi}=-\overline{\Phi}'$.
Using the transformation rule~(\ref{gt}),
the junction conditions are written in terms of the longitudinal gauge quantities as
\begin{eqnarray}
&&M_{\rm C}^4\left(\frac{1}{2}h_{\mu\nu}'|_{\rho_0}+
{\cal D}_{\mu\nu}\zeta_{\rm C}\right)
\nonumber\\&&\;
-M_{\rm B}^4\left(\frac{1}{2}h_{\mu\nu}'|_{r_0}+
{\cal D}_{\mu\nu}\zeta_{\rm B}\right)
=T_{\mu\nu},
\label{junc_munu}
\end{eqnarray}
and
\begin{eqnarray}
&&M_{\rm C}^4 \left(\Box\zeta_{\rm C}-4 \Psi'|_{\rho_0}+\frac{\phi}{\rho_0}\right)
\nonumber\\&&\;
-M_{\rm B}^4 \left(\Box\zeta_{\rm B}-4 \Psi'|_{r_0}+\frac{\phi}{r_0}\right)
=T_{\varphi}^{\;\varphi},
\label{junc_phiphi}
\end{eqnarray}
where ${\cal D}_{\mu\nu}:=\eta_{\mu\nu}\Box-\partial_{\mu}\partial_{\nu}$.
We used the background junction conditions and the perturbed Einstein equations
$\delta G_{\mu\rho}=\delta G_{\mu r}=0$ to make the expressions as compact as possible.
(The derivation and details are presented in Appendix~\ref{app:der}.)
Note that the perturbation of the brane-localized scalar field, $\delta\Sigma(x)$,
does not appear in the junction conditions as long as axisymmetric perturbations are considered.
The 4D trace of the $(\mu\nu)$ component of the junction conditions
is useful:
\begin{eqnarray}
M_{\rm C}^4\Box\zeta_{\rm C}
-M_{\rm B}^4\Box\zeta_{\rm B}=\frac{1}{3}T_{\mu}^{\;\mu}.\label{tr_junc_munu}
\end{eqnarray}
We now define
\begin{eqnarray}
M_4^2=\pi\rho^2_0M_{\rm C}^4,\quad r_c=\frac{\rho_0M_{\rm C}^4}{2M_{\rm B}^4},
\end{eqnarray}
and the energy-momentum tensor integrated along the $\varphi$ direction: $\overline{T}_{ab}=2\pi\rho_0 T_{ab}$.
Then, the above junction conditions are rewritten as
\begin{eqnarray*}
&&M_{\rm C}^4(\,\text{cap}\,)-M_{\rm B}^4(\,\text{bulk}\,)=T_{**}
\nonumber\\
&&\qquad\;\; \Leftrightarrow \;\;
\frac{2}{\rho_0}(\,\text{cap}\,)- \frac{1}{r_c}(\,\text{bulk}\,)=\frac{\overline{T}_{**}}{M_4^2}.
\end{eqnarray*}
Recall now that the brane position seen from the bulk side, $r_0$,
is related to the size of the disk-like cap, $\rho_0$, as $r_0=\rho_0/\beta$,
where $\beta$ controls the opening of the cone.
We assume that $\rho_0\ll r_c\lesssim r_0$. This requires
$\beta\ll 1$ (i.e., a very narrow cone) and $M_{\rm C}\gg M_{\rm B}$.
By $r_c\lesssim r_0$
we mean to allow for both $r_0<r_c$ and $r_c\ll r_0$, but we preclude $r_0\ll r_c$.
Let us focus on scales which lie between $\rho_0$ and $r_0$.
The approximation
$I_0(p\rho_0)\simeq1+(p\rho_0)^2/4$ yields
\begin{eqnarray}
\left. h_{\mu\nu}'\right|_{\rho_0} &\simeq&
\frac{\rho_0}{2}\int\hat h_{\mu\nu{\rm C}}(p)p^2e^{ip\cdot x}d^4p
\nonumber\\
&=&-\frac{\rho_0}{2}\left.\Box h_{\mu\nu}\right|_{\rho_0}.\label{h'=boxh}
\end{eqnarray}
Similarly, we have
\begin{eqnarray}
\left. \Psi'\right|_{\rho_0} \simeq -\frac{\rho_0}{2}\Box\Psi|_{\rho_0}\label{p'=boxp}
\end{eqnarray}
and
\begin{eqnarray}
\Psi|_{\rho_0}\simeq\Phi|_{\rho_0}, \label{p=1p}
\end{eqnarray}
where we used ${\cal I}(p\rho_0)\simeq 1$ in deriving the second relation.
Noting that 4D Ricci tensor of the metric $\eta_{\mu\nu}+\overline{\gamma}_{\mu\nu}$
is given by
\begin{eqnarray}
{\cal R}_{\mu\nu}=-\frac{1}{2}\Box h_{\mu\nu}|_{\rho_0}
-\eta_{\mu\nu}\Box\Psi|_{\rho_0}-2\partial_{\mu}\partial_{\nu}\Psi|_{\rho_0},
\end{eqnarray}
we find
\begin{eqnarray}
\frac{2}{\rho_0}\left(\frac{1}{2}h_{\mu\nu}'|_{\rho_0}+
{\cal D}_{\mu\nu} \zeta_{\rm C} \right)
\simeq {\cal R}_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}{\cal R}
+2{\cal D}_{\mu\nu} \phi.\label{cap_R=}
\end{eqnarray}
As for the bulk side, we make the approximation $pr_0\gg 1$ and so
${\cal K}(pr_0)\approx 3K_0(pr_0)$, leading to
\begin{eqnarray}
\Phi|_{r_0}\simeq 3\Psi|_{r_0}.\label{p=3pbulk}
\end{eqnarray}
Using this and the perturbed Einstein equation $\delta G_{rr}=0$,
we find
\begin{eqnarray}
\Psi'|_{r_0}\simeq 0.\label{p'bulk}
\end{eqnarray}
Now Eqs.~(\ref{con_psi}),~(\ref{con_phi}),~(\ref{p=1p}), and~(\ref{p=3pbulk}) yield
\begin{eqnarray}
\zeta_{\rm C}\simeq\rho_0\left(\phi+\Psi|_{\rho_0}\right),
\quad
\zeta_{\rm B}\simeq r_0\left(\phi+3\Psi|_{\rho_0}\right).
\label{zeta=p+p}
\end{eqnarray}
Substituting Eqs. (\ref{p'=boxp}),~(\ref{p'bulk}), and~(\ref{zeta=p+p})
into the junction conditions~(\ref{junc_phiphi}) and~(\ref{tr_junc_munu})
and solving for $\phi$ and $\Psi|_{\rho_0}$,
we obtain
\begin{eqnarray}
(3-2\alpha)\phi&\simeq& -\frac{\rho_0^2}{2M_4^2}\left(\overline{T}_{\mu}^{\;\mu}-
3\overline{T}_{\varphi}^{\;\varphi}\right)
\nonumber\\&&\quad
+\frac{\alpha\rho_0^2}{M_4^2}\left(\overline{T}_{\mu}^{\;\mu}-
\overline{T}_{\varphi}^{\;\varphi}\right),\label{sol-phi}
\\
(3-2\alpha)\Psi&\simeq&\frac{\rho_0^2}{6M_4^2}\left(\overline{T}_{\mu}^{\;\mu}-
3\overline{T}_{\varphi}^{\;\varphi}\right)
\nonumber\\&&\quad
-\frac{\alpha}{3M_4^2}\Box^{-1}
\overline{T}_{\mu}^{\;\mu}, \label{sol-psi}
\end{eqnarray}
with $\alpha:=r_c/r_0$.
Here, we
neglected terms suppressed by factors $\rho_0^2\Box$, $\rho_0^2/r_0^2$,
and $\rho_0^2/r_c^2$ relative to the others. However,
we keep the first term in the right hand side of Eq.~(\ref{sol-psi})
because $\alpha$ may also be small.
From Eq.~(\ref{sol-phi}) one finds that $\phi$ is algebraically
determined by the energy-momentum tensor on the brane and hence is a nonpropagating mode.
In what follows we do not consider the case in which $\alpha$ is equal to or very close to $3/2$.
Plugging Eqs.~(\ref{sol-phi}) and~(\ref{sol-psi}) into the expression~(\ref{zeta=p+p}),
we can write
the brane bending scalars in terms of the energy-momentum tensor as
\begin{eqnarray}
(3-2\alpha)\frac{\zeta_{{\rm C}}}{\rho_0}&\simeq& -\frac{\rho_0^2}{3M_4^2}\left(\overline{T}_{\mu}^{\;\mu}-
3\overline{T}_{\varphi}^{\;\varphi}\right)
\nonumber\\&&\quad
-\frac{\alpha}{3M_4^2}\Box^{-1}\overline{T}_{\mu}^{\;\mu},
\\
(3-2\alpha)\frac{\zeta_{{\rm B}}}{r_c}&\simeq&-\frac{1}{M_4^2}\Box^{-1}\overline{T}_{\mu}^{\;\mu}.
\label{sol-zb}
\end{eqnarray}
Using Eq.~(\ref{cap_R=}),
the junction conditions~(\ref{junc_munu}) are now written as
\begin{eqnarray}
{\cal R}_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}{\cal R}
\simeq\frac{\overline{T}_{\mu\nu}}{M_4^2}
-{\cal D}_{\mu\nu}\chi
+\frac{h_{\mu\nu}'|_{r_0}}{2r_c},\label{effeq}
\end{eqnarray}
where $\chi:=-\zeta_{\rm B}/r_c$.
Here, the term
${\cal D}_{\mu\nu}\phi$
was neglected because Eq.~(\ref{sol-phi}) implies that it is smaller
by a factor of $\rho_0^2\Box$ than the other terms.
Since $\phi=\overline{\delta q}_{\varphi\varphi}/2$,
this suppressed mode corresponds to the fluctuation of the size of the ring-like brane.
The same suppression is found also in~\cite{Peloso, Reg-TK, Takamizu}.
As $K_0'(pr)|_{r_0}\approx-pK_1(pr_0)\approx-pK_0(pr_0)$ for $pr_0\gg 1$,\footnote{Noting
that $h_{\mu\nu}'|_{r_0}\simeq [\overline{\gamma}_{\mu\nu}'-\eta_{\mu\nu}\overline{\gamma}']|_{r_0}$, the
last term in Eq.~(\ref{effeq}) implies Pauli-Fierz type massive gravity~\cite{PFterm} with
$m^2_{{\rm PF}}=\sqrt{-\Box}/r_c$, as in the 5D DGP model.
}
we have the estimate
\begin{eqnarray}
\frac{1}{r_c}h_{\mu\nu}'|_{r_0}\sim\frac{1}{r_c\mathtt{r}}h_{\mu\nu}|_{r_0},\label{dgpterm}
\end{eqnarray}
where $\mathtt{r}$ is the length scale under consideration,
so that the last term in the right hand side of Eq.~(\ref{effeq}) is negligible when $\mathtt{r}\ll r_c$.
The situation here is very similar to that of the DGP braneworld~\cite{DGP}.
From Eqs.~(\ref{effeq}) and~(\ref{sol-zb}), i.e.,
\begin{eqnarray}
\Box\chi\simeq \frac{1}{3-2\alpha}\frac{\overline{T}_{\mu}^{\;\mu}}{M_4^2},
\label{chitrace}
\end{eqnarray}
we conclude that linearized
gravity on scales smaller than $r_c$ and $r_0$
is effectively described by the 4D Brans-Dicke theory
with $\omega_{{\rm BD}}=-r_c/r_0$~\cite{BD}, and the Brans-Dicke scalar $(\chi)$ is identified
as the brane bending.
In order for this scalar not to be a ghost,
it is required that $3-2\alpha>0$~\cite{FujiiMaeda}.
Therefore, we preclude the case with $r_c\geq 3r_0/2\, (\gg\mathtt{r}\gg \rho_0)$.
As the volume of the bulk is infinite in the present model,
the zero mode is not normalizable and hence is removed from the spectrum.
Therefore, from the brane observer point of view,
Kaluza-Klein modes contribute to gravity on the brane.
Nevertheless, 4D gravity is recovered
because nearly massless modes are confined around the brane in the cap side.
These modes
play the role of 5D induced gravity term ``$R^{(5)}$'' with one dimension compactified to a small circle.
The fluctuation of the size of the Kaluza-Klein circle is suppressed, and hence
we can expect to recover a 4D tensor structure.
We may expect that gravity shows the higher dimensional nature
at longer distances, $\mathtt{r}\gtrsim r_c, r_0$.
From the higher dimensional perspective,
gravitons leak away into the extra dimensions at distances much larger than $r_c$.
However, for $r_c\ll\mathtt{r}\ll r_0$ one of the extra dimensions is
effectively compactified because the extra 2D space is just like a needle,
so that gravity will look 5D~\cite{Kaloper1}.
Working in the momentum space, one sees that gravity is indeed 5D because
the differentiation with respect to the bulk coordinate $r$
yields $p$ rather than $p^2$
and the last term in Eq.~(\ref{effeq}) is much bigger than ${\cal O}({\cal R}_{\mu\nu})$
in this regime.
On much larger scales, $\mathtt{r}\gg r_0$, a graviton will see the opening of the cone,
leading to the 6D behavior.
This picture is indeed true, as is shown by an explicit example in the next subsection.
Moreover, this is not the end of the story because,
even in the weak gravity regime, nonlinearities will set in below
a certain distance scale.
Although the Brans-Dicke behavior contradicts tests of gravity,
nonlinear effects help to remove the scalar degree of
freedom,
leaving 4D Einstein gravity on
the brane~\cite{Deffayet, Lue:2001gc, Gabadadze:2004iy, Gruzinov:2001hp, Tanaka, KScp}.
This is what will be discussed in Sec.~\ref{sec:nonlinear}.
\subsection{Case study: static ``point'' sources}\label{sec:pointsource}
In this subsection we compute the gravitational field produced by
a static point source.
Concerning gravity at short distances,
this is just a check of our general argument in the previous subsection.
This example also allows us to see explicitly how
gravity is modified at long distances.
Here we restrict ourselves to linear perturbations.
As in~\cite{Kaloper}, we model a static point source in our 4D world
as a loop of string along the ring-like brane.
The energy momentum tensor for a static string loop is given by
\begin{eqnarray}
\overline{T}_t^{\;t}=\overline{T}_\varphi^{\;\varphi}=
-{\cal M}\;\delta^{(3)}(\vec{x}),\quad
\overline{T}_{ij}=0,
\end{eqnarray}
with $i, j$ being 3-space indices.
Since the source and the gravitational field are static, we use the expression
\begin{eqnarray}
h_{\mu\nu}&=&
\begin{cases}
\displaystyle{
\int\hat h_{\mu\nu}(k)K_0(kr_0)I_0(k\rho)e^{i\vec{k}\cdot \vec{x}}d^3k
}\;\;\;\text{(cap)}\\
\displaystyle{
\int\hat h_{\mu\nu}(k)I_0(k\rho_0)K_0(kr)e^{i\vec{k}\cdot \vec{x}}d^3k
}\;\;\text{(bulk)}
\end{cases},\label{genk_h}
\\
\Psi&=&
\begin{cases}
\displaystyle{
\int\hat\psi(k)K_0(kr_0)I_0(k\rho)e^{i\vec{k}\cdot \vec{x}}d^3k
}\quad\;\text{(cap)}\\
\displaystyle{
\int\hat\psi(k)I_0(k\rho_0)K_0(kr)e^{i\vec{k}\cdot \vec{x}}d^3k
}\quad\text{(bulk)}
\end{cases},\label{genk_psi}
\\
\Phi&=&
\begin{cases}
\displaystyle{
\int\hat\psi (k) K_0(kr_0){\cal I}(k\rho) e^{i\vec{k}\cdot \vec{x}}d^3k
}\quad\text{(cap)}\\
\displaystyle{
\int\hat\psi(k)I_0(k\rho_0) {\cal K}(kr) e^{i\vec{k}\cdot \vec{x}}d^3k
}\quad\text{(bulk)}
\end{cases},\label{genk_phi}
\end{eqnarray}
where the continuity of $h_{\mu\nu}$ and $\Psi$ has been implemented.
In the following calculation, it is convenient to use the 4D Green function:
\begin{eqnarray}
{\cal G}_4(x)=\frac{1}{4\pi|\vec{x}|},\quad
\nabla^2{\cal G}_4 = -\delta^{(3)}(\vec{x}),
\end{eqnarray}
where $\nabla^2:=\delta^{ij}\partial_i\partial_j$.
The trace of the $(\mu\nu)$
junction conditions~(\ref{tr_junc_munu}) reads
\begin{eqnarray}
2\frac{\zeta_{\rm C}}{\rho_0}-\frac{\zeta_{\rm B}}{r_c}=\frac{{\cal M}}{3M_4^2}{\cal G}_4.
\label{2z-z}
\end{eqnarray}
Plugging this into~(\ref{junc_munu}), we obtain
\begin{eqnarray}
K_0(kr_0)\hat h_{00}(k)&\simeq& \frac{4{\cal M}}{3M_4^2}\hat{\cal G}(k),
\\
K_0(kr_0)\hat h_{ij}(k)&\simeq& \frac{2{\cal M}}{3M_4^2}
\left(\delta_{ij}-\frac{k_ik_j}{k^2}\right)\hat{\cal G}(k),
\end{eqnarray}
where
\begin{eqnarray}
\hat{\cal G}(k):=\frac{1}{(2\pi)^3}\left[k^2+\frac{k}{r_c}\frac{K_1(kr_0)}{K_0(kr_0)}\right]^{-1},
\label{kernel-G}
\end{eqnarray}
and we made use of the small argument expansion $I_0(k\rho_0)\simeq{\cal I}(k\rho_0)\simeq 1$
and $I_0'(k\rho)|_{\rho_0}\simeq k^2\rho_0/2$.
From the continuity of $\overline{\Phi}$ [Eq.~(\ref{con_phi})] we find
\begin{widetext}
\begin{eqnarray}
\left(1-2\alpha\right)\phi
=-\alpha\frac{{\cal M}}{3M_4^2} {\cal G}_4
+\int\hat\psi(k)\left[2\alpha K_0(kr_0)-{\cal K}(kr_0)\right]e^{i\vec{k}\cdot\vec{x}}d^3k,
\end{eqnarray}
The $(\varphi\varphi)$ component of the junction equations now reads
\begin{eqnarray}
\int\hat\psi(k)\left\{-\left(1-2\alpha\right)\left[k^2K_0(kr_0)+\frac{kK_1(kr_0)}{r_c}\right]
+\alpha \frac{K_0(kr_0)}{\rho_0^2}-\frac{{\cal K}(kr_0)}{2\rho_0^2}
\right\}e^{i\vec{k}\cdot\vec{x}}d^3k
\simeq\frac{{\cal M}}{6M_4^2}\left[
\frac{\alpha}{\rho_0^2}
+\left(1-2\alpha\right) \nabla^2\right]{\cal G}_4,\label{fourierJ}
\end{eqnarray}
\end{widetext}
where we neglected $\phi/(r_0r_c)\;(\ll\phi/\rho_0^2)$.
Looking at the small argument expansion
$K_0(x)\simeq -\gamma-\ln(x/2)$, $K_1\simeq 1/x$, and ${\cal K}(x)\simeq 4/x^2$ with
$\gamma=0.5772...\,$,
it turns out that the last two terms in the left hand side of Eq.~(\ref{fourierJ})
are much larger than the first two when $kr_0\ll 1$.
When $kr_0\gg 1$, the last two terms are again dominant.
Therefore, we have
\begin{eqnarray}
\left[2\alpha K_0(kr_0)-{\cal K}(kr_0)\right]\hat\psi\simeq
\frac{1}{(2\pi)^3}\frac{{\cal M}}{3M_4^2}\left(\frac{\alpha}{k^2}-\rho_0^2\right).
\label{kernel-psi}
\end{eqnarray}
Using this, it is easy to find $\phi\simeq 0$ and
\begin{eqnarray}
\frac{\zeta_{\rm C}}{\rho_0}\simeq\Psi|_{\rho_0},\quad
\frac{\zeta_{\rm B}}{r_c}\simeq 2\Psi|_{\rho_0}-\frac{{\cal M}}{3M_4^2}{\cal G}_4,
\label{ps_z}
\end{eqnarray}
where note that the last term in the right hand side of Eq.~(\ref{kernel-psi})
gives a delta function in the real space, which vanishes away from the source.
\subsubsection{$|\vec{x}|\ll r_0$}
This is nothing but the case studied in Sec.~\ref{genan}.
As we may approximate $K_0(kr_0)\approx K_1(kr_0)$ in this case,
we have
\begin{eqnarray}
\hat{\cal G}(k)\simeq \frac{1}{(2\pi)^3}\frac{1}{k^2+k/r_c}.
\end{eqnarray}
In the real space, the behavior of $h_{\mu\nu}$ is governed by
\begin{eqnarray}
\int\hat{\cal G}(k)e^{i\vec{k}\cdot\vec{x}}d^3k&\simeq&
\begin{cases}
{\cal G}_4(x)\quad\quad\quad\quad (|\vec{x}|\ll r_c)
\\
r_c/(2\pi^2|\vec{x}|^2) \quad\; (|\vec{x}|\gg r_c)
\end{cases},
\end{eqnarray}
i.e., this reduces to ${\cal G}_4$ at short distances and shows 5D behavior for $r_c\ll |\vec{x}|\ll r_0$.
As for the scalar-type perturbations,
we make the approximation
$\left[2\alpha K_0(kr_0)-{\cal K}(kr_0)\right]\hat\psi\approx- (3-2\alpha)K_0(kr_0)\hat\psi$
to obtain
\begin{eqnarray}
\Psi|_{\rho_0}\simeq
-\frac{\alpha}{3-2\alpha}\frac{{\cal M}}{3M_4^2}{\cal G}_4(x).\label{r<r0_psi}
\end{eqnarray}
Thus, gravity is 4D at short distances, $|\vec{x}|\ll r_c$, in agreement with the argument in Sec.~\ref{genan}.
It is easy to confirm that the parameterized post-Newtonian (PPN) parameter is given by that of
the Brans-Dicke theory with $\omega_{{\rm BD}}=-\alpha=-r_c/r_0$.
Despite the 4D behavior of $\Psi|_{\rho_0}$,
this is always suppressed relative to $h_{\mu\nu}$ if $r_c\ll r_0$.
Therefore, for $r_c\ll |\vec{x}|\ll r_0$
the dominant contribution comes from $h_{\mu\nu}$
and gravity on the brane looks 5D.
If $r_c\sim r_0$, there is no 5D regime from the beginning.
\subsubsection{$|\vec{x}|\gg r_0 \;(\gtrsim r_c)$}\label{psA2}
In this regime we have ${\hat {\cal G}}(k)\simeq(2\pi)^{-3}r_0r_cK_0(kr_0)$,
and hence\footnote{
We use the following formula:
$$
\int_0^{\infty}y K_0(y)\sin(\alpha y)dy=\frac{\pi}{2}\frac{\alpha^{-2}}{(1+\alpha^{-2})^{3/2}}
\quad (\alpha>0).
$
}
\begin{eqnarray}
h_{00}\simeq\frac{4{\cal M}}{3M_4^2} {\cal G}_6,
\;\;
h_{ij}\simeq \frac{2{\cal M}}{3M_4^2}
\left(\delta_{ij}-\partial_i\partial_j\nabla^{-2}\right){\cal G}_6,
\end{eqnarray}
where ${\cal G}_6(x):=r_0r_c/(4\pi|\vec{x}|^3)$.
Similarly, we find
$\hat\psi(k)\simeq-(2\pi)^{-3}r_0r_c ({\cal M} /12M_4^2)$,
leading to $\Psi|_{\rho_0}\simeq-({\cal M}/12M_4^2){\cal G}_6$.
Thus, we conclude that on the largest distance scales
gravity on the brane looks 6D.
\section{(Semi)nonlinear regime}\label{sec:nonlinear}
Based on the linear analysis, we have shown that for $\mathtt{r}\ll r_0$
the brane bending in the bulk side is given by
\begin{eqnarray}
\zeta_{\rm B}\sim r_c \frac{R_g}{\mathtt{r}},\label{z-rc}
\end{eqnarray}
where $R_g$ is the gravitational radius of the source [see Eq.~(\ref{sol-zb})]. (We
dropped the factor $(3-2\alpha)$ because we are assuming that
this always gives an ${\cal O}(1)$ coefficient.)
Since $r_c$ is a ``large'' parameter, nonlinearity of $\zeta_{\rm B}$ may be important.
Indeed, when
\begin{eqnarray}
(\partial_{\mu}\zeta_{\rm B})^2\sim \frac{R_g}{\mathtt{r}},\label{zz-rgr}
\end{eqnarray}
such nonlinear terms in the brane bending must be taken into account
even if gravity is weak, $R_g/\mathtt{r}\ll 1$.
Substituting Eq.~(\ref{z-rc}) into Eq.~(\ref{zz-rgr}) and using the estimate $\partial_{\mu}\sim\mathtt{r}^{-1}$,
we find that nonlinear terms become important when
\begin{eqnarray}
\mathtt{r} \lesssim R_*:= (r_c^2 R_g)^{1/3}.
\end{eqnarray}
The linear analysis in the previous section is reliable only for $\mathtt{r}\gtrsim R_*$.
In this section, we present a preliminary analysis of
weak gravity on even smaller scales, $(\rho_0\ll)\;\mathtt{r}\ll R_*$, taking into account
quadratic terms in $\zeta_{\rm B}$~\cite{Padilla, Lue:2001gc, Tanaka}.
Note that since $\zeta_{\rm C}$ is estimated to be
$\sim$ max$\left\{(\rho_0/\mathtt{r})^3R_g, (r_c/r_0)(\rho_0/\mathtt{r})R_g\right\}$
according to the linear analysis,
nonlinear terms in $\zeta_{\rm C}$ will never become relevant.
We are considering the weak gravity regime,
so that we will keep linear terms in the metric perturbations.
Although we are now
working in the quadratic terms in $\zeta_{\rm B}$,
we may still use the 6D linearized equations in the bulk.
Therefore,
the relations~(\ref{p=3pbulk}) and~(\ref{p'bulk}),
which were derived
only by using the general solution to the bulk Einstein equations,
hold as well as
the cap equations~(\ref{h'=boxh})--(\ref{cap_R=}).
However,
we must treat more carefully
the transformation from the longitudinal gauge to
the Gaussian normal gauge in the bulk side when we impose the boundary conditions.
We move from the longitudinal gauge to the Gaussian normal gauge by
making the infinitesimal coordinate transformation,
$x^{\mu}=\overline{x}^{\mu}-\delta x^{\mu}(\overline{x}, \overline{r})$,
$r=\overline{r}-\delta r(\overline{x}, \overline{r})$,
satisfying the conditions
\begin{eqnarray}
\delta x^{\mu}|_{r_0}=0 \quad\text{and}\quad \delta r|_{r_0}=-\zeta_{\rm B}(x).
\end{eqnarray}
Under the above transformation, the metric perturbation transforms as
\begin{eqnarray}
&&\overline{\delta g}_{AB}(\overline{x}, \overline{r})\simeq
\delta g_{AB}(\overline{x}-\delta x, \overline{r}-\delta r)
\nonumber\\&&\quad
-g_{AB,r}\delta r+\frac{1}{2}g_{AB,rr}(\delta r)^2
\nonumber\\&&\quad\quad
-\delta x^{C}_{\;,A}g_{CB}(\overline{x}-\delta x, \overline{r}-\delta r)-(A\leftrightarrow B)
\nonumber\\&&\quad\quad\quad
+\delta x^{C}_{\; ,A}\delta x^{D}_{\;,B}g_{CD}(\overline{x}-\delta x, \overline{r}-\delta r),
\label{2nd_trans}
\end{eqnarray}
where we neglected terms like ${\cal O}(\delta x^C_{\;,A})\times{\cal O}(\delta g_{BC})$.
Since $B=\overline{B}=\overline{\Gamma}=0$,
we require
\begin{eqnarray}
0&=&-\delta x_{\mu, r}-\delta r_{,\mu}+\delta r_{,\mu}\delta r_{,r}+\delta x^{\nu}_{\;,\mu}\delta x_{\nu,r},
\\
0&=&2\Gamma-2\delta r_{,r} +(\delta r_{,r})^2+\delta x_{\mu,r}\delta x^{\mu}_{\;,r}.
\end{eqnarray}
Evaluating these two equations at $\overline{r}=r_0$, we obtain
\begin{eqnarray}
\left. \delta x_{\mu,r}\right|_{r_0}&=&\partial_{\mu}\zeta_{\rm B},
\label{d_rx2}
\\
\left.\delta r_{,r}\right|_{r_0}&=&\Gamma|_{r_0}+\frac{1}{2}\partial_{\mu}\zeta_{\rm B}\partial^{\mu}
\zeta_{\rm B}.
\label{d_rr2}
\end{eqnarray}
Using Eqs.~(\ref{2nd_trans}),~(\ref{d_rx2}), and~(\ref{d_rr2}), we can compute the
metric perturbations and their $r$ derivatives evaluated on the brane:
\begin{eqnarray}
&&\overline{\Phi}=\Phi|_{r_0}-\frac{\zeta_{\rm B}}{r_0}-\frac{\zeta_{\rm B}^2}{2r_0^2},
\quad
\overline{\gamma}_{\mu\nu}=\gamma_{\mu\nu}|_{r_0}+\partial_\mu\zeta_{\rm B}\partial_{\nu}\zeta_{\rm B},
\nonumber\\
&&\overline{\Phi}'=\Phi'|_{r_0}+\frac{1}{r_0}\Gamma|_{r_0}+\frac{\zeta_{\rm B}}{r_0^2}+\frac{1}{2r_0}
\partial_{\mu}\zeta_{\rm B}\partial^{\mu}\zeta_{\rm B}+\frac{\zeta_{\rm B}^2}{r_0^3},
\nonumber\\
&&\overline{\gamma}_{\mu\nu}'=\gamma_{\mu\nu}'|_{r_0}-2\partial_{\mu}\partial_{\nu}\zeta_{\rm B}.
\label{trmet2}
\end{eqnarray}
The brane bending scalar associated with $\partial_{\mu}$ will become large at short distances,
but we may assume that $\zeta_{\rm B}/r_0$ and $\zeta_{\rm B}/r_c$
remain small (i.e., order of metric perturbations) even in the seminonlinear regime.
We therefore neglect $\zeta_{\rm B}^2/r_0^2$ in the first equation
and $\zeta_{\rm B}^2/r_0^3$ in the third equation.
Now the $(\mu\nu)$ component of the junction conditions reduces to
\begin{eqnarray}
&&{\cal R}_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}{\cal R}
-\frac{\overline{T}_{\mu\nu}}{M_4^2}
\nonumber\\&&\;=-2{\cal D}_{\mu\nu}\phi
+\frac{1}{r_c} {\cal D}_{\mu\nu}\zeta_{\rm B}
+\frac{1}{2r_cr_0}
\partial_\lambda\zeta_{\rm B}
\partial^{\lambda}\zeta_{\rm B}\eta_{\mu\nu},
\label{j2_mn}
\end{eqnarray}
where the last term in the right hand side is in fact negligible
because $\zeta_{\rm B}/r_c\gg (\zeta_{\rm B}/r_0)(\zeta_{\rm B}/r_c)$.
Thus, even though we
have taken into account the possible nonlinear terms
in the brane bending,
the $(\mu\nu)$ junction conditions turn out to be
the same as the linear result.
Note that we have already neglected the term $h_{\mu\nu}'|_{r_0}/r_c$
which is much smaller than $\Box h_{\mu\nu}|_{\rho_0}$.
The 4D trace of the $(\mu\nu)$ junction conditions reduces to
$
2\Box\zeta_{\rm C}/\rho_0-\Box\zeta_{\rm B} /r_c=
\overline{T}_{\mu}^{\;\mu}/(3M_4^2),
$
neglecting the nonlinear term for the reason stated above.
The $(\varphi\varphi)$ component of the junction conditions is given by
\begin{eqnarray}
\frac{2}{\rho_0}\left(\Box\zeta_{\rm C}+2\rho_0\Box\Psi|_{\rho_0}+\frac{\phi}{\rho_0}\right)
-\frac{1}{r_c}\Box\zeta_{\rm B}
=\frac{\overline{T}_{\varphi}^{\;\varphi}}{M_4^2},
\label{junc_phiphi_nl}
\end{eqnarray}
where we used
Eqs.~(\ref{p'=boxp}) and~(\ref{p'bulk}), and
dropped the term $\phi/(r_0r_c)$ which is much smaller than $\phi/\rho_0^2$.
Again, Eq.~(\ref{junc_phiphi_nl}) is the same as the linear result.
The continuity of the induced metric reads
$\phi\simeq-\Psi|_{\rho_0}+\zeta_{\rm C}/\rho_0\simeq-3\Psi|_{r_0}+\zeta_{{\rm B}}/r_0$.
We use the trace of the second equation in~(\ref{trmet2}),
$\Psi|_{\rho_0}=\Psi|_{r_0}+\partial_{\mu}\zeta_{\rm B}\partial^{\mu}\zeta_{{\rm B}}/8$,
to get
\begin{eqnarray}
\frac{\zeta_{\rm C}}{\rho_0}
-\frac{1}{3}\frac{\zeta_{\rm B}}{r_0}-\frac{2}{3}\phi
\simeq\frac{1}{8}\partial_{\mu}\zeta_{\rm B}\partial^{\mu}\zeta_{{\rm B}}.\label{quadra}
\end{eqnarray}
The nonlinear effect appears here and this will play a very important role
in screening away the scalar degree of freedom $\chi$ found in the linear analysis.
Suppose that the nonlinear effect dominates in the above equations:
$(\partial_{\mu}\zeta_{\rm B})^2\gg\zeta_{\rm B}/r_c, \zeta_{\rm B}/r_0$.
In this limit, one finds
\begin{eqnarray}
\frac{\zeta_{\rm C}}{\rho_0}&\simeq&\frac{1}{6M_4^2}\Box^{-1}\overline{T}_{\mu}^{\;\mu}
\,\simeq\,\Psi|_{\rho_0},
\\
\phi&\simeq& \frac{\rho_0^2}{2M_4^2}\left(\overline{T}_{\varphi}^{\;\varphi}
-\overline{T}_{\mu}^{\;\mu}\right),\label{2ndphi}
\end{eqnarray}
and
\begin{eqnarray}
\partial_{\mu}\zeta_{\rm B}\partial^{\mu}\zeta_{{\rm B}}\simeq\frac{4}{3M_4^2}
\Box^{-1} \overline{T}_{\mu}^{\;\mu}.\label{2ndz2}
\end{eqnarray}
Because of the large contribution from the quadratic term in Eq.~(\ref{quadra}),
the result obtained here is different from the linear calculation~(\ref{sol-phi})--(\ref{sol-zb}).
From Eq.~(\ref{2ndz2}) we obtain the estimate $\zeta_{\rm B}\sim\sqrt{R_g\mathtt{r}}$.
For $\mathtt{r}\ll R_*$ we indeed have
$(\partial_{\mu}\zeta_{\rm B})^2\sim$ (metric perturbations) $\gg\zeta_{\rm B}/r_c$,
and so the approximation is self-consistent.
Eq.~(\ref{2ndphi}) indicates that $\phi$ is suppressed
in much the same way as in the linear analysis.
Since ${\cal D}_{\mu\nu}\zeta_{\rm B}/r_c\ll{\cal O}({\cal R}_{\mu\nu})$,
all the terms in the right hand side of Eq.~(\ref{j2_mn}) are now found to be negligible.
Consequently, {\em 4D Einstein gravity is recovered on scales much smaller than $R_*$.}
In other words, the van Dam-Veltman-Zakharov discontinuity~\cite{vDVZ} disappears.
The analysis so far has neglected terms like
$ \gamma_{\mu\nu}\partial^{\nu}\zeta_{\rm B}$, $\gamma_{\mu\nu,r}\zeta_{\rm B}$,
$\gamma_{\mu\nu,rr}\zeta_{\rm B}^2, ...\,$.
Noting that $(\partial_r)^{n}\delta g_{AB}|_{r_0}\sim \delta g_{AB}|_{r_0}/\mathtt{r}^n$
and $\partial_{\mu}\zeta_{\rm B}\sim\zeta_{\rm B}/\mathtt{r}\sim\sqrt{R_g/\mathtt{r}}$,
one finds that such terms give higher order contributions.
\section{Summary and Conclusions}
Summarizing, we have studied weak gravity
in the 6D regularized braneworld having
the mechanism of long distance modification of gravity.
We resolved the codimension-2 singularity
by capping the apex of the cone-shaped bulk
and replacing the conical brane with a ring-like codimension-1 brane.
The brane contains a scalar field which cancels the pressure along the compact direction.
In regularizing the codimension-2 brane,
we have assumed the Planck scale in the cap to be much greater than the bulk Planck scale.
This assumption is along the line of the regularization of brane-induced gravity of~\cite{Reg-DGP}.
We have shown that the model {\em effectively} gives rise to DGP-type gravity
without introducing induced-gravity terms
on the brane. In the case of the sharp conical bulk $(\rho_0\ll r_c\lesssim r_0)$,
we have obtained the following result in the regime where linearization is justified:
\begin{itemize}
\item the 4D Brans-Dicke theory with $\omega_{{\rm BD}}=-r_c/r_0$ for $\mathtt{r}\ll r_c, r_0$;
\item 5D gravity for $r_c\ll\mathtt{r}\ll r_0$;
\item 6D gravity for $\mathtt{r}\gg r_c, r_0$.
\end{itemize}
(Of course, the second regime appears if $r_c\ll r_0$.)
The linear analysis is not valid below $R_*=(r_c^2R_g)^{1/3}$,
where $R_g$ is the gravitational radius of the source.
We have performed a seminonlinear analysis for $\mathtt{r}\ll R_*$ by
carefully taking into account quadratic terms in the brane bending.
Our finding is:
\begin{itemize}
\item 4D Einstein gravity is reproduced for $\mathtt{r}\ll R_*$.
\end{itemize}
The present model is very simple in that the background is given by
a locally flat spacetime.
In spite of this simplicity, the system is shown to have a rich and interesting structure.
It is easy to extend the model to more general background configurations
which are curved due to the cosmological constant and other fields living in the bulk.
The similar regularization scheme can be applied to such cases as well.
Bulk-cap asymmetry will then help to reproduce 4D gravity on the brane~\cite{Padilla, KoyamaKoyama}.
As mentioned in the Introduction, long distance modification of gravity
is often motivated by explaining the current acceleration of the Universe.
It would be interesting to construct cosmological solutions and
explore whether
the accelerating expansion is possible\footnote{
It has been known that the Lovelock term in the bulk leads to induced gravity on the
brane~\cite{Ig, Lovelock, Lovelock2}.
Self-accelerating solutions were found in the framework of 6D Gauss-Bonnet braneworlds
in~\cite{Charmousis}.
} due to gravity leakage in the present model.
We hope to address this issue in the near future.
\acknowledgments
I would like to thank Tetsuya~Shiromizu for a careful reading of the manuscript
and Kei-ichi~Maeda for helpful comments on the scalar-tensor theories of gravity.
I am supported by the JSPS under Contact No.~19-4199.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,336 |
{"url":"http:\/\/fsaraceno.wordpress.com\/tag\/ecb\/","text":"### Archive\n\nPosts Tagged \u2018ECB\u2019\n\n## Walls Come Tumbling\u00a0Down\n\nYesterday I quickly commented\u00a0the disappointing growth data for Germany and for the EMU as a whole, whose GDP Eurostat splendidly defines \u201cstable\u201d. This is bad, because the recovery is not one, and because we are increasingly dependent on the rest of the world for that growth that we should be\u00a0able to generate domestically.\n\nHaving said that, the real bad news did not come from Eurostat, but from the August 2014 issue of the ECB monthly bulletin, published on Wednesday. Thanks to Ambrose Evans-Pritchard\u00a0I noticed the following chart ( page 53):\n\nThe interesting part of the chart is the blue dotted line, showing that the forecasters\u2019 consensus on longer term inflation sees more than a ten points drop of the probability that inflation will stay at 2% or above. Ten points in just a year. And yet, just a few pages above we can read:\n\nAccording to Eurostat\u2019s flash estimate, euro area annual HICP inflation was 0.4% in July 2014,\u00a0after 0.5% in June. This reflects primarily lower energy price inflation, while the annual rates of\u00a0change of the other main components of the HICP remained broadly unchanged. On the basis of\u00a0current information, annual HICP inflation is expected to remain at low levels over the coming\u00a0months, before increasing gradually during 2015 and 2016. Meanwhile, inflation expectations for\u00a0the euro area over the medium to long term continue to be firmly anchored in line with the aim of\u00a0maintaining inflation rates below, but close to, 2%\u00a0(p. 42, emphasis added)\n\nThe ECB is hiding its head in the sand, but expectations, the last bastion against deflation, are obviously not firmly anchored. This can only mean that private expenditure will keep tumbling down in the next quarters. It would be foolish to hope otherwise.\n\nSo we are left with good old macroeconomic policy. I did not change my mind since\u00a0my latest piece on the ECB. Even if the ECB inertia is appalling, even if their stubbornness in claiming that everything is fine (see above) is more than annoying, even if announcing mild QE measures in 2015 at \u00a0the earliest is borderline criminal, it remains that I have no big faith in the capacity of monetary policy to trigger\u00a0decent growth.\u00a0\u00a0The latest issue\u00a0of the ECB bulletin also\u00a0reports the results of the latest Eurozone Bank Lending Survey. They\u00a0show a slow easing of credit conditions, that proceed\u00a0in parallel with a pickup of credit demand from firms and households. While for some countries credit constraints may play a role in keeping private expenditure down (for example, in Italy), the overall picture for the EMU is of demand and supply proceeding in parallel. Lifting constraints to lending, in this situation, does not seem likely to boost credit and spending. It\u2019s the liquidity trap, stupid!\n\nThe solution seems to be one, and only one: expansionary fiscal policy, meaning strong increase in government expenditure (above all for investment) in countries that can afford it (Germany, to begin with); and delayed consolidation for countries with struggling public finances. Monetary policy should accompany this fiscal boost with the commitment to maintain an expansionary stance until inflation has overshot the 2% target.\n\nFor the moment this remains a mid-summer dream\u2026\n\n## Praising the Bundesbank\n\nI am puzzled by Wolfgang M\u00fcnchau\u2019s latest piece in the Financial Times. Let me start by quoting the end:\n\n[...] The ECB should have started large-scale asset purchase a year ago. It certainly should do so now. The EU should allow governments to overshoot their deficit targets this year, and suspend the fiscal compact, which will result in further fiscal pain from 2016.\n\nEven a casual reader of this blog will quickly realize that it would be hard for me to agree more with these statements. The macroeconomic stance at the EMU level has been seriously inappropriate since 2010, with fiscal policy globally restrictive (thank you austerity), and monetary policy way too timid.\nSo, what is the problem? The problem is the first part of M\u00fcnchau\u2019s editorial, in which he attacks the Bundesbank for its plea in favour of faster wage growth in Germany (the Buba asked for an average wage increase of 3%).\nThis is frankly hard to understand. The eurozone problems, and it\u2019s flirting with deflation, stem from the victory of the Berlin View, that laid the burden of adjustment on the shoulders of peripheral countries alone.\nThe call for wage increases in Germany signals, and it was about time, that even conservative German institutions are beginning to realize the obvious: there will be no rebalancing, and therefore no robust recovery, unless German domestic demand recovers. This means a fiscal expansion, as well as private expenditure recovery. Unsurprisingly, the Buba rules out the former, but it is nice to see that at least the latter has become an objective. Faster wage growth may not make a huge difference in quantitative terms, but it still marks an important change of attitude. This is a huge step away from the low-wage-high-productivity-export-led model that the Bundesbank and the German government have been preaching (and imposing to their partners).\nM\u00fcnchau is right in calling for a different policy mix in the EMU. But this is complementary, not alternative, to a change in the German growth model. I would have expected him to applaud a small but potentially important change in attitude. Instead I have read a virulent attack. Puzzled, puzzled\u2026\n\nCategories: EMU Crisis\n\n## ECB: Great Expectations\n\nAfter the latest disappointing data on growth and indeflation in the Eurozone, all eyes are on\u00a0today\u2019s\u00a0ECB meeting. Politicians and commentators speculate about the shape that QE, Eurozone edition, will take. A bold move to contrast lowflation would be welcome news, but a close look at the data suggests that the messianic expectation of the next \u201cwhatever it takes\u201d may be misplaced.\n\nFaced with mounting deflationary pressures, policy makers rely on the probable loosening of the monetary stance. While necessary and welcome, such loosening may not allow embarking the Eurozone on a robust growth path. The April 2014 ECB survey on bank lending confirms that, since 2011, demand for credit has been stagnant at least as much as credit conditions have been tight. Easing monetary policy may increase the supply for credit, but as long as demand remains anemic, the transmission of monetary policy to the real economy will remain limited. Since the beginning of the crisis, central banks (including the ECB) have been very effective in preventing the meltdown of the financial sector. The ECB was also pivotal, with the OMT, in providing an insurance mechanism for troubled sovereigns in 2012. But the impact of monetary policy on growth, on both sides of the Atlantic, is more controversial. This should not be a surprise, as balance sheet recessions increase the propensity to hoard of households, firms and financial institutions. We know since Keynes that in a liquidity trap monetary policy loses traction. Today, a depressed economy, stagnant income, high unemployment, uncertainty about the future, all contribute to compress private spending and demand for credit across the Eurozone, while they increase the appetite for liquidity. At the end of 2013, private spending in consumption and investment was 7% lower than in 2008 (a figure that reaches a staggering 18% for peripheral countries). Granted, radical ECB moves, like announcing a higher inflation target, could have an impact on expectations, and trigger increased spending; but these are politically unfeasible. It is not improbable, therefore, that a \u201csimple\u201d quantitative easing program may amount to pushing\u00a0on a string. The ECB had already accomplished half a miracle, stretching its mandate to become de facto a Lender of Last Resort, and defusing speculation. It can\u2019t be asked to do much more than this.\n\nWhile monetary policy is given almost obsessive attention, there is virtually no discussion about the instrument that in a liquidity trap should be given priority: fiscal policy. The main task of countercyclical fiscal policy should be to step in to sustain economic activity when, for whatever reason, private spending falters. This is what happened in 2009, before the hasty and disastrous fiscal stance reversal that followed the Greek crisis. The result of austerity is that while in every single year since 2009 the output gap was negative, discretionary policy (defined as change in government deficit net of cyclical factors and interest payment) was restrictive. In truth, a similar pattern can be observed in the US, where nevertheless private spending recovered and hence sustained fiscal expansion was less needed. Only in Japan, fiscal policy was frankly countercyclical in the past five years.\n\nAs Larry Summers recently argued, with interest rates at all times low, the expected return of investment in infrastructures for the United States is particularly high. This is even truer for the Eurozone where, with debt at 92%, sustainability is a non-issue. Ideally the EMU should launch a vast public investment plan, for example in energetic transition projects, jointly financed by some sort of Eurobond. This is not going to happen for the opposition of Germany and a handful of other countries. A second best solution would then be for a group of countries to jointly announce that the next national budget laws will contain important (and coordinated) investment provisions , and therefore temporarily break the 3% deficit limit. France and Italy, which lately have been vocal in asking for a change in European policies, should open the way and federate as many other governments as possible. Public investment seems the only way to reverse the fiscal stance and move the Eurozone economy away from the lowflation trap. It is safe to bet that even financial markets, faced with bold action by a large number of countries, would be ready to accept a temporary deterioration of public finances in exchange for the prospects of that robust recovery that eluded the Eurozone economy since 2008. A change in fiscal policy, more than further action by the ECB, would be the real game changer for the EMU. But unfortunately, fiscal policy has become a ghost. A ghost that is haunting Europe\u2026\n\n## ECB: One Size Fits\u00a0None\n\nEurostat just released its flash estimate for inflation in the Eurozone: 0.5% headline, and 0.8% core. We now await comments from ECB officials, ahead of next Thursday\u2019s meeting, saying that everything is under control.\n\nJust this morning, Wolfgang M\u00fcnchau in the Financial Times rightly said that EU central bankers should talk less and act more. M\u00fcnchau also argues that quantitative easing is the only option. A bold one, I would add in light of todays\u2019\u00a0deflation inflation data. Just a few months ago, in September 2013, Bruegel estimated the ECB interest rate to be broadly in line with Eurozone average macroeconomic conditions (though, interestingly, they also highlighted that it was unfit to most countries taken individually).\n\nIn just a few months, things changed drastically. While unemployment remained more or less constant since last July, inflation kept decelerating until today\u2019s very worrisome levels. I very quickly extended the Bruegel exercise to encompass the latest data (they stopped at July 2013). I computed the target rate as they do as\n\n$Target=1+1.5\\pi_{core}-1(u-\\overline{u})$.\n\n(if you don\u2019t like the choice of parameters, go ask the Bruegel guys. I have no problem with these). The computation gives the following:\n\nUsing headline inflation, as the ECB often claims to be doing, would of course give even lower target rates. As official data on unemployment stop at January 2014, the two last points are computed with alternative hypotheses of unemployment: either at its January rate (12.6%) or at the average 2013 rate (12%). But these are just details\u2026\n\nSo, in addition to being unfit for individual countries, the ECB stance is now unfit to the Eurozone as a whole. And of course, a negative target rate can only mean, as M\u00fcnchau forcefully argues, that the ECB needs to get its act together and put together a credible and significant quantitative easing program.\n\nTwo more remarks:\n\n\u2022 A minor one (back of\u00a0 the envelope) remark is that given a core inflation level of 0.8%, the current ECB rate of 0.25%, is compatible with an unemployment gap of 1.95%. Meaning that the current ECB rate would be appropriate if natural\/structural unemployment was 10.65% (for the calculation above I took the value of 9.1% from the OECD), or if current unemployment was 11.5%.\n\u2022 The second, somewhat related but more important to my sense, is that it is hard to accept as \u201cnatural\u201d an unemployment rate of 9-10%. If the target unemployment rate were at 6-7%, everything we read and discuss on the ECB excessively restrictive stance would be significantly more appropriate. And if the problem is too low potential growth, well then let\u2019s find a way to increase it\n\n## Mario Draghi is a Lonely\u00a0Man\n\nI just read an interesting piece by Nicol\u00f2 Cavalli on the ECB and deflationary risks in the eurozone. The piece is in Italian, but here is a quick summary:\n\n\u2022 Persisting high unemployment, coupled with inflation well below the 2% target, put deflation at the top of the list of ECB priorities.\n\u2022 Mario Draghi was adamant that monetary policy will remain loose for the foreseeable horizon.\n\u2022 As we are in a liquidity trap, the effect of quantitative easing on economic activity has been limited (in the US, UK and EMU alike).\n\u2022 Then Nicol\u00f2 quotes studies on quantitative easing in the UK, and notices that, like the Bank of England, the ECB faces additional difficulties, linked to the distributive effects of accommodating monetary policy:\n\u2022 Liquidity injections inflate asset prices, thus increasing financial wealth, and the value of large public companies.\n\u2022 Higher asset prices increase the opportunity costs of lending for financial institutions, that find it more convenient to invest on stock markets. This perpetuates the credit crunch.\n\u2022 Finally, low economic activity and asset price inflation depress investment, productivity and wages, thus feeding the vicious circle of deflation.\n\nNicol\u00f2 concludes that debt monetization seems to be the only way out for the ECB. I agree, but I don\u2019t want to focus on this. Read more\n\n## Of Actions and Words in\u00a0Frankfurt\n\nLast Thursday the ECB cut rates, somewhat unexpectedly. This shows that it takes the risk of deflation very seriously. Good news, I\u2019d say. But unfortunately, press conferences follow ECB Council meetings. And I say unfortunately, because Mr Draghi words often fail to match his actions. Here is what he said on Thursday (I could not resist adding some bold here and there):\n\nIf you look at the euro area from a distance, you see that the fundamentals in this area are probably the strongest in the world. This is the area that has the lowest budget deficit in the world. Our aggregate public deficit is actually a small surplus. We have a small primary surplus of 0.7%, compared with, I think, a deficit of 6 or 7% deficit in US, \u2013 6 I think \u2013 and 8 % in Japan. This is the area with the highest current account surplus. And it is also the area, as we said before, with one of the lowest \u2013 if not the lowest \u2013 inflation rate.\n\nFascinating. Truly fascinating. I will pass on the fact that one of the strong \u201cfundamentals\u201d Mr Draghi quotes, low inflation, is actually the main source of worry for economists and policymakers worldwide, including the ECB, that had to rush into a rate cut that was not planned at least until December! I will also pass on his praise of high current account surpluses while the Commission itself is considering opening an infraction procedure against Germany, for perpetuating an important source of imbalances within the eurozone and worldwide.\n\nNo, what I find more shocking is the list of fundamentals Draghi gives: public debt and deficit; inflation; current account balance. Now, it dates back a little, but I remember all of those, in Econ 101, to be defined as instruments of economic policy, supposed to serve the final objectives of growth and employment. It is true that we do rather well in what Draghi calls fundamentals, but I continue preferring to call instruments. Look at this table:\n\nI have reported, for ease of comparison, data from the IMF World Economic Outlook (October 2013), therefore they are not the latest (quarterly or monthly) data. Also, I have highlighted in red the worst performer, and in green the best. And boy, Draghi is right! (Notice incidentally that eurozone inflation was 2.5 percent on average in 2012. With the latest data at 0.7 percent, this suggests that \u00a0we are running, not walking, towards deflation.)\n\nBut if we look at the supposed objectives of economic policy (how would Draghi call these?), the picture changes, quite a bit:\n\nNo other major advanced economy is doing nearly as badly as the eurozone in terms of unemployment and GDP. But according to the ECB President we have \u201cthe strongest fundamentals in the world\u201d. Does this means that Draghi did not take Econ 101? No, I know for sure that he did take it, and \u00a0he actually had excellent mentors. To understand Draghi\u2019s claim, it may be useful to read his whole sentence. After arguing that the eurozone has strong fundamentals he goes on:\n\nThis does not translate automatically into a galloping recovery. But, actually, it gives you the fundamentals upon which you can pursue the right economic policies. Structural reforms are the necessary and sufficient condition for this to happen. In the absence of that, unfortunately, we are going to stay here for quite a long time.\n\nHere is the answer. The only and one answer. Focusing on instruments instead of targets is the strategy of those who do not believe that a role exists for active economic policies. It is a pity that one of these guys is heading the second most important central bank of the world. And it is paradoxically reassuring that the situation is currently so bad that he is forced to abandon his creed and implement active monetary policies.\n\nAdvice for the next episodes: praise Mario Draghi actions, and avoid reading the transcripts of his press conferences.","date":"2014-09-20 09:58:21","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\": 0, \"img_math\": 1, \"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\": 0.36313578486442566, \"perplexity\": 3022.673617777195}, \"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-2014-41\/segments\/1410657133078.21\/warc\/CC-MAIN-20140914011213-00034-ip-10-196-40-205.us-west-1.compute.internal.warc.gz\"}"} | null | null |
\section{Introduction}
A central problem in cosmology today is the nature and abundance
of the matter in the universe. While a given matter content
uniquely determines the shape of the power spectrum
of density fluctuations (up to variations in the primeval shape),
the amplitude of mass fluctuations remains unspecified theoretically and its
value must be sought from observations.
After the {\sl COBE} discovery of primeval fluctuations
(\cite{smoot92}) it became possible
to use temperature fluctuations at the surface of last
scattering to
compute the small scale normalization $\sigma_8$ (the rms relative mass
fluctuation in a sphere of radius $8\,h^{-1}\,{\rm Mpc}$,
$H_0=100$ h km s$^{-1}$ Mpc$^{-1}$) for any particular
model (\cite{wright92}; \cite{efstathiou92}; \cite{adams93}).
This allowed researchers to shift
their emphasis from the amplitude determination to the study of the spectral
shape. Any additional independent amplitude determinations would
enable one to discriminate between different models of structure formation.
The {\sl COBE} results are able to provide a constraint on the shape of the
primordial power spectrum, but not on the matter content of the universe,
because on the scales probed by {\sl COBE} different matter contents
produce similar temperature fluctuations. Small scale CMB anisotropy
experiments
are just beginning to be useful for determining the spectral shape
(e.g., Bond et al. 1991;
Dodelson \& Jubas 1993; G\'orski, Stompor \& Juszkiewicz 1993).
An alternative approach is to use measured redshift-distance samples
to determine the amplitude of mass fluctuations. This
method has the advantage that the reconstructed peculiar velocities
are directly sensitive to the underlying mass distribution.
Most work to date has been based on estimates of the bulk flow averaged
over large volumes (Bertschinger et al. 1990; Courteau et al. 1993).
These statistics have several limitations. First, the statistical
distribution of spectral amplitude estimates based on the bulk flow
is broad ($\chi^2$ with only 3 degrees of freedom) and consequently
bulk flows can only weakly constrain the models. Second, bulk flows
are particularly sensitive to the systematic errors introduced by
nonuniform sampling (the sampling gradient bias of Dekel, Bertschinger,
\& Faber 1990, hereafter DBF). Finally, the scales contributing to the
bulk flow estimates are large ($\gsim40$--60 $h^{-1}$ Mpc), whereas
most of the peculiar velocity measurements come from smaller distances
(10--30 $h^{-1}$ Mpc). It should be possible to place additional
independent model constraints on the smaller scales alone.
Several groups have combined different peculiar velocity data to
probe models on a range of scales (e.g., Del Grande \& Vittorio 1992;
Muciaccia et al. 1993; Tormen et al. 1993), but all except
Tormen et al. use bulk flow estimates with their associated uncertainties.
In this paper we estimate the amplitude of mass fluctuations on
intermediate scales by applying the maximum-likelihood method directly to the
POTENT
reconstruction of density perturbations from the peculiar velocities
(Bertschinger \& Dekel 1989; DBF).
This analysis assumes that the measured velocities are a fair tracer
of the underlying velocity field, which was induced by the underlying
gravitational field.
Use of a large sample of peculiar velocity data reduces
both the statistical and systematic errors, provided that care is
exercised concerning nonlinear corrections and Malmquist bias effects.
We test several models: standard cold dark matter (CDM), CDM plus a
cosmological constant(CDM+$\Lambda$), and CDM plus massive neutrinos
(CDM+HDM), with several different choices for the Hubble constant and
the primeval spectral index.
Because we use the relatively sparse 1990 dataset we do not
attempt to discriminate between the models using the velocity data alone.
Instead,
we compare our results to the {\sl COBE} normalization to derive some
conclusions about the viability of the models. We also briefly
comment on the
agreement with other methods of amplitude estimation.
\section{Method and data analysis}
Assuming a potential flow the present velocity field can be extracted from
observed radial peculiar velocities of galaxies (e.g. Lynden-Bell et al.
1988) by integrating along radial rays. This is the essence of the
POTENT reconstruction method (Bertschinger \& Dekel 1989; DBF).
Furthermore, in linear perturbation theory there is a simple relation
between velocity and density fields (Peebles 1980): $\delta({\bf
r})=-(H_0f)^{-1}\nabla\cdot {\bf v(r)}\equiv
f^{-1}\widetilde{\delta}({\bf
r})$, where $\delta({\bf r})$ is
the density fluctuation at position ${\bf r}$, $H_0$ is the Hubble
constant and $f$ is the the growing mode logarithmic growth rate.
For CDM and CDM+$\Lambda$ models at low redshifts f is well approximated
by $f(\Omega_m)=\Omega_m^{0.6}$ (Peebles 1980;
Lahav et al. 1991), where $\Omega_m$ is the total matter content in the
universe. For the CDM+HDM model $f$ is in general a function of the
wavenumber $k$
(see section 3.2). We introduce $\widetilde{\delta}({\bf r})$
as the measured quantity to be compared with the theoretical predictions.
The input data and their treatment are described in Bertschinger et al.
(1990). We discard all points with $r > 50h^{-1}$ Mpc
and with the distance to the fourth nearest neighbor $R_4>10h^{-1}$ Mpc,
thus keeping only the points with small sampling and measurement errors.
At the end we are left with $N=111$ data points $\widetilde{\delta_i}$
on a grid of spacing $10h^{-1}$ Mpc (with gaussian smoothing
radius $12h^{-1}$ Mpc),
which we use for comparison with the theoretical predictions.
Despite the many
data points there are only about 10 independent degrees of freedom in the
sample (Dekel et al. 1993), which currently prevents us from extending
the analysis beyond the amplitude determination.
Given the data one can estimate the power spectrum parameters using the
maximum likelihood method. The initial density perturbations are assumed to
constitute a gaussian random field in all the models that we study here.
Nonlinear effects and nontrivial coupling between signal and
noise (sampling and measurement errors)
in the data affect the distributions and in general the resulting data are
not normally distributed. However, one can still define a gaussian
likelihood function and use its maximum as a statistic
to estimate the unknown parameters. Moreover, if the deviations from
a normal distribution are small, the increase in variance owing to the use
of the wrong likelihood function should be small. We use Monte Carlo
simulations with the correct distributions to estimate the bias and variance
of the estimated parameters.
We define the likelihood function as
\begin{equation}
L(\sigma_8)={1 \over \sqrt{(2\pi)^N \vert M\vert}}
\exp\left[ -{1\over 2}\sum_{i=1}^N \sum_{j=1}^N
M_{ij}^{-1}(\widetilde{\delta_i}-\langle \widetilde{\delta_i} \rangle)(
\widetilde{
\delta_j}-\langle \widetilde{\delta_j}
\rangle) \right],
\label{likest}
\end{equation}
where $M_{ij}^{-1}$ and $M$ are the inverse and determinant of
the correlation matrix $M_{ij}=\langle (\widetilde{\delta_i}-\langle
\widetilde{
\delta_i}
\rangle)(\widetilde{\delta_j}-\langle \widetilde{\delta_j} \rangle) \rangle$.
Here $\langle \rangle$ denotes averaging over the random field
ensemble (signal) and distance errors (noise).
For a given theoretical model, the correlation matrix depends on the
parameters of the power spectrum, most notably on its amplitude. An
estimate of the amplitude $\sigma_8$ is given by the value that maximizes
$L$. In general this estimate is biased, in part because the maximum
likelihood estimator is only asymptotically unbiased,
but also because of our assumption of normality and owing to the presence
of sampling gradient and other biases in the $\{\widetilde{\delta_i}\}$.
We estimate the statistical bias by Monte Carlo simulations
and correct for it as described below.
To apply the method in practice, one needs to compute the correlation matrix
$M_{ij}$. The matrix has contributions from both the noise and the true
underlying signal. The signal contribution is given by
\begin{equation}
M_{ij}^s=\int d^3k\,f^2(\Omega_m,k)\,W^2(k)\,P(k)\,e^{i{\bf k}\cdot
({\bf r}_i-{\bf r}_j)}\ ,
\end{equation}
where $P(k)$ is the density fluctuation power spectrum and $W(k)$
is the smoothing window function in $k$-space.
Note that we have retained a possible $k$ dependence of $f$.
Measurement noise arises from errors in the galaxy
distance estimates and from the sparse spatial sampling of the data. A detailed
analysis, performed in the appendix of DBF, shows that the noise contribution
is correlated with the signal contribution in $M_{ij}$. Thus,
one cannot compute each of the two contributions separately
and then add them together with the appropriate amplitude, rather, one must
compute $M_{ij}$ directly for each amplitude.
We computed $M_{ij}$ using Monte Carlo simulations as described by
DBF and Bertschinger et al. (1990), with input peculiar velocities given
by a random sample of the linear velocity field from a particular gaussian
theory (e.g., CDM). We evaluated the radial velocities at the positions
of real galaxies and added noise assuming a lognormal distribution of
distance errors (in contrast with the normal distribution used in the
earlier work). For each model and for a dense grid of amplitudes,
500 POTENT reconstructions were averaged over the signal and noise
ensembles to compute $\langle \widetilde\delta_i\rangle$ and $M_{ij}$.
Before discussing the results we have to address the possible effects
of bias $\langle \widetilde\delta_i \rangle$. As discussed in DBF,
the two main contributions are from
the sampling gradient bias and the Malmquist bias.
The first one is adequately taken into account by the Monte Carlo
simulations,
because they use similar sampling of space as the real data.
On the other hand, Malmquist bias computed by the Monte Carlo simulations
is not exactly the real bias present in the data.
Since the real data had the homogeneous Malmquist bias subtracted,
the only bias left
is the density gradient bias $(\partial \ln n /\partial \ln r)
(\sigma^2/ r) $. However, the bias in the Monte Carlo samples is
$(3.5+\partial \ln nP /\partial \ln r)
(\sigma^2/ r) $, where $P(\vec r\,)$ is a poorly known selection function.
Without knowledge of $P$, one cannot properly correct the simulations
for the Malmquist
bias. To test the importance of inhomogeneous Malmquist bias on the
parameters we wish to
estimate, we compare the results for two simple cases. In first case, we
correct the real data
for the Monte Carlo bias, while in the second we do not.
This is a correct treatment of the bias if the
density gradient bias is negligible in the first case or the
volume and selection function contributions to the bias
exactly cancel in the second case.
We find that the final mass amplitudes
differ from each
other by about 10\%. This is significantly smaller than the statistical
errors of our amplitude estimates.
Another possible source of error are nonlinear effects, which tend to
change the densities and velocities compared to their linear values.
The real velocities are nonlinear while our simulations are strictly
linear.
Due to the large smoothing radius, the differences are small in the
case studied here, even if the unsmoothed density fluctuations are large.
To test this one can also compute the true density field
by applying the following correction
in the quasi-linear regime ($-0.8\lower.5ex\hbox{\ltsima}\delta\lsim4.5$):
$f^{-1}\widetilde{\delta}=\delta/(1+0.18\delta)$
(Nusser et al. 1991).
This changes our final estimated amplitudes by 5--10\%, depending on the value
of $f$.
Note that inhomogeneous Malmquist bias tends to increase the amplitude
of the measured peculiar velocities, while nonlinear effects decrease the
linear $\widetilde\delta$. Because both effects are small ($\sim10\%$)
and opposite in sign, we will neglect them.
\section{Results}
Our analysis was restricted to the simplest
generalizations of standard CDM model that decrease the power on
small scales relative to that on large scales and have been proposed recently
as viable models of large scale structure.
We computed transfer functions for the models by integrating the coupled
linearized relativistic Einstein, Boltzmann, and fluid equations for
baryons, CDM, photons, and neutrinos.
In all cases, we fix the baryon contribution to
$\Omega_B=0.0125h^{-2}$, as given by the nucleosynthesis constraint
(Walker et al. 1991).
For each model considered, we estimated the maximum-likelihood value
of the amplitude, which we denote $\sigma_{8,v}$.
The error distribution of estimated $\sigma_{8,v}$ from the Monte Carlo
samples is somewhat asymmetric with longer
tails toward larger values (fig. \ref{fig1}).
For all the models,
the relative one-sigma errors are well approximated by $(+0.3/-0.25)$.
To this one must add the sum of systematic errors due to the Malmquist
bias in the data and to residual nonlinear effects, whose sign is unclear
but whose magnitude is, at most, about 10\%.
The results for different models
are summarized in Table \ref{table1}. Our
estimates of $\sigma_{8,v}$ from POTENT include a small statistical bias
correction ($\sim5\%$). In addition, in Table \ref{table1}
we include $\sigma_{8,l=2}$,
the {\sl COBE} normalization based on assuming a value of
$Q_{\rm rms-PS}=15.7\exp[0.46(1-n)]\,\mu$K
(\cite{smoot92}; \cite{seljak93}).
We also include the age of the universe as a possible
additional constraint for these models.
Despite the complicated window function,
we find that there is only a weak dependence of $\sigma_{8,v}$ on the shape
of the power spectrum. This is not surprising considering that the
POTENT sample is sparse at large distances and has been smoothed
with a gaussian of radius $12\,h^{-1}\,{\rm Mpc}$.
We find that $\sigma_{8,v}$ depends mainly on $\Omega_m$ through
$f(\Omega_m)$. An approximate value valid through most of the parameter
space is thus
\begin{equation}
\sigma_{8,v}\,\Omega_m^{0.6}=1.3^{+0.4}_{-0.3}\ .
\label{eqn1}
\end{equation}
95\% confidence limit intervals give $\sigma_8\,\Omega_m^{0.6}\sim 0.7-2.3$.
For a given model (and thus, for a given value of $n$), the relative
1$\sigma$ uncertainty of $\sigma_{8,l=2}$ is only $17\%$ (\cite{seljak93};
Scaramella \& Vittorio 1993), which is 2 times smaller than the uncertainty
of $\sigma_{8,v}$. As a first approximation
one can thus neglect the uncertainty of $\sigma_{8,l=2}$ when
comparing it with $\sigma_{8,v}$.
\begin{table}[p]
$$
\begin{array}{llllllll}
\Omega_{CDM+B} & \Omega_{\Lambda} & \Omega_{\nu} & h & n & \sigma_{8,v} &
\sigma_{8,l=2} & {\rm age [Gyr]}\\
\hline
1.0 & 0.0 & 0.0 & 0.5 & 1.0 & 1.4 & 1.05 & 13.1\\
0.2 & 0.8 & 0.0 & 0.8 & 1.0 & 3.0 & 0.67 & 13.2\\
0.7 & 0.0 & 0.3 & 0.5 & 1.0 & 1.3 & 0.66 & 13.1\\
0.8 & 0.0 & 0.2 & 0.75 & 1.0 & 1.3 & 1.20 & 8.7\\
1.0 & 0.0 & 0.0 & 0.5 & 0.75& 1.3 & 0.55 & 13.1\\
\hline
\end{array}
$$
\caption{$\sigma_{8,v}$, $\sigma_{8,l=2}$ and age of the universe
for various models discussed
in the text. Relative errors are $^{+0.4}_{-0.3}$ for $\sigma_{8,v}$
and $\pm 0.17$ for $\sigma_{8,l=2}$.}
\label{table1}
\end{table}
\subsection{CDM model}
The first model we tested is
the CDM model. We performed Monte Carlo simulations for the standard
case with $h=0.5$. The $\sigma_{8,v}$ and $\sigma_{8,l=2}$
amplitudes agree well with each other (see Table \ref{table1}). In fact,
the amplitude predicted by the
standard CDM model is consistent with the {\sl COBE} normalization over most
of the allowed range of $h$. Thus, for example, $\Omega_m=1$ and $h=0.75$
(not shown in the Table) gives
$\sigma_{8,l=2}=1.5$, which is still compatible within the uncertainties
with $\sigma_{8,v}=1.3$.
The CDM model cannot be ruled out based on the comparison
between the velocity data and {\sl COBE}. This conclusion
agrees with the previous
comparisons based on the bulk flow estimates on somewhat larger scales
(Bertschinger et al. 1990; Efstathiou et al. 1992; Courteau et al. 1993),
but the present analysis
gives smaller uncertainty in the amplitude and thus places more
stringent limits on the models.
\subsection{CDM+$\Lambda$ models}
We studied a family of generalized CDM models adding
a cosmological constant $\Lambda$ with $\Omega_m+\Omega_{\Lambda}=1$.
Monte Carlo simulations were performed for
the model with $h=0.8$ and
$\Omega_{\Lambda}=0.8$. This model is a representative of the models
which match the recent determinations
of large scale galaxy clustering power (Maddox et al. 1990;
Kofman, Gnedin \& Bahcall 1993)
and the value of $h$ (Jacoby et al. 1992),
yet is compatible with globular cluster ages.
Our values for $\sigma_{8,l=2}$ are lower than the corresponding values
given by Efstathiou et al. (1992) owing to a different baryon content (which
affects the density transfer function)
and because we include the contribution
from the time derivative of the potential integrated along the line of sight
when $\Lambda \ne 0$ (\cite{kofman85}; \cite{gorski92}).
The agreement with the low $\Omega_m$ model is
not very good. While decreasing $\Omega_m$ increases $\sigma_{8,v}$
[due to the $f(\Omega_m)$ factor],
the time derivative of the potential integrated along the line of
sight tends to decrease $\sigma_{8,l=2}$ for a given $\Delta T/T$
quadrupole.
Decreasing $h$
even further decreases $\sigma_{8,l=2}$. We find that the results
strongly constrain these models toward small
$\Omega_{\Lambda}$ values.
95\% upper limits on $\Omega_{\Lambda}$ are 0.6 for $h=0.8$ and 0.4
for $h=0.5$.
\subsection{CDM+HDM models}
A mixed dark matter model with $\Omega_{CDM+B}=0.7$
and $\Omega_{\nu}=0.3$ has recently emerged as one of the best
candidates to explain the large-scale structure measurements
(Schaefer, Shafi, \& Stecker 1989;
Davis, Summers, \& Schlegel 1992; Taylor \& Rowan-Robinson 1992;
Klypin et al. 1993).
In this model, the
growth factor $f(k)$ depends on wavenumber because free-streaming damps
small wavelengths; $f(k)$ ranges between 1 on large scales to
${1\over 4}[(1+24\Omega_{CDM+B})^{1/2}-1]$ on small scales
(Bond, Efstathiou, \& Silk 1980; Ma 1993).
For $h=0.5$, which corresponds to $m_{\nu}=7$ eV, $f=1$ on large scales
and $f\approx 0.8$ on small scales, with a transition at
$k\approx1\,{\rm Mpc}^{-1}$. Because this is a relatively small scale,
the effect on our estimate of $\sigma_{8,v}$ is small.
The results for this model in Table \ref{table1} imply that the CDM+HDM
model is not strongly constrained,
although the estimated $\sigma_{8,v}$ is
somewhat high compared to the $\sigma_{8,l=2}$.
Decreasing the $\Omega_{\nu}$
contribution or increasing the value of $h$ increases
$\sigma_{8,l=2}$ and thus reduces the discrepancy.
We find excellent agreement between the two normalizations
for a particular model of $\Omega_{\nu}=0.2$ and $h=0.75$, but
of course other parameter values with smaller $\Omega_{\nu}$
and/or larger $h$ give acceptable results as well. In general,
the difference between CDM and CDM+HDM power spectra is small on large
scales ($>8 h^{-1}$Mpc) and consequently the {\sl COBE}
normalized $\sigma_{8,l=2}$ differs little for the two models.
\subsection{Tilted models}
A third way to decrease small scale power relative to that on large scales
is to tilt the
primordial power spectrum $P(k)\propto k^n$ by decreasing $n$
(\cite{adams93}; \cite{muciaccia93}).
The power spectrum for tilted models
differs from its standard CDM counterpart ($n=1$) on all scales,
not just on small scales as for the CDM+HDM power spectra.
Because the {\sl COBE}
scale is about 3 orders of magnitude larger than the $\sigma_8$ scale,
a given $Q_{\rm rms-PS}$ normalization drastically
changes $\sigma_{8,l=2}$ even for modest changes in $n$. This is
true despite the fact that
the best-fit value of $Q_{\rm rms-PS}$ increases when $n$ is decreased
(\cite{seljak93}). For the CDM transfer function with $h=0.5$ we find
\begin{equation}
\sigma_{8,l=2}=1.05\,(1\pm0.17)e^{-2.48(1-n)},
\label{adams}
\end{equation}
similar to the expression based on the $10^{\circ}$ {\sl COBE} normalization
(Adams et al. 1993).
Low values of $n$ generally imply an excessively small value of
$\sigma_{8,l=2}$ relative to $\sigma_{8,v}$.
For the particular case $n=0.75$ and $h=0.5$, the discrepancy between
$\sigma_{8,l=2}$ and $\sigma_{8,v}$ is more than a factor
of 2 (see Table \ref{table1}; the {\sl COBE} normalization assumes
no gravitational wave contribution). Including a possible
gravitational wave contribution (Lucchin, Matarrese, \& Mollerach 1992;
Davis et al. 1992) further decreases $\sigma_{8,l=2}$,
by a factor of $\sqrt{(3-n)/(14-12n)}$.
Higher values of $h$ allow somewhat lower values of $n$,
but in general $n$ cannot differ from 1 by more than $\sim 0.1$-$0.2$.
95\% confidence limits give $n>0.85$ with no gravitational wave contribution
and $n>0.94$ with a gravitational wave contribution, assuming $h=0.5$.
\section{Summary}
The analysis presented here gives the amplitude of mass
fluctuations $\sigma_{8,v}\approx1.3\,\Omega_m^{-0.6}$ in different models.
Comparing the $\sigma_{8,v}$ with the {\sl COBE} normalized
$\sigma_{8,l=2}$, one
can constrain different models of structure formation,
due to the fact that the two normalizations work on very different scales.
In general, all the usual extensions of CDM, which decrease the power on
small scales relative to that on
large scales, decrease the $\sigma_{8,l=2}$ value
relative to the standard CDM value, but leave $\sigma_{8,v}$ almost
unchanged.
While the estimated $\sigma_{8,v}$
agrees well with the {\sl COBE} value for standard CDM, the agreement
becomes worse for the extensions of CDM.
This particularly strongly challenges
the non-zero cosmological constant models and the tilted
models. It also points to a somewhat lower value of
$\Omega_{\nu}$ or a higher value of $h$
than have been assumed by most workers studying the mixed dark matter models.
Recently, using the constraints from the masses and abundances
of rich clusters, Efstathiou, White, \& Frenk (1993) obtained
$\sigma_8=0.57\,\Omega_m^{-0.56}$. This result is inconsistent with our
result at more than the 2$\sigma$ level, although including all the
sources of systematic errors could bring the two results into better
agreement. Nevertheless, there is increasing evidence that the
estimate of $\sigma_8$ from cluster abundances give lower values
than estimates based on peculiar velocities (Lilje 1992;
Henry \& Arnaud 1991; Evrard 1989).
The discrepancy may point to a significant
systematic error in either of the two methods. Further investigations
are needed to resolve this issue.
\acknowledgments
We thank Avishai Dekel for allowing us the use of the POTENT results.
This work was supported by grants NSF AST90-01762 and NASA NAGW-2807.
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import logging
logger = logging.getLogger(__package__)
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{"url":"https:\/\/www.iamcdocumentation.eu\/index.php\/Capital_and_labour_markets_-_COFFEE-TEA","text":"Capital and labour markets - COFFEE-TEA\n\n${\\displaystyle K{_{r,t}}=I{_{r,t}}+(1-\\delta {_{r}})K{_{r,t-1}}}$\nwhere: ${\\displaystyle K{_{r,t}}}$ is the capital stock in region r and time t; ${\\displaystyle I{_{r,t}}}$ is the investment in new capital goods in region r and time t; ${\\displaystyle \\delta {_{r}}}$ is the depreciation rate of capital in region r.","date":"2020-02-27 10:03:11","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 4, \"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\": 0.27876994013786316, \"perplexity\": 1015.2900314803998}, \"config\": {\"markdown_headings\": false, \"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-2020-10\/segments\/1581875146681.47\/warc\/CC-MAIN-20200227094720-20200227124720-00401.warc.gz\"}"} | null | null |
Indie Folk duo, Gwyneth and Monko are on the menu at Knoxville's WDVX. They will perform for a live audience as part of the radio station's daily show Blue Plate Special on November 8th. Knoxville, TN is but one of many stops on Gwyneth and Monko's tour of the US tour supporting their recently released EP, Good Old Horse. Featuring soulful vocals by Gwyneth with intricate instrumentation by Monko on both electric and acoustic guitars as well as mandolin, this energetic duo blends old timey folk sentiments with modern indie-folk sounds. Laura Meyer will join them. Listen to the show LIVE ONLINE!
The WDVX Blue Plate Special is a live performance each weekday at noon. Monday thru Friday at the Knoxville Visitors Center on the corner of Gay Street and Summit Hill Drive in Downtown Knoxville. Past guests include: BR-549, Marty Stuart, and Mary Gauthier, The WDVX Blue Plate Special is broadcast and webcast around the world. Can't make it down for a show? Listen live on-line. | {
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Jarim-Lim III – słabo znany amorycki król syryjskiego państwa Jamhad, panujący w 2 połowie XVIII wieku p.n.e., syn Nikmepy, brat Irkabtum, ojciec Hammurapi II.
Bibliografia
hasło Yarim-Lim III, [w:] Gwendolyn Leick, Who's Who in the Ancient Near East, London and New York 2002, s. 179.
Władcy Jamhadu | {
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\section{Introduction}
The randomization of a complete first order theory $T$ is the complete continuous theory $T^R$
with two sorts, a sort for random elements of models of $T$, and a sort for events in an underlying probability space.
The aim of this paper is to investigate algebraic independence relations in randomizations of first order theories.
We will use results from our earlier papers [AGK1], which characterizes definability in randomizations, and [AGK2], where it is shown that the randomization of every
o-minimal theory is real rosy, that is, has a strict independence relation.
We focus on the independence axioms introduced by Adler [Ad2] (see Definition \ref{d-adler} below).
In first order model theory, algebraic independence is anti-reflexive and satisfies all of Adler's axioms except perhaps base monotonicity,
and also satisfies \emph{small local character}, a property that implies local character with the smallest possible bound $\kappa(D)=(|D|+\aleph_0)^+$.
It was shown in [BBHU] and [EG] that for any complete continuous theory, the algebraic independence relation satisfies all of the Adler's axioms except perhaps base
monotonicity, extension, and finite character, and also satisfies countable character (a weakening of finite character), has local character with
bound $\kappa(D)=((|D|+2)^{\aleph_0})^+$, and is anti-reflexive. We show here that if the underlying first order theory
$T$ has $\operatorname{acl}=\operatorname{dcl}$ (that is, algebraic closure coincides with definable closure), then algebraic closure in $T^R$ also satisfies extension and small local character.
However, for every $T$, algebraic independence in $T^R$ never has finite character and never satisfies base monotonicity.
Another relation on models of $T^R$ is \emph{pointwise algebraic independence}, which was introduced in [AGK2] and roughly means algebraic
independence almost everywhere. We show that for arbitrary $T$ (rather than just when $T$ has $\operatorname{acl}=\operatorname{dcl}$), pointwise algebraic independence
in $T^R$ satisfies all of Adler's axioms except perhaps base monotonicity and extension. In particular,
it does have finite character. Moreover, pointwise algebraic independence satisfies extension for countable sets, has small local character, and satisfies base monotonicity
if and only if algebraic extension in $T$ satisfies base monotonicity. However, pointwise algebraic independence is never anti-reflexive.
This paper is organized as follows. In Section 2 we review Adler's axioms for independence relations and some general results from the literature about algebraic
independence in first order and continuous model theory. Section 3 contains some notions and results about the randomization theory $T^R$ that we will need from the papers
[AGK1] and [AGK2]. Section 4 contains the proofs of the negative results that in $T^R$, algebraic independence never has finite character and never satisfies
base monotonicity. To better understand why this happens, we take a closer look at the example of dense linear order. Section 5 contains the proof of the result
that if $T$ has $\operatorname{acl}=\operatorname{dcl}$ then algebraic independence in $T^R$ satisfies the extension axiom. In Section 6 we prove that if $T$ has $\operatorname{acl}=\operatorname{dcl}$ then
algebraic independence in $T^R$ has small local character. On the way to this proof, we introduce the pointwise algebraic independence relation in $T^R$,
and show that it has small local character whether or not $T$ has $\operatorname{acl}=\operatorname{dcl}$. Finally, in Section 7 we prove the other results stated in the
preceding paragraph about pointwise algebraic independence in $T^R$. We also show that in $T^R$, pointwise algebraic independence never implies algebraic independence,
and algebraic independence implies pointwise algebraic independence only in the trivial case that the models of $T$ are finite.
For background in continuous model theory in its current form we refer to the papers [BBHU] and [BU].
We assume the reader is familiar with the basics of continuous model theory, including the notions of a theory, model, pre-model, reduction, and completion.
For background on randomizations of models we refer to the papers [Ke] and [BK].
We follow the terminology of [AGK2]. A continuous pre-model is called \emph{pre-complete} if its reduction is its completion. The set of all finite tuples in a set $A$
is denoted by $A^{<\mathbb{N}}.$ We assume throughout this paper that $T$ is a complete first order theory with countable signature $L$ and models of cardinality $>1$, and
that $\upsilon$ is an uncountable inaccessible cardinal that is held fixed. We let $\cu M$ be the \emph{big model} of $T$, that is,
the (unique up to isomorphism) saturated model $\cu M\models T$ that is finite or of cardinality $|\cu N|=\upsilon$.
We call a set \emph{small} if it has cardinality $< \upsilon$, and \emph{large} otherwise.
\section{Independence}
\subsection{Abstract Independence Relations}
Since the various properties of independence are given some slightly different names in
various parts of the literature, we take this opportunity to declare that we are following the terminology established in [Ad2], which is
repeated here for the reader's convenience. In this paper, we will sometimes write $AB$ for $A\cup B$, and write $[A,B]$ for
$\{D\colon A\subseteq D \wedge D\subseteq B\}$
\begin{df}[Adler] \label{d-adler}
Let $\cu N$ be the big model of a continuous or first order theory. By a \emph{ternary relation over $\cu N$} we mean
a ternary relation $\ind$ on the small subsets of $\cu N$. We say that $\ind$
is an \emph{independence relation} if it satisfies the following \emph{axioms for independence relations} for all small sets:
\begin{enumerate}
\item (Invariance) If $A\ind_CB$ and $(A',B',C')\equiv (A,B,C)$, then $A'\ind_{C'}B'$.
\item (Monotonicity) If $A\ind_C B$, $A'\subseteq A$, and $B'\subseteq B$, then $A'\ind_C B'$.
\item (Base monotonicity) Suppose $C\in[D, B]$. If $A\ind_D B$, then $A\ind_C B$.
\item (Transitivity) Suppose $C\in[D, B]$. If $B\ind_C A$ and $C\ind_D A$, then $B\ind_D A$.
\item (Normality) $A\ind_CB$ implies $AC\ind_C B$.
\item (Extension) If $A\ind_C B$ and $\hat B\supseteq B$, then there is $A'\equiv_{BC} A$ such that $A'\ind_C \hat B$.
\item (Finite character) If $A_0\ind_CB$ for all finite $A_0\subseteq A$, then $A\ind_CB$.
\item (Local character) For every $A$, there is a cardinal $\kappa(A)<\upsilon$ such that, for any set $B$, there is a subset $C$ of $B$ with $|C|<\kappa(A)$ such that $A\ind_CB$.
\end{enumerate}
If finite character is replaced by countable character (which is defined in the obvious way), then we say that $\ind$ is a
\emph{countable independence relation}. We will refer to the first five axioms (1)--(5) as the \emph{basic axioms}.
\end{df}
\begin{df}
An independence relation $\ind$ is \emph{strict} if it satisfies
\begin{itemize}
\item[(9)] (Anti-reflexivity) $a\ind_B a$ implies $a\in \operatorname{acl}(B)$.
\end{itemize}
\end{df}
There are two other useful properties to consider when studying ternary relations over $\cu N$:
\begin{df}
\noindent\begin{enumerate}
\item[(10)] (Full existence) For every $A,B,C$, there is $A'\equiv_C A$ such that $A'\ind_C B$.
\item[(11)] (Symmetry) For every $A,B,C$, $A\ind_CB$ implies $B\ind_CA$.
\end{enumerate}
\end{df}
\begin{fact} \label{f-fe-vs-ext} (Remarks 2.2.4 in [AGK2]).
\noindent\begin{enumerate}
\item[(i)] Whenever $\ind$ satisfies invariance, monotonicity, transitivity, normality, full existence, and symmetry,
then $\ind$ also satisfies extension.
\item[(ii)] Any countable independence relation is symmetric.
\end{enumerate}
\end{fact}
\begin{df}
\noindent\begin{enumerate}
\item[(i)] We say that $\ind$ has \emph{countably local character} if for every countable set $A$ and every small set $B$, there is a countable subset $C$ of $B$ such that $A\ind_CB$.
\item[(ii)] We say that $\ind$ has \emph{small local character} if for all small sets $A, B, C_0$
such that $C_0\subseteq B$ and $|C_0|\le|A|+\aleph_0$, there is
a set $C\in[C_0,B]$ such that $|C|\le|A|+\aleph_0$ and $A\ind_C \ B$.
\end{enumerate}
\end{df}
\begin{fact} \label{f-countably-localchar} (Remark 2.2.7 in [AGK2]).
\noindent\begin{enumerate}
\item[(i)] If $\ind$ has small local character, then $\ind$ has local character with bound $\kappa(D)=(|D|+\aleph_0)^+$ (the smallest possible bound).
In the presence of monotonicity, the converse is also true.
\item[(ii)] If $\ind$ has local character with bound $\kappa(D)=(|D|+\aleph_0)^+$, then $\ind$ has countably local character.
\item[(iii)] If $\ind$ has invariance, countable character, base monotonicity, and countably local character, then
$\ind$ has local character with bound $\kappa(D)=((|D| + 2)^{\aleph_0})^+$.
\end{enumerate}
\end{fact}
We say that $\ind[J]$ is \emph{weaker than} $\ind[I]$, and write $\ind[I]\Rightarrow \ind[J]$, if $A\ind[I]_C B\Rightarrow A\ind[J]_C B$.
\begin{rmk} \label{r-weaker} Suppose $\ind[I]\Rightarrow\ind[J]$. If $\ind[I]$ has full existence, local character,
countably local character, or small local character. Then $\ind[J]$ \ has the same property.
\end{rmk}
\subsection{Algebraic Independence}
\begin{df} In first order logic, a formula $\varphi(u,\vec v)$ is \emph{functional} in $T$ if
$$T\models(\forall \vec v)(\exists ^{\le 1} u)\varphi(u,\vec v).$$
$\varphi(u,\vec v)$ is \emph{algebraical} in $T$ if there exists $n\in\mathbb{N}$ such that
$$T\models(\forall \vec v)(\exists ^{\le n} u)\varphi(u,\vec v).$$
\end{df}
The \emph{definable closure} of $A$ in $\cu M$ is the set
$$\operatorname{dcl}^{\cu M}(A)=\{b\in M\mid \cu M\models\varphi(b,\vec a) \mbox{ for some functional } \varphi \mbox{ and } \vec a\in A^{<\mathbb{N}}\}.$$
The \emph{algebraic closure} of $A$ in $\cu M$ is the set
$$\operatorname{acl}^{\cu M}(A)=\{b\in M\mid \cu M\models\varphi(b,\vec a) \mbox{ for some algebraical } \varphi \mbox{ and } \vec a\in A^{<\mathbb{N}}\}.$$
We refer to [BBHU] for the definitions of the algebraic closure $\operatorname{acl}^{\cu N}(A)$ and definable closure $\operatorname{dcl}^{\cu N}(A)$ in a continuous structure $\cu N$.
If $\cu N$ is clear from the context, we will sometimes drop the superscript and write $\operatorname{dcl}, \operatorname{acl}$ instead of $ \operatorname{dcl}^\cu N, \operatorname{acl}^\cu N$.
We will often use the following facts without explicit mention.
\begin{fact} \label{f-algebraic-cardinality} (Follows from [BBHU], Exercise 10.8)
For every set $A$, $\operatorname{acl}(A)$ has cardinality at most $( |A|+2)^{\aleph_0}$. Thus the algebraic closure of a small set is small.
\end{fact}
\begin{fact} \label{f-definableclosure} (Definable Closure, Exercises 10.10 and 10.11, and Corollary 10.5 in [BBHU])
\begin{enumerate}
\item If $ A\subseteq\cu N$ then $\operatorname{dcl}( A)=\operatorname{dcl}(\operatorname{dcl}( A))$ and $\operatorname{acl}( A)=\operatorname{acl}(\operatorname{acl}( A))$.
\item If $ A$ is a dense subset of the topological closure of $B$ and $ B\subseteq\cu N$, then $\operatorname{dcl}(A)=\operatorname{dcl}( B)$ and $\operatorname{acl}(A)=\operatorname{acl}( B)$.
\end{enumerate}
\end{fact}
It follows that for any $ A\subseteq\cu N$, $\operatorname{dcl}( A)$ and $\operatorname{acl}( A)$ are topologically closed.
\noindent In any complete theory (first order or continuous), we define the notion of \emph{algebraic independence}, denoted $\ind[a]$,
by setting $A\ind[a]_CB$ to mean $\operatorname{acl}(AC)\cap \operatorname{acl}(BC)=\operatorname{acl}(C)$. In first order logic, $\ind[a]$ satisfies all axioms for a strict
independence relation except for perhaps base monotonicity.
\begin{prop} \label{p-alg-indep}
In any complete continuous theory, $\ind[a]$ satisfies symmetry and all axioms for a strict countable independence relation except perhaps for
base monotonicity and extension.
\end{prop}
\begin{proof}
The proof is exactly as in [Ad2], Proposition 1.5, except for some minor modifications. For example, countable character of $\operatorname{acl}$ in
continuous logic yields countable character of $\ind[a]$. Also, in the verification of local character, one needs to take
$\kappa(A):=((|A|+2)^{\aleph_0})^+$ instead of $(|A|+\aleph_0)^+$.
\end{proof}
\begin{question} \label{q-a-full-existence}
Does $\ind[a]$ always have full existence (or extension) in continuous logic?
\end{question}
The proof that $\ind[a]$ has full existence in first order logic uses the negation connective, which is not available in continuous logic.
\section{Randomizations}
\subsection{The Theory $T^R$}
The \emph{randomization signature} $L^R$ is the two-sorted continuous signature
with sorts $\BK$ (for random elements) and $\BB$ (for events), an $n$-ary
function symbol $\llbracket\varphi(\cdot)\rrbracket$ of sort $\BK^n\to\BB$
for each first order formula $\varphi$ of $L$ with $n$ free variables,
a $[0,1]$-valued unary predicate symbol $\mu$ of sort $\BB$ for probability, and
the Boolean operations $\top,\bot,\sqcap, \sqcup,\neg$ of sort $\BB$. The signature
$L^R$ also has distance predicates $d_\BB$ of sort $\BB$ and $d_\BK$ of sort $\BK$.
In $L^R$, we use ${\sa B},{\sa C},\ldots$ for variables or parameters of sort $\BB$. ${\sa B}\doteq{\sa C}$
means $d_\BB({\sa B},{\sa C})=0$, and ${\sa B}\sqsubseteq{\sa C}$ means ${\sa B}\doteq{\sa B}\sqcap{\sa C}$.
A structure with signature $L^R$ will be a pair $\cu N=(\cu K,\cu E)$ where $\cu K$ is the part of sort $\BK$ and
$\cu E$ is the part of sort $\BB$.
The following fact, which is a consequence of Proposition 2.1.10 of [AGK1], gives a model-theoretic characterization of $T^R$.
\begin{fact} \label{f-neat}
There is a unique complete theory $T^R$ with signature $L^R$ whose big model $\cu N=(\cu K,\cu E)$ is the reduction of a
pre-complete-structure $\cu P=(\cu J,\cu F)$
equipped with a complete atomless probability space $(\Omega,\cu F,\mu )$ such that:
\begin{enumerate}
\item $\cu F$ is a $\sigma$-algebra with $\top,\bot,\sqcap, \sqcup,\neg$ interpreted by $\Omega,\emptyset,\cap,\cup,\setminus$.
\item $\cu J$ is a set of functions $a\colon\Omega\to M$.
\item For each formula $\psi(\vec{x})$ of $L$ and tuple
$\vec{a}$ in $\cu J$, we have
$$\llbracket\psi(\vec{a})\rrbracket=\{\omega\in\Omega:\cu M\models\psi(\vec{a}(\omega))\}\in\cu F.$$
\item $\cu F$ is equal to the set of all events
$ \llbracket\psi(\vec{a})\rrbracket$
where $\psi(\vec{v})$ is a formula of $L$ and $\vec{a}$ is a tuple in $\cu J$.
\item For each formula $\theta(u, \vec{v})$
of $L$ and tuple $\vec{b}$ in $\cu J$, there exists $a\in\cu J$ such that
$$ \llbracket \theta(a,\vec{b})\rrbracket=\llbracket(\exists u\,\theta)(\vec{b})\rrbracket.$$
\item On $\cu J$, the distance predicate $d_\BK$ defines the pseudo-metric
$$d_\BK(a,b)= \mu \llbracket a\neq b\rrbracket .$$
\item On $\cu F$, the distance predicate $d_\BB$ defines the pseudo-metric
$$d_\BB({\sa B},{\sa C})=\mu ( {\sa B}\triangle {\sa C}).$$
\end{enumerate}
\end{fact}
\begin{fact} \label{f-glue} (Lemma 2.1.8 in [AGK1])
In the big model $\cu N=(\cu K,\cu E)$ of $T^R$, for each $\bo a ,\bo b \in\cu K$ and ${\sa B}\in\cu E$,
there is an element $\bo c \in\cu K$ that agrees with $\bo a $ on ${\sa B}$ and agrees with $\bo b $ on $\neg{\sa B}$,
that is, ${\sa B}\sqsubseteq\llbracket \bo c =\bo a \rrbracket$ and $(\neg{\sa B})\sqsubseteq\llbracket \bo c =\bo b \rrbracket$.
\end{fact}
\begin{df} A first order or continuous theory has $\operatorname{acl}=\operatorname{dcl}$ if $\operatorname{acl}(A)=\operatorname{dcl}(A)$ for every set $A$ in every model of the theory.
\end{df}
For example, any first order theory $T$ with a definable linear ordering has $\operatorname{acl}=\operatorname{dcl}$.
\begin{fact} \label{f-acl=dcl} ([AGK1], Proposition 3.3.7, see also [Be2])
In the big model $\cu N$ of $T^R$, $\operatorname{acl}_\BB( A)=\operatorname{dcl}_\BB( A)$ and $\operatorname{acl}( A)=\operatorname{dcl}( A)$. Thus $T^R$ has $\operatorname{acl}=\operatorname{dcl}.$
\end{fact}
As a corollary, we obtain the following characterization of algebraic independence in $\cu N$.
\begin{cor} \label{c-alg-indep} In the big model $\cu N$ of $T^R$, $ A\ind[a]_C B$ if and only if
$$[\operatorname{dcl}(AC)\cap\operatorname{dcl}(BC)=\operatorname{dcl}(C)] \wedge [\operatorname{dcl}_\BB(AC)\cap\operatorname{dcl}_\BB(BC)=\operatorname{dcl}_\BB(C)].$$
\end{cor}
\begin{proof} By the definition of algebraic independence in the two-sorted metric structure $\cu N$ and Fact \ref{f-acl=dcl}.
\end{proof}
From now on we will work within the big model $\cu N=( {\cu K}, {\cu E})$ of $T^R$.
By saturation, $\cu K$ and $\cu E$ are large. Hereafter, $A, B, C$ will always denote small subsets of $\cu K$, and $\cu N_A$ will denote
the expansion of $\cu N$ formed by adding a constant symbol for each $\bo a\in A$.
We will write $\operatorname{dcl}, \operatorname{acl}$ for $\operatorname{dcl}^{\cu N}, \operatorname{acl}^{\cu N}$, and $\ind[a]$ will denote the algebraic independence relation in $\cu N$.
For each element $\bo b\in {\cu K}$, we will also choose once and for all a \emph{representative} $b\in\cu J$ such that
the image of $b$ under the reduction map is $\bo b$. It follows that for each first order formula $\varphi(\vec v)$,
$\llbracket\varphi(\vec{\bo a})\rrbracket$ in $\cu N$ is the image of $\llbracket\varphi(\vec a)\rrbracket$ in $\cu P$ under the reduction map.
Note that any two representatives of an element $\bo b\in\cu K$ agree except on a set of measure zero.
For any small $A\subseteq\cu K$ and each $\omega\in\Omega$, we define
$$ A(\omega)=\{a(\omega)\mid \bo a\in A\},$$
and let $\operatorname{cl}( A)$ denote the closure of $ A$ in the metric $d_\BK$. When $\cu A\subseteq {\cu E}$,
$\operatorname{cl}(\cu A)$ denotes the closure of $\cu A$ in the metric $d_\BB$, and
$\sigma(\cu A)$ denotes the smallest $\sigma$-subalgebra of $ {\cu E}$ containing $\cu A$.
Since the cardinality $\upsilon$ of $\cu N$ is inaccessible, whenever $A\subseteq\cu K$ is small, the closure $\operatorname{cl}(A)$ and
the set of $n$-types over $A$ is small. Also, whenever $\cu A\subseteq\cu E$ is small, the closure $\operatorname{cl}(\cu A)$ is small.
\subsection{Definability in $T^R$}
In this section we review some notions and results about definability that we will need from the paper [AGK1].
We write $\operatorname{dcl}_\BB( A)$ for the set of elements of sort $\BB$ that are definable over $ A$ in $\cu N$,
and write $\operatorname{dcl}( A)$ for the set of elements of sort $\BK$ that are definable over $ A$ in $\cu N$.
Similarly for $\operatorname{acl}_\BB( A)$ and $\operatorname{acl}( A)$.
\begin{df} We say that an event $\sa E$ is \emph{first order definable over $A$}, in symbols $\sa E\in\operatorname{fdcl}_\BB(A)$, if
$\sa E=\llbracket\theta(\vec{\bo a})\rrbracket$ for some formula $\theta$ of $L$ and some tuple $\vec{\bo a}\in A^{<\mathbb{N}}$.
\end{df}
\begin{df}
We say that $\bo b$ is \emph{first order definable over $ A$}, in symbols $\bo b\in\operatorname{fdcl}( A)$, if there is a functional formula
$\varphi(u,\vec v)$ and a tuple $\vec{\bo a}\in {A}^{<\mathbb{N}}$ such that
$\llbracket \varphi(b,\vec{a})\rrbracket=\top$.
\end{df}
\begin{fact} \label{f-separable} ([AGK1], Theorems 3.1.2 and 3.3.6)
$$\operatorname{dcl}_\BB( A)=\operatorname{cl}(\operatorname{fdcl}_\BB( A))=\sigma(\operatorname{fdcl}_\BB(A))\subseteq\cu E,\qquad \operatorname{dcl}( A)=\operatorname{cl}(\operatorname{fdcl}( A))\subseteq\cu K.$$
\end{fact}
It follows that whenever $A$ is small, $\operatorname{dcl}(A)$ and $\operatorname{dcl}_\BB(A)$ are small.
\begin{rmk} \label{r-dcl-B} For each small $A$,
$$\operatorname{fdcl}_\BB(\operatorname{fdcl}(A))=\operatorname{fdcl}_\BB(A),\quad\operatorname{dcl}_\BB(\operatorname{dcl}(A))=\operatorname{dcl}_\BB(A).$$
\end{rmk}
We will sometimes use the $\llbracket \ldots\rrbracket$ notation in a general setting. Given a property $P(\omega)$, we write
$$\llbracket P\rrbracket=\{\omega\in\Omega\,:\,P(\omega)\}.$$
\begin{df} \label{d-pointwise-def}
We say that $\bo b$ is \emph{pointwise definable over $A$}, in symbols $\bo b\in\operatorname{dcl}^\omega(A)$, if
$$\mu(\llbracket\bo b\in\operatorname{dcl}^{\cu M}(A_0)\rrbracket)=1$$
for some countable $A_0\subseteq A$.
We say that $\bo b$ is \emph{pointwise algebraic over $A$}, in symbols $\bo b\in\operatorname{acl}^\omega(A)$, if
$$\mu(\llbracket \bo b\in\operatorname{acl}^{\cu M}(A_0)\rrbracket)=1$$
for some countable $A_0\subseteq A$.
\end{df}
\begin{rmk} \label{r-dcl-omega}
$\operatorname{dcl}^\omega$ and $\operatorname{acl}^\omega$ have countable character, that is,
$\bo b\in\operatorname{dcl}^\omega(A)$ if and only if $\bo b\in\operatorname{dcl}^\omega(A_0)$ for some countable $A_0\subseteq A$, and similarly for $\operatorname{acl}^\omega$.
\end{rmk}
The next result is a useful characterization of $\operatorname{dcl}(A)$.
\begin{fact} \label{f-dcl3} ([AGK1], Corollary 3.3.5) For any element $\bo b\in {\cu K}$,
$\bo{b}$ is definable over $ A$ if and only if:
\begin{enumerate}
\item $\bo b$ is pointwise definable over $A$;
\item $\operatorname{fdcl}_\BB(\bo b A)\subseteq\operatorname{dcl}_\BB(A).$
\end{enumerate}
\end{fact}
\begin{cor} \label{c-pointwise-alg-def}
In $\cu N$ we always have
$$\operatorname{acl}(A)=\operatorname{dcl}(A)\subseteq\operatorname{dcl}^\omega(A)=\operatorname{dcl}^\omega(\operatorname{dcl}^\omega(A))\subseteq\operatorname{acl}^\omega(A)=\operatorname{acl}^\omega(\operatorname{acl}^\omega(A)).$$
\end{cor}
\subsection{Algebraic Independence in the Event Sort}
The ternary relation $\ind[a\BB] \ \ $ on the big model $\cu N$ of $T^R$ was introduced in the paper [AGK2] and will be useful here. It is the
analogue of algebraic independence obtained by restricting the algebraic closures of sets to the event sort.
\begin{df} \label{d-event-indep}
For small $A, B, C\subseteq \cu K$, define
$$A\ind[a\BB]_C \ \ B\Leftrightarrow\operatorname{acl}_\BB(AC)\cap\operatorname{acl}_\BB(BC)=\operatorname{acl}_\BB(C).$$
\end{df}
\begin{rmk} \label{r-acl=dcl}
By Fact \ref{f-acl=dcl}, for small $A, B, C\subseteq \cu K$, we have
$$A\ind[a\BB]_C \ \ B\Leftrightarrow\operatorname{dcl}_\BB(AC)\cap\operatorname{dcl}_\BB(BC)=\operatorname{dcl}_\BB(C).$$
By Corollary \ref{c-alg-indep}, we also have
$$A\ind[a]_C B \Leftrightarrow (\operatorname{dcl}(AC)\cap\operatorname{dcl}(BC)=\operatorname{dcl}(C))\wedge A\ind[a\BB]_C \ B.$$
\end{rmk}
\begin{fact} \label{f-event-aB} (Proposition 6.2.4 in [AGK2]).
In $T^R$, the relation $\ind[a\BB] \ \ $ satisfies all the axioms for a countable independence relation except base monotonicity.
It also has symmetry and small local character.
\end{fact}
We will also need the following fact, which is given by Lemma 6.1.6, Corollary 6.1.7, and Lemma 6.2.3 of [AGK2], and is a consequence of a result in [Be].
\begin{fact} \label{f-event-dB}
There is a countable independence relation $\ind[d\BB] \ $ (dividing independence in the event sort) that has small local character over $\cu N$
and is such that $\ind[d\BB] \ \Rightarrow \ind[a\BB] \ .$
\end{fact}
\section{Negative Results: Finite Character and Base Monotonicity}
In this section we show that for \emph{every} $T$, algebraic independence in $T^R$ satisfies neither finite character nor base monotonicity.
The following lemmas and notation will be useful for these results.
\begin{lemma} \label{l-char-function}
There exists a pair $Z=\{\bo{0},\bo{1}\}\subseteq\cu K$ such that $\llbracket\bo 0\ne \bo 1\rrbracket=\top$, and $\operatorname{dcl}_\BB(Z)=\{\top,\bot\}.$
\end{lemma}
\begin{proof}
This follows from Fact \ref{f-neat} and the fact that $\cu N$ is saturated.
\end{proof}
By Fact \ref{f-glue}, for each event $\sa B\in\cu E$, there is a unique element
$1_{\sa B}\in\cu K$ that agrees with $1$ on $\sa B$ and agrees with $0$ on $\neg\sa B$.
Given a set $\cu A\subseteq\cu E$, let $\bo 1_{\cu A}=\{\bo 1_{\sa B} \mid \sa B\in\cu A\}.$
\begin{lemma} \label{l-char-functions}
If $\cu A\subseteq\cu E$, then $\operatorname{dcl}_\BB(\bo 1_{\cu A} Z)=\sigma(\cu A).$
\end{lemma}
\begin{proof} By definition, $\sa E\in\operatorname{fdcl}_\BB(\bo 1_{\cu A}Z)$ if and only if $\sa E=\llbracket\theta(\vec{\bo a},\bo 0,\bo 1)\rrbracket$
for some formula $\theta\in L$ and tuple $\vec{\bo a}\subseteq\bo 1_{\cu A}.$ For each $\bo b\in\bo 1_{\cu A}$, we have $\mu(\llbracket\bo b\in Z\rrbracket)=1.$
It follows that $\llbracket\theta(\vec{\bo a},\bo 0,\bo 1)\rrbracket\in\sigma(\cu A)$. By Fact \ref{f-separable}, $\operatorname{dcl}_\BB(\bo 1_{\cu A} Z)\subseteq\sigma(\cu A).$
For each $\sa B\in\cu A$ we have $\sa B=\llbracket \bo 1_{\sa B} = \bo 1\rrbracket,$ so $\cu A\subseteq \operatorname{fdcl}_\BB(\bo 1_{\cu A} Z).$
Then by Fact \ref{f-separable} again, $ \sigma(\cu A)\supseteq\operatorname{dcl}_\BB(\bo 1_{\cu A} Z).$
\end{proof}
\subsection{Finite Character}
\begin{prop} \label{p-no-finite-character}
For every $T$, $\ind[a]$ in $T^R$ does not satisfy finite character.
\end{prop}
\begin{proof} Since $\mu$ is atomless, there is an event $\sa B$ and a sequence of events $\<\sa B_n\>_{n\in\mathbb{N}}$ such that for each $n$,
$$ \sa B_n\sqsubseteq\sa B_{n+1}\sqsubseteq\sa B, \quad \mu(\sa B\setminus\sa B_n)=2^{-(n+1)},\quad \mu(\sa B)=1/2.$$
Let $\bo b=\bo 1_{\sa B}$, $\cu A_n=\{\sa B_m\mid m<n\}$, $\cu A=\{\sa B_m\mid m\in\mathbb{N}\}.$
Then $\bo 1_{\cu A}=\bigcup_n \bo 1_{\cu A_n}.$ By Lemma \ref{l-char-functions},
$$\operatorname{dcl}_{\BB}(\bo 1_{\cu A_n} Z)=\sigma(\cu A_n), \quad \operatorname{dcl}_{\BB}(\bo b Z)=\sigma(\{\sa B\})\subseteq\sigma(\bo 1_{\cu A} Z)=\operatorname{dcl}_{\BB}(\cu A).$$
Note that every element of $\cu K$ that is pointwise definable from $\bo 1_{\cu A} Z$ is pointwise definable from $Z$.
Then by Fact \ref{f-dcl3}, we have
$$ \operatorname{dcl}( Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid \operatorname{fdcl}_\BB(\bo x Z)\subseteq \{\top,\bot\}\},$$
$$ \operatorname{dcl}(\bo 1_{\cu A} Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid \operatorname{fdcl}_\BB(\bo x \bo 1_{\cu A} Z)\subseteq \sigma(\cu A)\},$$
$$ \operatorname{dcl}(\bo 1_{\cu A_n} Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid \operatorname{fdcl}_\BB(\bo x \bo 1_{\cu A_n} Z)\subseteq \sigma(\cu A_n)\},$$
$$ \operatorname{dcl}(\bo b Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid \operatorname{fdcl}_\BB(\bo x \bo b Z)\subseteq \sigma(\sa B)\}.$$
But $\sigma(\cu A_n)\cap\sigma(\{\sa B\})=\{\top,\bot\}$, so by Fact \ref{f-acl=dcl},
$$\operatorname{acl}(\bo 1_{\cu A_n} Z)\cap\operatorname{acl}(\bo b Z)=\operatorname{dcl}(\bo 1_{\cu A_n} Z)\cap\operatorname{dcl}(\bo b Z)=\operatorname{dcl}(Z)=\operatorname{acl}(Z),$$
and hence $\bo 1_{\cu A_n}\ind[a]_Z \bo b.$
However, $\sa B\in\sigma(\sa A)$, so by Lemma \ref{l-char-functions} we have
$\operatorname{dcl}_\BB(\bo b \bo 1_{\cu A} Z)=\sigma(\cu A).$ Moreover, $\bo b\in\operatorname{dcl}^\omega(Z)$. Therefore
$$\bo b\in \operatorname{acl}(\bo 1_{\cu A} Z)\cap\operatorname{acl}(\bo b)\setminus \operatorname{acl}(Z),$$
so $\bo 1_{\cu A} \nind[a]_Z \bo b$ and finite character fails.
\end{proof}
\subsection{Base Monotonicity}
By Proposition 1.5 (3) in [Ad1], for any complete first order theory $T$, $\ind[a]$ satisfies base monotonicity if and only if the lattice of algebraically closed sets is modular.
The argument there shows that the same result holds for any complete continuous theory.
We show that for $T^R$, $\ind[a]$ never satisfies base monotonicity, and thus is never modular and is never a countable independence relation.
\begin{prop} \label{p-a-nobase-monotonicity} For every $T$,
$\ind[a]$ in $T^R$ does not satisfy base monotonicity.
\end{prop}
\begin{proof} Since $\mu$ is atomless, there are two
independent events $\sa D, \sa F$ in $\cu E$ of probability $1/2$. Let $\sa E=\sa D\sqcap\sa F$. $\bo a=1_{\sa D}$, $\bo b = 1_{\sa E}$, and $\bo c=1_{\sa F}$. Then
$$\operatorname{dcl}_\BB(\bo a)=\sigma(\{\sa D\}),\quad
\operatorname{dcl}_\BB(\bo c)=\sigma(\{\sa F\}),$$
$$ \operatorname{dcl}_\BB(\bo a\bo c)=\sigma(\{\sa D,\sa F\}),\quad
\operatorname{dcl}_\BB(\bo b \bo c)=\sigma(\{\sa E,\sa F\}).$$
As in the proof of Proposition \ref{p-no-finite-character}, we have
$$ \operatorname{dcl}( Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid \operatorname{fdcl}_\BB(\bo x Z)\subseteq \{\top,\bot\}\},$$
$$ \operatorname{dcl}(\bo a Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid \operatorname{fdcl}_\BB(\bo x\bo a Z)\subseteq\sigma(\{\sa D\})\},$$
$$ \operatorname{dcl}(\bo c Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid\\operatorname{fdcl}_\BB(\bo x\bo c Z)\subseteq\sigma(\{\sa F\})\},$$
$$ \operatorname{dcl}(\bo a \bo c Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid \operatorname{fdcl}_\BB(\bo x\bo a\bo c Z)\subseteq\sigma(\{\sa D,\sa F\})\},$$
$$ \operatorname{dcl}(\bo b \bo c Z)=\{\bo x\in\operatorname{dcl}^\omega(Z)\mid\operatorname{fdcl}_\BB(\bo x\bo b\bo c Z)\subseteq\sigma(\{\sa E,\sa F\})\}.$$
It follows that $\bo a\ind[a]_Z \bo b\bo c Z$. But
$$\sa E\in\sigma(\{\sa D,\sa F\})\cap\sigma(\{\sa E,\sa F\})\setminus\sigma(\{\sa F\}),$$
so
$$\bo b\in\operatorname{acl}(\bo a\bo c Z)\cap \operatorname{acl}(\bo b\bo c Z)\setminus \operatorname{acl}(\bo c Z).$$
Therefore $\bo a\nind[a]_{\bo c Z} \bo b\bo c Z$, and base monotonicity fails.
\end{proof}
Recall from [Ad1] that the ternary relation $\ind[M] \ $ is defined by
$$A\ind[M]_C \ B \ \Leftrightarrow (\forall D\in[C,\operatorname{acl}(BC)]) A\ind[a]_D B.$$
$\ind[M] \ $ is the weakest ternary relation that implies $\ind[a]$ and satisfies base monotonicity.
So in both first order and continuous model theory, if $\ind[a]$ satisfies base monotonicity then $\ind[M] \ =\ind[a]$
and hence $\ind[M] \ $ satisfies symmetry. Thus the following corollary is an improvement of Proposition \ref{p-a-nobase-monotonicity}.
\begin{cor} \label{c-no-symmetry} For every $T$, $\ind[M] \ \ $ in $T^R$ does not satisfy symmetry.
\end{cor}
\begin{proof}
We use the notation introduced in the proof of \ref{p-a-nobase-monotonicity}. Since $\bo a\nind[a]_{\bo c Z} \bo b\bo c Z$ and
$\ind[M]\Rightarrow\ind[a]$, we have $\bo a\nind[M]_{\bo c Z} \ \bo b\bo c Z.$ However, it follows from the proof of \ref{p-a-nobase-monotonicity} that
$\bo b\bo c Z\ind[M]_Z \ \bo a,$ so $\ind[M] \ $ does not satisfy symmetry.
\end{proof}
As an example, we look at the relations $\ind[a]$ and $\ind[M] \ $ in the continuous theory $\operatorname{DLO}^R$, the randomization of the theory of dense linear order without endpoints.
We will see that these relations are much more complicated in $\operatorname{DLO}^R$ than they are in $\operatorname{DLO}$.
\begin{ex} Let $T=\operatorname{DLO}$. In the big model $\cu M$ of $\operatorname{DLO},$ we have $\operatorname{acl}(A)=\operatorname{dcl}(A)=A$ for every set $A$. Thus in $\cu M$ the lattice
of algebraically closed sets is modular, $\ind[a]=\ind[M] \ $, and $\ind[a]$ is a strict independence relation.
In the big model $\cu N$ of $\operatorname{DLO}^R$, $\ind[a]$ does not satisfy base monotonicity by Proposition \ref{p-a-nobase-monotonicity}, and $\ind[M] \ $ does not satisfy symmetry by
Corollary \ref{c-no-symmetry}. Proposition 4.2.3 of [AGK1] shows that for every finite
set $A\subseteq\cu K$, $\operatorname{dcl}(A)$ is the smallest set $D\supseteq A$ such that whenever $\bo a, \bo b, \bo c, \bo d\in D$, the element of $\cu K$
that agrees with $\bo c$ on $\llbracket \bo a <\bo b\rrbracket$ and agrees with $\bo d$ on $\neg\llbracket \bo a < \bo b\rrbracket$ belongs to $D$.
Let $\bo a\vee\bo b$ and $\bo a\wedge\bo b$ denote the pointwise maximum and minimum, respectively.
We leave it to the reader to work out the following characterizations
of $A\ind[a]_C B$ and $A\ind[M]_C B$ in the simple case that $A, B, C$ are singletons in $\cu N$.
\begin{enumerate}
\item $\bo{a}\ind[a]_\emptyset \bo{b}\Leftrightarrow\bo{a}\not=\bo{b}.$
\item $\operatorname{acl}(\bo a\bo b)=\{\bo a,\bo b,\bo a\vee\bo b,\bo a\wedge\bo b\}$.
\item $\bo{a}\ind[a]_{\bo{c}}\bo{b}\Leftrightarrow
\{\bo{a},\bo{c},\bo{a}\vee\bo{c},\bo{a}\wedge\bo{c}\}\cap \{\bo{b},\bo{c},\bo{b}\vee\bo{c},\bo{b}\wedge\bo{c}\}=\{\bo{c}\}.$
\end{enumerate}
To see where base monotonicity fails for $\ind[a]$, let $\sa E$ be an event with $0<\mu(\sa E)<1$ and take $\bo a, \bo b,\bo c$ so that
$\bo a=\bo b<\bo c$ on $\sa E$ and $\bo c<\bo a<\bo b$ on $\neg\sa E$. Then use (1) and (3) to show that $\bo a\ind[a]_{\emptyset} \bo b$ but
$\bo a\nind[a]_{\bo c} \bo b$.
\begin{itemize}
\item[(4)] If $\bo b\in\{\bo b\vee\bo c,\bo b\wedge\bo c\}$, then $\bo{a}\ind[M]_{\bo{c}}\bo{b}\Leftrightarrow\bo{a}\ind[a]_{\bo{c}}\bo{b}$.
\item[(5)] If $\bo b\notin\{\bo b\vee\bo c,\bo b\wedge\bo c\}$, then $\bo{a}\ind[M]_{\bo{c}}\bo{b}$ if and only if:
$$\bo{a}\ind[a]_{\bo{c}}\bo{b}, \quad \bo{b}\notin \operatorname{dcl}(\{\bo{a},\bo{c},\bo{b}\wedge\bo c\}), \quad \bo{b}\notin \operatorname{dcl}(\{\bo{a},\bo{c},\bo{b}\vee\bo c\}).$$
\end{itemize}
To see where symmetry fails for $\ind[M] \ $, partition $\Omega$ into three events $\{\sa D, \sa E, \sa F\}$ of positive
measure. Take $\bo a, \bo b, \bo c$ so that $\bo a=\bo b<\bo c$ in $\sa D$, $\bo a<\bo c<\bo b$ in $\sa E$, and $\bo c<\bo a<\bo b$ in $\sa F$.
Use (5) to show that $\bo a\ind[M]_{\bo c} \ \bo b$ but $\bo b\nind[M]_{\bo c} \ \bo a.$
\end{ex}
\section{Full Existence and Extension} \label{s-alg-t^R}
By Proposition \ref{p-alg-indep}, $\ind[a]$ in continuous model theory satisfies symmetry and all axioms for a strict countable independence relation except for base monotonicity and extension.
\begin{rmk}
\label{r-stable-extension} If $T$ is stable, then the relation $\ind[a]$ in the theory $T^R$ satisfies full existence and extension.
\end{rmk}
\begin{proof} by Theorem 5.1.4 in [BK], $T^R$ is stable, so it has a unique strict independence relation. This relation satisfies full existence and is stronger than $\ind[a]$.
Then by Remark \ref{r-weaker}, $\ind[a]$ satisfies full existence. By Fact \ref{f-fe-vs-ext},
$\ind[a]$ in $T^R$ satisfies extension.
\end{proof}
Our main result in this section is another sufficient condition for algebraic independence in $T^R$ to satisfy full existence and extension
\begin{thm} \label{t-fe} Suppose $T$ has $\operatorname{acl}=\operatorname{dcl}$. Then the relation $\ind[a]$ in $T^R$ satisfies full existence and extension.
\end{thm}
\begin{proof}
By Fact \ref{f-fe-vs-ext} and Proposition \ref{p-alg-indep}, if $\ind[a]$ over $\cu N$ has full existence, then it has extension.
By Remark \ref{r-acl=dcl}, to prove full existence
we must show that for all small $A,B,C$, there is $A'\equiv_{C} A$ such that
$$[\operatorname{dcl}(A'C)\cap\operatorname{dcl}(BC)=\operatorname{dcl}(C)] \wedge A'\ind[a\BB]_C \ B.$$
In view of Fact \ref{f-definableclosure} and Remark \ref{r-dcl-B}, we may assume without loss of generality that
$C=\operatorname{acl}(C)$, $A=\operatorname{acl}(AC)\setminus \operatorname{acl}(C)$, and $B=\operatorname{acl}(BC)\setminus \operatorname{acl}(C)$.
Then $C=\operatorname{dcl}(C)$, $A=\operatorname{dcl}(AC)\setminus \operatorname{dcl}(C)$, and $B=\operatorname{dcl}(BC)\setminus \operatorname{dcl}(C)$. By Fact \ref{f-event-aB}, the relation
$\ind[a\BB] \ \ $ over $\cu N$ has full existence. Therefore we may also assume that $ A\ind[a\BB]_C \ B.$ By Remark \ref{r-acl=dcl},
$$\operatorname{dcl}_\BB(AC)\cap\operatorname{dcl}_\BB(BC)=\operatorname{dcl}_\BB(C).$$
So it suffices to show that there is $A'\equiv_CA$ such that
$$A'\cap B=\emptyset \wedge \operatorname{dcl}_\BB(A'C)=\operatorname{dcl}_\BB(AC).$$
For each element $\bo a\in A$, we
define $\varepsilon(\bo a)$ as the infimum of all the values $1-\mu(\llbracket a\in\operatorname{dcl}^{\cu M}(D)\rrbracket)$ over all countable $D\subseteq C$.
Note that $\varepsilon(\bo a)=0$ if and only if $\bo a$ is pointwise definable over some countable subset of $C$.
Add a constant symbol for each $\bo a\in A, \bo b\in B$, and $\bo c\in C$. For each $\bo a\in A$, add a variable $\bo a'$.
Consider the set $\Gamma$ of all conditions of the form
$$\llbracket\theta(\vec{\bo a},\vec{\bo c})\rrbracket=\llbracket\theta(\vec{\bo a}',\vec{\bo c})\rrbracket\wedge
\bigwedge_{i\le|\vec{\bo a}|}d_\BK({\bo a}'_i,\bo b)\ge \varepsilon({\bo a}_i)$$
where $\theta$ is an $L$-formula, $\vec{\bo a}\in A^{<\mathbb{N}}, \vec {\bo c}\in C^{<\mathbb{N}}$, and $\bo b\in B$.
\
\emph{Claim 1}. For every finite subset $\Gamma_0$ of $\Gamma$, there is a set $A'=\{{\bo a}'\colon \bo a\in A\}$ that satisfies $\Gamma_0$
in $\cu N_{ABC}$.
\
\emph{Proof of Claim 1}: Let $A_0, B_0, C_0$ be the set of elements of $A, B, C$ respectively that occur in $\Gamma_0$.
Then $A_0, B_0, C_0$ are finite. If $A_0$ is empty, then $\Gamma_0$ is trivially satisfiable in ${\cu N}_{ABC}$,
so we may assume that $A_0$ is non-empty. Let
$$A_0=\{{\bo a}_0,\ldots,{\bo a}_n\},\vec{\bo a}=\<{\bo a}_0,\ldots,{\bo a}_n\>,
C_0=\{{\bo c}_0,\ldots,{\bo c}_k\},\vec{\bo c}=\<{\bo c}_0,\ldots,{\bo c}_k\>.$$
Let $\Theta_0$ be the set of all sentences that occur on the left side of an equation in $\Gamma_0$. Then $\Theta_0$ is finite. By combining tuples,
we may assume that each sentence in $\Theta_0$ has the form $\theta(\vec{\bo a},\vec{\bo c})$.
Since the algebraic independence relation over $\cu M$ satisfies full existence, and $T$ has $\operatorname{acl}=\operatorname{dcl}$, for each $\omega\in\Omega$ there exists
$$G_0(\omega)=\{g_0(\omega),\ldots,g_n(\omega)\}\subseteq M$$
such that
$$\operatorname{tp}^{\cu M}(G_0(\omega)/C_0(\omega))=\operatorname{tp}^{\cu M}(A_0(\omega)/C_0(\omega))$$
and
$$ G_0(\omega) \cap B_0(\omega)\subseteq\operatorname{dcl}^{\cu M}(C_0(\omega)).$$
Let $i\le n$. Whenever $a_i(\omega)\notin\operatorname{dcl}^{\cu M}(C_0(\omega))$, we have $g_i(\omega)\notin\operatorname{dcl}^{\cu M}(C_0(\omega))$, and hence
$g_i(\omega)\notin B_0(\omega)$.
Let $Z=\{\bo 0, \bo 1\}$ be as in Lemma \ref{l-char-function}.
For each $i\le n$ let
$$\sa E_i=\llbracket a_i\in\operatorname{dcl}^{\cu M}(C_0)\rrbracket.$$
By Fact \ref{f-glue}, for each $i$ there exists a unique element $\bo 1_{\sa E_i}\in\cu K$ that agrees with $\bo 1$ on $\sa E_i$
and agrees with $\bo 0$ on $\neg\sa E_i$. By applying Condition (5) in Fact \ref{f-neat} to the formula
$$ \bigwedge_{\theta\in\Theta_0}(\theta(\vec u,\vec{\bo c})\leftrightarrow\theta(\vec{\bo a},\vec{\bo c}))\wedge
\bigwedge_{i=0}^n \bigwedge_{\bo b\in B_0} (1_{{\sa E}_i}=\ti 0\rightarrow u_i\ne\bo b),$$
we see that there exists a set
$$G_0=\{{\bo g}_0,\ldots,{\bo g}_n\}\subseteq \cu K$$
such that for each $\omega\in \Omega$, $\theta(\vec{\bo a},\vec{\bo c})\in\Theta_0$, $i\le n$, and $\bo b\in B_0$:
\begin{itemize}
\item
$\cu M\models \theta(\vec g(\omega),\vec c(\omega))\leftrightarrow\theta(\vec a(\omega),\vec c(\omega));$
\item if $a_i(\omega)\notin\operatorname{dcl}^{\cu M}(C_0(\omega))$, then $g_i(\omega)\ne b(\omega)$.
\end{itemize}
It follows that $\llbracket\theta(\vec{\bo g},\vec {\bo c})\rrbracket=\llbracket\theta(\vec{\bo a},\vec {\bo c})\rrbracket$ for each
$\theta(\vec{\bo a},\vec {\bo c})\in\Theta_0$, and that $d_\BK({\bo g}_i,{\bo b})\ge\varepsilon({\bo a}_i)$ for each $i\le n$ and $\bo b\in B_0$.
Therefore $\Gamma_0$ is satisfied by $G_0$ in $\cu N_{ABC}$, and Claim 1 is proved.
\
By saturation, $\Gamma$ is satisfied in $\cu N_{ABC}$ by some set $A'$. $\Gamma$ guarantees that $A'\equiv_C A$ and $\operatorname{dcl}_\BB(A'C)=\operatorname{dcl}_\BB(AC).$
It remains to show that for each $\bo a\in A$, $\bo a'\notin B$. Let $\bo a\in A$. By hypothesis
we have $\bo a\notin\operatorname{dcl}(C)$.
By Fact \ref{f-dcl3}, either $\bo a$ is not pointwise definable over a countable subset of $C$ and thus $\varepsilon(\bo a)>0$,
or there is a formula $\theta(u,\vec v)$ and a tuple $\vec{\bo c}\in C^{<\mathbb{N}}$ such that
$$\llbracket\theta(\bo a,\vec{\bo c})\rrbracket\in\operatorname{fdcl}_\BB(\{\bo a\}\cup C)\setminus\operatorname{dcl}_\BB(C).$$
$\Gamma$ guarantees that $d_\BK(\bo a',B)\ge\varepsilon(\bo a)$, so in the case that $\varepsilon(\bo a)>0$ we have $\bo a'\notin B$.
$\Gamma$ also guarantees that
$$\llbracket\theta(\bo a',\vec{\bo c})\rrbracket=\llbracket\theta(\bo a,\vec{\bo c})\rrbracket,$$
so in the case that $\varepsilon(\bo a)=0$, we have
$$\llbracket\theta(\bo a',\vec{\bo c})\rrbracket=\llbracket\theta(\bo a,\vec{\bo c})\rrbracket\in\operatorname{dcl}_\BB(AC)\setminus\operatorname{dcl}_\BB(C).$$
But we are assuming that
$$\operatorname{dcl}_\BB(AC)\cap\operatorname{dcl}_\BB(BC)=\operatorname{dcl}_\BB(C),$$
so
$$\llbracket\theta(\bo a',\vec{\bo c})\rrbracket\notin\operatorname{dcl}_\BB(BC),$$
and hence $\bo a'\notin B$. This completes the proof.
\end{proof}
\section{Small Local Character}
In this section we show that if $T$ has $\operatorname{acl}=\operatorname{dcl}$, then algebraic independence in $T^R$ has small local character.
In order to do this, we need the pointwise algebraic independence relation $\ind[a\omega] \ \ ,$ which is of interest in its own right
and will be studied further in the next section.
In the following, $\forall^c D$ means ``for all countable $D$'', and $\exists^c D$ means ``there exists a countable $D$''.
\begin{df} Let $I$ be a ternary relation over $\cu M$ that has monotonicity. The ternary relation $\ind[I\omega] \ \ $ over $\cu N$ (called \emph{pointwise $I$-independence}) is defined as follows. For all small $A,B,C$, $ A\ind[I\omega]_C \ B$ if and only if
$$ (\forall^c A'\subseteq A)(\forall^c B'\subseteq B)(\forall^c C'\subseteq C)(\exists^c D\in[C',C])A'\ind[I\omega]_{D} \ B'.$$
\end{df}
\begin{fact} \label{f-pointwise-a} (Consequence of Lemma 4.1.4 in [AGK2].)
If $I$ be a ternary relation over $\cu M$ that has monotonicity, then
for all countable $A,B,C$,
$$ A\ind[I \omega]_C \ B\Leftrightarrow \mu(\llbracket A\ind[I]_C B\rrbracket)=1.$$
\end{fact}
We recall a definition from [AGK2].
\begin{df} In $T$, $A\ind[c]_C \ B$ (read ``$C$ covers $A$ in $B$''), is the relation that holds
if and only if for every first order formula
$\varphi(\bar x,\bar y,\bar z)\in[L]$ and all tuples $\bar{ a}\in A^{|\bar x|}$, $\bar{ b}\in B^{|\bar y|}$ and $\bar{ c}\in C^{|\bar z|}$,
there exists $\bar{ d}\in C^{|\bar y|}$ such that
$$ \cu M\models\varphi(\bar{ a},\bar{ b},\bar{ c})\Rightarrow\varphi(\bar{ a},\bar{ d},\bar{ c}).$$
\end{df}
\begin{fact} \label{f-covering-omega-small} (Lemma 7.2.4 in [AGK2].)
In $T^R$, the relation $\ind[c\omega] \ \ $ has small local character.
\end{fact}
\begin{lemma} \label{l-ind[c]-implies-ind[a]}
In $T$, $\ind[c]\Rightarrow\ind[a].$
\end{lemma}
\begin{proof} Suppose $A, B, C$ are small and $A\ind[c]_C B$ in $\cu M$. Let $e\in\operatorname{acl}^{\cu M}(AC)\cap\operatorname{acl}^{\cu M}(BC)$. Then there are algebraical formulas
$\varphi(u,\vec x,\vec z), \psi(u,\vec y,\vec w)$ and tuples $\vec a\in A^{<\mathbb{N}}, \vec b\in B^{<\mathbb{N}}, \vec c,\vec {c'}\in C^{<\mathbb{N}}$ such that
$$\cu M\models\varphi(e,\vec a,\vec c)\wedge\psi(e,\vec b,\vec {c'})$$
and
$$(\forall u\in M)[\cu M\models \varphi(u,\vec a,\vec c)\Rightarrow \operatorname{tp}(u/AC)=\operatorname{tp}(e/AC)].$$
Then
$$\cu M\models(\exists u)[\varphi(u,\vec a,\vec c)\wedge\psi(u,\vec b,\vec {c'})].$$
Since $A\ind[c]_C B$, there exists $\vec d\in C^{<\mathbb{N}}$ such that
$$\cu M\models(\exists u)[\varphi(u,\vec a,\vec c)\wedge\psi(u,\vec d,\vec {c'})].$$
Therefore
$$\cu M\models\psi(e,\vec d,\vec {c'}),$$
so $e\in\operatorname{acl}^{\cu M}(C).$
\end{proof}
\begin{prop} \label{p-a-omega-small}
In $T^R$, $\ind[a\omega] \ \ $ has small local character.
\end{prop}
\begin{proof} By Lemma \ref{l-ind[c]-implies-ind[a]}, for all countable $A,B,C\subseteq\cu K$, we have
$$\llbracket A\ind[c]_C B\rrbracket\sqsubseteq\llbracket A\ind[a]_C B\rrbracket.$$
It follows easily that $\ind[c\omega] \ \ \Rightarrow \ind[a\omega] \ \ .$ $\ind[c\omega] \ \ $ has small local character by Fact \ref{f-covering-omega-small},
so by Remark \ref{r-weaker}, $\ind[a\omega] \ \ $ has small local character.
\end{proof}
\begin{prop} \label{p-I-small}
In $T^R$, $\ind[a\omega] \ \wedge \ind[a\BB] \ $ has small local character.
\end{prop}
\begin{proof} By Fact \ref{f-event-dB}, $\ind[d\BB] \ \Rightarrow\ind[a\BB] \ \ ,$ so $\ind[a\omega] \ \wedge \ind[d\BB] \ \Rightarrow\ind[a\omega] \ \wedge \ind[a\BB] \ .$
Then by Remark \ref{r-weaker}, it suffices to show that $\ind[a\omega] \ \wedge \ind[d\BB] \ $ has small local character.
Let $A, B, C_0$ be small subsets of $\cu K$ such that $C_0\subseteq B$ and $|C_0|\le|A|+\aleph_0$. By Fact \ref{f-event-dB}, $\ind[d\BB] \ $ has small local character, so
there is a set $C_1\in[C_0,B]$ such that $|C_1|\le|A|+\aleph_0$ and $A\ind[d\BB]_{C_1} \ B.$ By Proposition \ref{p-a-omega-small},
there is a set $C_2\in[C_1,B]$ such that $|C_2|\le|A|+\aleph_0$ and $A\ind[a\omega]_{C_2} \ B.$ By Fact \ref{f-event-dB}, $\ind[d\BB] \ $ has base monotonicity,
so $A\ind[d\BB]_{C_2} \ B.$ Therefore $\ind[a\omega] \ \wedge \ind[d\BB] \ $ has small local character.
\end{proof}
\begin{prop} \label{p-I-implies-a-omega} The following are equivalent:
\begin{itemize}
\item[(i)] $T$ has $\operatorname{acl}=\operatorname{dcl}$.
\item[(ii)] In $T^R$, $\ind[a\omega] \ \wedge \ind[a\BB] \ \Rightarrow \ind[a].$
\end{itemize}
\end{prop}
\begin{proof}
Suppose that (i) fails. Then in $\cu M$ there is a finite set $C$ and an element $a\in\operatorname{acl}^{\cu M}(C)\setminus\operatorname{dcl}^{\cu M}(C)$.
By Fact \ref{f-neat} and saturation, there is an element $\bo b$ and a finite set $D$ in $\cu K$ such that for each first order formula $\varphi(u,\vec v)$,
if $\cu M\models \varphi(a,C)$ then $\mu(\llbracket\varphi(\bo b,D)\rrbracket)=1$ in $\cu N$. Therefore in $\cu N$ we have
$\bo b\ind[a\omega]_{D} \ \bo b\wedge \bo b\ind[a\BB]_{D} \ \bo b,$ but $\bo b\notin \operatorname{acl}(D).$ Then $\bo b\nind[a]_D \bo b$, so (ii) fails.
Now suppose (i) holds, and assume that $A\ind[a\omega]_C \ B\wedge A\ind[b\BB]_C \ B.$ We prove that $A\ind[a]_C B.$ By Remark \ref{r-acl=dcl},
it suffices to show that $\operatorname{dcl}(AC)\cap\operatorname{dcl}(BC)\subseteq\operatorname{dcl}(C).$ Let $\bo d\in\operatorname{dcl}(AC)\cap\operatorname{dcl}(BC).$ By Fact \ref{f-dcl3},
\begin{equation} \label{eq a}
\bo d\in\operatorname{dcl}^\omega(AC),\quad \bo d\in\operatorname{dcl}^\omega(BC),
\end{equation}
and
$$\operatorname{fdcl}_{\BB}(\bo d AC)\subseteq \operatorname{dcl}_\BB(AC),\quad \operatorname{fdcl}_{\BB}(\bo d BC)\subseteq \operatorname{dcl}_\BB(BC).$$
By Fact \ref{f-separable},
$$\operatorname{dcl}_{\BB}(\bo d AC)\subseteq \operatorname{dcl}_\BB(AC),\quad \operatorname{dcl}_{\BB}(\bo d BC)\subseteq \operatorname{dcl}_\BB(BC).$$
Then
$$\operatorname{dcl}_{\BB}(\bo d C)\subseteq \operatorname{dcl}_\BB(AC)\cap \operatorname{dcl}_\BB(BC).$$
Since $A\ind[a\BB]_C \ B,$ we have
\begin{equation} \label{eq-b}
\operatorname{dcl}_{\BB}(\bo d C)\subseteq \operatorname{dcl}_\BB(C).
\end{equation}
We next show that
\begin{equation} \label{eq-c}
\bo d\in\operatorname{dcl}^\omega(C).
\end{equation}
By Fact \ref{f-dcl3}, it will then follow that $\bo d\in\operatorname{dcl}(C)$, as required.
By (\ref{eq a}),
there are countable sets $A_0\subseteq A, B_0\subseteq B, C_0\subseteq C$ such that
$$\mu(\llbracket \bo d\in\operatorname{dcl}^{\cu M}(A_0 C_0)\rrbracket)=\mu(\llbracket\bo d\in\operatorname{dcl}^{\cu M}(B_0 C_0)\rrbracket)=1.$$
Since $A\ind[a\omega]_C \ B$, there is a countable set $C_1\in[C_0,C]$ such that
$$\mu(\llbracket A_0\ind[a]_{C_1} B_0\rrbracket)=1.$$
Then
$$\mu(\llbracket \operatorname{dcl}^{\cu M}(A_0 C_0)\cap\operatorname{dcl}^{\cu M}(B_0 C_0)\subseteq\operatorname{acl}^{\cu M}(C_1)\rrbracket)=1,$$
so $\mu(\llbracket\bo d\in\operatorname{acl}^{\cu M}(C_1)\rrbracket)=1$, and hence $\bo d\in\operatorname{acl}^\omega(C)$. By (i), $\operatorname{acl}^\omega(C)=\operatorname{dcl}^\omega(C)$, so (\ref{eq-c}) holds.
\end{proof}
\begin{thm} \label{t-small-local-char}
Suppose $T$ has $\operatorname{acl}=\operatorname{dcl}$. Then the relation $\ind[a]$ in $T^R$ has small local character.
\end{thm}
\begin{proof} By Proposition \ref{p-I-small}, $\ind[a\omega] \ \wedge \ind[a\BB] \ \ $ has small local character.
By Remark \ref{r-weaker}, Proposition \ref{p-I-implies-a-omega}, and the hypothesis that $T$ has $\operatorname{acl}=\operatorname{dcl}$,
it follows that $\ind[a]$ in $T^R$ has small local character.
\ \ \end{proof}
Here is a summary of our results about algebraic independence in $T^R$:
For any $T$, algebraic independence in $T^R$ does not satisfy finite character and does not satisfy base monotonicity.
If $T$ has $\operatorname{acl}=\operatorname{dcl}$, then algebraic independence in $T^R$ satisfies all the axioms
for a strict countable independence relation except base monotonicity, and also satisfies finite character and small local character.
\section{Pointwise Algebraic Independence}
In the preceding sections we obtained results about the algebraic independence relation $\ind[a]$ in $T^R$ under the assumption that the underlying first
order theory $T$ has $\operatorname{acl}=\operatorname{dcl}$. In the general case where $T$ is not assumed to have $\operatorname{acl}=\operatorname{dcl}$,
the pointwise algebraic independence relation $\ind[a\omega] \ \ $ may be an attractive alternative to the algebraic independence
relation $\ind[a]$ in $T^R$. In this section we investigate the properties of $\ind[a\omega] \ \ $ in $T^R$ when the underlying
first order theory $T$ is an arbitrary complete theory with models of cardinality $>1$. We first recall some results from [AGK2].
\begin{fact} \label{f-pointwise-axioms} (Special case of Proposition 7.1.4 in [AGK2].)
In $T^R$, $\ind[a\omega] \ \ $ satisfies symmetry and all the axioms for a countable independence relation except perhaps base monotonicity and extension.
Also, if $\ind[a]$ in $T$ has base monotonicity, then so does $\ind[a\omega] \ $ in $T^R$.
\end{fact}
\begin{fact} \label{f-pointwise-small} (Corollary 7.2.5 in [AGK2].) In $T^R$, $\ind[a\omega] \ \ $ has small local character.
\end{fact}
\begin{df} A ternary relation $\ind[I]$ has the \emph{countable union property} if whenever
$A, B, C$ are countable, $C=\bigcup_n C_n$, and $C_n\subseteq C_{n+1}$ and $A\ind[I]_{C_n} B$
for each $n$, we have $A\ind[I]_C B$.
\end{df}
\begin{fact} \label{f-omega-union} (Special case of Proposition 7.1.6 in [AGK2].)
If the relation $\ind[a]$ in $T$ has monotonicity, finite character, and the countable union property, then the relation $\ind[a\omega] \ \ $ in $T^R$
has finite character.
\end{fact}
\begin{thm} \label{t-finite-char}
In $T^R$, the relation $\ind[a\omega] \ \ $ has finite character.
\end{thm}
\begin{proof} It is well-known that $\ind[a]$ in $T$ has monotonicity and finite character. We show that $\ind[a]$ in $T$ has the countable union property.
Suppose $A, B, C$ are countable, $C=\bigcup_n C_n$, and $C_n\subseteq C_{n+1}$ and
$A\ind[I]_{C_n} B$ for each $n$. Let $d\in\operatorname{acl}^{\cu M}(AC)\cap\operatorname{acl}^{\cu M}(BC)$. Then for some $n$ we have $d\in\operatorname{acl}^{\cu M}(AC_n)\cap\operatorname{acl}^{\cu M}(BC_n)$.
Since $A\ind[a]_{C_n} B$, $d\in\operatorname{acl}^{\cu M}(C_n)$, so $d\in\operatorname{acl}^{\cu M}(C)$. Therefore $A\ind[a]_C B$, and hence $\ind[a]$ has the countable union property.
So by Fact \ref{f-omega-union}, $\ind[a\omega] \ \ $ has finite character.
\end{proof}
$\llbracket A\ind[I]_C B\rrbracket\in\cu F$ for all countable $A,B,C\subseteq\cu K$.
\begin{lemma} \label{l-a-measurable} For all countable sets $A, B, C\subseteq\cu K$, the set $\llbracket A\ind[a]_C B\rrbracket$ belongs to $\cu F$, and thus is
measurable in the underlying probability space $(\Omega,\cu F,\mu)$.
\end{lemma}
\begin{proof} Let $\{\varphi_i(u,\vec x)\mid i\in\mathbb{N}\}$, $\{\psi_j(u,\vec y)\mid j\in\mathbb{N}\}$, and $\{\chi_k(u,\vec z)\mid k\in\mathbb{N}\}$ enumerate
all algebraical formulas over the indicated variables. Then the set $\llbracket A\ind[a]_C B\rrbracket$ is equal to
$$ \bigcap_{i\in\mathbb{N}}\bigcap_{\vec a\subseteq AC} \bigcap_{j\in\mathbb{N}}\bigcap_{\vec b\subseteq BC}\bigcup_{k\in\mathbb{N}}\bigcup_{\vec c\subseteq C}
\llbracket \forall u[\varphi_i(u,\vec {\bo a})\wedge\psi_j(u,\vec{\bo b})\Rightarrow \chi_k(u,\vec{\bo c})]\rrbracket.$$
\end{proof}
\begin{thm} \label{t-aa-omega-existence}
The relation $\ind[a\omega] \ \ $ over $\cu N$ satisfies extension and full existence for all countable sets $A, B, \widehat{B}, C$..
\end{thm}
\begin{proof}
We first prove full existence for countable sets. Let $A, B, C$ be countable subsets of $\cu K$. By Fact \ref{f-event-aB}, the relation
$\ind[a\BB] \ \ $ over $\cu N$ has full existence. Therefore we may assume that $ A\ind[a\BB]_C \ B.$ By Fact \ref{f-acl=dcl},
$$\operatorname{dcl}_\BB(AC)\cap\operatorname{dcl}_\BB(BC)=\operatorname{dcl}_\BB(C).$$
Since $\ind[a]$ has full existence in $\cu M$,
for each $\omega\in\Omega$ there exists a set $A'_0\subseteq M$ such that $A'_0\equiv_{C(\omega)} A(\omega)$ and $A'_0\ind[a]_{C(\omega)} \ B(\omega)$
in $\cu M$.
Let $\varphi_i(u,A,C)$, $\psi_i(u,B,C)$, and $\chi_i(u,C)$ be enumerations of all algebraical formulas
over the indicated sets (with repetitions) such that for each pair of algebraical formulas $\varphi(u,A,C)$ and $\psi(u,B,C)$ there
exists an $i$ such that $(\varphi_i,\psi_i)=(\varphi,\psi)$. Whenever $\omega\in\Omega$, $A'_0\subseteq\cu M$, and $A'_0\ind[a]_{C(\omega)} \ B(\omega)$ in $\cu M$,
for each $i\in \mathbb{N}$ there exists $j\in\mathbb{N}$ such that
\begin{equation} \label{eq-ind}
\cu M \models \forall u [\varphi_i(u,A'_0,C(\omega))\wedge\psi_i(u,B(\omega),C(\omega))\rightarrow\chi_j(u,C(\omega))].
\end{equation}
Let $\mathbb{N}^0=\{\emptyset\}$ and $\sa E_\emptyset=\Omega$.
For each $n>0$ and $n$-tuple $s=\langle s(0),\ldots,s(n-1)\rangle$ in $\mathbb{N}^n$, let $\sa E_s$ be the set of all $\omega\in\Omega$ such that
for some set $A'_0\subseteq\cu M$, $A'_0\equiv_{C(\omega)} B(\omega)$ and (\ref{eq-ind}) holds whenever $i<n$ and $j=s(i)$.
Let $L'$ be the signature formed by adding to $L$ the constant symbols
$$\{ k_a, k_b, k_c \ : \ a\in A, b\in B, c\in C\}.$$
For each $\omega\in\Omega$, $(\cu M,A(\omega),B(\omega),C(\omega))$ will be the $L'$-structure where $k_a, k_b, k_c$ are interpreted by
$a(\omega), b(\omega), c(\omega)$. Form $L''$ by adding to $L'$ countably many additional constant symbols $\{k'_a \, : \, a\in A\}$ that will be used for elements of
a countable subset $A'_0$ of $\cu M$.
Then for each $n>0$ and $s\in \mathbb{N}^n$, there is a countable set of sentences
$\Gamma_s$ of $L''$ such that for each $\omega$, $\omega\in \sa E_s$ if and only if $\Gamma_s$ is satisfiable in $(\cu M,A(\omega),B(\omega),C(\omega))$.
Since $\cu M$ is $\aleph_1$-saturated, $\Gamma_s$ is satisfiable if and only if it is finitely satisfiable in $(\cu M,A(\omega),B(\omega),C(\omega))$.
It follows that the set $\sa E_s$ belongs to the $\sigma$-algebra $\cu F$. Moreover, since $\ind[a]$ has full existence in $\cu M$, for each $n$ and $s\in \mathbb{N}^n$ we have
$$\Omega\doteq\bigcup\{\sa E_t \colon t\in \mathbb{N}^n\},\quad \sa E_s \doteq\bigcup\{ \sa E_{sk}\colon k\in\mathbb{N}\},$$
where $sk$ is the $(n+1)$-tuple formed by adding $k$ to the end of $s$. We now cut down the sets $\sa E_s$ to sets $\sa F_s\in\cu F$ such that:
\begin{itemize}
\item[(a)] $\sa F_\emptyset=\Omega$;
\item[(b)] $\sa F_s\subseteq\sa E_s$ whenever $s\in \mathbb{N}^n$;
\item[(c)] $\sa F_s\cap\sa F_t=\emptyset$ whenever $s,t\in \mathbb{N}^n$ and $s\ne t$;
\item[(d)] $\sa F_s\doteq\bigcup\{\sa F_{sk}\colon k\in\mathbb{N}\}$ whenever $s\in\mathbb{N}^n$.
\end{itemize}
This can be done as follows. Assuming $\sa F_s$ has been defined for each $s\in\mathbb{N}^n$. we let
$$ \sa F_{sk}=\sa F_s\cap(\sa E_{sk}\setminus\bigcup_{j<k} \sa F_{sj}).$$
Now let $\theta_i(A,C)$ enumerate all first order sentences with constants for the elements of $AC$. Let $\Sigma$ and $\Delta$ be the following countable sets of sentences of $(L'')^R$:
$$\Sigma=\{\llbracket \theta_i(A',C)\rrbracket\doteq\llbracket\theta_i(A,C)\rrbracket\colon i\in\mathbb{N}\}.$$
$$\Delta=\{\sa F_s\sqsubseteq \llbracket\forall u[\varphi_i(u,A',C))\wedge\psi_i(u, B, C))\rightarrow \chi_{s(i)}(u,C))]\rrbracket\colon s\in\mathbb{N}^{<\mathbb{N}}, i<|s|\}.$$
It follows from Fact \ref{f-neat} (5) and conditions (a)--(d) above that $\Sigma\cup\Delta$ is finitely satisfiable in $\cu N_{ABC}$. Then by saturation, there is a set $A'$
that satisfies $\Sigma\cup\Delta$ in $\cu N_{ABC}$. Since $A'$ satisfies $\Sigma$, we have $A'\equiv_C A$.
The sentences $\Delta$ guarantee that $A'\ind[a\omega]_C \ B$.
By the proof of Fact \ref{f-fe-vs-ext} (1) (see the Appendix of [Ad1]), invariance, monotonicity, transitivity, normality, symmetry,
and full existence for all countable sets implies extension for all countable sets. Then by the preceding paragraphs and Fact \ref{f-pointwise-axioms},
$\ind[a\omega] \ \ $ satisfies extension for all countable sets.
\end{proof}
\begin{question} Does $\ind[a\omega] \ \ $ satisfy extension for countable $A,B,C$ and small $\widehat{B}$?
\end{question}
\begin{question} Does $\ind[a\omega] \ \ $ satisfy full existence and/or extension?
\end{question}
We conclude by showing that the relations $\ind[a]$ and $\ind[a\omega] \ \ $ are incomparable except in trivial cases.
\begin{prop} \label{p-pointwise-algebraic-vs-algebraic}
\noindent\begin{enumerate}
\item[(i)] $\ind[a\omega] \ \ $ is not anti-reflexive.
\item[(ii)] $\ind[a\omega] \ \Rightarrow \ind[a]$ always fails in $\cu N$.
\item[(iii)] $\ind[a]\Rightarrow\ind[a\omega] \ \ $ holds in $\cu N$ if and only if the models of $T$ are finite.
\end{enumerate}
\end{prop}
\begin{proof} (i) and (ii) Let $Z=\{\bo 0, \bo 1\}$ be as in Lemma \ref{l-char-function}. Let $\sa D$ be a set in $\cu F$
of measure $\mu(\sa D)=1/2$, and let $\bo 1_{\sa D}$ agree with $\bo 1$ on $\sa D$ and agree with $\bo 0$ on $\neg \sa D$.
Then $\bo 1_{\sa D}\ind[a\omega]_Z \ \bo 1_{\sa D}$, but $\bo 1_{\sa D}\notin\operatorname{acl}(Z).$ Therefore $\bo 1_{\sa D}\nind[a]_Z \bo 1_{\sa D}$
and $\ind[a]$ is not anti-reflexive.
(iii) If $\cu M$ is finite, then $\operatorname{acl}^{\cu M}(\emptyset)=M$, so $A\ind[a]_C B$ always holds in $\cu M$. Therefore $A\ind[a\omega]_C \ B \ $ always holds in $\cu N$,
and hence $\ind[a]\Rightarrow\ind[a\omega] \ \ $ holds in $\cu N$.
For the other direction, assume $\cu M$ is infinite. By saturation of $\cu M$, there exist elements $0,1,a,b\in M$ such that
$$ 0\ne 1,\quad a\notin\operatorname{acl}^{\cu M}(01), \quad\operatorname{tp}(a/\operatorname{acl}^{\cu M}(01))=\operatorname{tp}(b/\operatorname{acl}^{\cu M}(01)), \quad a\ind[a]_{01} \, b.$$
By a routine transfinite induction using Fact \ref{f-neat} and the saturation of $\cu N$, there is a mapping $a\mapsto\ti a$ from $M$ into $\cu K$ such that
for each tuple $a_0,a_1,\ldots$ in $M$ and formula $\varphi(v_0,v_1,\ldots)$ of $L$, if $\cu M\models \varphi(a_0,a_1,\ldots)$ then
$\mu(\llbracket\varphi(\ti a_0,\ti a_1,\ldots)\rrbracket)=1$ in $\cu N$. Let $\ti M=\{\ti a\mid a\in M\}.$
To simplify notation, suppose first that $T$ already has a constant symbol for each element of $\operatorname{acl}(01)$. Then
$\operatorname{acl}^{\cu M}(01)=\operatorname{acl}^{\cu M}(\emptyset)$, so
$$ 0\ne 1, \quad a\notin\operatorname{acl}^{\cu M}(\emptyset),\quad\operatorname{tp}(a)=\operatorname{tp}(b), \quad a\ind[a]_{\emptyset} \, b \quad\mbox{ in } \cu M,$$
$$ \mu(\llbracket {\ti 0}\ne {\ti 1}\rrbracket)=1,\quad {\ti a}\notin\operatorname{dcl}(\emptyset),\quad\operatorname{tp}({\ti a})=\operatorname{tp}({\ti b}), \quad {\ti a}\ind[a]_{\emptyset} \, {\ti b} \quad\mbox{ in } \cu N.$$
By Results \ref{f-acl=dcl} and \ref{f-separable}, for each $A\subseteq \ti M$,
$$\operatorname{acl}(A)=\operatorname{dcl}(A)=\operatorname{cl}(\operatorname{fdcl}(A))=\operatorname{fdcl}(A)\subseteq \ti M.$$
Let $\sa E\in\cu E$ be an event of measure $\mu(\sa E)=1/2$.
Let $ \bo c$ agree with $\ti 1$ on $\sa E$ and $\ti 0$ on $\neg\sa E.$ Let $\bo d$ agree with $\ti a$ on $\neg \sa E$ and with $\ti b$ on $\sa E$
(see the figure).
\begin{center}
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\put(30,0){\line(0,1){20}}
\put(50,0){\line(0,1){20}}
\put(70,0){\line(0,1){20}}
\put(10,0){\line(1,1){20}}
\put(50,0){\line(1,1){20}}
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\put(9,10){\makebox(0,0){$\ti 0$}}
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\put(49,10){\makebox(0,0){$\ti a$}}
\put(69,10){\makebox(0,0){$\ti b$}}
\put(18,10){\makebox(0,0){$\bo c$}}
\put(58,10){\makebox(0,0){$\bo d$}}
\end{picture}
\end{center}
\emph{Claim 1}: $\ti a\ind[a]_{\emptyset} \bo{cd} $ in $\cu N$.
\emph{Proof of Claim 1}: Suppose $ x\in\operatorname{acl}({\ti a})\cap\operatorname{acl}(\bo{cd})$ in $\cu N$. Then $x\in \operatorname{dcl}(\ti a)$,
so $x=\ti z$ for some $z\in\operatorname{dcl}^{\cu M}(a)$.
Moreover, $x\in\operatorname{dcl}(\bo c\bo d)$, so $x\in\operatorname{dcl}^\omega(\bo c\bo d)$, and hence $x(\omega)\in\operatorname{dcl}^{\cu M}(1b)=\operatorname{dcl}^{\cu M}(b)$ for all $\omega\in\sa E$.
Therefore $z\in\operatorname{dcl}^{\cu M}(b)$. Since ${\ti a}\ind[a]_{\emptyset} \, {\ti b}$ in $\cu N$, we have
$x\in\operatorname{acl}(\ti a)\cap\operatorname{acl}(\ti b)=\operatorname{acl}(\emptyset)$.
\medskip
\emph{Claim 2}: $\ti a\nind[a\omega]_{\emptyset} \ \bo{cd} $ in $\cu N$.
\emph{Proof of Claim 2}:
For all $\omega\in\neg\sa E$ we have $\bo d(\omega)=\ti a(\omega)$, so
$$\ti a(\omega)\in\operatorname{acl}^{\cu M}(\ti a(\omega))\cap \operatorname{acl}^{\cu M}(\bo{cd}(\omega))\setminus \operatorname{acl}^{\cu M}(\emptyset),$$
and hence $\omega\notin\llbracket \ti a\ind[a]_\emptyset \bo c\bo d\rrbracket$.
Therefore $\mu(\llbracket \ti a\ind[a]_\emptyset\bo c\bo d\rrbracket)\le 1/2$, so $\ti a\nind[a\omega]_{\emptyset} \ \bo{cd} $.
By Claims 1 and 2, $\ind[a]\Rightarrow\ind[a\omega] \ \ $ fails in $\cu N$.
We now turn to the general case where $T$ need not have a constant symbol for each element of $\operatorname{acl}(01)$.
Our argument above shows that $\ti a\ind[a]_{\ti 0 \ti 1} \bo{cd}$ but $\ti a\nind[a\omega]_{\ti 0 \ti 1} \ \bo{cd}$ in $\cu N$, so
$\ind[a]\Rightarrow\ind[a\omega] \ \ $ still fails in $\cu N$.
\end{proof}
\section*{References}
\vspace{2mm}
[Ad1] Hans Adler. Explanation of Independence. PH. D. Thesis, Freiburg, AxXiv:0511616 (2005).
[Ad2] Hans Adler. A Geometric Introduction to Forking and Thorn-forking. J. Math. Logic 9 (2009), 1-21.
[AGK1] Uri Andrews, Isaac Goldbring, and H. Jerome Keisler. Definable Closure in Randomizations.
To appear, Annals of Pure and Applied Logic. Available online at www.math.wisc.edu/$\sim$Keisler.
[AGK2] Uri Andrews, Isaac Goldbring, and H. Jerome Keisler. Randomizing o-minimal Theories. Submitted. Available online at www.math.wisc.edu/$\sim$Keisler.
[Be] Ita\"i{} Ben Yaacov. On Theories of Random Variables. Israel J. Math 194 (2013), 957-1012.
[BBHU] Ita\"i{} Ben Yaacov, Alexander Berenstein,
C. Ward Henson and Alexander Usvyatsov. Model Theory for Metric Structures. In Model Theory with Applications to Algebra and Analysis, vol. 2,
London Math. Society Lecture Note Series, vol. 350 (2008), 315-427.
[BK] Ita\"i{} Ben Yaacov and H. Jerome Keisler. Randomizations of Models as Metric Structures.
Confluentes Mathematici 1 (2009), pp. 197-223.
[BU] Ita\"i{} Ben Yaacov and Alexander Usvyatsov. Continuous first order logic and local stability. Transactions of the American
Mathematical Society 362 (2010), no. 10, 5213-5259.
[CK] C.C.Chang and H. Jerome Keisler. Model Theory. Dover 2012.
[EG] Clifton Ealy and Isaac Goldbring. Thorn-Forking in Continuous Logic. Journal of Symbolic Logic 77 (2012), 63-93.
[GL] Rami Grossberg and Olivier Lessman. Dependence Relation in Pregeometries. Algebra Universalis 44 (2000), 199-216.
[Ke] H. Jerome Keisler. Randomizing a Model. Advances in Math 143 (1999), 124-158.
\end{document}
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The Abu Dhabi Fund for Development (ADFD) today formalized a Memorandum of Understanding (MoU) with the Government of the Comoros, which stipulates that the ADFD will offer AED 184 million (US$50 million) to fund the Comorian government's various development programs and projects.
Comoros President Azali Assoumani attended the signing of the MoU, which was signed by Mohammed Saif Al Suwaidi, Director-General of the ADFD and H.E. Mr. Mohamed El Amine Souef, Comoros Minister of Foreign Affairs, International Cooperation and the Francophonie, in charge of Comorians abroad. Government officials and senior representatives of the two entities also attended the signing ceremony.
The signing ceremony took place on the sidelines of the UAE delegation visit to Comoros' Capital city Moroni, to discuss ways to promote bilateral cooperation between the two countries, in all areas related to economic, trade, investment and development.
Mohammed Saif Al Suwaidi, Director-General of the ADFD, said that the Fund's top priority is to provide development funding to allow Comoros to achieve sustainable growth and has already disbursed significant grants to help the Comorian government to overcome economic and developmental challenges in key sectors such as health, education, transport, water and electricity.
H.E. Mr. Mohamed El Amine Souef expressed his appreciation for the UAE's ongoing support for the Comoros, stressing that the grant will enable the government to further drive economic and social growth in the country.
The Abu Dhabi Fund for Development (ADFD) activities in the Comoros date back to 1979, where the Fund has financed several development projects to the tune of about Dh253 million in strategic sectors. | {
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namespace $.$$ {
export class $mol_app_inventory_enter extends $.$mol_app_inventory_enter {
event_submit() {
this.domain().credentials({
login : this.login() ,
password : this.password() ,
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Last summer, the Literature group went outdoors for three events in the open air – one in Hyde Gate, one in Hyde Abbey Garden and one on the move, walking around Winchester. One of the group's regulars, Kevin Barrett, has written two poems inspired by these events. He read them out for the first time at December's 'Third Thursday literature night in the Hyde Tavern. He's kindly given us permission to reproduce them here.
We hear again the widow's cry.
K.J. Barrett. 17th December. 2012. | {
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Government needs more input from veterans on pension program
Canadian Veterans Support Forum :: Miscellaneous ll
by SniperGod on Fri 15 Mar 2019, 11:09 am
War pensioners rep. says government needs more input from veterans on pension program
By Tom Sasvari -March 15, 2019
SUDBURY—The chief executive officer of the Manitoulin-North Shore War Pensioners of Canada (WPC) is perplexed as to why the federal government didn't garner more input from veterans before making changes in laws to enact its new Pension for Life program.
"I shake my head as to why the government of Canada will not pay more attention to the input of veterans before making changes in law to enact a new Pension for Life Program, which will come into effect on April 1, 2019, April Fool's Day," stated Colin Pick.
"As stated in a Globe and Mail article (January 2, 2019), it's a systematic attack upon veterans' rights to be denied the chance to participate in the very democracy that they were willing to die to defend," said Mr. Pick, quoting Sean Bruyea, a veterans' rights advocate. Mr. Bruyea is in the process of suing then-Veterans Affairs Minister Seamus O'Regan for defamation of character, as a result of comments made in response to his criticism of the Pension for Life Program, explained Mr. Pick.
"The problems stem from government actions portraying the same old story that whatever the government does for the veterans, it turns out to be more about saving the government money on the backs of the veterans, instead of genuinely improving the financial lot for Canada's most seriously disabled veterans. Once again, the new proposals fall far short of ensuring financial stability, equality or some semblance of fairness between different generations of veterans for the remainder of their lives," continued Mr. Pick.
He said, "in the words of retired Major Mark Campbell, who lost both legs in Afghanistan, this validates and vindicates everything we are trying to do. Meaning, trying to bring back the previous disability pension afforded to veterans prior to 2006, under the old pension act, then when the problematic components of the New Veterans Charter came into effect, this was another cost savings measure on the backs of the veterans."
The Globe and Mail reported on February 21, 2019, that most disabled veterans will receive more financial support under the federal government's new pension program, though some of the most severely impaired will lose out on hundreds of thousands of dollars, according to a new report.
The parliamentary budget officer (PBO) study also concluded that the benefits promised by the soon-to-be-implemented Pension for Life program will be less than what veterans would have reaped in a previous disability pension program scrapped in 2006.
In its report released Thursday, the PBO found that all current recipients of disability benefits will receive an "equal or greater amount" of funding in the Liberals' Pension for Life regime, set to go into effect on April 1.
But it warned that three percent of new veteran entrants would have received around $300,000 more in financial support from the existing program, in part because of the elimination of a supplement for "veterans with severe and permanent impairment and diminished earning capacities."
Known as the Career Impact Allowance Supplement, it offered more than $21,000 to the most severely affected veterans in 2017. While current recipients can continue to access the supplement, it will not be available to new applicants, the PBO study notes.
Relying on data from Veterans' Affairs, the PBO predicted the costs to government for the existing cohort of beneficiaries and projected new applicants over 2019 and 2023. It found that the pre-2006 program, instituted in the Pension Act, would be the most generous for the veterans and also most expensive for the federal government, projecting Ottawa would need to cough up $40 million to cover those costs.
That program offered regular pension payments for disabled veterans while the New Veterans Charter in 2006 largely replaced those with lump-sum payments that topped out at $365,400, according to reporting by the Globe and Mail. Critics argued that this was far less than what veterans would have received in the pre-2006 program.
The PBO determined that the Liberals' new package was "slightly more generous" than the Veteran Well-Being Act regime introduced in 2006. According to the PBO, the government would have to pay $22 billion under the existing system in veterans pensions and $25 billion under the new Pension for Life program.
"The bottom line being that the pre-2006 old pension act paid out 1.5 times more over the lifetime of a veteran some 13 years ago than it does now," said Mr. Pick. "Now as we approach the third attempt at a Pension for Life overhaul, this government still cannot get it right because when these changes come into place on April 1, 2019, some of the seriously disabled veterans are going to get less than before because the government has dropped the supplementary benefit designed to compensate for those seriously disabled veterans who are no longer able to work because of their injuries. This is nothing short of another kick in the teeth for veterans, which is absolutely wrong on moral grounds and equality for all veterans' pensions."
https://www.manitoulin.ca/war-pensioners-rep-says-government-needs-more-input-from-veterans-on-pension-program/
SniperGod
CF Coordinator
Re: Government needs more input from veterans on pension program
by Trooper on Fri 15 Mar 2019, 6:42 pm
The government did not garner info from Veterans on the Pension For Life because the government did not create the Pension For Life. Walt Natynczyk and his team of bureaucrats created the Pension For Life. Hehr, O'Regan, and now MacAulay have the exact same knowledge as the general public has on the Pension For Life. As soon as the Ministers are appointed Walt gives them the breakdown as Walt see's it. The Ministers do not have the smarts nor the knowledge to question what Walt creates. This is why we end up with a broken promise Pension For Life. The Ministers will say they look forward to hearing from Veterans, but the fact is this is just for show, and public relations. You will notice when something shows up in the media for example a Veterans being denied access to Camp Hill the Ministers run and hide. The Ministers are suppose to stand for disabled Veterans in this Country. I have yet to witness this. Instead what we witness is Ministers taken orders from Walt Natynczyk!!! This is fact of the way in which the Veterans file in Canada is run. The Veterans file is run by Walt Natynczyk, and his bureaucratic team. The Ministers, and quite frankly the government is useless at handling the Veterans file. My advice for those reading this is to look at the history behind the Veterans file, in particular since 2006. Those who understand the pre 2006 system understands exactly where I'm coming from with my insights. The government is no friend to it's disabled Veterans, and this goes for all political parties in Canada. The government does not have enough knowledge or even the initiative to run the Veterans file. It is to complex for them so they turn to the deputy Minister to run the file in full. So in my advice I would caution anyone especially disabled Veterans in this Country against anything, including promises by any political party to be authentic!!!
I'm sorry Mr. Colin Pick but unfortunately you will continue to shake your head moving forward regarding the incompetence in our government towards their handling of the Veterans file.
Sean Bruyea has an impeccable understanding, and knowledge with the Veterans file. He scares Ministers of Veterans affairs, and the bureaucrats that create such mess, and chaos within our file. We are lucky to have Sean advocating for us not only because of his competence, also because he is old system which shows his good character fighting for those post 2006 Veterans. I hold a lot of respect for Sean Bruyea he truly cares for his fellow Veterans!!!
In closing I would say that our government is allowing the bureaucrats to run the show, a show that shows the Veterans file getting worse step by step. Those post 2006 disabled Veterans don't come close to what pre 2006 Veterans are getting. We have both the Conservatives, and the Liberals to blame for that. And this is a fact that is incontestable.
» Input-box says in "username"
» BOER WAR TRIBUTE - CITY OF OTTAWA
» Improvements needed in compensation for veterans, says Ombudsman
» McCanns will not take legal action against Portuguese Government
» MILITARY PENSION FOR LIFE | {
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I begged them to kill me, says Xinjiang detainee
Mihrigul Tursun says she was put in a 40-square-foot cell with more than 40 other women. Photo: RTHK
Maggie Ho reports
Even as Beijing continues to reject claims of torture and illegal detention in camps across Xinjiang, three former detainees have told RTHK that they were subjected to various mental and physical abuse in such places.
On Monday, news reports had circulated quoting Xinjiang officials as saying the camps were set up to re-educate people and clamp down on terrorism and extremism.
A number of foreign media representatives were taken around some of the camps to see for themselves these claims.
But the former detainees tell a different story.
"They beat me up badly. I begged them to kill me. They tortured us, and some people died," said 30-year-old Mihrigul Tursun.
In the summer of 2015, the Uighur, who is married to an Egyptian, brought her newborn triplets back to her hometown in Xinjian. She was taken away by mainland officials once she landed at the airport, accused of conducting subversive activities.
Tursun said inmates had to "make confessions" every day before they were given meals, which they had to eat while squatting.
They were also forced to sing Communist songs and recite passages to praise the party "long live Xi Jinping", she said.
She remembers being put in a 40-square-foot cell with more than 40 other women locked up for various reasons, such as keeping a Koran at home, attending an Islamic wedding, or wearing a face veil in public.
Among them was a village girl who was locked up for listening to an Arabic song on her phone, said Tursun.
"She told me she heard this Arabic song, she didn't understand the language but it was nice to listen to. She said that's why they detained her. She had just finished primary school, what could she have done?"
Tursun said she was released 13 months later, after the intervention of the Egyptian Embassy.
Another former camp inmate, Omir Bekali, grew up in Xinjiang but had become a citizen of Kazakhstan. He returned to China to visit his relatives last year and was locked up for allegedly threatening national security.
He also rejected the authorities' claims that people are in the camp for vocational training and Chinese language classes.
"There were doctors, post graduates, teachers, and lawyers. Why did they have to be educated? They were university graduates. The authorities are covering up the truth, they are lying to the world. And they have the audacity to say they're doing it for human rights."
He said people who disobeyed instructions were punished. Some were not given food or water for 24 hours, and others were forced to stand outside naked, in extreme weather.
The 43-year-old said his father died in the camp, and his mother, siblings, as well as his wife's relatives are all still incarcerated. He was freed after eight months, with the assistance of the embassy of Kazakhstan.
Jalilova Gulbakhar, born and raised in Kazakhstan, had been doing business in China for two decades when she was locked up in May 2017, for transferring 17,000 yuan to a Turkish company. Officials charged her with financing terrorists.
She said she wrote down the name of more than 200 people she lived with in the camps, most of them Uighurs, adding she believes they are all innocent.
"I have to speak up for them. I have to tell the world that China is growing strong, but in Xinjiang there's a dark side to it," she said. "I can't sleep thinking of the girls every night. I won't stop what I'm doing even if it will cost me my life".
She too, was only released after the intervention of the Kazakhstan Embassy.
All three told RTHK that they have no animosity towards Han Chinese people and it is the government's policies that caused their suffering.
More on this in Hong Kong Today
E-mail: enews@rthk.hk | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,485 |
{"url":"https:\/\/math.stackexchange.com\/questions\/4107427\/convexity-and-solution-of-a-quadratic-program","text":"# Convexity and solution of a quadratic program\n\n$$\\begin{array}{ll} \\underset{x_1,x_2,x_3}{\\text{minimize}} & x_1 x_2 +\\frac{1}{2} x_1^2 + \\frac32 x_2^2 + 2 x_3^2 + 2 x_1 + x_2 + 3x_3\\\\ \\text{subject to} & x_1 + x_2 + x_3 = 1\\\\ & x_1 - x_2=0\\\\ & x_1, x_2, x_3 \\geq 0\\end{array}$$\n\nFirst, I want to check if the objective function is convex. I did this by finding the eigenvalues and they were all positive so this is positive definite and is strictly convex.\n\nNow, I am trying to show that $$x^*=(\\frac12, \\frac12, 0)$$ is an optimal solution to this problem by finding vectors $$y$$ and $$s$$ that satisfy the optimality conditions jointly with $$x^*$$. I think this should be done via Primal-Dual Interior-Point Method but I am not much familiar with this approach and am pretty confused.\n\n\u2022 What have you tried? Apr 18 at 20:13\n\u2022 Do you mean $x_1x_2+ x_1^2+ (3\/2x_2^2)+ 2x_2^2+ 2x_3^2+ 2x_1+ x_2+3x_1$ or $(x_1x_2+ x_1^2+ 3)\/(2x_2^2+ 2x_2^2+ 2x_3^2+ 2x_1+ x_2+3x_1)$? And should that last \"$3x_1$\" be \"$3x_3$\"? If not, $2x_1+ 3x_1= 5x_1$! Apr 18 at 20:30\n\u2022 @IgorRivin I don't know how to approach it tbh. Apr 18 at 23:22\n\u2022 Yes, I just edited it, thanks for letting me know. @user247327 Apr 18 at 23:23\n\nWell, first, from $$x_1- x_2= 0$$, $$x_1= x_2$$ of course so we can write the expression to be minimized as $$x_1^2+ x_1^2+ \\frac{3}{2}x_1^2+ 2x_3^2+ 2x_1+ x_1+ 3x_3= \\frac{7}{2}x_1^2+ 3x_1+ 3x_3$$\n\nAnd from $$x_1+ x_2+ x_3= 2x_1+ x_3= 1$$, $$x_3= 1- 2x_1$$\n\nSo we can write the expression as a function, $$\\frac{7}{2}x_1^2+ 3x_1+ 3- 6x_1= \\frac{7}{2}x_1^2- 3x_1+ 3$$, of the single variable $$x_1$$!\n\nThat will have an extremum where $$7x_1- 3= 0$$ so at $$x_1= \\frac{3}{7}$$, $$x_2= x_1= \\frac{3}{7}$$, and $$x_3= 1- 2x_1= 1- \\frac{6}{7}= \\frac{1}{7}?$$.\n\nThe second derivative is 7, positive, so that is a minimum. (We could also have observed that the function in $$x_1$$ is a parabola opening upward.)\n\nLet $$f$$ be the cost function and note that $${\\partial f(x^*) \\over \\partial x} = (3,3,3)$$. Let $$g(x) = x_1+x_2+x_3-1$$ and note that $$g(x^*) = 0$$ and $${\\partial f(x^*) \\over \\partial x} + \\lambda {\\partial g(x^*) \\over \\partial x} = 0$$ with $$\\lambda = -1$$. Hence we see that $$x^*$$ solves the convex problem $$\\min \\{ f(x) | g(x) = 0 \\}$$ and hence is a solution for the more restrictive problem in the question.","date":"2021-09-26 23:18:22","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\": 25, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9063001871109009, \"perplexity\": 115.19975676833619}, \"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\/1631780057973.90\/warc\/CC-MAIN-20210926205414-20210926235414-00498.warc.gz\"}"} | null | null |
Produced by David Edwards, Emmy and the Online Distributed
Proofreading Team at http://www.pgdp.net (This file was
produced from images generously made available by The
Internet Archive)
Transcriber's Note: Symbols are used in the text to indicate =bold= and
_italic_ text.
THE BOY SCOUTS OF BOB'S HILL
[Illustration: BE PREPARED]
CHARLES PIERCE BURTON
OFFICERS OF THE NATIONAL COUNCIL
Honorary President, THE HON. WOODROW WILSON
Honorary Vice-President, HON. WILLIAM H. TAFT
Honorary Vice-President, COLONEL THEODORE ROOSEVELT
President, COLIN H. LIVINGSTONE, Washington, D. C.
Vice-President, B. L. DULANEY, Bristol, Tenn.
Vice-President, MILTON A. McRAE, Detroit. Mich.
Vice-President, DAVID STARR JORDAN, Stanford University, Cal.
Vice-President, F. L. SEELY, Asheville, N. C.
Vice-President, A. STAMFORD WHITE, Chicago, Ill.
Chief Scout, ERNEST THOMPSON SETON, Greenwich, Connecticut
National Scout Commissioner, DANIEL CARTER BEARD, Flushing, N. Y.
NATIONAL HEADQUARTERS
BOY SCOUTS OF AMERICA
THE FIFTH AVENUE BUILDING, 200 FIFTH AVENUE
TELEPHONE GRAMERCY 545
NEW YORK CITY
FINANCE COMMITTEE
John Sherman Hoyt, Chairman
August Belmont
George D. Pratt
Mortimer L. Schiff
H. Rogers Winthrop
GEORGE D. PRATT, Treasurer
JAMES E. WEST, Chief Scout Executive
ADDITIONAL MEMBERS OF THE EXECUTIVE BOARD
Ernest P. Bicknell
Robert Garrett
Lee F. Hanmer
John Sherman Hoyt
Charles C. Jackson
Prof. Jeremiah W. Jenks
William D. Murray
Dr. Charles P. Neill
George D. Porter
Frank Presbrey
Edgar M. Robinson
Mortimer L. Schiff
Lorillard Spencer
Seth Sprague Terry
July 31st, 1913.
TO THE PUBLIC:--
In the elecution of its purpose to give educational value and moral
worth to the recreational activities of the boyhood of America, the
leaders of the Boy Scout Movement quickly learned that to effectively
carry out its program, the boy must be influenced not only in his
out-of-door life but also in the diversions of his other leisure
moments. It is at such times that the boy is captured by the tales of
daring enterprises and adventurous good times. What now is needful is
not that his taste should be thwarted but trained. There should
constantly be presented to him the books the boy likes best, yet always
the books that will be best for the boy. As a matter of fact, however,
the boy's taste is being constantly vitiated and exploited by the great
mass of cheap juvenile literature.
To help anxiously concerned parents and educators to meet this grave
peril, the Library Commission of the Boy Scouts of America has been
organized. EVERY BOY'S LIBRARY is the result of their labors. All the
books chosen have been approved by them. The Commission is composed of
the following members: George F. Bowerman, Librarian, Public Library of
the District of Columbia, Washington, D. C.; Harrison F. Graver,
Librarian, Carnegie Library of Pittsburgh, Pa.; Claude G. Leland,
Superintendent, Bureau of Libraries, Board of Education, New York City;
Edward F. Stevens Librarian, Pratt Institute Free Library, Brooklyn, New
York; together with the Editorial Board of our Movement, William D.
Murray, George D. Pratt and Frank Presbrey, with Franklin K. Mathiews,
Chief Scout Librarian, as Secretary.
In selecting the books, the Commission has chosen only such as are of
interest to boys, the first twenty-five being either works of fiction or
stirring stories of adventurous experiences. In later lists, books of a
more serious sort will be included. It is hoped that as many as
twenty-five may be added to the Library each year.
Thanks are due the several publishers who have helped to inaugurate this
new department of our work. Without their co-operation in making
available for popular priced editions some of the best books ever
published for boys, the promotion of EVERY BOY'S LIBRARY would have been
impossible.
We wish, too, to express our heartiest gratitude to the Library
Commission, who, without compensation, have placed their vast experience
and immense resources at the service of our Movement.
The Commission invites suggestions as to future books to be included in
the Library. Librarians, teachers, parents, and all others interested in
welfare work for boys, can render a unique service by forwarding to
National Headquarters lists of such books as in their Judgment would be
suitable for EVERY BOY'S LIBRARY.
Signed
James E. West
Chief Scout Executive.
"DO A GOOD TURN DAILY."
[Illustration: "I HAVE LOST THE CAMP. HELP!"--_Page 132._]
EVERY BOY'S LIBRARY--BOY SCOUT EDITION
THE BOY SCOUTS OF BOB'S HILL
_A Sequel to "The Bob's Hill Braves"_
BY
CHARLES PIERCE BURTON
AUTHOR OF
THE BOYS OF BOB'S HILL, THE BOB'S CAVE BOYS,
AND THE BOB'S HILL BRAVES
WITH ILLUSTRATIONS BY
GORDON GRANT
[Illustration]
NEW YORK
GROSSET & DUNLAP
PUBLISHERS
COPYRIGHT, 1912,
BY
HENRY HOLT AND COMPANY
_Published October, 1912_
=To=
THE RAVENS,
Patrol 1, Troop 3, of Aurora, Illinois,
BOY SCOUTS OF AMERICA
CONTENTS
CHAPTER PAGE
I. "THE BAND" AND THE CAVE 1
II. RAVEN PATROL HITS THE TRAIL 20
III. TRACKING THE ROBBERS 37
IV. "DANGER--COME" 53
V. A CAMPFIRE ON BOB'S HILL 67
VI. A FOURTEEN-MILE HIKE 82
VII. "BILL HASN'T COME BACK" 102
VIII. SMOKE SIGNALS ON THE MOUNTAIN 120
IX. FOUND AT LAST 135
X. A MAIDEN IN DISTRESS 146
XI. TREED BY A BEAR 162
XII. WHAT HAPPENED TO THE BEAR 174
XIII. EAGLE PATROL JOINS THE SCOUTS 191
XIV. PLANNING A CAMPING TRIP 206
XV. SCOUTING IN THE GREAT NORTHWEST 219
XVI. CLOUDBURST ON GREYLOCK 233
XVII. ON THE WAY AT LAST 246
XVIII. SCOUTING THROUGH A WILDERNESS 262
XIX. ON HISTORIC GROUND 278
XX. SCOUTS TO THE RESCUE 295
ILLUSTRATIONS
PAGE
"I HAVE LOST THE CAMP. HELP!" _Frontispiece_
WITH SKINNY LEADING, WE STARTED, DODGING FROM
TREE TO TREE 13
"IT GIVES ME PAIN," SHE SAID, "TO INFORM YOU
THAT THE WOODBOX IS EMPTY" 206
AS WE RAN, WE HEARD A YELL OF PAIN, OR FRIGHT,
AND IT WAS NOT A BEAR'S VOICE AT ALL 261
THE BOY SCOUTS OF BOB'S HILL
CHAPTER I
"THE BAND" AND THE CAVE
BLACKINTON'S barn is exactly at the foot of Bob's Hill. Phillips's is,
too, and so is our garden; but I am not telling about those now. Beyond
the barns are apple orchards, reaching halfway up the hill, as you know,
if you have read about the doings of the Band.
When they built Blackinton's barn they cut into the hill, so that the
roof of the stable <DW72>s clear down to the ground, on the hill side in
the orchard. It makes a fine place for us boys to sit and talk about
things.
Mrs. Blackinton, who owns the barn, says that maybe climbing around on a
roof isn't the best thing in the world for shingles but boys have got to
do something and she is willing to take a chance; only to be as careful
as we can, and not to eat any more apples than are necessary to our
happiness and well being.
Anyhow, seven of us Bob's Hill boys sat there one Saturday afternoon in
May, planning what to do in the long vacation. Every member of the Band
was there, not counting Tom Chapin, except Skinny Miller; and we were
expecting him every minute.
He was late then, and every little while one of us would stick his head
around the edge of the barn to see if he wasn't coming up the driveway
from Park Street. We might as well have sat still, for you never can
tell which way he will come.
Pa says that Skinny is like the wind, which bloweth whither it listeth.
I don't exactly know what he meant but that is what he said, or
something like that.
It was quiet in the orchard. There was hardly a sound except the buzzing
of insects in the sunshine, and somehow that only seemed to make it more
quiet and dreamy.
Suddenly Bill Wilson stood up on the sloping shingles and gave such a
warwhoop that it almost made the bark rattle on the trees. When Bill
turns his voice loose it is something awful.
We looked up to see what it all was about. He had grabbed Benny Wade by
the hair and, giving another yell louder than the first, was pretending
to scalp him. Bill always likes to play Indian.
Benny didn't want to be scalped. Although he is two years younger and
not nearly so big, he grabbed Bill around the legs and held on until
they both slipped and went tumbling down the steep roof to the ground,
where they sat, with the rest of us laughing down at them.
Just then we heard another warwhoop, sounding from up the hill
somewhere, beyond the orchard. Bill and Benny scrambled to their feet,
and we all looked and listened.
We saw nothing for a minute or two. Then something darted through the
gate, which leads into the orchard from the hill; dropped down out of
sight behind the fence, and commenced crawling backward toward the
nearest apple tree. Every few seconds, it would raise up long enough to
point something, which looked like a gun, at the enemy.
"Great snakes!" whispered Bill. "What's that?"
But we could tell in a minute without asking, for when it reached the
tree it stood up and peered around the trunk, aiming a stick and
pretending to fire. We knew then that Skinny was on the way.
"It's Skinny!" shouted Benny, throwing a stick at him.
Skinny waved one arm for us to be quiet, then began to wriggle back to
the next tree. Making his way slowly from tree to tree, with a quick
dash he finally reached the roof, where he felt safe.
"That was a close call, Skinny," said Bill. "I heard a bee buzzin'
around out there in the orchard, a few minutes ago."
"Bee, nothin'!" Skinny told him, still pointing with his gun and looking
around in every direction. "They pretty near had me surrounded."
That was the beginning of this history, which tells all about the doings
of the Band, that set all the people talking about us for miles around.
Perhaps you never heard about the Band; how we found a cave at Peck's
Falls, part way up the mountain, and had all kinds of fun playing there
and on Bob's Hill. There are eight of us in all. Skinny is captain. His
folks call him Gabriel but we don't like that name. Skinny is a good
name for him, he is so fat. He can run though, even if he is heavy, and
you would think that he could fight some if you had seen him once, when
the Gingham Ground Gang got after us.
Benny Wade is the littlest fellow in the bunch but he feels just as big
as anybody and sometimes that is almost as good as being big. Besides
these there are Harry, Wallie, Chuck, Bill Wilson, Hank Bates,--Oh, yes,
I most forgot,--and myself.
My name is John Alexander Smith. The boys call me Pedro, and I have been
secretary ever since Tom Chapin found the cave. It's up to me to write
the doings of the Band and the minutes of the meetings.
Tom Chapin was our first captain and he meets with us now, whenever he
is in town.
The village where we live is in a long, narrow valley, with little
Hoosac River flowing north through the center of it, until it gets
beyond the mountain range. Then it turns west and hurries down into the
Hudson.
Bob's Hill stands just west of the village and looks down upon the
highest steeples. Over the brow of the hill and a little south are
Plunkett's woods. West, straight back, a mile or more, begins the
timbered <DW72> of old Greylock, which, everybody knows, is the highest
mountain in Massachusetts. And in the edge of the first woods, a little
back from the road, is the prettiest place you ever sat eyes upon.
Grown-up folks call it "the glen," but we boys just say "Peck's Falls."
I don't know why, only there is a waterfall there, which begins in a
brook, somewhere up on the mountainside, and plays and tumbles along,
until finally it pours down from a high cliff into a pool a hundred feet
below; then dashes off to join Hoosac River.
A queer-shaped rock, with a high back and narrow ledge, which we call
the "pulpit," bridges the ravine in front of the falls, fifty feet and
maybe more, above the rushing water. A little farther down the ravine,
at the edge of the stream, is another rock. It will do no harm now to
say that our cave is under that rock, because folks have found out about
it, although not many know about there being two entrances.
All these things that I have told about belong to us boys. Mr. Plunkett
thinks that he owns Plunkett's woods and Bob's Hill. I mean the very top
of it. And somebody has been cutting trees off from Greylock, until it
looks like a picked chicken in spots. But we call them all ours because
we have more fun with them than anybody else does, and it seems to us
that things belong to those who get the most out of them.
We knew from the way Skinny was acting that he had something on his
mind, so we sat down and waited for him to tell us.
"Fellers," said he, after a while, "we've been Injuns and we've been
bandits, and we have had fun, good and plenty. I ain't sayin' that
Injuns and bandits are not all right sometimes but----"
"Guess what!" broke in Benny. "We've been 'splorers, too. Don't you
remember 'sploring out in Illinois last summer? About LaSalle and that
other guy and What's-her-name who fell over the cliff?"
"That was all right, too," said Skinny, "and I couldn't forget it in a
thousand years, but I tell you those things are back numbers. They are
out of date."
"Never mind about the date," said Hank, "but hurry and get it out of
your system. We've got to be something, haven't we? If we ain't Injuns
and we ain't bandits, what are we?"
"We are Scouts," shouted Skinny, aiming with his gun and dodging so
quickly that he almost slid down the roof.
We all looked at one another in surprise, wondering what he meant. Benny
spoke up first.
"What are those things, Skinny?" he asked.
"Why," said Skinny, "haven't you been readin' about 'em? They
are--er--they are--er--they're just Scouts, that's all.--They scout
around, you know, and do all kinds of stunts."
"Scoot around, you mean," I told him.
"Well, it's the same thing, ain't it?"
"Not for mine," said Bill, shaking his head. "Scouts may be all right,
but Injuns and bandits are good enough for me."
"Here's the book, anyhow," said Skinny.
He pulled out of his pocket a little book, which told all about "The Boy
Scouts of America."
"That's what we are going to be, the Boy Scouts of America, or part of
them. They have members all over the country. We'll call ourselves 'The
Boy Scouts of Bob's Hill,' when we have our meetings."
Say, it looked good to the Band, except Bill, after Skinny had read the
book to us a little, sitting there on the roof. It was a good deal like
what we had been doing, only more so. Even Bill said it was almost as
good as being Injuns and when Benny heard about the uniforms he hardly
could wait.
"How are we going to do it?" I asked, after we had talked until we were
tired.
"That is what I came to tell you about," said Skinny. "Mr. Norton, who
teaches my class in Sunday school, is getting one up."
"One what, Skinny?" asked Benny, his eyes bulging out like saucers, he
was so interested.
"Something he called a 'patrol.' You see, the Boy Scouts are almost like
an army, with all kinds of officers, only they call them different
names, and the different companies are called patrols. He is getting up
a patrol in the Sunday school and wanted me in that, but when I told him
about the Band he said that we could have a patrol of our own, if we
wanted to. There are eight of us, you know, and that is just enough. I
don't know much about it yet, but Mr. Norton wants me to bring the Band
up to his house Monday night and talk it over. He's going to have ice
cream; I heard him say so to Mrs. Norton."
When he said that last, he looked at Bill, because Bill liked ice cream,
although he didn't seem to think much of the Scout business.
"Will you go?" asked Skinny. "I've got to tell him to-morrow, so he'll
know how much ice cream to make."
Benny looked at me and I could see by the way his eyes were shining
that he wanted to go. But Bill never likes to change his mind.
"I think we ought to vote on it," he said, "and have Pedro put it in the
minutes of the meetin'."
"Shall I put it down in invisible ink," I asked, "or in the kind that
shows?"
We always write our most secret doings in invisible ink, made of lemon
juice, so that nobody can read about them. We don't need to read it
ourselves, because we know all about it anyway. If we want to, by
holding the writing up to a fire we can make the letters show.
"Write it with chalk," said Skinny, "and make the letters a foot high.
This is something we want folks to know about."
"Uniforms wouldn't be so very much good," said Benny, "if folks couldn't
see us with them on."
Skinny nodded his head; then took a piece of chalk out of his pocket,
and commenced to mark on the clapboards, back of the sloping roof.
I thought at first that he was going to write the minutes of the meeting
before it happened and was going to kick about it, being secretary.
Instead of that, however, he made a big circle, and in the center of
the circle he drew a picture of a tomahawk. Then, after looking at a
watch which his folks gave him for Christmas, he put the figures 18
above the tomahawk, and 16 below.
That was our Indian sign. The circle meant our cave at Peck's Falls,
that being sort of round. The figures said for us to meet on the
eighteenth day of the month, at the sixteenth hour, which would be at
four o'clock that very afternoon. We had half an hour in which to get to
the cave.
When we saw the Sign we all gave a yell, Bill Wilson louder than
anybody, and were going to start for the cave on a jump, but Skinny
hissed like a snake and held up one hand for us to keep still.
"My braves," said he, after he had made up a lot of Indian words, which
we couldn't understand, only they sounded fierce, "do you want to lose
your scalps? You don't know what is waitin' for us on yonder hill."
We didn't, either. If we had, maybe we shouldn't have gone.
[Illustration: WITH SKINNY LEADING, WE STARTED, DODGING FROM TREE TO
TREE.]
"Follow me," said he. "Keep behind the trees until we get out of the
forest, and mum's the word!"
So, with Skinny leading, we started, dodging from tree to tree on the
hillside, until we came to the orchard fence. After that there were no
trees except on the very top.
There is a sort of road leading out of the orchard and winding around
the hill, where the walking is easy, but on that side Bob's Hill itself
rises almost straight up from the orchards, and the <DW72> is covered
with slippery grass, with now and then a big stone sticking its nose out
of the ground. To climb it you have to dig in with the sides and heels
of your shoes and work hard.
Skinny started straight up and we after him, except Bill, who can climb
faster than anybody. He soon was ahead.
As Bill neared the top, forgetting all about danger, Skinny gave a
warning hiss. Bill looked back; then dropped to the ground and began to
crawl slowly up, pulling at the grass and stones to help him along. The
rest of us waited to see what would happen to Bill.
In a few minutes we saw him stick his head up carefully above the brow
of the hill. Then he dodged down out of sight and slid back part way
toward us, motioning for us to come on and not to make any noise.
I didn't know what to think of it, for I hadn't really supposed anybody
would be there. Skinny is 'most always careful that way because, he
says, you never can tell what may happen.
"Gee!" said he, when Bill motioned. "Didn't I tell you they pretty near
had me surrounded? Steady now, and mum's the word!"
Slowly we crawled up toward Bill. When we had come up even with him,
without a word he crept toward the top of the hill, we crawling along
after him, and my heart was pounding like a trip-hammer, partly from the
work of climbing and partly because it was scary.
Pretty soon we began to hear voices. The eight of us put our heads up at
about the same time; then sank down again out of sight, and I heard
Skinny whisper, "Jerusalem!" and Bill saying "Great snakes!" to himself.
We lay there for a moment, looking at each other and not knowing what to
do. Then Benny spoke up.
"Come on, fellers," said he. "Who's afraid of them? It's only a lot of
girls."
That's what it was. About twelve high-school girls were sitting there
under a tree, with lunch baskets around, looking at Greylock and waiting
for it to be time to eat. There was no way for us to pass without being
seen except to go back and around through Plunkett's woods, and we
didn't want to do that.
"Let's scare 'em," said Skinny at last. "We'll yell the way we did on
Greylock that time we scared the wild cat."
"It's all right to scare 'em," said Hank, "for they haven't any business
on our hill. But a girl ain't a wild cat or anything like it, and you
never can tell what she will do. They may not scare worth a cent."
"I'll tell you what," I said. "If we all yell, they'll know that it
must be the Band. So let's have only one yell. Give Bill a chance and
there will be something doing."
We left Bill and crawled up to where we could see them and they couldn't
see us. Then he commenced.
Say, I've heard Bill Wilson a lot of times, but I never heard anything
like that. Although I knew what was doing it, shivers chased up and down
my back, until I 'most forgot about the girls.
He started with a moan like he was in pain. Then for a minute it sounded
as if a whole menagerie had been turned loose, with a dog fight in the
middle. From the midst of the dog fight came a blood-curdling screech
which died away again in a moan and sob, and then all was still while
Bill was getting his breath for another.
It was awful to hear, and the girls didn't wait for another, or even for
the sob part. At the first moan they started to their feet, looking
around with scared faces, and when the menagerie turned loose away they
went on a run.
"Charge, my braves!" cried Skinny, as soon as he could stop laughing
long enough to speak. "Let's surround 'em."
With a yell, we charged across the top of the hill, down the <DW72>
beyond and into a field which rose gently up to Plunkett's woods.
Just before the girls reached the woods one of them looked back, saw us,
and told the others. I thought they would run harder than ever when they
saw us coming, but it was just as Hank said about not knowing what they
would do. They turned and stood there, the whole twelve of them, looking
so mad that we stopped running and waited to see what would happen.
"We know who you are, Skinny Miller," said the one who had seen us
first, "and you ought to be ashamed of yourself. We'll fix you for
this."
She said something to the others, which we couldn't hear, and pointed
toward us. Then they stooped and each one grabbed a stick from the edge
of the woods.
"Great snakes!" said Bill. "I wish I hadn't come."
"Fellers," said Skinny, looking at his watch. "It's 'most four o'clock.
We'll have to run like sixty if we get to the cave in time for the
meetin'."
There are a lot of boys who never saw a mountain, and the Band, even,
never saw the Rockies and big mountains like those. But Greylock is big
enough for us. On a summer day, with fleecy clouds chasing over his head
like great, white butterflies; sunshine resting on the pine trees, and
the mountain smiling down on us with arms outstretched, as if he would
gather in all of Massachusetts and a part of Vermont, and the cawing of
crows in the Bellows Pipe, and no school to call us back--say, that's
living; that is!
Soon we came to the woods and followed along a path until we could hear
the rushing and roaring of Peck's Falls in front of us, sounding as if
old Greylock himself was talking.
We stopped at Pulpit Rock a minute to see the falls and the foaming pool
below; then followed Skinny down the side of the steep ravine to our
cave at the edge of the stream.
"The meetin' will come to order," said Skinny, after we had crawled in
and were sitting on the floor. "Are we all here?"
"I am," said Benny, "and I," "and I," "and I," said the others, faster
than I could count them.
"All the fellers that want to go to Mr. Norton's," said Skinny, as soon
as he had found that everybody was there, "to see about this Scout
business--and eat ice cream," he added, looking at Bill when he said it,
"mark a cross on the floor of the cave with your knives."
Everybody marked except Bill. He didn't have his knife with him.
"It's all right," said he. "I'll go, anyhow, knife or no knife. I'd
rather be an Injun than a Scout any day in the week, but there ain't any
use letting that ice cream go to waste."
"'Tis well," said Skinny. "We have spoken."
CHAPTER II
RAVEN PATROL HITS THE TRAIL
WHEN Monday night came, the Band met at Skinny's and went from there to
Mr. Norton's. He seemed glad to see us and started in for a good time
without saying a word about the Scout business. I was just going to ask
him about it when Mrs. Norton brought in the ice cream. After that we
were too busy to ask anything.
When at last we had eaten all that we wanted and Bill had put away three
dishes, Mr. Norton gathered us around him and said that he would tell us
a story, if we wished to hear it.
We told him to go ahead, and, after thinking a moment, he began.
"You boys probably do not remember the Boer war in Africa. You were too
young at the time. During that war the Boers surrounded a town called
Mafeking. All the able-bodied men were needed for fighting in order to
defend the city and could not be spared for the work of carrying
despatches and things like that.
"They had some lively lads in that town. As soon as the boys found out
the situation they made up their minds that they could do that kind of
work just as well as the men could. They did, too. Back and forth they
hurried on bicycles, through a rain of bullets, from fort to fort,
carrying messages and scouting. I tell you, those English boys were
heroes. I don't see how they escaped being killed. They must have dodged
the bullets."
When Skinny heard Mr. Norton speak of their being English boys he looked
troubled, because Skinny thinks a lot of the United States of America.
"Is this an English story, Mr. Norton?" he asked. "Because if it is I
don't know about it. How about George Washington, Bunker Hill, seeing
the whites of the enemy's eyes, and all those things? We named our boat
out on Fox River in Illinois, the 'Paul Revere.'"
"Guess what!" put in Benny, laughing at something he was thinking.
"Skinny couldn't dodge any bullets? 'Cause why? He's too fat. They
couldn't miss him."
"Aw, what's the matter with you?" said Skinny. "I could dodge as many as
you could, I guess. If a bullet hit you there wouldn't be anything left
of you; that's what. Why, I----"
"A hero is a hero," said Mr. Norton, before Skinny had time to finish,
"and a boy is a boy, I guess, no matter in what country he happens to
live. I have heard all about the Band, and I know that if you had been
in Mafeking that time you would have been among the first to volunteer
for scout service, bullets or no bullets, and Washington or no
Washington."
"Hurrah!" yelled Bill, forgetting where he was. "That's the stuff. Injun
or no Injun, too. I knew an English boy once, and he was all right. Say,
you ought to have seen him in a scrap."
Mr. Norton laughed and went on with his story.
"A few years later Gen. Robert Baden-Powell, who had been colonel in
command of the English forces at Mafeking, got to thinking about those
boys in South Africa and how manly it made them to help in the scouting.
He liked boys and he made up his mind that if scouting had been good for
those boys it would be good for any boys. Not the fighting part, I mean,
but the outdoor life, learning to take care of themselves in the
wilderness, make camps, build fires, find their way through the forest,
follow a trail, and such things. So he called a meeting of a lot of boys
and talked to them and showed them how to do it. They played at being
Indians mostly."
"They don't have Injuns in England," said Bill, shaking his head,
"unless it's in a Wild West show, and that doesn't count."
"You are stopping the story, Bill," Skinny told him. "What's the
difference?"
"Well, they don't," grumbled Bill.
"Anyhow," Mr. Norton went on, "the boys enjoyed the play, and the idea
spread like wildfire, until now there are Boy Scouts all over the world.
In America here Ernest Thompson Seton had much the same idea. He was
teaching the boys woodcraft, camp life, and such things by organizing
the Seton Indians that you may have heard about. Then he went to
England, where he and General Baden-Powell put their heads together and
worked out the Boy Scout idea. In this country the boys are known as
'the Boy Scouts of America,' but nearly every civilized nation has its
Boy Scouts under some name or other, and the movement is very popular
among the boys.
"I invited you up here to-night to get acquainted with the Band. Skinny,
I mean Gabriel, tells me that you are all live wires. I want to know if
you will join the Scouts. You can have a patrol of your own, select your
own patrol leader and your own patrol animal."
"What's a patrol animal?" we asked.
"Patrol animal? Why, each patrol is named after some animal, and the
Scouts all have to be able to imitate its call, so that they can let
each other know where they are hiding."
When Mr. Norton told us that you hardly could have heard yourself think
for a minute. Mrs. Norton didn't know what had broken loose and came
running in from the next room. Skinny was hissing like a snake; Bill
croaked like a frog; Benny cawed like a crow; Hank barked like a dog,
and the other boys did something else, and nobody could tell what they
were doing.
"You seem to have the right idea," smiled Mr. Norton.
There was a lot more to it, uniforms and rules and signs and all that
sort of thing, but that doesn't belong in this history. It didn't take
us long to decide that we would go in. Bill Wilson was the craziest one
in the bunch.
Mr. Norton thought that we ought to decide on a patrol leader before we
went home. We told him that there was nothing to decide.
"Skinny is captain, all right," said Benny, "and the Band is the Band, I
guess, whether we are Scouts or Injuns."
"Yes, I'm captain of the Band," Skinny told him, when Mr. Norton waited
to see what he had to say about it, "but I don't know about this patrol
business. It wouldn't do to vote on it here, anyway. The cave is where
we meet. We ought to vote in the cave, seeing it is summer time. If it
was winter we could meet in Pedro's barn."
We left it that way and were so busy during the closing days of school
that we didn't have time to think much more about it until Friday. When
we came in from afternoon recess, there was the Sign, as big as life,
drawn with chalk on the blackboard.
I saw teacher looking at it, sort of puzzled, as if she was wondering
what it all was about, and some of the girls were giggling at it. They
seemed to think it was a joke of some kind, instead of something
important. Anyhow, the Sign said for us to meet at the cave, Saturday,
at ten o'clock.
Saturday morning, long before ten, every boy was at our house, that
being nearest to the cave. Each one carried a lot of good things to eat,
so we should not have to go home for dinner unless we wanted to.
Besides his dinner Hank had with him a little camera, which his folks
had given to him on his birthday because he promised not to make any
more awful smells with chemicals in the cellar. Hank was always mixing
things to see what would happen and he pretty near blew his house up at
one time. He is an inventor, too, and says that when he grows up he is
going to make a flying machine. He nearly made one once. He made a kite
that would pull us uphill on our sleds.
One time he made a spanking machine which worked with a crank, and when
teacher wanted us to lick Bill we spanked him with it. Only we laid a
horse hair across the seat of his pants to see what it would do and it
broke the machine. Of course, he didn't make the camera, but he had a
place down cellar where he developed and printed his pictures after the
camera had taken them.
"Gee, fellers," said Skinny, "Hank is goin' to take our pictures.
Everybody look pleasant."
"Not on your life," Hank told him. "You'd break the machine; that's
what."
We went up through Blackinton's orchard and followed the road around to
the top of the hill.
In a field, a little west of the top, the same field where we chased the
high-school girls, stand what we call the "twin stones." They are big
ones, six feet high and maybe more. One of these we use for a
fireplace. It is near Plunkett's woods, where it is always easy to find
dry sticks to burn. A piece of the rock has been split off in such a way
that it makes a kind of hearth, with a place between for a fire.
"Let's come back here for dinner," I said. "When we build a fire in the
cave the smoke makes our eyes smart. What do you say?"
So we went into the woods and hid our lunch and some potatoes, which we
had carried in our pockets to cook, but Hank wouldn't leave his camera.
He said it cost too much to let it lie around in the woods. His folks
paid three dollars for it.
Then we hurried on to the cave.
"Open sesame!" said Skinny, pounding the outside of the cave with a
club, like the robber did in "Arabian Nights."
"Is she open?" asked Bill, who was in a hurry to get in.
Skinny didn't answer. He was peering up and down the ravine to see if
anybody was looking. When he found that no one was in sight he motioned
for us to go in.
"Old Long Knife will guard the pass," said he.
And he did, for when I put my head out of the cave a little later to
find out why he did not come, he was fighting like sixty. He swung his
club and jumped around for a minute; then gave a fearful whack and drew
himself up with his arms folded, like an Injun or a bandit.
"Lie there, villain!" he hissed. "Sick semper turn us, and don't you
forget it."
After that he came in with his face all red, he had been working so
hard. We already had the candle lighted and were ready to begin.
"Fellers," said Skinny, when we all had sat down on the floor in front
of him and I had called the roll. "I don't know whether this is the Band
or the patrol, or whether we are bandits, or Injuns, or Scouts, and I
don't know that it makes much difference. I am captain of the Band, but
what we want to find out is, who is leader of the patrol. We could fight
for it, perhaps, only I hate to muss my clothes."
Some looked at Bill, for we knew that he kind of wanted to be leader. He
would make a good one, too, only it seemed to belong to Skinny.
Nobody said a thing for 'most a minute. Then Benny stood up, bumped his
head against the roof of the cave, and sat down again.
"Mighty chief," said he, when we were through laughing at him, "may I
speak and live?"
He never had said that before and it surprised us.
"You may," said Skinny, looking fierce and swinging his club.
"Fellers," began Benny, "Skinny was a good enough leader when we went
'sploring out in Illinois last summer and I 'most got drowned in Fox
River, and he was a good enough leader when we found a tramp in this
'ere cave and smoked him out. He lassoed the robber, that time, didn't
he, when the guy was stealin' Hank's pearl, and--and--lots of things? I
guess that anybody who could do that is good enough to be patrol
leader."
That was a long speech for Benny to make, and we all patted him on the
back except Bill, who sat thinking and getting ready to say something.
All of a sudden he spoke up.
"Fellers," said he, "three cheers for Skinny Miller, who is always there
with the goods."
"You're out of order," Skinny told him, but nobody could hear.
I shouldn't wonder if they heard us voting clear down in the village.
We also had to have an assistant patrol leader, called a corporal, and
we elected Bill Wilson. Bill is great at such things. As corporal he
would be in command whenever Skinny was away. That didn't count for
much, though, for Skinny is almost always around when anything is going
on.
The next thing to do was to decide upon our patrol animal, like the book
said.
At first we couldn't agree very well on that. Nearly every one wanted a
different animal. Skinny wanted us to choose a snake because he liked
the hissing part and a picture of a snake would be easy to draw on our
signs.
Hank and Bill thought a dog would be best.
"A dog," said Bill, "is man's best friend, and that is what Scouts are
for."
Hank could bark like a dog. That was why he wanted it.
Benny thought a crow would be the thing, but it seemed to me that the
American eagle would be better. We heard one once on Greylock and it was
great.
Skinny liked the eagle pretty well, especially the American part, but
when he found that Benny Wade wanted a crow he said he was for a crow,
too. That was because Benny had made the speech.
"A snake is all right for some things," he said, "and you don't want to
step on them or on us. Don't you remember that old flag which had a
rattlesnake on it and the words, 'Don't tread on me'? The hissing is all
right, too, when we are close together and can hear, but how about it
when we are not? What if I was hiding in Plunkett's woods and you were
on the way to the cave and I should be attacked by Injuns or something.
I might hiss until I was black in the face and who'd hear me? You could
hear me caw almost to Peck's Falls."
"Yes, that's so about snakes," I told them. "I don't think much of
snakes myself. But I don't know about crows. The eagle is such a noble
bird."
"Noble nothin'!" said he. "What did an eagle ever do that was noble any
more than a crow? Besides a crow can talk if you split its tongue. I
read it in a book. You can't draw an eagle. You'd have to write under it
what it was."
"So you would under a crow," I told him.
"Anyhow," he went on, "I'll bet nobody here can make a noise like an
eagle. Let's hear you do it, Pedro. Cawing is easy."
That ended the eagle business. Skinny was right. Not one of us could
make a noise like an eagle.
"What makes you want it a crow, Benny?" asked Hank.
"I don't know how to tell it," said Benny, sort of bashful like. "I
wasn't thinking about drawing it. A crow would be hard to draw, I
guess, but we could make something that looked like a bird and we boys
would know what bird was meant. I wasn't thinking either whether it was
noble or not. Maybe a crow ain't exactly noble, but somehow when I see a
big fellow soaring around in the Bellows Pipe, between the mountains, it
makes me feel kind of noble myself and as if I ought to soar, too. And
when I hear the cawing of a crow, no matter where I am, even in North
Adams or Pittsfield, I can see Bob's Hill and old Greylock and the
Bellows Pipe, and big crows flying around in the air as if they owned
them all. We are Bob's Hill boys and Greylock boys. That's why I want it
a crow. They sort of belong together."
We never had thought of that before, but when we came to talk it over it
seemed that way to us, too. So we chose the crow for our patrol animal,
only we didn't call ourselves "the crows" but "the ravens," because it
sounded so much nobler. While we can't draw a very good one when we make
our signs, it looks some like a bird and we all know what kind it is, as
Benny said.
By that time we were getting hungry and so we made a bee-line for
Plunkett's woods, sounding as if a whole flock of crows were starting
south.
"Everybody scatter for wood," shouted Skinny, when we had come to the
big stone where we build our fires. "I'll get the grub."
We ran to different parts of the woods where we knew there were dead
branches lying on the ground, trying to see which would get a fire going
first. Then, just as Bill and I met at the stone, with arms full of
sticks, and the others close behind, we heard a terrible cawing over in
the woods, only it didn't sound so much like a crow as it did like
Skinny.
We looked at one another, wondering what it all meant, for the Scout
business was new to us. Besides it sounded as if something had happened.
"'Tention, Scouts," said Bill, in a hurry to get in his work as corporal
while Skinny was away. "Everybody caw!"
We made a great racket. In a moment there came an answering caw from the
woods; then Skinny stepped out into the clearing in plain sight and
motioned for us to come.
We knew something was the matter and started for the woods on a jump,
the corporal in the lead.
"It's gone!" shouted Skinny, when we had come near. "Some guy has stolen
our dinner."
"Great snakes!" groaned Bill. "And I'm starving to death."
We all gathered around the place where we had hidden the things under
some bushes. Skinny was right; they were gone. I tell you he was mad.
"I don't know whether we are Scouts or bandits or Injuns," said he, "and
I don't care, but I'd like to get hold of the critter that stole our
dinner. We wouldn't do a thing to him. Oh, no. Maybe not."
"Everybody scatter," he shouted. "Look for signs and tracks. We'll
follow him to the ends of the earth."
CHAPTER III
TRACKING THE ROBBERS
WE didn't have any idea who took our things and there didn't seem to be
any way of finding out. The ground in the woods was carpeted with pine
needles, which left no trace of footprints.
We thought that maybe those girls that we had chased had taken our
dinner to get even, and it might have been the Summer Street boys, or
maybe the Gingham Ground Gang.
We scattered, like Skinny told us, and gradually worked out from the
center, crawling on our hands and knees, and watching every inch of the
ground and the bushes.
We didn't get any trace at all until I found a potato. Then Skinny, who
was a little ahead of me and at one side, gave a groan and yelled:
"Here's my wishbone. They've eaten all my fried chicken."
It always makes Skinny mad to have somebody eat his fried chicken.
Farther on we found pieces of eggshell and then more, as if somebody had
peeled an egg while walking and thrown the shells on the ground.
We knew then that there was no chance of getting our dinners back, but
we followed the trail, just the same.
After a time we came to the queerest looking tracks, where somebody had
stepped on a soft piece of ground. Benny found them first.
"The spoor!" he yelled. "The spoor! I've found the spoor."
"Well, don't tell the whole town about it," said Skinny. "Keep quiet and
we'll surround 'em."
"But the chicken and eggs are gone," he added, after a moment. "I was
going to give you some of that chicken, Bill."
We stopped and had a long look at the tracks. There were four footprints
and a hole, which looked as if it had been made with a stick, or cane.
Three of the prints were like those which any man would make in walking
and one was the print of a bare foot, only it had a queer look that we
couldn't understand.
"We've got 'em," whispered Skinny. "We'll know that footprint again
anywhere we find it. Forward, and mum's the word!"
Twice after that we found the same queer footprint; once in the dust of
a road that runs along the south side of Plunkett's woods, and again on
the edge of a brook which comes down from the mountain somewhere.
Then we lost the trail and didn't know where to go. Just because we
didn't know what else to do, we followed the brook up, until we came to
a gully out of sight from the road.
Skinny was ahead, aiming with his stick and saying what he would do if
he should catch the fellow that stole his chicken. All of a sudden we
saw him drop behind a bush and lie still. We dropped, too. We didn't
know what for, but I've noticed that it is 'most always a good thing to
drop first and find out why afterward. Then we crawled slowly up to him
to see what had happened.
There, sitting on the ground in a grassy ravine, near the brook, were
two men, and they were eating what remained of our lunch. One of them
had his left shoe off and his foot done up in a bandage. That was what
had made the track look so queer.
Now that we had caught them we didn't know what to do with them, for
they were too big for us to tackle.
"I believe we could get away with the lame one," whispered Skinny, "only
they have about eaten it all up; so what's the use? Besides, the other
one looks as big as a house."
"If we only had a rope, Skinny," said Benny, "you could creep up behind
and lasso them, the same as you did the robber out near Starved Rock."
"Bet your life I could," he replied, "but we haven't got one. Fellers,
don't you ever go out again without a rope. You can't ever tell when you
will need it."
"Great snakes!" said Bill, thinking of the chicken Skinny had been going
to give him. "I'm starving to death. Let's heave some rocks at 'em,
anyhow, and then run."
He picked up a big stone as he spoke and was going to throw it, when
Hank caught his arm.
"Wait," said he. "I know a trick worth two of that. I'm going to shoot
'em."
"Shoot them?" I gasped in surprise. "What with?"
"With my camera. You fellows stay here out of sight and caw like a crow
if they make any move before I am ready for them. If I can only get
behind that clump of bushes back of them without their seeing me, I'll
take their picture."
"Aw, cut it out," said Bill.
But Hank was gone, and after a little we could see him running through a
field out of sight of the men, so as to come into the ravine from the
other end. Pretty soon we saw him crawling in, creeping from bush to
bush, in sight only for a second at a time.
There was not a sound except the voices of the men, who were talking
about something, and the ground might have opened and swallowed Hank for
all we could see of him.
We waited a long time and began to get nervous, not knowing what had
happened, and I saw Bill feeling around for another stone.
Then all of a sudden Hank stood up above the bushes he had told us
about. He looked toward where he knew we were hiding and put one finger
to his lips. Then he tossed a stone toward the men and dropped down out
of sight again before it could fall.
"Great snakes!" whispered Bill. "If he's goin' to throw, why don't he do
it, and not give a baby toss like that?"
Skinny held up one hand warningly as the pebble fell into the brook
right back of the men, making a little splash and gurgle, as if a frog,
or maybe a trout, had leaped out after a fly.
When they heard it both men jumped up and stood there in the sunshine,
looking toward the sound. We couldn't see Hank, but knew that he was
somewhere in the bushes taking their picture.
You almost could have heard our hearts beat for a minute, not knowing
what would happen. Then the men sat down again and went on talking.
We waited five minutes to give Hank a chance to get away, and crawled
back the way we had come. When we reached the road we heard a crow
cawing in the woods and knew that he was safe.
"You answer, Benny," said Skinny. "You do it best."
He gave three caws so real that I almost thought it was a sure enough
crow. Hank joined us and we hurried down the road toward home, hoping
that the dinner would not be all eaten up.
"Did you get the picture?" I asked.
He nodded. "I think so, but I can't be sure until it has been developed.
I had a splendid chance. They stood just right and there was a fine
opening through the bushes."
"It took you a long time," grumbled Bill. "I could have hit them with a
rock easy."
"I was trying to hear what they were saying. I couldn't hear very well,
but I think they are robbers or something."
"You bet they are robbers," said Skinny. "Didn't they steal my fried
chicken?"
We didn't think much more about the men because we had important work on
hand. The first thing we had to do was to eat dinner. That is always
important, especially when your mother knows how to cook beefsteak that
makes you crazy just to smell. After that came a ball game. Our nine,
the "Invincibles," played a picked nine from Summer Street. We beat, 25
to 19.
I didn't see any of the boys again until in church, Sunday morning. When
I went in Bill Wilson was there, looking so dressed up that I hardly
knew him.
He saw me and motioned for me to come into his pew, but Ma wouldn't let
me do it. Bill had something on his mind. It was easy to tell that. He
looked excited, and every time I turned around he went through with all
sorts of motions with his mouth, trying to make me understand what he
wanted to say.
It bothered me. Every time the minister twisted up his face, trying to
make us understand how important it was what he was saying, I'd think of
Bill's mouth going back of me. I couldn't help it.
When at last we went into Sunday school he told me.
"Great snakes, Pedro!" said he, grabbing me by one arm. "Haven't you
heard about it?"
"How can I tell whether I have or not, when I don't know what it is?" I
told him.
"They robbed Green's store last night; stole him blind."
"Who did?"
"The guys that we saw yesterday. Our robbers."
When Bill told me that you could have knocked me down with a feather. It
made me almost as excited as he was. He didn't have time to say any more
because teacher made him sit at the end of the line away from me so that
he wouldn't whisper so much.
But after Sunday school was over he told me all about it. Burglars had
broken into Green's store during the night. They blew open the safe and
took all the money, nearly one hundred dollars, and they carried off a
lot of knives and revolvers. There is an alley back of the store. They
broke into the basement from there and then made their way upstairs.
"How do you know that it was our robbers who did it?" I asked.
Bill drew himself up and swelled out his chest, just like Skinny does
sometimes.
"I'm a Boy Scout, ain't I?" he said. "A corporal, too."
"You are only a Tenderfoot," I told him.
That was true. You have to be a Tenderfoot before you can get to be a
real Scout.
"It's the same thing," he said, winking one eye. "One of the robbers has
a tender foot, anyhow."
"Look here, Bill," I told him. "You are getting to be worse than Skinny.
What are you talking about?"
"Pedro," he said, "you'll never make a Scout. You're a good bandit and a
good secretary, but this Scout business is too much for you. I saw their
tracks; that's what."
"In the alley?"
He nodded. "Come on and I'll show you."
We hurried down to Center Street and turned into the alley back of the
stores. The ground in the alley was hard and didn't show any tracks
except wagon ruts.
Bill looked up and down the alley to make sure that nobody was watching;
then tiptoed over to one side, and lifted up a big piece of wrapping
paper, which lay there as if it had been blown out of the store. Under
the paper there was the same kind of footprint which we had followed
from Plunkett's woods the day before.
There was no doubt about it. The man with a bandaged foot must have been
in the alley back of the store which had been robbed.
Bill was the proudest fellow you ever saw over that footprint. When I
had finished looking at it he put the paper back again and we went out
into the street.
"What do you think of that?" said he. "I guess Skinny ain't the whole
thing--on Sundays."
"Does the marshal know?"
"I haven't told a soul except you, Pedro. I am saving it for the Band--I
mean the patrol. This is our chance. What's the good of bein' a Scout if
you don't do any scoutin'?"
"Anyhow, I think we ought to tell the marshal about this," I said.
"Those robbers are not going to wait for the Scouts to get busy. They
probably jumped a freight last night and are in New York by this time.
But maybe the marshal could do something."
Bill was bound to tell the other Scouts about it first. So after dinner
we got the boys together and all went over and took a look at the
footprint.
Skinny was even more excited than Bill was.
"We are hot on the trail, fellers," said he. "The thing to do is to
surround them. We ought to have captured them yesterday. Bet your life
we'll take a rope next time."
But when Pa found us talking it over on our woodpile, and we told him
about it, he said for us to go to the marshal's at once, and if we
didn't he would.
It being Sunday, we went to the marshal's house and found him sitting on
the front porch dressed in his best clothes. He was some surprised when
he saw the eight of us walk into his yard. It made us wish that we had
uniforms on.
"To what do I owe the honor of this visit?" said he. "Is this a
committee of distinguished citizens to ask me to run for mayor or
something?"
Bill was bursting with the news, but Skinny was the first to speak.
"We want you to run for those burglars," he said, "and we can tell you
who they are."
When he heard that the marshal began to get interested.
"Well, who were they? Maybe," he went on, smiling at us, "you youngsters
have come to give yourselves up."
"We didn't do it," put in Bill. "We wouldn't do such a thing, but we
know who did. We don't know his name, but we know his track. We could
have caught him yesterday if we'd wanted to. I wish we had now."
Then we told him about losing our dinners and following the robbers
through Plunkett's woods, and about the queer looking track made by the
bandaged foot.
"I'd know that footprint in China," said Bill, "and I found one just
like it in the alley back of Green's store. The man with the lame foot
made it. I 'most know he did."
"Say, William, you are a regular sleuth," said the marshal. "I have a
notion to put you on the force."
But he didn't guy us any more after that. He put on his coat and walked
downtown with us.
After he had looked at the footprint he covered it up again so that
nobody would step on it.
"That's the one all right," Hank told him. "There were two of them. I
heard them say something about robbing, when I was taking their
pictures."
"Taking their pictures! They don't go around breaking into stores with
an official photographer along, do they?"
"I don't know what they go around with," Hank said, "but I crept up
close behind them and lay back of a bush where I could hear them
talking, although I couldn't understand much of what they said. I
thought it would be fun to take their pictures when they didn't know
anything about it."
"They stood up when Hank threw a stone and looked right at the camera,
only they didn't know it was there," Benny explained.
"Great Scott, boy! Do you mean to tell me that you took a photograph of
the rascals?"
"I snapped them all right," Hank told him, "but I won't know whether I
got a good picture or not until I develop the roll. I haven't done it
yet."
"Well, you develop it right away, or, better still, get your camera and
we'll have Marsh, the photographer, do it and make sure of things. He'll
do it, if it is Sunday."
Hank hung back. "Can't you wait a while?" he asked. "I've got five shots
left in the camera and don't want to waste them. They cost money."
The marshal looked disgusted. "Waste them! How much did they cost?"
"Twenty-five cents a roll; six in a roll."
The marshal pulled a quarter out of his pocket and handed it to him.
"You'll be a rich man some day," said he. "Now that roll of films
belongs to me and that picture is going to be developed before you are
an hour older. Can you do the job or shall I look up Marsh?"
"I can do it all right, if there is any picture to develop."
"Very well, go ahead with it and bring it down to my office just as soon
as you can. And I'll tell you further, young fellow, if we catch those
burglars through your help, you'll get part of the reward."
Hank looked at us a moment with his eyes shining. Then he drew himself
up.
"I'm a Scout," said he, "and Scouts are not looking for rewards. 'A
Scout's duty is to be useful and to help others.' The book says so."
It made us all feel proud to have Hank say that. The marshal gave a
surprised whistle.
"If that is the case," said he, laughing, "give me back my quarter."
But Hank wouldn't do that, although Skinny nudged him. I don't suppose
you can learn to be a Scout all at once.
CHAPTER IV
"DANGER--COME"
IT was anxious work, standing around while Hank ran the film from his
camera through some kind of machine which he had, to bring out the
picture. After what seemed like a long time he took it out and looked
through it toward the light.
"Hurrah!" he yelled. "We've got 'em."
We all crowded around to look, and sure enough at one end of the film we
could see as plain as day two men standing up and looking toward us. And
there was the brook, too, and the ravine, so real that we almost could
hear the water pouring over the stones, which we think is the sweetest
music in the whole world. Away back in the picture was the bush, behind
which we boys were hiding when Hank took it. Only you couldn't see us at
all, for we had been careful to keep out of sight.
It is wonderful, isn't it? I don't know how it is done and I don't
believe that anybody else knows, but I know that it is so because I saw
it with my own eyes.
Hank washed the film, and after it was dry put it in a frame with some
paper which he had, and held it up to the gas jet. In a few seconds the
picture showed up on the paper fine, just like our writing does when we
do it in invisible ink and hold it up to a blaze.
We could tell who it was, all right. The big one had a scowl on his
face, as if he had put it there when Hank tossed the stone and hadn't
had time to smooth it out again.
"This picture is for the marshal," Hank told us. "Now I'll print another
for the patrol. We'll let them soak and wash a while, and then dry them
out. It'll take quite a long time, but we've got 'em all right."
When we finally went down to the marshal's it was evening. He was
tickled when he saw the picture. It made Skinny feel real chesty and we
all of us were proud.
"I tell you, Mr. Michael," said he, "the Band's the stuff. I mean the
patrol is. They don't get away from us very often. I only wish we'd had
a rope with us that time."
"You boys certainly did the trick," said the marshal, examining the
picture. "I don't know those men myself, but I know where they will know
them, and that is the next best thing. That is, if they are old crooks,
as I suspect they are."
"Where's that?" asked Skinny.
"At police headquarters in New York. They have a rogues' gallery there
that would surprise you. It contains the pictures and records of nearly
every crook in the country. If these men are among them they'll pretty
near know where to put their hands on them. I'll mail this down
to-night. I've telegraphed already. Come around to-morrow and I'll tell
you if I hear anything."
He met us with a broad grin the next afternoon and showed us a telegram.
This is what it said, for I put it down. Skinny thought it ought to be
in the minutes of the meeting.
"Men well-known crooks. Are under arrest. Got the goods and most of the
money."
* * * * *
"More than ten words are in that telegram," said Hank, counting them.
"There you go again," laughed the marshal. "I'll have to call the New
York chief down for being so careless. Anyhow, your robbers will go to
the penitentiary as sure as preaching."
"I don't know about it," Benny told us afterward, when we were talking
it over. "I'm 'most sorry that we did it. I shall always be thinking
that if it hadn't been for us those men wouldn't be locked up away from
birds and grass and trees. Maybe they didn't have such good folks as
we've got. You know that guy out in Illinois didn't have."
But after we saw Pa we felt better about it.
"I'm glad you feel that way," said he. "Still you did the right thing
after you found out about the robbery. I wouldn't advise you, however,
to go around taking photographs of burglars. You might get into trouble
another time. It surely is an awful thing to be in state's prison, but
being away from the trees and grass is not the worst thing about it. The
worst thing is being so bad that you have to be locked up in order to
make other people safe. It is a terrible thing to be a criminal, whether
you are in prison or not."
He was quiet for a minute; then went on:
"I can't think of a worse prison for a human soul than a human body that
does mean things, lies and steals or is vile in any way."
A few days later when Skinny and I went to the post-office together the
postmaster handed him a letter.
"I say," said he, "you have been promoted, haven't you?"
On the envelope was written, "Captain Gabriel Miller, Patrol Leader,
Raven Patrol, Boy Scouts of America."
It made us both excited.
"It's for the whole patrol," said Skinny, trying to look through it. "I
don't think I ought to open it until we are all together, and I hardly
can wait."
He rushed to the door as he spoke and whistled through his teeth, for he
saw Bill and Hank passing on the other side of the street, going to my
house.
"I could have cawed," he explained when they had come across, "but I
didn't think that I ought to when folks were looking."
We went over to Benny's and found him piling wood and glad enough to
quit.
"Never mind about the other boys," I told them. "They will be along
pretty soon. Whatever it is, we'll want to read it twice, anyhow."
Skinny opened the letter and looked at the writing.
"Jee-rusalem, fellers!" he shouted. Then he commenced to caw like some
crow that was crazy with the heat.
Bill cawed, too, but he didn't know what for. Then he tried to snatch
the letter out of Skinny's hand.
"Aw, cut it out, can't you?" said he, when Skinny dodged out of the way.
"Read it."
"I am readin' it," said Skinny. "It's great."
"Well, read it out loud."
Then Skinny started to read, and this is what the letter said, only it
doesn't tell how Skinny's eyes shone, nor how he stopped every few lines
to punch the enemy.
"_To the Boy Scouts of Bob's Hill:_
"I want to thank every boy in Raven Patrol, and
especially Henry Bates, for the recovery of my
property. But for you I should never have seen it
again and the burglars would still be at large. I
offered a reward for the capture of the thieves
and it rightfully belongs to you, but the marshal
has told me that, being Boy Scouts, you do not
want to be rewarded for good deeds. What I wish to
say is this: I like the Boy Scout idea and want to
help it along. Not as a reward but just because I
like boys, will you let me buy uniforms for your
patrol?
"Sincerely your friend,
"ROBERT GREEN."
That is how we happen to have such fine uniforms that make folks turn
around and look every time we pass.
On the day we first wore the uniforms we were made real Scouts; not
First class ones but Second class. You see, there are three kinds. First
you have to be a Tenderfoot. That doesn't mean that your feet are
tender, but that you are new to the business. To get to be a Second
Class Scout, you have to do all kinds of stunts and you have to be a
Tenderfoot at least a month.
We knew how to build fires and cook things out in the woods and things
like that, which Scouts have to do, and the way we tracked the burglars
showed that we knew something about that.
The hardest things we had to do were to learn the Morse alphabet of dots
and dashes for signaling and to learn what to do when folks get hurt,
how to put on bandages and things like that and how to bring folks back
to life when they are nearly drowned. We learned them all right, and it
is a good thing we did.
Signaling was the most fun of all. We could do it with flags like they
do in the army; by waving our arms like a semaphore, and by smoke from
fires like the Indians do. We also could spell out things with smoke in
the Morse alphabet, which was something the Indians couldn't do, by
making the smoke go up in puffs like dots and dashes.
Part of us would go up on Bob's Hill and part on the hill opposite,
beyond the Basin where we go swimming, build fires, and signal to each
other. It was hard at first, but after a while we could spell out 'most
anything and understand some of it.
It came in handy, too, because one afternoon, after we had been playing
in our yard, we decided to practise our signaling. Just after all the
boys had started for the east hill, except Skinny and me, who were going
up on Bob's Hill, Ma came out and wanted to know where the other boys
were.
"It is too bad that they have gone," said she. "I was going to ask them
to stay to supper."
"Maybe they'll come back," said Skinny, winking at me.
"We are not going to have much, but I thought you boys would enjoy
eating together and we should like it, too. We do not often have the
honor of sitting down to the table with young gentlemen who have
uniforms on."
"We'll stay," said Skinny, "if you will let us do something to help.
According to Scout law, a Scout must try his best to do somebody a good
turn every day. I haven't done it now for 'most two days."
"If that is the case," Ma told him, "my woodbox seems to be getting
empty."
That is the greatest woodbox I ever saw for getting empty. We filled it
so full that the wood fell off all over the floor; then started for the
hill.
"Now is our chance," said Skinny. "We've just got to make them
understand this time. We never have had anything much to tell the boys
before, but this is important."
We climbed to the very top of Bob's Hill and soon had a fire going. When
it was well started we threw on some green stuff that made a big smoke.
Pretty soon we saw smoke going up across the valley and knew that the
other boys were ready.
"They are there," I said. "Now we'll tell them."
"Wait," said Skinny. "First let's give the danger signal. That'll fetch
'em."
"But there ain't any danger," I told him. "What's the use of lying, even
with smoke?"
"You bet there's danger," said he. "There's danger of losing your
mother's supper, ain't there?"
So I gave him one end of a wet blanket which I was carrying, and I
grabbed hold of the other end. We covered the fire with it, stopping all
of the smoke; then took it off and let a big puff go up; then covered it
again and sent up a little puff, and kept doing that until I was sure
the boys would be most crazy, for that sign means danger.
After we had done it a while, we spelled out the word "come." We did
that by using the blanket to make a short puff of smoke for a dot and a
long puff for a dash, like this:
... C .. O -- M . E
We waited and spelled it out twice more to make sure, and then went down
the hill to the house.
"Shall I set the table for the others?" Ma asked, when she saw us
coming.
"They will be here in a few minutes," said Skinny, looking at his watch.
We were not sure of it, but we hoped they would and, as Skinny said, it
wouldn't do any hurt to get the table ready.
We were beginning to be afraid that they had not understood and were not
coming, when we heard a faint cawing, a long way off somewhere. It
seemed from beyond Summer Street.
Skinny answered, while I ran into the house to tell the folks that it
was all right. Then we went out in front and waited.
The first we saw of them was when Bill Wilson turned into Park Street in
a cloud of dust and came tearing up the middle of the road on a jump.
The other boys were close behind, running to beat the band, and every
mother's son of them was carrying a big club.
They didn't even yell when they saw us, they were so nearly winded, but
Bill, being corporal, ran up to Skinny, gave the Scout salute, and then
whirled his club around his head three times.
It was great to see them come up that way, every Scout whirling his
club and all out of breath. Skinny's eyes shone like stars, he was so
proud, and I saw Ma looking out of a window, surprised some, I guess.
"Show 'em to us!" yelled Bill, as soon as he could speak. "We'll eat 'em
up."
"You'll get all the eating you want in about five minutes," Skinny told
him.
"Where are they?" yelled Bill again, while the other boys marched up and
stood in a row, each with his club in the air.
"You are crazy," said Skinny. "Where's who?"
"The Gingham Ground Gang. Didn't you tell us the Gang was after you and
for us to come quick?"
"Not much. I said supper was ready and that if you didn't get a move on
yourselves you would lose out."
"Ain't there going to be a fight?"
Just then Ma came out and it was a good thing she did, because there
might have been a fight, after all.
"Boys," said she, smiling at us, "you are all invited to stay to supper,
and you will just about have time to wash up and cool off a little. We
are having supper early to-night. I was so disappointed when I found out
that you had gone that your patrol leader, Captain Miller, told me that
he would signal to you and that Corporal Wilson would get you here on
time if he had to run his legs off. I don't exactly see how he did it
but you are here, that is certain. I've let your folks know, so you can
stay just as well as not, unless you don't like my cooking."
When she said that the boys set up a shout, for they knew all about Ma's
cooking.
"I wish you would tell me how you do it," she added, turning back as she
was going into the house. "If your secretary would come like that when I
call him, I should be the proudest woman in the village."
CHAPTER V
A CAMPFIRE ON BOB'S HILL
"JEE-RUSALEM, fellers," said Skinny a few days later, "we're going to
have a campfire to-night on Bob's Hill. Mr. Norton, the Scoutmaster, is
going to be there, and he says for us not to eat too much supper because
there will be something doing along about eight o'clock. It will beat
the Fourth of July."
We hardly could wait for evening to come. The folks thought that I must
be sick because I didn't want much supper, until I told them about the
campfire.
"You'd better eat a bowl of bread and milk, anyhow," said Ma. "If I know
anything about boys, and I have seen a few in my day, you will be ready
for another meal by eight o'clock."
I don't know how it is, but things always seem to happen just as Ma says
they will. Long before eight o'clock came we were waiting for Mr.
Norton at our house, as hungry as bears.
After a while he came along, lugging a big basket and wearing a smile
that would have made us warm to him if we never had before.
"Captain," said he to Skinny, "if you will detail two of your men to
bring some water, we'll get started. Of course, if we were going to make
a regular camp we should see that there was water near. We'll have to
carry it this time, but it isn't far to the top of the hill. One of you
might help me with this basket; there seems to be something in it."
Fifteen minutes later we were all at the top of the hill and had brought
some sticks from Plunkett's woods for a fire and a curl of birch bark to
kindle it with.
"I understand that you boys came near burning up the woods and village
once with a fire up here," said Mr. Norton. "We must be careful about
that. Fire is a good servant but a very hard master. We do not need a
big blaze for a campfire, so hot that we cannot sit around it. All we
need is just enough to look cheerful, to heat our coffee, and furnish
enough hot coals for cooking this beefsteak."
He was unpacking the basket while he talked, and Skinny was lighting the
fire.
"I don't know that I can tell you anything about making fires and
cooking. You boys just about live out of doors in summer, so far as I
have observed. You are in great luck to have your homes in a small
village. If you should play some of your pranks in a city, I am afraid
that you might become unpopular and the police might get after you. Boys
in great cities, like Chicago or New York, know little of the freedom
and sweetness of country life."
He went over to a little clump of trees and came back with a small
branch, from which he stripped the leaves and twigs. When he had
finished he had what he called a "pot hanger" of green wood, about four
feet long and with a kind of crotch at the smaller end. He put the big
end under a stone, the right distance from the fire, and drove a short,
crotched stick into the ground to hold the pot hanger over the blaze at
the right angle. When that was done all we had to do was to hang a pail
of water on the end of the pot hanger and wait for the water to boil.
"I thought that we wouldn't bother with potatoes this time," said he,
"although they make good eating when baked in hot ashes, as you boys
probably know. Mrs. Norton put in a whole stack of bread and butter
sandwiches and some other things, which we must get rid of somehow, and
Mrs. Smith gave me this bag as we were leaving the house. I don't know
what is in it, and she told me not to open it until the feast was
ready."
We all kept our eyes on the bag and wondered what was in it. I thought
that I could make a good guess, being better acquainted with Ma than the
other boys were, but I couldn't be sure.
By the time the water was boiling the fire had burned down to red-hot
coals. Mr. Norton poured the water over the coffee and set the pot in a
hot place. Then he began to get busy with the meat, using a broiler
which he had brought in the basket. The delicious smell of the beefsteak
and the coffee almost drove us crazy, and we began to be afraid that it
would bring the whole village up the hill to us.
It seems as if every meal that we eat out of doors that way is better
than any which we ever have had before. It grew dark before we had
finished Ma's doughnuts, which we found on opening the bag. As we sat
there we could see lights begin to glow all up and down the valley and
back of us from an occasional farmhouse, up toward Greylock. Stars came
out overhead, and after a little we saw a light in the sky above the
East mountain and knew that in a few minutes the moon would come up.
After we had eaten all that we wanted, we threw some wood on the coals
to make a little blaze, and then lay around and talked.
Finally Benny said, "I wish you would tell us a story, Mr. Norton, like
Mr. Baxter did out in Illinois last summer."
"I am going to tell you a whole lot of stories before we get through
with our meetings," he replied, "but let us discuss this Scout business
a little more first. When you took the Scout's oath and were enrolled
in the Tenderfoot class, you pledged your word of honor that you would
do your duty to God and your country, that you would help other people
at all times, and that you would obey the Scout law. That Scout law is
important. Suppose we talk it over. Gabriel, you are leader, can you
tell us what the first law is?"
Skinny stood up and folded his arms.
"A Scout is trustworthy," said he.
"It is a great thing to be trustworthy; to be dependable," said Mr.
Norton. "In a few years, you boys and others like you will be running
this country and the other countries which make up what we call the
civilized world. To you doubtless that time seems far off. Let me tell
you that it will be here almost before you know it. It seems only
yesterday when I myself was a youngster like you."
"I'm going on twelve," Benny told him, "and I have begun to grow again."
"The Band is dependable all right," said Skinny, stabbing around in the
air with his fork. "I mean the patrol is. Bet your life, when they
monkey with the Band they run up against a buzz saw."
Bill didn't say a word, but he cawed three times; then flapped his arms
and crowed, and ended by standing on his hands and kicking his feet in
the air. Bill didn't have to talk. He could do things that made us know
what he meant, without saying a word.
"To be dependable," went on Mr. Norton, "means more than to fight for
your rights, or for your country's rights. It means that in all walks of
life you must be ready to 'deliver the goods.' When a Scout gives his
word of honor that settles it. That which he says is true, is true; you
can depend upon it, and he will do exactly what he says he will do. That
is a quality which we greatly need in men as well as in boys, who soon
will be men."
"Corporal, what is the second law?"
Bill thought a minute and then said:
"A Scout is loyal."
"Right you are. You must be loyal to your country, to your parents, to
your officers, to your employers, when you get to work. Loyalty is a
great thing. It means to stick together. One boy, or one man, alone,
cannot accomplish much. Several working loyally together for a single
object, are a power. You and the Gingham Ground Gang used to have
considerable trouble, didn't you?"
"We do now," we told him, "except with Jim Donavan. Jim is square and
we'd like to have him join us, but he won't leave the Gang; says it
wouldn't be right."
"That is the kind of boy we want for a Scout. He is loyal and his honor
is to be trusted. You must help me to organize the Gang, as you call
them, into another patrol. But what I was going to say is this: When you
and the Gang were enemies, which I hope you never will be again, what
would have happened if one of you had ventured alone down near the
gingham mills?"
"They would have done him up."
"Exactly. Now suppose the eight of you had stood together, back to back,
shoulder to shoulder, working against a common enemy?"
"We did once," said Benny, "and they licked us, anyhow, but there were
more of them than there were of us."
"Bet your life they didn't lick us very bad," put in Skinny. "It was a
snowball fight. They drove us from their hill, but afterward they asked
us to come back and slide with them, and we did. We had a fine time."
"It seems to me that in that case both sides won a victory. The greatest
victory a boy or man can win is one over himself, over his own passions,
his selfishness and meanness. The greatest enemy that he or his country
can have will be found right inside his own heart. There is where we all
have a fight on hand continually. But, remember, you are Scouts and a
Scout's honor is to be trusted."
"Benny, what is the next law?"
"A Scout is helpful."
"There you have it. The highest type of man is the useful one. There was
once an old philosopher who said that he counted that day lost in which
he did no good deed. A Scout ought to feel the same way. You must try to
do something for somebody every day."
"They don't have giants and dragons, any more," said Skinny. "I wish
they did; we'd paralyze 'em."
"Henry, what is the next one?"
"I am not quite sure whether it comes next or not, but I think it does.
The law says, 'A Scout is a friend to all and a brother to every other
Scout.' Does that mean that we must be brothers to the Gingham Ground
Gang when they get to be Scouts?"
"Surely it does. Why not? Your folks may have a little more money than
their folks and not so much as some one else. What of it? There is
something better than money, and that something is manhood. Don't be
snobs, whatever you are."
"Now, Mr. Secretary, it is your turn."
"A Scout is courteous," I told him.
"Politeness is a great thing. If he lives up to his pledge, a Scout will
be courteous, especially in his treatment of women and children who are
younger than he is, and of old people and those who are feeble or
handicapped in some way by being crippled or sick. Don't forget that old
men started as boys and that you boys, if you live, will become old
men. Now for number six."
"A Scout is kind and a friend to animals," Harry said.
"And the next?"
"A Scout is obedient," said Chuck.
"Now we are getting down to business. The first duty of a soldier is to
obey, and it is so important that he should obey in time of war that a
soldier, or scout, who refused to obey orders would be shot. You are
supposed to obey orders without question. Obey your parents especially.
Obey me as Scoutmaster. Obey your patrol leader; that is your duty as
Scouts. If the order does not suit you, do your kicking afterward, not
before. First deliver the goods; then you will be in a position to
criticise, if necessary."
"We haven't heard from you, Wallie. Let's have number eight."
"A Scout is cheerful."
"That's the idea. Don't grumble or whine. That will never get you
anywhere, or the world anywhere.
"I want to say a few words about the next law, 'A Scout is thrifty.'
Thrift is of the greatest importance. Save your money. Save your
pennies. Put them in the bank. I think they ought to teach thrift and
the importance of saving in the public schools. It does not mean that
you should be stingy. When you boys worked hard one winter and gave a
purse of money to an unfortunate stranger, you were living up to the
highest ideals of a Scout. It doesn't mean that money is the most
important thing in the world, for it is far from it. But remember this:
a man's first duty to his country is to be self-supporting, and to be
self-supporting in his old age he must be thrifty in his youth. He must
make hay while the sun shines. He must learn to save his money. That is
why a Tenderfoot must have one dollar in the bank before he can become a
Second Class Scout, and a Second Class Scout must have two dollars
before he becomes a First Class Scout. The habit of thrift is very
important. When you grow older and go to work, no matter what you earn,
I want you to save a part of it.
"There are three more laws," he went on, after a minute, "and they speak
for themselves: 'A Scout is brave,' 'A Scout is clean,' 'A Scout is
reverent.' I need not tell you to be brave in the presence of danger. Do
you understand that sometimes it takes greater courage to stand up for
the right? Keep yourselves clean; not only your bodies but your thought
and speech. And be reverent, boys, toward God, who made old Greylock and
these beautiful hills for you to enjoy."
When he had finished Skinny started to throw some wood on the fire, but
Mr. Norton stopped him.
"Never go away," he said, "leaving a fire where it possibly can do any
damage. We'll be going home in a few minutes, and before we go this fire
must be put out. If the wind should come up in the night the flames
might spread into Plunkett's woods."
We saw in a minute that he was right, and, taking sticks, beat out what
little fire there was; then started down the hill.
"I'll tell you what I have been thinking," said Mr. Norton, when we
were going through Blackinton's orchard. "We have had so much fun
to-night that I should like to go camping with you boys for a week, some
time this summer. These mountains and woods are just the places for
scouting and we could have a campfire every night. What do you say?"
"We say yes," said Skinny, "if our folks will let us, and I know they
will."
"Can we play Indian, Mr. Norton?" asked Benny.
"We certainly can. I think everybody likes to get out into the woods and
be an Indian once a year. You boys have something to do first, however.
I want every one of you to be able to show a First Class Scout badge."
"We can do most of the stunts now," I told him, "only we haven't been
seven miles and back."
The book says that before becoming a First Class Scout a boy must go on
foot to a point seven miles away and return again, and afterward to
write a short account of the trip. It says, too, that it would be
better to go one day and come back the next, and that means to camp out
all night.
That last was a hard thing to do because our mothers did not want us to
go off that way alone. Mothers always seem to think a boy is going to
get hurt or something. Mr. Norton finally talked them into it, all
except Benny's mother. She wouldn't stand for it. Benny cried, he felt
so badly about it.
"Do it in one day, then," Mr. Norton told him. "Remember that the law
says for you to obey your parents without question. That is more
important than to do the stunt."
CHAPTER VI
A FOURTEEN-MILE HIKE
SCHOOL let out Thursday, June 22, and it had seemed to us as if the day
never would come. Not that we don't like school because we
do--sometimes; but when the sap drips from the maples and bees buzz
around the pussywillows on the river bank and all the trees take on a
different look, as if there was going to be something doing right away,
then the time has come for us to get out our marbles and tops and to fix
up the cave for the summer.
Pretty soon the buds begin to throw off their overcoats, and Bob's Hill
grows green again in the warm sunshine; the woods are bright with wild
flowers, and the songs of birds and smell of spring fill the air.
Then the mountains and hills tease us away from our books, when we look
out of the window. The river, all swelled up with joy and melting
snows, shouts for us to come on, every time we cross the bridge. On
Saturdays the brook at Peck's Falls, grown big and noisy, roars out a
welcome and tries to say how glad it is to have us back at the cave
again.
Say, how can a boy sit quiet in school when all those things are going
on?
Last day finally came. It always does, no matter how slowly the time
seems to pass. The very next morning the Ravens met to do the final
stunts that would make us First Class Scouts.
For more than a week we had thought of little except the fourteen-mile
hike. It took several meetings before we could decide where to go. Our
first idea was to tramp up into the mountains somewhere, but that scared
our folks and we had to give it up.
"It isn't as if you were all going together," said Pa. "In that case, if
one should get hurt the others could take care of him and go for help.
If one of you alone should break your leg on the mountain we might never
be able to find you. I think you'd better stick to civilization and the
beaten paths. You are not mollie-coddles and probably would come out all
right, anyhow. At the same time, I should sleep better nights if I knew
that my boy wasn't off on the mountain somewhere, alone."
That left us only two directions to go, north and south, because on the
east and west there are mountains and the valley between is narrow.
South near Cheshire Harbor it narrows down so much that there is room
only for a wagon road, the river, and the railroad, side by side, but
there is another road part way up the hill on the east.
On that account we decided that all should not go on the hike the same
day, but to go four at a time, each taking a different road. There are
two roads leading north to North Adams, one on each side of the river,
and two leading south. One goes through Maple Grove and Cheshire Harbor
to Cheshire, where a lot of swell folks from New York spend their summer
vacations. The other, as I have said, is part way up the east hill and
goes through a place, called Pumpkin Hook. It's a queer name but we
didn't name it.
The plan that we finally decided on was for each to follow one road one
day for seven miles; then go up into the hills somewhere to make camp
for the night, and the next day to go back again by the other road. In
that way we should stand a chance of meeting two Scouts some time during
the trip, one on the morning of the second day, when we would be
crossing over to take the other road, and one when the first boys on
their way home would pass the second boys on the way out.
We drew cuts to see who should be the first four to go. Skinny, Harry,
Wallie, and Bill won the first chance. They were to start the next
morning at seven o'clock sharp from the bridge, two going north and two
south. Hank, Benny, Chuck, and myself were to wait until seven o'clock,
the second day, and then start. When we all had come back, we planned to
meet Mr. Norton and tell him about where we had been and what we had
seen and done.
Benny and I live nearest to the bridge. My house is only a stone's throw
north of it; Benny's is a little north of mine and on the other side of
Park Street. That made it easy for us to get to the bridge first, but
pretty soon the others began to come.
"Has anybody seen Skinny?" I asked, looking at Mr. Norton.
Skinny's house is near Mr. Norton's, and we had thought that maybe they
would come together.
"I stopped in as I passed," said he. "Mrs. Miller told me that he had
started."
Just then we heard a caw, sounding from over toward Plunkett's woods
somewhere. It didn't take us long to answer. Then we watched down the
railroad track, where it curves into town between the wooded hillside
and the river.
We didn't have long to wait. In a few minutes we saw Skinny put his head
out between the trees which line a high bank, fifteen or twenty feet
above the track. He looked carefully in every direction; waved one arm,
when he saw that we were watching, and then dodged back again out of
sight.
"He's surrounding something," said Bill, giving a caw so loud it must
have almost scared the crows up in the Bellows Pipe.
"There are only four minutes left before leaving time."
Mr. Norton was looking at his watch. He had hardly spoken, when, with a
whoop and yell, Skinny slid down the embankment and was running like mad
up the track toward us, waving his hatchet in one hand and swinging a
rope around his head with the other.
"One minute to spare," said Mr. Norton, smiling as he put his watch back
into his pocket. "That's the way to do it. Be prompt. If you say that
you'll be somewhere at a certain time, be there."
"Say, Skinny," said Bill, winking at me and giving the Scout salute,
"did you get 'em surrounded?"
Skinny wouldn't answer, or even look at him except to return the salute.
He pulled out his own watch, held it a moment; then pounded on the
bridge with his hatchet.
"The meetin' will come to order?" said he.
As he spoke, the bell on the woolen mill began to ring and we knew that
it was seven o'clock and time to start.
Quite a little crowd had gathered by that time and there was a cheer
when the boys started, Skinny and Harry marching south on Center Street,
side by side, and Bill and Wallie, north on Park Street.
Pretty soon their ways branched off. They turned and waved to us; then
were gone. Once after that we heard some crows cawing in the distance,
and a little later I heard Bill yell from somewhere down the river. I
knew that he was doing his best, but I hardly could hear him.
It wasn't easy to wait until the next day, with the other boys gone and
knowing that we should have to do it, too, in the morning.
Pa said that maybe the time would pass more quickly if I'd hoe in the
garden a spell, but it didn't seem to make any difference. My mind was
following the boys, especially Skinny, on his long walk over a hilly
road to Pumpkin Hook.
"Scout's law says that we must be useful and help others," he had told
us, "and, bet your life, I am going to do things."
"Maybe," said he, after a minute, "I can rescue some fair damsel in
distress, like the knights used to do, even if there ain't any dragons
now-a-days. The road goes too far from the river for me to save anybody
from drowning; unless I come back by the river road."
In the evening Benny and I sat out on the woodpile, talking about it. We
wondered where the boys were making their camps, if anything would
happen to them and if Skinny had rescued anybody yet.
That night I dreamed that I was on the way. I met a little, old woman,
going to market, and carried her basket for her.
"Noble boy," said she. "Because of your kind act I'll change shoes with
you. Mine hurt my feet."
I didn't like to do it very well because her shoes were old and shabby,
but Scout law says to be courteous. So I thanked her as well as I could
and put them on.
And, say, they were magic shoes. I got to North Adams in about three
jumps and liked it so well that I went on to Boston. I was just going to
sleep on Boston Common when a big policeman grabbed me by one shoulder
and gave me a shake.
"Quit!" I said. "A Scout's honor is to be trusted."
"John! John!" came a voice. "It's time to be up and away."
I opened my eyes and there was Pa, laughing down at me.
"You're a pretty Scout," said he. "It's after six o'clock and you have
to start at seven."
Ma hated to see me go, knowing that I'd be out all night, but Pa didn't
care, or pretended that he didn't.
"He's all right," he said. "What's going to hurt him, I'd like to know?"
Before seven o'clock the four of us were at the bridge and, say, we
looked fine in our uniforms. Each one carried a little pan to cook in,
some bacon and other things to eat, and a blanket strapped on his back.
We also carried "first aid to injured" things, to be ready if we should
find somebody getting hurt.
When the bells rang for seven o'clock we started. This time it was
Benny and I who went north on Park Street, and Hank and Chuck, south.
"You watch my smoke," whispered Hank to me, when we were ready to start.
"I've got a new invention and I'm going to try it on somebody."
When we were passing Benny's house Mrs. Wade came out and waved to us.
"Benny Wade," she shouted, "if you are not home by nine o'clock
to-night, your mother will have a fit."
I knew from the look on Benny's face how hard it was for him to be
cheerful, when he wanted to stay out all night, like the rest of us.
"All right, Ma," said he. "Don't worry. I'll come back, if I live."
"If you live!" I heard her yell; but Benny was turning the corner to
take the east road and in another second was out of sight.
At first I hardly could believe that I really was on the way. I took Mr.
Norton's message out of my pocket and looked at it, to make sure,
several times. He had given each of us a message to some one at the end
of the line and told us to bring back a receipt or an answer. Mine was
to a man in North Adams.
The Bob's Hill boys are used to walking. That didn't bother me any. But
somehow this was different from any other walk that I ever had taken. I
suppose it was because it was so important and because I was all alone.
I walked along at pretty good speed until I had almost reached the
Gingham Grounds. Then I slowed down and kept my eyes open for the Gang,
hoping that I should see Jim Donavan somewhere. Jim was their captain
and one of our best friends, but some of the others had it in for us.
I had begun to think that I was going to get through all right, without
any trouble, when I saw one of them coming toward me. He was one of the
best fighters in the Gang, too, and he had a dog with him. Jim was
nowhere in sight.
Isn't it queer what things will come into your head when you are scared?
Pa says that I can't remember twenty-five cents' worth of groceries from
our house to the store; but that is something else.
I was scared, all right, and wanted to run, because fighting always is
scary until after you get started. Then, all of a sudden, I thought of
something that Pa had once read to me about General Grant. Grant was
marching up a hill once, expecting to find the enemy on the other side
and wanting to run all the time, only he was too proud. Then when he
reached the top, where he could see down into the enemy's camp, he found
that they had been more scared than he was and not so proud, for they
had run away.
"So," said he, or something like it, "no matter how frightened you are,
or how much you want to run, remember that the other fellow probably is
just as badly scared as you are."
When I thought of that I braced up and walked along fast, pretending
that I was in a hurry and didn't see him, but keeping one eye on him,
just the same, and the other on a stone which lay in the road, near
where the dog stood whining. The boy was patting his head and trying to
coax him along.
He pretended that he didn't see me, too, until I was passing. Then he
spoke.
"Hello, you village guy," said he.
"Hello, yourself," I said, stopping and edging toward the stone.
"Where do you think you are going?"
"North Adams."
"What for?"
"Oh, just for fun."
"Huh!" said he. "Ain't the trains runnin'?"
"I've got something that's better than trains. It's legs."
"What's the uniform for?"
"Anything the matter?" I asked, after I had told him that I was a Boy
Scout, for I could see that he was feeling badly about something.
"It's my dog," he told me, rubbing his sleeve across his eyes. "Somebody
broke his leg with a stone and I've got to kill him. He's all I have."
"A Scout should be kind to animals," I said to myself. "A Scout is a
friend to all." "A Scout should be useful."
Then I answered myself back.
"What's the use? This ain't any damsel-in-distress business, like Skinny
is going to do. Besides, if I hurry maybe I'll get a chance to signal to
Benny from the turn in the road on ahead."
"Come on and help me kill him," said he.
Just then the dog gave such a pitiful whine that I couldn't stand it,
Benny or no Benny. So I took out my bandage.
"I think I can fix his leg, if you'll help me," I told him. "Get me a
couple of sticks."
I told him what I wanted, and when he had brought them and I had
whittled them into shape to use as splints, I fitted the broken bones in
place and bandaged the leg, just as Mr. Norton had taught us, while the
boy held the dog. The dog yelped a little, but seemed to know that I was
doing it to help him.
"It will soon grow together," I said, when I had finished, "and then it
will be almost as good as new."
It made me feel kind of queer and happy to see how glad he was. The dog
licked my hand, too, and seemed to be trying to say something. I wish
dogs could talk.
"How did you come to know so much?" he asked. "Is your father a doctor?"
Then I told him all about the Scouts and our hike and what Mr. Norton
had said about wanting the Gang to join.
"Bully!" said he. "We'll do it. The others went up on the mountain this
morning after berries. I'd have gone, too, only for the dog. But I'll
tell them when they get home to-night."
"Say," I called out, after I had started on. "You know Benny Wade, don't
you?"
"The kid what always goes around with youse?"
I nodded.
"Yes, I know him when I see him. Why?"
"He'll come through here this evening some time, on his way back from
North Adams. Let him look at the dog and see if he is all right. He
knows as much about those things as I do. Bill Wilson ought to be along
some time during the day on his way back. He started yesterday. Say, you
ought to see Bill do up a leg."
Nothing happened after that, although I kept close watch of the river,
hoping that I might find somebody drowning. Some boys were in swimming
at one place, but they were not drowning nor anywhere near it.
I could have reached North Adams easily long before noon, if I had
wanted to, but I had all day to do it in, so loafed along, expecting to
meet Bill every minute. I rested in the shade whenever I felt like it.
But although I did a lot of cawing every few minutes and kept a sharp
look-out, I didn't see Bill, and I didn't hear him, which I couldn't
understand, unless he had taken the east road home to keep away from the
Gingham Grounds.
At noon I went down by the river, cut a pole, and fished a little,
although I didn't catch anything. I didn't build a fire and cook because
I had a good lunch in my pack. It seemed sort of lonesome, being there
so far away and knowing I couldn't go home when night came.
After a long rest I walked on until I came to a bridge, and then,
feeling sure Benny must be in North Adams by that time, I crossed over
to the east road, where I knew some folks, and went up into the hills to
where Hoosac Tunnel begins. It was fun to see the trains dart in and out
of that great hole which reaches four miles through the mountain, and I
sat there a long time watching.
Four o'clock came before I found my man in North Adams and delivered the
message. By that time I was tired enough to go into camp for the night.
He smiled when he saw me coming in my Scout uniform.
"This letter," said he, when he had read it, "says for me to buy you a
life size ice cream soda? Do you want it?"
There isn't anything in Scout law, is there, which says a Scout mustn't
eat ice cream soda? And the tireder and hotter you are the better it
tastes, doesn't it? I guess yes. Only I wished that Benny was there,
eating one with me.
That night I camped on the bank of a brook, part way up the mountain and
a mile or more beyond the city. The water was clear as crystal and
seemed kind of company, for it gurgled as it poured over the stones,
making music that was great.
I hardly could wait to build a fire and fry my bacon, I was so hungry.
But what is the use of carrying bacon and a pan seven miles, unless you
fry the stuff after you get there? I tell you it tasted good and so did
the wild strawberries that I picked afterward for dessert.
But when it began to grow dark and lights shone out down in the city and
in the sky above, and queer sounds came from the mountain and woods back
of me, I'd have given fifteen cents to have been at home, or at any
rate, to have had somebody with me.
After a while I heard a voice say:
"A Scout should smile and look pleasant."
"Who--who--is that talking?" I asked.
"It's your friend, the brook," came back the answer, in a sweet, gurgly
voice. "I'm a Scout, too. Hear me sing."
"So am I," came the deep voice of the mountain back of me. "A Scout
should be brave. Sleep, my brother. I'll watch over you."
"So are we Scouts," came in whisperings from every side, through the
darkness, and I knew that the trees were talking to me. "We'll take care
of you."
Then I grew brave all in a minute and started up to go to them. As I did
so, the darkness fled, leaving me there lying on the ground in broad
daylight, while the brook sang its loudest and all the trees waved
good-morning. Would you believe it? I had slept all night long and
dreamed that about the brook and the mountain.
On the way home, I came in sight of the houses of the village before ten
o'clock, tired but happy because I had done the last test and now could
be a First Class Scout.
Benny met me outside the village, and he looked scared when he saw that
I was alone.
"Have you seen Bill Wilson?" he shouted, as soon as he could make me
hear.
"I missed him somewhere," I called. "He must have come back by the east
road. Why? What's the matter?"
He already was hurrying home so fast that I hardly could catch up with
him. As he ran he shouted back over his shoulder something that set my
heart to beating and made me forget how tired I was.
"Bill hasn't come back."
CHAPTER VII
"BILL HASN'T COME BACK"
ALL it meant to say that Bill hadn't come back did not come over me
until I found myself hurrying after Benny down Park Street. Bill had
left home on the morning of the second day before, intending to camp out
one night and come back the next day. Two nights had passed and he was
still away. What had become of him?
I hurried along faster and faster, thinking of all the things that might
have happened. Mr. Norton and Bill's folks reached the house almost as
soon as I did. I don't know how they found out that I had come back.
Bill's folks were nearly crazy about him. The first night out, they
expected him to be away, of course, and so did not worry much. When
dinner time came the next day and he hadn't showed up, they began to
wonder what was keeping him, for the other boys who had started at the
same time were home.
When night came again and he still was away, they began to grow very
anxious and sent for Mr. Norton.
"I can't understand it," said he. "I supposed that he had come home long
ago, and have been too busy to find out. The other three are back, I
understand."
"Yes, they came back in time for dinner."
"I am surprised that William is still out, but do not feel alarmed, Mrs.
Wilson. Something has detained him, but it cannot be anything serious.
Both roads to North Adams are well traveled and the farmhouses are near
together. As likely as not he stopped to help somebody out of a
difficulty and it has taken longer than he expected. One of our laws,
you know, says that a Scout's duty is to be useful and to do somebody a
good turn every day. I'll run over and talk with Wallace. They started
together and may have met when they crossed over from one road to the
other."
Mr. Norton was more anxious than he pretended. Wallie said that he
hadn't seen him and hadn't heard him, which was worse, for Bill usually
could be heard a long way off. Wallie said that he had called to him
every few rods when crossing over to the west road beyond North Adams
but hadn't heard a thing. It would have been easy for them to miss each
other, unless they happened to take the same crossroad.
"I might get track of him in North Adams," said Mr. Norton, after a
little. "You see, I gave him a message to deliver to a friend of mine
there. He surely will know something about him, but he hasn't a
telephone and I think is out of town to-day, anyhow. Maybe I'd better
drive up. The boy probably will get back before I do, but it will make
me feel better to be doing something."
By that time everybody was getting scared. I mean all our folks were.
Mrs. Wade was sure that Benny never would come home again, although it
wasn't quite nine o'clock, the time when he said he would come.
Mrs. Wade is all right most of the time, only she can think of more
trouble for Benny to get into than he could find in a week, if he
looked for it. Mothers are often that way. I guess it is because they
like us so well.
* * * * *
"He said he would come back, if he lived. Those were his last words. And
he hasn't come."
She told that to Ma, over and over again.
"He'll come back all right," said Ma, "and so will John, when the time
comes."
But she was worried about me, just the same, all on account of Bill. Of
course, I didn't know about it at the time. I found out afterward.
No one ever made better time driving the six miles to North Adams than
Mr. Norton did that night. Just outside the village he met Benny, coming
on a run, and stopped long enough to ask him if he had seen Bill.
"No," said he. "I missed him. The Gang held me up at the Gingham Ground
and almost made me late. I told Ma that I would be home by nine o'clock
if I lived. I'm 'most dead, but guess I can hold out until I get there.
She'll be having a fit pretty soon if I don't hurry. What time is it,
anyhow?"
Mr. Norton whipped up his horse before Benny finished.
"William hasn't come back!" he shouted over his shoulder, just as Benny
called to me in almost the same place. Then he tore down the road toward
the Gingham Ground.
It was after midnight when he came back. There was a light burning in
our house and he stopped.
"He has not been there!" was all that he could say, when Pa met him at
the door.
"Hasn't been there!"
"No, I found Jenks, to whom I had sent the message, and he said that he
had seen nothing of him, although he had been expecting him. You see, I
told him that the boy was coming. The message has not been delivered."
"Mr. Smith," he went on, after a moment, "I can't face Mrs. Wilson with
that news. You go to her, while I get the marshal started and see if
something cannot be done. I tell you something has happened. I am
convinced of that. Young Wilson would have delivered that message if he
possibly could have reached the place, and it would have taken a great
deal to stop him. There isn't a yellow streak in that boy anywhere."
"Did you make any inquiries?"
"Yes, I stopped at every house along the road where there was a light
burning. Not a person had seen him, although several had seen your boy
on the way out. At North Adams I notified the police, but I don't know
what they can do."
"I'll go to Mrs. Wilson right away," Pa told him. "This certainly is bad
business, but we can't do much until morning. As soon as it is daylight
we'll send out a search party. There are only two roads, unless he went
up through the Notch, which is not at all probable. It ought not to be a
difficult matter to get some trace of him."
"I'll tell you where he is," he went on, after thinking a minute. "He
met my John and went back to camp all night with him. They will come
home together to-morrow; you see if they don't. John is a pretty safe
boy. He's full of pranks, like the others, but he is more cautious.
He'll come home all right and bring Bill with him."
Mr. Norton shook his head.
"I sincerely hope so," he said, "but it is not at all probable. Mr.
Smith, I never will forgive myself if anything has happened to that
boy."
"You are not to blame at all," Pa told him. "Depend upon it, if anything
has happened, and we don't know that there has, the boy himself is to
blame. He is a fine lad, but is a little reckless and thoughtless at
times. Cheer up. It might be a lot worse. Now, if the boys had gone up
into the mountains as they talked of doing at first, there would be real
cause for worry."
That was why Benny waited for me outside the village the next day, and
why Mr. Norton and Mr. and Mrs. Wilson met me at the house and why
Skinny and the other boys came in a few minutes afterward.
Mrs. Wilson knew by my face that I had not seen anything of Bill and
burst out crying.
"There couldn't have anything happened to him, Mrs. Wilson," I told her,
sort of choking up in my throat, myself, because she was feeling so
bad. "I mean anything much. Maybe a tramp locked him up somewhere when
he was asleep, or some gipsies stole him. I saw some gipsies up above
North Adams and they were going west to beat the band. But he'll get
away from them. I'll bet on Bill every time."
When I spoke of gipsies to make Mrs. Wilson feel better it seemed to
scare her worse than ever.
"Nonsense!" said Pa. "Gipsies don't go around stealing thirteen-year-old
boys, who can make as much noise as Bill can."
"Well, I saw some, anyhow," I told him.
Just then Skinny jumped out in front of the rest of us, with his eyes
shining and his cheeks redder than I ever had seen them before, and
stood there with his arms folded, like a bandit, or a Scout, I don't
know which.
"Fellers," said he, "Scouts, I mean. We got Bill into this scrape and we
will get him out again. This is a job for us, not for the police. If
anybody can find Bill, bet your life we can. We know the call of the
Ravens. We know the signs and we know Bill better than his own folks
know him. We'll track him. We'll follow him to the ends of the earth.
Will you go with me?"
We sprang up with a cheer, forgetting how tired we were, those of us who
had just come home from the long walk.
"Everybody scatter and look for signs."
"Wait a minute, boys," said Ma. "It's almost dinner time. You must not
start without something to eat. There is no telling when you will get
back. Let me give you a bite in the kitchen first."
That was just like Ma. We saw in a minute it was the thing to do and
hurried in for a quick lunch.
"The boy is right," we heard Pa saying. "They'll find him, depend upon
it. I never knew those boys to get into a scrape yet that they couldn't
pull out of. But it won't hurt if the rest of us look around a little,
too."
"Who saw him last?" asked Skinny, after we had started.
"I did," said Wallie. "We walked together until I turned off to take
the east road. He kept straight on toward the Gingham Ground and I heard
him yell some time afterward."
"You don't suppose that the Gang got after him, do you, and locked him
up or something?"
"I'll bet that's what they did," said Benny. "That is just what
happened. They got after me, too. I was scared half to death and didn't
want to go through the Grounds, but it was getting late and I knew that
Ma would be worried, so I braced up and started through on a run. In a
minute two of them ran out and grabbed me by the collar."
"'It's one of them village kids,' said one of them. 'Let's call the Gang
and duck him. He needs it to cool off.'
"Then he whistled and a lot of the others came and they hustled me down
to the river. Gee, I was mad and I was scared. Then, just as I had about
given up, another boy came chasing after us.
"'Is this Benny Wade?' said he.
"'It's all that is left of me,' I told him.
"With that he jumped in and took hold of me.
"'Youse ain't a goin' to duck this kid,' said he, 'unless you duck me
along with him. His partner came through here this morning and fixed my
dog's broken leg and he told me to watch out for Benny Wade and have him
look at the bandage, to see if it was all right. Now, kid, you come
along with me and look at my dog.'
"'Duck 'em both,' said some one.
"I guess maybe they would have done it, too, if Jim Donavan hadn't come
along just in time."
"Maybe it was Bill who fixed up the dog," said Hank.
"No, I did it," I told them.
We had been walking along while Benny was talking. What he said
surprised us some and would have made us mad at any other time. Benny
had been so worried about Bill that he hadn't said anything about
himself before, and neither had any of us.
"The first thing to do," said Skinny, "is to go to Jim's house and start
from there. If Bill went through the Gingham Ground I'll bet that some
of the Gang saw him."
The place which we call the Gingham Ground is a settlement near some
big gingham mills. There are two long rows of brick tenement houses with
a street between. We knew that Skinny was right, because Bill would have
had to walk down that street between the rows of houses, and some one
would have been sure to see him. He might have stopped at Jim's, or,
anyhow, would have called to him when he passed.
It didn't take us long to get there, and as we came near we could see
the Gang getting together. You see, they thought we were after them on
account of what they had done to Benny.
We didn't pay much attention to them but went straight to Jim's house
and found him eating dinner. He was surprised to see us and was glad.
"Wait until I call the Gang," said he, after we had told him about Bill.
In a few minutes they had all come up, as friendly as could be when they
found out that we were not looking for a fight.
Not one of them had seen Bill. They all knew him and they felt sure that
if he had gone through in daylight some of them would have seen him.
"I'll tell you what we'd better do," said Jim, finally. "I don't believe
that he came this way, but, to make sure, the Gang will work north from
here and ask at every house. You go back and look between here and the
village. If he left there and didn't get as far as this, then he must
have turned off somewhere."
We went back, stopping at every house we came to, on each side of the
road. We couldn't find a person who remembered having seen him or any
one like him. You see, if he passed at all, it must have been soon after
seven o'clock in the morning. The men had gone to work in the mills and
the women were busy in the back parts of the houses.
Then we started back again, not knowing what to do next. There was one
house, larger than the others, which we had not visited, because it
stood high above the road on a hillside and could be reached only by a
long driveway. It was about halfway between the Gingham Ground and our
house in the village. We couldn't think of anything else to do, so we
went up there.
"I don't remember seeing any one," said the lady who met us at the
door. "Of course, there are boys passing at all hours of the day. I
might have seen him."
We looked at Skinny in despair.
"This one," said he, "was probably making a noise. Maybe he was cawing
like a crow."
"I saw him, Mama," shouted a little girl, who had come up and stood
listening. "I saw a boy go past, making an awful racket, and it sounded
something like a crow."
"Was he carrying anything?" I asked.
"Yes, he had a rolled-up blanket on his back. I remember thinking he
looked funny and wondering what he was going to do with it. Oh, yes, he
had on a uniform, too."
"It was Bill, all right," said Skinny. "We've struck the trail at last."
We went down to the road and talked it over.
"He passed here," said Skinny, "on time and going north, and he didn't
pass through the Gingham Ground. We feel sure of that much. He must have
turned off somewhere in the next half-mile."
"We know something else," I told him. "He couldn't have turned east,
because the river is in the way and there isn't any bridge."
We made up our minds to separate, one party to work north from where we
were standing; one to work south from the Gingham Ground, and the others
to work in between, to see if we could find where he had left the road.
"Look for a sign," said Skinny, "and look on the west side. There isn't
much chance for finding footprints."
Hank was the one who found it. We heard him yell and went to him on a
run.
He came out to the roadside and waited for us, waving his hat in the
air, he was so excited; then, when we had come up, took us back from the
road through a sort of lane, which pretty soon turned south and wound
off through the woods.
Just at the turn stood a big stone, out of sight from the road. That is
why we had not seen it before. On the stone was something which set us
all yelling.
It was a circle and in the circle was the picture of a crow and there
was an arrow. It was the Scout sign for "I took this path." The crow
meant that whoever drew the sign belonged to Raven Patrol. We knew then
that it was Bill.
"We've got him," shouted Skinny. "He went through this way so as not to
meet the Gang."
It did look like that, but although we examined every inch of the way
between there and the Gingham Ground, we couldn't find another sign of
any kind. And we couldn't understand why he had not delivered the
message to Mr. Jenks and come back home.
Sorrowfully we made our way out to the sign again and sat down to rest
and talk about what to do next.
"Guess what!" said Benny, after a little. "That arrow doesn't point
toward the Gingham Ground at all. It points straight back from the road.
Let's go that way and see."
There didn't seem to be much use in doing it, but we had to do
something.
"Come on," said Skinny, springing up. "He is somewhere; that's a cinch,
and we know that he was all right when he drew that sign."
We hurried along and soon struck a little path, up which we ran as fast
as we could, for it was growing late.
"Look for another sign," warned Skinny. "Scouts and Injuns always mark
the paths they take."
"Hurrah, here it is!" he shouted, a little farther on.
When we had come up, he pointed to a stone, which had been placed in the
middle of the path, with a smaller stone on top of it. It was the Indian
sign for "This is the trail."
We couldn't understand it, for it was leading away from North Adams.
We hurried on, calling every now and then, but not a sound could we
hear, except the birds and squirrels, and not another sign or track
could we find.
All that time we were going uphill and away from North Adams. At last,
we came out of the woods on top of the hill, where we could see up and
down the valley, and Greylock over beyond. Feeling too disappointed to
speak we threw ourselves down on the grass.
Suddenly Skinny gave a yell and we thought for a moment that he had gone
crazy.
"Look! Look! Look there!" he shouted, pointing back at the mountain.
We looked; then, when the full meaning of what we saw came to us, grew
as excited as he was, threw our hats in the air, and danced around and
cheered ourselves hoarse.
From the very top of Greylock, two columns of smoke were going almost
straight up, for there happened to be no wind to speak of. If it was
Bill, and we felt sure that it was, those two columns of smoke meant:
"I have lost the camp. Help."
CHAPTER VIII
SMOKE SIGNALS ON THE MOUNTAIN
BEFORE Bill started on his trip he made up his mind that he would walk
farther and do a bigger stunt than any of us. When Bill Wilson is for
anything, he is for it. There is no halfway doings with him. He didn't
take to the Scout business very well at first because he didn't know
much about it and thought that Indians or bandits would be better. But
as soon as he had joined he cared more than anybody.
Trying to do more than the other Scouts did was what got him into
trouble. He started for North Adams, the same as Wallie, Benny, and
myself, and he took with him a message for Mr. Jenks, as I have said.
But a seven-mile walk and back again the next day was not good enough
for Bill. He made up his mind that he would deliver the message first
and then go on as far as Williamstown and stay all night there.
Williamstown is five or six miles west of North Adams. There is a big
college there, called Williams College. I guess it was the name that
made Bill think of going there.
Our valley runs north and south until it gets to North Adams and then
turns west. Hoosac River turns with it. After flowing north all the
time, which everybody knows is no way for a river to flow, it turns
west, and so finally reaches the Hudson. Then, of course, its waters
flow south in the Hudson and at last reach the Atlantic Ocean at New
York.
After Bill had left Wallie the first morning of his trip, he walked
along lively, knowing that he had a long way to go to Williamstown, and
he did a lot of cawing on the road, just as Skinny thought. Nothing
happened to him at all until he found himself almost to the Gingham
Ground. Then he saw five or six members of the Gang playing ball near
where he would pass.
That made him stop. Bill is brave, all right, but what is the good of
being brave when they are six to your one, and the whole six have it in
for you?
That is what Bill thought, anyhow, and he started to leave the road and
try to work around out of sight through the woods and fields. Then he
thought of something to do, which scared him at first, but the more he
thought about it, the more he wanted to do it.
Hoosac Valley, as I have said, swings off toward the west at North
Adams. That brings Williamstown on the opposite side of Greylock from
where we live.
We found that out once when we went up on the mountain and came near
getting lost, which you know if you have read about the doings of the
Band. Almost straight down in front of us, on the east, was our village,
with Bob's Hill back of it, looking flat and not like a hill at all. We
could tell that it was Bob's Hill because we could see the twin stones,
standing there like tiny thimbles on a table. Looking north, we could
see North Adams; looking south, Cheshire, and on the west side of the
mountain and a little north, was Williamstown.
Bill thought of that when he was wondering how he could pass the
Gingham Ground without the Gang's seeing him.
"What's the use of going that way at all?" he said to himself. "What's
the matter with going straight back over the hills, climbing Greylock,
and then, after seeing exactly where Williamstown is, making a bee line
for it? I can deliver the message on the way back."
Say, that would be a great stunt! We are going to do it some time, when
we get bigger and our folks get over being scared.
He wanted to prove to us that he had done it; so made signs at different
places on the way, beginning where he turned off the road. We struck the
trail at the second sign.
Bill can beat us all climbing and he went along fast, having a lot of
fun all by himself. There is a path which leads up on Greylock from the
Gingham Ground; he followed that.
Before he had gone far he found a couple of bottles, which some one had
thrown away, and he hung those around his neck with a string. He took
them both so that one would balance the other. You see, he knew that
there was no water on Greylock. It has to be carried there from some
spring part way up. The day was hot, and he was thirsty, already.
When the sun grew hotter he took it easy along, picking berries and
lying around in the shade. He didn't get to the spring, where he was
going to fill his bottles, until almost noon. After that there was a
hard climb to get to the top, as steep as Bob's Hill, maybe steeper in
places.
He stopped at the spring to rest and eat his lunch; also to fix some
signs.
At last he stood on the very top of Greylock, which, as you probably
know, is the highest mountain in the State of Massachusetts, and it has
all kinds of mountains. Our geography says that it is 3,505 feet high.
Those last five feet seemed a mile to Bill, and they would to you, if
you were climbing the mountain on a hot day, with a pack on your back
and two bottles of water hanging from your neck.
I guess there never had been so much cawing on the top of Greylock as
when Bill stood there, after his hard climb, looking down on the hills,
which did not seem like hills, he was so much higher.
The air was so clear that Williamstown seemed close. So, after resting a
few minutes and drawing the sign on a flat rock to show which way he had
gone, he started down the west side of the mountain on a run, whooping
and yelling like an Indian at every jump.
Then, just as he was thinking how easy it was and what fun he would have
bragging to us boys about what he had done, he caught his foot in a root
or something, fell headlong, rolled down until he struck a tree; then
lay still.
How long he had lain there, when he finally came to life again, he
couldn't tell. At first he didn't know where he was or what had
happened. Then he remembered and tried to get on his feet and go on.
With a cry of pain, he sank back again. He had sprained his ankle and
hardly could move it without yelling.
When Robinson Crusoe was shipwrecked on an island he wrote on a piece
of paper the good things and the bad things that had happened to him. To
start with, he wrote on one side, "I am shipwrecked on an island," or
something like that, and on the other, "but I am alive."
Bill did the same, only he didn't write it. He thought it.
"I've busted my ankle," he said to himself, "but I didn't break my
bottles or spill my water.
"I can't walk a step, but I can yell to beat the band.
"I can't get to Williamstown and I can't get home, but I have something
to eat in my pack and plenty of matches in my pocket.
"Nobody knows where I am, but----"
That last "but" was to much for Bill. He couldn't find anything to go
with it, for he began to think of what Pa had told us, that if a person
should get hurt on the mountain he might die there and not be found for
weeks or years. His ankle was aching fearfully, too.
He tried yelling for a while and Bill is the best yeller that I ever saw
or heard.
"Help! Help!" he cried. "HELP!"
He might as well have saved his breath for all the good it did.
Then he lay still for a long time, trying to think what to do. That was
what Mr. Norton had told us.
"If anything happens," said he, "don't lose your heads. Think it over
calmly. Decide what is best to do and then do it."
"I'm a Scout," said Bill to himself, "and, bet your life, I ain't a
going to stay here and die on no mountain."
He took off his shoe and stocking and bathed his ankle in water from one
of the bottles--not much water because he couldn't spare it, and he took
a little sip himself. Then he thought of his "first aid to the injured"
package.
"What's the matter with bandaging myself?" said he. "It will be good
practice."
When he had finished and had rested a few minutes, he found that his
ankle did not hurt him quite so much and that he could move around a
little, if he didn't bear any weight on it.
He thought at first that he would crawl on his hands and knees to
Williamstown, or until he came to some house, but when he tried he found
that he couldn't do it.
"I'll tell you what I can do," he said at last, because he liked to hear
somebody talking, even if it was only himself. "Maybe I can crawl back
to the top of Greylock. Nobody ever would find me here and folks
sometimes go up there."
The Boy Scouts of Raven Patrol think that it took grit to crawl up the
steep and rough mountainside, with his ankle hurting at every move so
badly that it made him feel faint.
It wasn't far to the top, but Bill thought he never would get there, he
had to stop so many times to rest and wait for the pain to go away. An
hour or more passed before he finally crawled out into the clearing,
with nothing but the blue sky above him.
It was then getting late in the afternoon. Skinny was at Pumpkin Hook by
that time, probably surrounding the enemy. Wallie was somewhere in
North Adams or beyond. I was hoeing the garden at the very foot of
Greylock, little thinking that Bill was in so much trouble on top.
The summit of Greylock is almost level and is not very large. On the
east side Bill saw a lot of brush which somebody had cut and piled up,
probably to make a big fire; then for some reason had not lighted it.
He crawled over to that after the sun went down, built a little fire,
and cooked a small piece of bacon for his supper, which he ate with a
piece of bread and butter. It tasted good, but it made him thirsty and
he didn't dare drink much water.
Then, being tired out and more comfortable, he said his prayer and
repeated all of the Scout laws, from being loyal to being reverent,
wondering what good it was doing him to have two dollars in the bank
down in the village, and went to sleep.
When he awoke it was broad daylight. Benny and I were just starting on
our hikes, down in Park Street, but he couldn't see us, Bob's Hill being
in the way. By standing upon his one good foot, he could see the
village down below, and thought he could make out the very house he
lived in. He was as hungry as a bear and his ankle seemed a little
better, although it was still swollen so much that he couldn't get his
shoe on and he couldn't step on the foot.
He had plenty of food for breakfast, but he didn't know how many meals
he would need before he could get away; so he ate only a little and
waited, hoping every minute that somebody would come up on the mountain
and find him.
When the day at last dragged around and the sun was going down again in
Hudson River, Bill knew that he would have to spend another night on the
mountain and he felt pretty bad.
There were only a few mouthfuls of food left. One bottle of water was
all gone and the other nearly so. He knew that by that time his folks
would feel sure that something had happened and would begin to look for
him. That was some comfort.
Far down below, lights shone out from the houses, one by one. Down there
was his home. One of those lights was shining out of his window,
shining for him, while his mother sat and waited--waited for her boy who
never would come back again.
He sobbed aloud and stretched out his hands into the darkness.
"Mother, mother," he whispered, "I wish I hadn't come."
When he awoke in the morning he was frightened to find that the little
food which he had saved for his breakfast was gone. Some animal had
stolen it in the night.
His ankle was still badly swollen but it did not pain him so much except
when he tried to stand on it.
He was hungry and looked around for something that he could eat. A
little below the edge of the mountain stood a birch tree. He dragged
himself down to it and cut off long strips of the bark. This he chewed
for his breakfast, washing it down with a few sips of water, which
seemed hardly to wet his parched throat.
"I'll crawl down to the spring, if I can, and die there," he thought.
"Maybe they will find me sometime."
Then, as he was starting, something came to him.
Smoke signals! Perhaps one of the Scouts would see them and know what
they meant.
He was too weak and lame to spell out a message, like we did on Bob's
Hill. Instead, he built two fires, throwing on grass and leaves to make
a thick smoke. There was no wind and the smoke went straight up. That
was one of the signals, which Mr. Norton had taught us. It meant:
"I have lost the camp. Help."
He hadn't lost any camp, of course, but he didn't know what else to
send. He hoped it would let us know where he was and that something had
happened.
All day long he tended his fires, his ankle aching horribly because he
had to move around so much. Between times he sat on the mountain,
looking down at Bob's Hill and Plunkett's woods and the village beyond,
chewing birch bark and moistening his lips with the few drops of warm
water that were left.
Late that afternoon he gave up and made up his mind that he would crawl
down to the spring before dark and die there, he was so thirsty. He
turned to look down at his home, perhaps for the last time, and to see
Bob's Hill once more.
There were Plunkett's woods, and there, the twin stones, like thimbles,
they were so far away. And there--what was that?
From the ground close to one of the stones, the one where we build our
fires, a great column of smoke went up and he saw some things moving
around it, like flies or ants, they looked so small. Then the column of
smoke broke into long and short puffs. It was a signal.
Slowly he spelled the words:
"I-S, Is; I-T, it; Y-O-U, you; B-I-L-L, Bill?"
Jumping to his feet, although he almost screamed with pain, Bill grabbed
his blanket and held it down over one of the fires, which was still
sending out a big smoke; then pulled it off. Again and again he sent up
the puffs of smoke. His blanket was blazing; his hands were burned to a
blister; he was almost strangled with the smoke; but Bill kept on,
until he had spelled out something which could be seen from the top of
Bob's Hill, far below:
.... H
. E
-- L
..... P
Then he fainted away.
CHAPTER IX
FOUND AT LAST
WHEN we saw the smoke signal on Greylock, the first thing we thought of
was to signal back. But Skinny said:
"Come on. He won't be looking for us here. Bob's Hill is the place. He
can see us there."
We started on a run across the fields, getting more excited every
minute.
"I don't see how Bill could lose any camp," exclaimed Benny.
"And I don't see what he is doing on Greylock when he started for North
Adams," Hank said.
"Maybe it isn't Bill, at all," I told them. "I've seen smoke on Greylock
more than once."
"It's Bill all right," Skinny said. "I can almost hear him. We don't
know how he got there, but he's there and he can't get back. Something
has happened."
"Anyhow, we'll soon find out," we all thought, when we came in sight of
the twin stones.
"I'll run down home and get a blanket," I told them, "while the rest of
you make a fire."
Our house is right at the foot of the hill and it didn't take me long.
The old horse blanket which we used in signaling was in the woodshed. I
only stopped long enough to wet it and call to Ma that Bill was up on
Greylock signaling.
She was almost as excited as I was.
"Hurry!" said she. "Don't wait for me. I'll come as soon as I can."
I hadn't thought of waiting for anybody.
She grabbed a pair of field glasses off the shelf and rushed after me. I
heard her calling to Mrs. Blackinton when she went through the yard and
I had to go some to keep ahead.
By the time we had climbed the hill, the boys had a big fire going and
were piling on green branches and leaves to make it smoke. Then we
caught hold of the blanket by the corners, ready to shut off the smoke.
"Ask if it's Bill," Skinny told us, watching the two smokes on the
mountain.
Then we signaled, "Is it you, Bill?" and repeated it. Before we had
finished the second time Skinny gave a shout.
"It's Bill," said he. "He's signaling."
We could see one column of smoke break up into puffs, but couldn't see
very plain because the smoke was so thin and far away.
"Here, take this glass," said Ma, handing the field glass to Skinny.
"Hurrah," he cried, after he had looked through them. "I can see real
good."
Then he held up one hand and we waited while he called off the letters.
"H-E-L-P."
That was all. We waited for more but nothing came.
Before we had turned to go Ma was halfway down the hill and running to
beat the band. I knew that if Bill didn't get help it wouldn't be her
fault.
"See if you can get hold of Mr. Wilson," she called, as soon as we came
in sight. "I'll telephone his house. If you can't get him, get somebody.
Your father has gone to hitch up and he will be ready to start in a few
minutes."
In five minutes it seemed as if the whole town knew about it and were
out in front of our house, or else climbing the hill to see the smoke.
Mr. Wilson came on a run and was in the wagon before Pa could stop the
horse.
"I want one of you boys to go with us," said Pa. "We may need some more
signaling. Benny Wade, you are the lightest. Can you stand the climb?"
"Can I?" said he. "You watch me."
The marshal chased up with a light stretcher and another lantern.
"You can't have too many," he said. "It will be dark before you get up
there."
Ma came running out with a basket of bread and butter and some meat.
"We'll light a big fire on the mountain, if all is well," they told
her.
"The water!" called Skinny. "Pedro, get them a big bottle."
In another minute they were off, while the others went home to wait,
which is the hardest part.
I found out afterward what happened. They couldn't drive all the way up
Greylock from our side. There was a road from North Adams and another
from Cheshire but those were too far.
Pa planned to drive as far as they could and then to leave the horse
tied and walk up the rest of the way. They went around the road by the
Quaker Meeting House to Peck's Falls. From there a road goes part way up
the mountain, steep and winding. It was hard pulling for the horse.
I don't believe Greylock ever was climbed so fast before, although it
seemed slow enough to poor Bill waiting on top, thirsty and faint. He
knew that his signal had been seen and that was something.
The first thing that he heard was a call of a crow, over to the south
and far down the mountainside.
"Caw, caw, caw," came the sound, and it seemed to be Benny's voice.
Bill stood up on one foot and listened.
"Caw, caw, caw," it came again, this time nearer.
Then Bill braced himself and seemed to grow stronger, all in a minute.
"Caw," he yelled. "Caw, caw!"
The sound went floating down into the gathering darkness, until it
reached two men and a boy, toiling up the mountainside.
"That's Bill!" cried Benny.
"Thank God!" said Mr. Wilson. "He's alive. We know that."
Twenty minutes later he had Bill in his arms and Benny was building the
biggest fire that had been seen on Greylock since I could remember. We
were watching for it down below and knew that everything was all right.
"Now," said Pa, "let's have some supper. I don't know about William, but
I feel hungry."
It was late at night when they finally brought Bill home. Mrs. Wilson
nearly had a fit again when she saw them carrying him into the yard on
a stretcher.
"Speak to her, son," said his father, "so that she will know you are
alive."
Bill propped himself up on one elbow and gave such a yell that it scared
the neighbors, and ended with a caw. Then she knew that it was all right
and felt better.
Skinny was the proudest fellow you ever saw because we had found Bill.
It made him real chesty and we all felt good about it.
"Say, we're the stuff," said he. "If you don't believe it, watch our
smoke. That's all I've got to say. Hurry up and get well, Bill, so we
can have a meeting and tell about our hikes. I want to see a First Class
Scout badge on my manly bosom."
We were sitting in Bill's house at the time, to cheer him up a little
because he couldn't go out without a crutch.
"What's the matter with having the meeting here?" said Bill. "I don't
suppose Mr. Norton will give me a badge because I haven't delivered his
message yet, but I'd like to hear what the rest of you did. I can't get
out for a few days. When I do, I'm going to North Adams and back, if it
takes a whole leg. Believe me."
"You did more than any of us," Benny told him, "badge or no badge."
"I guess you won't chase over the mountain the next time," I said. "When
you stick to the roads there don't anything happen."
"Oh, there don't, don't they?" exclaimed Skinny. "Say, you fellers ought
to have been with me. There was something doing every minute. Ma says
it's a wonder that I'm alive. I've had awfully hard work to keep from
telling about it."
"Tell us about it now."
"Not much, you wouldn't be able to sleep to-night. Besides, it might
make Bill's ankle worse."
"Great snakes!" said Bill. "There ain't anything the matter with me,
only it hurts me to step on my foot. Come on, Skinny. Let's have it."
"No-p. We've got to have a meetin' first."
"Suppose that you have your meeting here to-night," said Mrs. Wilson,
who had come into the room in time to hear what we were talking about.
"Willie is a great deal better and I can have him take a nap to brace
him for the story. If you boys will come around after supper you can
meet right in this room, and perhaps, I don't say for sure, perhaps the
neighbors will bring in some ice cream to quiet your nerves and make you
sleep."
"May we bring Mr. Norton?" I asked. "He is our Scoutmaster and he ought
to be with us when we tell about the doings of the patrol."
"Surely you can. He is coming, anyway. He sent word this morning that he
would call to-night."
We met at Skinny's a little before eight o'clock and went over in a
bunch. On the way Skinny told us what to do.
"When we get to the gate," said he, "let's stop and each one caw three
times."
"What for?" I asked. "We know that he is there; don't we? Besides Bill
is sick. Maybe we'd better keep quiet."
"Sick nothin'! He ain't any more sick than I am. He said so himself.
He's hurt his ankle a little, that's all. Ankles can't hear, can they?"
"Maybe it will cheer him up to hear us," I told him. "He can't get out,
you know. It is hard to be cooped up in the house that way, and Fourth
of July coming."
"Anyhow," said Benny, "let's not all caw at once. We can take turns and
it will not make so much noise."
That was what we did, standing just outside the gate, where we could see
a light streaming through an open window in Bill's room.
Skinny led off with three. I followed, and the others in turn, ending
with Benny. Skinny said that it sounded like the booming of minute guns
in some battle or other, that he read about in a book.
Say, it surprised the folks living around there. Before we were half
through, they came running out of their houses to see what was going on.
It made us feel proud and we were just going to do it over again, when
we heard Bill cawing in the house and Mrs. Wilson threw the door open
and stood there laughing.
"I judge by the sound," said she, "that the Ravens have arrived and are
in good voice."
We found Bill sitting in a big chair, with his foot propped up and his
eyes shining.
At first we didn't know just how to act, until in a few minutes Mr.
Norton came and then Mrs. Wilson brought in some ice cream and some
clusters of strawberries, with dishes of powdered sugar to dip them
into.
We knew how to act then, all right, and for a few minutes we were too
busy to talk.
I am not going to tell what all the Scouts did on that hike. I already
have told what happened to some of us. There didn't much happen to most
of them, anyhow, any more than there did to me. It was different with
Skinny. Something almost always happened to him.
CHAPTER X
A MAIDEN IN DISTRESS
"FELLERS," Skinny had told us, when we were getting ready to start on
the hike, "you always ought to carry a rope. Something happens every
time when you don't have a rope along."
"It happens when you do," Benny said. "Anyhow, a rope is too much
bother. A blanket and a frying pan and things like that are all I want
to carry."
"A rope is the thing, just the same. Didn't I lasso the robber last
summer out on Illinois River, at Starved Rock? How could I lasso
anything without a rope? And didn't we let you down into Horseshoe
Canyon with a rope and pull Alice What's-her-name up again?"
"Bet your life we did," Bill put in. "You need a rope when you are
camping out or are in a boat on the river, but what good is it in
walking seven miles?"
"Maybe it is and maybe it isn't; but, just the same, you'll be sorry if
you don't take one along."
He was right, too, for Bill told us afterward that he would have given a
good deal for a rope when he was sitting on top of Greylock. He didn't
need it for anything, only, he said, it would have been sort of company
for him.
Skinny was bound to carry a rope. When he marched down Center Street
with it coiled around his shoulders, over his blanket, and with his
tomahawk in his belt, people ran out of the stores to look at him.
The road that he took is uphill a good part of the way. It goes up
through the foothills of the east mountain and isn't easy walking. We
slide down that road sometimes in winter. When the coasting is good we
can slide nearly a mile, clear into the village; then hitch on to a bob
and ride back again for another.
There were no bobs for Skinny. It was warm in the sun and he loafed
along, taking it easy and looking for somebody to rescue. Once he
stopped to help a man in a field. Along about ten or eleven o'clock he
began to get hungry and tired. No matter where he looked there didn't
anything happen, so he made up his mind to take a long rest the next
time he came to some good shade, and maybe to cook his dinner.
A half-mile farther on he came to a real shady spot by the roadside,
under a tree which stood in a corner of a pasture on the other side of a
fence. A tiny stream crossed the road, and ran down through the pasture.
This was the place he had been looking for and, after drinking, he threw
himself down on the ground and went to sleep.
He didn't know how long he slept but he felt first rate when he woke up,
only hungrier than ever. Over in the pasture stood a cow with her back
to him, looking at something and growing real excited about it.
"I wonder what ails the critter," said Skinny to himself. "She looks mad
about something, snorting and shaking her head that way."
Just then he heard a girl's voice singing. She sang real loud, like boys
whistle sometimes to keep up their courage, when they are half scared.
Then in a few minutes she came in sight, walking across the pasture and
keeping one eye on the cow.
Skinny hadn't seen her before because the cow had stood in the way.
"Jerusalem!" said he. "Here's luck. She's got a fire-red sunbonnet and
cows don't like red sunbonnets a little bit."
On came the girl, singing louder than ever, trying to edge off away from
the cow but not daring to run.
Skinny could see that the cow was getting madder all the time. He knew
that something was going to happen at last, and he began to uncoil his
rope.
"Run, you little fool," said he. "Run."
He meant the girl and not the cow. He said it under his breath so she
wouldn't hear, for he didn't want to lose the chance to do the rescue
act and have something to tell us boys about afterward.
The girl was scared. Any one with half an eye could have seen that. The
cow hadn't quite made up its mind what to do, and Skinny was beginning
to be afraid that the girl would get across without giving him a chance
to get in his work. Then what did she do but take off her sunbonnet and
swing it around by one string, just to let the cow know that she wasn't
afraid of any animal that walked on four legs.
She hadn't seen Skinny yet, on account of his being back of the cow. The
cow didn't know he was there, either, until about four seconds
afterward. It knew then, all right.
Maybe the cow wasn't mad when she saw that red sunbonnet whirling around
in the air. She tore up the sod with her horns, gave a big snort, and
started, head down.
Say, it was Skinny's busy day about that time. Before the cow could get
fairly going he had crawled under the fence and run up behind, whirling
his lasso around his head. Then he gave a yell like a wild Indian and
threw it.
I think the yell scared the girl worse than the cow did. Anyhow, between
the cow and the Indian she was scared stiff; just stood there
paralyzed. And she didn't do any more singing.
If that lasso had caught there would have been a paralyzed cow all
right. Skinny threw it in great shape. It went straight for her horns,
but when he yelled she lifted her head suddenly. The loop struck against
one of the horns, instead of going over it, and then fell off to the
ground.
"Gee!" groaned Skinny. "Missed!"
There wasn't time to say anything more, and he knew that he would have
to get mighty busy or there wouldn't be any rescuing done.
When something happens that way and you have to do something first and
think about it afterward, the mind seems to work like chain lightning.
There was only one thing to do and it didn't take Skinny long to do
that. He dropped the rope, grabbed hold of the cow's tail with both
hands, and dug his feet into the ground.
"Run!" he yelled. "Run for the fence! I've got her."
When Bill heard about it he said that it seemed to him as if the cow had
Skinny. Anyhow, she was surprised some and she was mad. She will think
twice next time before she does any chasing, when anybody from Raven
Patrol is around, I guess.
Skinny had a good hold and she couldn't get away. First she stopped
running and tried to get at whatever it was back of her, with her horns,
chasing herself around in a circle.
Skinny hung on like a good fellow. He had to. If he had let go once it
would have been all up with him. She never touched him. Every time the
cow stopped, there was a hundred pounds of boy hanging to the end of her
tail.
It was like playing crack the whip, he told us afterward, "and being the
littlest fellow on the tail end."
Then for a few moments it was hard to tell which was the cow and which
was Skinny, for she started on a run for the other side of the pasture,
Skinny sliding and bumping behind, and both of them scared half to
death. Skinny was so excited he couldn't think to let go of the tail.
Hank said that he would have given a quarter if he could have taken a
picture of it with his camera.
All this didn't take so long as it does to tell about it. The girl had
reached the fence, crawled under, and was yelling for help.
Just then it seemed to Skinny as if the tail had come off in his hands,
for he went tumbling along, heels over head, until he struck with a jar
that almost loosened his teeth.
What really happened was that he stumbled on a stone and his hands were
jerked loose. In another minute the cow was out of sight in a hollow.
Skinny scrambled to his feet and went back after the rope, trying not to
limp because he could see the girl looking at him through the fence.
He felt pretty chesty to think that he had rescued a maiden, only he
didn't know what to do with her, now that he had saved her.
She spoke first, as he stood there sort of brushing his clothes off.
"Are you hurt, boy?"
"What, me?" said Skinny. "Me hurt? Say, didn't you see the critter run
when I got after her?"
"I should say I did, only I was scared. Wasn't you scared?"
"I don't scare worth a cent," he told her. "I ain't afraid of any cow
a-livin'. You don't suppose I'd 'a' chased her all over the pasture, if
I'd been scared, do you?"
"N-no, but----"
"Say, if my lasso hadn't slipped, there would have been something doing.
It's lucky for you that I got hold of her tail. That's the way to do it.
When you twist a cow's tail, it scares 'em."
It's just as Hank says, you never can tell what a girl will do. That
girl tried to say something; then choked up and went off into a fit of
laughing that made the tears roll down her cheeks and left her so weak
that she had to hang on to the fence.
Skinny grinned a little to be polite, but he didn't like it very well.
"Oh," said she, as soon as she could speak, "it was too--too funny for
anything to see you sailing along behind the cow."
"It wouldn't have been so funny if the cow had been running toward you,
instead of away from you. You would have laughed out of the other side
of your mouth, I guess."
She saw that he was mad about it.
"You mustn't mind my laughing," said she, stuffing her handkerchief into
her mouth. "I can't help it. It's a disease."
"A disease?"
"Yes, it's high strikes. When folks have them they can't stop laughing.
They laugh when they ought to cry, maybe."
"Sounds like a ball game," said Skinny.
"It's something like that," she told him. "Maybe that isn't it exactly
but it's something. I'm better now."
"Oh, well, if it's something that ails you, I suppose it's all right.
I'd laugh, too, only I am all out of breath from chasing the cow."
When he said that the girl burst out laughing again, and Skinny laughed
with her. That made them feel acquainted.
"I guess I've got 'em, too," said he. "They must be catching. Well, I
must be going now."
"My name is Mary Richmond," she told him, "I live in Holyoke and I am
visiting over where you see that red barn."
"Mine is Gabriel Miller. I don't like the name very well. Gabe isn't so
bad. The boys call me Skinny. I live down in the village and I am on a
hike. I guess I'd better be going now."
"I don't see any."
"Any what?"
"What you said you were on, a hike."
"You will see one in about a minute. I am out for a long walk. I belong
to the Boy Scouts and I've got to walk seven miles, camp out to-night,
and come back to-morrow."
"My," said she, "you must be hungry--all that walking and--and--chasing
the cow, too."
"I am," said Skinny, bracing up. "I believe I'll eat my lunch right here
in the shade. Wish you'd stay and eat with me. I can cook some bacon."
Wasn't that a nervy thing to say? Skinny is brave when he gets started.
"It would be fine," she told him, "only Ma is expecting me at the house.
She is visiting, too. Wouldn't it be nicer for you to come with me?
They will be glad to see you because you saved me from the cow. I am
awfully hungry and Grandma is the best cook. We're going to have
lemonade. She told me so. Come on, do."
"Lemonade would taste good," he said, "if I only dast."
"Huh!" said she, tossing her head. "I thought that you were not afraid
of anything."
"I ain't of a cow. This is different. Say, that was a swell song you
were singing. I wish I knew it."
"I'll teach it to you after dinner, if you will come. If you don't
you're a 'fraid cat."
"All right. I'll go if it kills me."
Skinny says that he never ate a dinner that tasted any better than that
one did. Mrs. Richmond was scared when she heard about the cow and she
couldn't say enough about how he had saved her little girl from a
terrible death.
"That wasn't anything," he told her. "Scouts are always doing those
things. I'm going to try to save somebody from drowning when I come back
along the river to-morrow."
"I'll tell you a better stunt than that," said Mary's grandfather,
winking one eye at the rest of the folks. "Why don't you go up to Savoy
on the east mountain. That would make a walk of about seven miles from
the village. You won't find anybody drowning up there, but several deer
have been seen around there lately."
"Gee!" said Skinny, his eyes sticking out when he thought of the deer.
"If I only had a gun!"
"It's against Massachusetts law to shoot deer. That's why they are
getting so common. You have your rope. Maybe you can lasso one. There is
no law against that, I guess."
"I'll do it," Skinny told him. "Bet your life the boys will be surprised
when they see me bringing home a deer. Maybe I'll get two or three. Mr.
Norton didn't give me a message to anybody, so it won't make any
difference which way I go."
"Don't get too many. We'd like to save a few. And be careful that some
bear doesn't get you," went on Mr. Richmond, laughing to see how excited
Skinny was. "They are not very common, but once in a while one is seen
on the mountain."
"How do you get up there?"
"Go back to Pumpkin Hook. It isn't far, and then follow the road which
turns east. It will take you right to Savoy. You will find a pretty good
road all the way, and you won't have any more trouble than you would
going to Cheshire--unless," he added in a fierce voice that made Skinny
jump, "unless A BEAR GETS YOU!"
"Now, father, don't scare the boy to death," said Mary's mother. "You
know well enough there are no bears and the road to Savoy is a
well-traveled one."
"Of course it is, or I shouldn't have suggested his going there. But
there have been bears seen on the Savoy Mountain. I saw one myself, last
year."
"Huh! I ain't afraid of no bear," put in Skinny, drawing himself up and
looking fierce. "I tracked one once on Bob's Hill. It went up to Peck's
Falls and hid in our cave. We smoked it out. I didn't have a gun or
knife or anything, but I hit it with a snowball."
You could have hung a hat on Mary's eyes when Skinny told them that.
"Was it a really and truly bear?" she asked. "And did it stand on its
hind legs like in the circus pictures over at the Hook?"
"It stood on its hind legs, all right," he told her, "but it wasn't
really a bear. We thought it was. It made tracks in the snow just like
bear's tracks, but when we had smoked it out we found that it wasn't
anything but a man."
"It was Jake Yost, a foolish feller," he explained, turning to Mr.
Richmond. "He had his boots on the wrong feet and wouldn't change them
back for fear of changing his luck. That was what made his tracks look
like bear's tracks."
It tickled them to hear about that, but it didn't tickle us boys much
when it happened. It was too scary.
"If you will stop here on your way back to-morrow," said Mary's grandma,
"we'll give you a nice dinner. I think you will be wanting one about
that time. Mary may walk with you as far as the Hook, if you like, and
show you the road."
"I think maybe I'd better go along, too, with my gun," said Mr.
Richmond, "on account of the bears."
"Don't you mind his nonsense," she said. "You run along."
So off they went together, Skinny with his rope and tomahawk and Mary
with her red sunbonnet, but they kept away from the pasture.
From Pumpkin Hook Skinny went on alone, up the mountain road, whirling
his tomahawk around his head and every little while pretending to lasso
the enemy, because he knew that Mary was watching him from below.
Then pretty soon he came to a bend in the road. He turned and waved to
her, and in a minute was out of sight.
CHAPTER XI
TREED BY A BEAR
I AM writing what happened to Skinny as if we found out all about it at
once, which we didn't. He told us some of it the first time, with Bill
sitting up and listening and Mr. Norton asking questions whenever Skinny
began to run down. But every time we saw him after that for several days
he would think of something more to tell, or something a little
different, so that it took a long time before we felt sure that we knew
all about it.
For instance, he didn't say much at first about Mary Richmond, the
Holyoke girl, except the rescue part. He was afraid that the boys would
make fun of him for walking down the mountain with a girl--but I haven't
told about that yet. I am going to put everything in just when it
happened, so that you can understand it better.
There didn't much happen, anyhow, while he was going up to Savoy. The
road was steep and winding, and climbing it kept Skinny busy and made
him wish more than once that he had gone in some other direction.
What Mr. Richmond had said about bears made him nervous. Every time he
saw a stump of a tree, he was sure it was a bear, and every time he came
to a part of the woods where the trees stood very close together and it
looked dark inside, he had to whistle and sing louder than Mary did when
she was afraid of the cow.
Whenever he felt real scared he would caw like a crow, and that made him
feel almost brave again, for sometimes when you just pretend you are
brave and act as if you are, all of a sudden you get brave. I don't know
why it is but I have noticed it.
He kept a sharp eye out for deer, for he wanted to bring us one, but he
didn't see a thing all the way up that looked like a wild animal except
a calf, which ran when he threw a stick at it, and the birds, which
don't count.
It was hot work but the air was fine, and he could see all up and down
Hoosac Valley, and that is worth seeing any time. If he had taken a
spy-glass with him, perhaps he could have seen the other Scouts on the
way to North Adams and Cheshire.
Once in a while he came to a mountain brook, gurgling and singing over
the stones. Then he would throw himself down to rest and listen to the
pouring water, which we boys think is the sweetest music in all the
world, unless it is the cawing of a crow away off somewhere, on the
mountainside.
Late in the afternoon he came to Savoy and stopped in a field to cook
himself a good supper.
That night he slept in a barn, cuddling down in the haymow, where he
could hear some horses stirring in their stalls. They seemed sort of
like company for him, although they couldn't talk any.
"Were you not afraid up there, all alone?" Mr. Norton asked, when Skinny
was telling about the horses.
"What, me?" said he. "Anyhow, I wouldn't have been, only there were all
kinds of noises in the night and once I heard something scratching at
the door. I think it was a bear; maybe, two bears."
"Great snakes!" said Bill, and we all thought so, too. But Skinny waved
one hand, as if that wasn't anything worth mentioning, and went on.
When morning finally came and the sun shone in through a cobwebby window
across the haymow he slipped out of the barn on the side away from the
house, so that the folks wouldn't see him.
Just the same, they saw him cooking his breakfast, and were going to set
the dog on him. But when the farmer's wife found out that it was a Boy
Scout and not a tramp she told him to come right into the house and eat
with them. He went, too, because he could smell the breakfast cooking
and it 'most made him crazy.
"How about it, Mr. Norton?" said Bill. "That makes two meals Skinny had
given to him, not counting the dinner at Richmond's the next day, which
he hasn't told about yet. That makes three. Didn't he have to cook them
himself on account of the Scout business?"
Before Mr. Norton could answer Skinny spoke up.
"Aw, g'wan!" said he. "I cooked enough to make up for it, I guess. Why,
I stopped two or three times and cooked something. You don't suppose a
feller can climb mountains without eatin', do you?"
"I didn't eat much," said Bill with a grin, "but I wanted to."
"I think Gabriel is right," laughed Mr. Norton. "Besides it sometimes is
harder to work folks for a meal than it is to cook it, yourself."
"Anyhow," Skinny told him, "I didn't get to Richmond's in time for that
dinner and I paid for those other meals. I rescued the girl the first
time, didn't I? That ought to be good for a dinner. And to pay for my
breakfast I carried in a lot of wood for the farmer's wife. She liked it
so well that she said she would be glad to have me stay to dinner. There
wasn't any chance to do any rescuing in Savoy, so I had to do something
else."
"That's business!" exclaimed Mr. Norton. "Pay as you go. Gabriel, my
boy, you showed yourself a true Scout and I'm proud of you."
He reached over and fastened a First Class Scout badge to Skinny's coat.
"Maybe I am a little ahead of the game," said he, "but Gabriel is leader
and I think that he has earned a badge. This seems to be the
psychological moment to present it."
Benny spoke up before we could stop him.
"What's a skological moment?" said he.
Say, that stumped Mr. Norton. He couldn't tell us.
"I'd like very much to give you one, William," he went on, after a
little, turning to Bill. "You showed yourself a hero and you have done
everything except the hike. How would it do to give you the badge now,
with the understanding that you will make good on the hike later, when
you get well?"
Skinny swelled all up when Mr. Norton gave him the badge, and I guess
anybody would. He didn't know what to do or say at first, but in a
minute he came to his senses. He jumped to his feet and gave the Scout
salute. It was great to see him.
"Fellers," said he, turning to us with his arms folded, while Mr. Norton
looked on, wondering what was going to happen.
"Who are going to be the best Boy Scouts in America, or England,
either?"
"We are!" we shouted.
"Who is the best Scoutmaster that ever happened?"
"Mr. Norton!" we yelled.
"Who is great stuff, if he did sprain his ankle on Greylock?"
"Bill Wilson!"
"'Tis well. Everybody caw. Now!"
There was some racket around that room when we turned ourselves loose.
Bill sat there smiling and with his face all flushed up, he was so
tickled over what Mr. Norton and Skinny had said.
Then Mr. Norton pulled another badge out of his pocket and started to
pin it on Bill's clothes. Bill stopped him.
"It wouldn't be fair, Mr. Norton," said he. "I started out to do my
hike and I didn't do it. I know that I did something which was harder
but I didn't do that. I wouldn't feel right about wearing the badge
until after I had made good."
"What do you say, boys?" asked Mr. Norton, his eyes shining because he
was so proud of Bill.
"Bill's all right," said Hank. "We all know that he can do the stunt and
that he will do it, but he hasn't done it yet."
Then Benny spoke up.
"Guess what!" said he. "Let's all wait until Bill gets well and does it,
before getting our badges. Except Skinny; he's got his."
"Bet your life I'll wait, too," said Skinny.
He started to take the badge off, but we wouldn't let him.
"Forget it," said Bill, "and go on with the story. You stopped in an
interesting place. I don't believe much happened, anyhow, except the
cow, and you've told us about that."
"I don't like to tell the rest. It will make you walk in your sleep and
that will hurt your foot. But I'm willing to risk it if you are."
You see, when Skinny started toward home from Savoy, he made up his mind
that he would lasso a deer, or know the reason why, because it would
look fine to have one stuffed and standing in front of our cave at
Peck's Falls. So, when he had found a place that looked wild and sort of
scary, he left the road and, getting his rope in shape to throw, made
his way in through the brush, as still as he could, so as not to
frighten the deer away.
He didn't see any deer, but after a while he found a big patch of wild
strawberries, so thick he couldn't step without tramping on some. That
made him forget all about his deer for 'most an hour.
Then, all of a sudden, he heard a crackling in the bushes on the other
side of a clearing, and he felt sure that his chance had come.
Skinny dropped on his hands and knees and crawled toward the sound. It
was slow work because he had to be careful not to make any noise, and he
grew more excited every moment.
At last he was crouching down behind some big bushes, and on the other
side he could hear the deer real plain, tramping around like a horse.
"Gee!" thought he. "It's a big one and will look great up by our cave."
He didn't say it out loud because he knew that although the deer could
not smell him on account of the wind blowing the other way, he would
hear him, unless he was very careful.
Then, getting the rope ready to throw, with the slip noose working
easily, he parted the bushes gently and crept through.
There was a great crashing as some big animal broke his way through the
bushes in front of him. Then came a snarl and a growl that made Skinny's
heart almost stop beating. And there he stood, paralyzed, looking
straight into the eyes of a bear!
It wasn't any Jake Yost with his boots on wrong, either. It was the real
thing, looking as big as the Quaker Meeting House to Skinny, although it
was really only a cub, about half grown.
I guess the bear wasn't expecting anybody to call, for he stood there,
sort of paralyzed himself, his eyes looking right into Skinny's and one
big paw raised to take another step.
Skinny gave a howl and started for the nearest tree, one that was too
small for a bear to climb.
Say, if tree climbing had been one of the Scout stunts, Skinny would
have won two badges.
It isn't any fun to sit in a tree on a mountain, with a real live bear
sniffing around at the bottom and you both getting hungrier every
minute.
Skinny knew he was safe as long as he stayed in the tree, but he didn't
dare get down while the bear was in sight, and the cub wouldn't go away
more than a few rods. I guess Skinny looked good to him, he was so fat.
Dinner time came and went. He was still in the tree and the bear was
still fooling around below.
Skinny called for help until he was hoarse, but there wasn't anybody
passing at that time of day. Then he began to get mad, and when Skinny
gets mad, look out!
"You think you're smart," said he, "but old Long Knife will show you a
thing or two."
First he let down his rope and found that it would reach the ground.
Then he fixed the noose up in good shape, tied the other end around a
limb and waited.
By and by the bear came smelling around that rope to see what it was,
and that was exactly what Skinny had been waiting for. He leaned down
and tried to swing the noose over the cub's head. The bear didn't know
what to make of it and every time the rope would hit his nose he would
growl and strike it away with his paw.
Skinny saw that he would have to get closer. He climbed down to a lower
limb; then held on with one hand, swung out over the bear, and tried to
lasso him with the other.
He almost did it, too, but just as he leaned still farther down, all of
a sudden there was a cracking noise and the limb broke.
With an awful scream of despair, Skinny fell.
CHAPTER XII
WHAT HAPPENED TO THE BEAR
THE Band, I mean the Ravens, don't know so very much about bears. That
was the only bear we ever had come across and we had been berrying all
over those mountains, although mostly on the Greylock side. Pa says that
they usually keep away from the road, the few that are left, because
they are afraid of folks.
Anyhow, it isn't any picnic to fall out of a tree at any time,
especially when there is a bear at the bottom.
When the limb began to crack, Skinny knew that he was a goner. He yelled
so loud that it surprised the bear and it looked up into the tree to see
what was going on. Just at that second the leader of Raven Patrol landed
on the cub's nose, like a thousand of brick. Boy and bear both went
sprawling, one in one direction and the other in another.
Skinny was the first to get on his feet and the way he shinned up the
tree again was a caution. He didn't stop to look until he had reached
the limb where the rope was tied. Then he felt safe.
The bear had picked himself up and was standing close to the foot of the
tree, looking up and whining, as if he didn't like being hit in the head
by a boy very well.
It was the chance which Skinny had been waiting for. He gathered the
rope up in his hands and opened the noose wide. Then, leaning down as
far as he dared, until he was right over the bear, he dropped it. The
noose fell as straight as a die and, spreading out around the cub's
head, lay across his shoulders with the side nearest the tree almost
touching the ground.
Just as the bear stepped one foot over the loop, Skinny grabbed the rope
with both hands and gave a quick jerk. The noose tightened; and there
was the most surprised bear you ever saw, tied fast to the tree! Skinny
stood on the limb above like a big crow, cawing to beat the band and so
excited that he came near falling again.
"Gee, but that bear was mad," said Skinny, when he was telling us about
it. "He growled and he snapped and he rolled on the ground; then he ran
around and around the tree, until he had wound himself up short, but he
couldn't get away. It was great, only I didn't dare jump on him again.
He was too crazy."
"Great snakes, Skinny!" exclaimed Bill. "You always have all the fun."
"I guess you wouldn't have thought it so much fun if you had been up in
the tree and couldn't get down. I'd 'a' choked him with the rope, if he
hadn't got his feet tangled up in it so that I couldn't."
"How did you get down, Skinny?" asked Benny, because Skinny had a way of
stopping at the most interesting places and pretending that he was
through telling about it.
In order to tell about that I'll have to go back a little in this
history.
When Mr. Richmond told Skinny to go up to Savoy and to be careful not to
let the bears get him, he was trying to scare a Boy Scout. He says that
he hadn't any idea there would be a bear or deer around, or he shouldn't
have let him go. But the next morning a man from Savoy drove past the
house and told about seeing a bear on the way down. He didn't have his
gun along and besides the bear ran into the woods when he saw him.
That made Mr. Richmond feel uneasy.
"I wish I hadn't let the boy go up the mountain," he said. "I don't
suppose anything will happen to him, but I'd feel better if he hadn't
gone. I guess, of the two, the bear would be the most scared if they
should meet."
"He told me that he'd surely come in time for dinner," said Mary.
When dinner time came she put a plate on for him. He didn't show up, of
course. He was up in the tree about that time, wondering how he ever
would get down. After that Mr. Richmond grew real anxious and went to
the house several times to see if Skinny had come.
"That boy looked to me," he said at last, "as if he wouldn't be guilty
of missing a good dinner if he could help it. I am going after him. He
may be all right, but I'm going to find out for sure."
With that, he hitched up a horse, took down his gun, and started.
"Let me go, too," Mary called after him. "I can hold the horse while you
are looking."
"All right. Jump in. We'll probably meet him on the road somewhere."
The first they saw or heard of him was the yell which Skinny gave when
the limb broke. It scared them.
"Take the reins," said Mr. Richmond. "There is trouble over there. Turn
around and if anything comes run the old horse down the road."
Say, he was paralyzed, when he found the bear tied to a tree and Skinny
standing on a limb, cawing.
"I was that flabbergasted," said he afterward, "that I hardly could pull
the trigger."
But he pulled it, all right, and that was the end of Mr. Bear.
Skinny didn't like it because Mr. Richmond killed the bear. He wanted to
tame it and give a show in our barn. He was bound to take it home,
anyhow, so as to save the skin.
It took a lot of pulling and hauling to get the cub out to the road, and
Mary had to help before they could lift him into the wagon.
"Jump in," said Mr. Richmond, when everything was ready. "It is time
that I was getting home."
"I can't," said Skinny. "You see, I am doing a stunt for the Scouts and
I have to walk."
Just before they started Mary thought of something.
"Say," said she, "maybe I'd walk, too, if anybody asked me; that is, if
Grandpa would let me and it wouldn't make any difference with the
Scouts."
"Come on, do," said Skinny. "May she, Mr. Richmond?"
"Well," said he, "seein' as how you've got a rope and it ain't very far,
I'm willin'. But it will be mighty lonesome for me."
I never saw Skinny so chesty as he was over catching that bear. And he
had a right to be, for everybody was talking about it and there was a
long piece in the paper. He even wanted to change the name of Raven
Patrol to the Bears, but we wouldn't stand for that. We didn't know how
to make a noise like a bear, anyway.
After that the folks told us to keep away from Savoy Mountain, rope or
no rope, and we had to do it. But Skinny wanted to go back and get a
bear for each of us.
"I think that our patrol leader has made good," said Mr. Norton, when
Skinny had finished. "What I'm wondering is, who was the most
frightened, Gabriel or the bear?"
"The bear was," said Skinny; "anyhow, after I jumped on him. Say, I'll
bet you fellers wouldn't dast jump on a live bear, when he was growling
and showing his teeth. It was great, just like jumping on a cushion,
only the bear didn't like it very well."
The other boys didn't have much to tell, much that was exciting, I mean,
but Mr. Norton made us all report what we did. Hank came last of all.
"Well, Henry," said Mr. Norton, "what have you to say for yourself? You
went to Cheshire by the river road, I believe?"
"How about that new invention, Hank?" I asked. I'd forgotten all about
it until then.
"Have you a new invention, Henry? Tell us about it."
"'Tain't nothin'," said Hank, squirming in his chair. "It didn't work
just right. I guess I'll have to go home now. Ma said to get in by ten
o'clock."
"We'll have time for your report," Mr. Norton told him.
Hank kept nudging me, trying to get me to go with him, but I wouldn't do
it, so after a while he began.
You see his invention, the one he spoke to me about just before we
started, was a Life Saver. When we were learning to be Scouts Mr. Norton
taught us how to bring drowned people back to life again; that is, if
they haven't been in the water too long. What Hank wanted to do was to
invent something that would keep them from getting drowned in the first
place.
"It's all right to bring them to life," he told me, "but it would be a
heap better not to have 'em drown at all."
After doing a lot of thinking, he made a sort of balloon of oiled silk,
with the mouth fastened to a hollow reed and a piece of potato to put
over the end of the reed, instead of a cork. Hanging from the mouthpiece
were two pieces of stout cord.
"What's it for, Hank?" asked Skinny, when he was showing it to us. "It
looks like a bagpipe."
"It's a Life Saver," he said. "You carry it in your pocket when the air
is out of it and look along the river until you find somebody drowning.
Then you throw him the Life Saver, if he hasn't got one in his own
pocket. He ties it around his neck, puts the mouthpiece to his lips, and
blows the bag full of wind. Then he puts the potato on the end to keep
the air from leaking out. He can't sink, can he? The balloon will hold
him up."
"Great snakes, Hank!" said Bill. "You've got a great head--like a tack."
"A tack's head is level, just the same."
"Guess what," said Benny. "Let's go swimming up to the Basin, to-morrow,
and try it."
"We can go swimming if we want to," Hank told him, "but I did try it. It
worked and it didn't work."
"What's the answer?" I asked.
"Well, you see, I walked all the way to Cheshire Harbor, looking for a
chance to use the Life Saver and I couldn't find anybody even in
swimming, let alone drowning. The water isn't deep enough for drowning
in most places, anyhow. But when I got to Cheshire Harbor I found a kid
sitting on the bank of the race, fishing.
"'What you got?' he asked, when he saw me fooling with the Life Saver.
"'Jump in,' I said, after I had told him about it. 'I'll show you how it
works.'
"'Jump in yourself,' he said. 'I don't want to get my feet wet. Let's
see the old thing, anyway.'
"I handed it to him and he blew up the bag until I thought it would
bust, and then tied it on with the strings.
"'Say, that's great stuff,' said he. 'I'll bet it will work all right.'
"When he said that, I don't know why I did it, but it seemed as if I
couldn't help it. I felt as if I just had to save him. I pushed him in,
balloon and all."
"Gee-e-ewhilikens!" shouted Skinny.
"You mutt!" said Bill.
Mr. Norton was too surprised to say anything, but he had the funniest
look on his face.
"Did it work?" Benny asked.
"It worked all right, but----"
"But what?" I said, beginning to get mad because Hank kept stopping at
the most interesting parts.
"He had tied it on to one ankle, instead of around his neck. It made his
ankle float, but his head went under, and he couldn't swim. I rescued
him, but I had to jump in after him and pull him out. It was hard work
because he kept trying to hit me all the time. Then, after I'd got him
out, I had to lick him before he would let me go on and do my stunt."
"I hardly think that was according to Scout law," said Mr. Norton, when
the rest of us had finished laughing and pounding Hank on the back.
"I rescued somebody, just the same. Only it wasn't a maiden."
"We still have a few minutes," said Mr. Norton. "Suppose that we play a
new game which I have here. It is a kind of invention of my own and is
called baseball."
"Seems as if I'd heard of that game somewhere," said Skinny, poking me
in the ribs.
"Not this one. This is parlor baseball and is brand new," replied the
Scoutmaster.
He brought out a chart, marked off in squares to represent different
plays, and laid it flat on the floor, about six inches from the wall, at
the end of the room.
"Now," said he, "we'll choose sides, then stand off about ten feet and
toss silver dollars at the squares. That is the same as going to bat. I
mention silver dollars because I brought some with me. Any disk, or
ring, about the same size and weight would do as well and might be more
convenient. The square on which the disk rests gives the result of your
play. If the disk rolls off the chart it counts as a strike, and three
strikes are out. Usually the Scoutmaster or Scout leader acts as umpire,
calls off each play as made and keeps the score. To-night, however, as
William is not able to play, we will make him umpire and I will take
part in the game to make even sides."
HOME RUN STRIKE THREE BASE
HIT
FLY CATCH BATTER HIT OUT ON
FIRST
SINGLE BALL TWO BASE
HIT
FOUL PASS BALL BALK
"Let me illustrate," he went on. "We will suppose that the first man up
throws three disks and all of them roll off the chart. That counts as
three strikes and he is out. The second player may throw a two-bagger or
a single. He then returns to his seat and the third player, by throwing
a three-bagger, brings the second man home and gains third base for
himself. The runners are advanced each time as many bases as the batter
makes. They also are advanced one base by a pass ball, a fly catch or an
out-on-first. The first two fouls count as strikes, of course, and four
balls entitle the batter to first base. The arrangement of these squares
is important. The home run is guarded on three sides by strikes and in
front by a fly catch. The three-base hit is as carefully guarded."
"Say, that game is all right," said Skinny, after we had finished
playing. "Three caws for Mr. Norton, our 'stinguished and celebrated
Scoutmaster."
As soon as he could make himself heard, Bill spoke up.
"I think the secretary," said he, "ought to put how to play that game in
the minutes of the meetin'."
"There ain't goin' to be any," I told him. "It's too much work."
"I think that William's suggestion is a good one," Mr. Norton said, "and
I also appreciate the force of your secretary's objection. How would it
be if I should do the work? I'll have typewritten copies of the rules of
the game struck off, so that each of you can have one."
That is what he did, the very next day. I am going to put the rules into
this history right here, just as he wrote them, because other Scouts may
want to play the game.
_Scouts' Parlor Baseball.--Rules for Play._
Divide the patrol into two equal groups and arrange them in batting
order on opposite sides of the room. Place the baseball chart six or
eight inches from one end of the room on the floor and indicate a mark
ten feet from the chart for the "batter" to stand on. The Scouts having
their inning then take turns at tossing a silver dollar (another
metallic disk or ring of equal size will suffice) at the chart. Each
player's record at bat is told by the square on which the dollar rests,
off the chart entirely counting as a strike. If the dollar rests
squarely across a line it is tossed again.
The rules of baseball govern the game. After a player finishes his turn,
he takes position at the farther end of his side, and the next in line
takes his turn, thus preserving the batting order. When three players
have been declared out, that side is retired and the other side takes
its inning. If time permits, a nine-inning game is played; otherwise the
number of innings to be played should be decided before beginning.
When a "batter" wins a position on a base he is advanced at each play as
many bases as the next player earns at the "bat." He also advances one
base on out-on-first, fly-catch, balk, and pass-ball plays, and when
forced. He must keep track of his supposed position on the bases and
report to the official when making a score.
The official, usually the patrol leader or Scoutmaster, decides the
plays and tosses the dollars back to the players. He also keeps the
score, and may correct a player, if necessary, for being noisy, or for
leaving his seat when not playing. In fact, he is in control of the
game, but is not allowed to play except when there is present an odd
number without him.
The chart should be made of stiff paper so as to lie flat on the floor,
or of cloth, in order to be tacked down. Each square should be 9 x 9
inches, but a smaller size may be used if the room is not large. In that
case the players should stand less than ten feet from the chart. The
squares must be labeled as in the diagram. Young Scouts, or beginners,
are sometimes allowed to stand eight, or even six, feet from the chart,
in order to make the sides more equal. This and any other questions that
may arise are decided by the official.
CHAPTER XIII
EAGLE PATROL JOINS THE SCOUTS
YOU must not think, when you read this history, that something all the
time was happening to us Scouts. I am only telling about what did
happen. Pa says that when it comes to starting things we have them all
beaten to a frazzle and Ma told us that it would be a mercy if we ever
lived to grow up, without losing any of our hands or feet. But we don't
think so. Boys have to be doing something all the time, don't they? If
they didn't they would get into mischief.
Anyhow, there didn't much of anything happen after Skinny lassoed the
bear, for a long time, unless you count the Fourth of July. Nobody can
help having the Fourth of July. It's part of the year. It is for our
country.
One Fourth of July, long ago, even before Pa was born, they rang old
Liberty Bell in Philadelphia, to beat the band, and they fired off
guns. 'Cause why? 'Cause there was a paper signed on that day, which
said that the United States of America should be free and independent.
But England was like old Pharaoh, with the Hebrew children, that the
Bible tells about. They didn't want to let us go. I don't blame them
much for it, either, but Skinny does.
Anyhow, I guess God must have meant for us to go free, just as He did
the Children of Israel because, although England was the greatest Nation
in the world and the best one, too, it seems to me, and we were only a
few scattering colonies without much money or anything, we came out
ahead. That is why Skinny thinks that George Washington could have
licked Napoleon Bonaparte with one hand tied behind his back.
So we have the Fourth of July, and we boys ring the church bells at four
o'clock in the morning, when they don't catch us at it, just like old
Liberty Bell was rung so many years ago.
One of Skinny's ancestors was killed in the battle of Bunker Hill. That
is what makes him so fierce against the Britishers. Every Fourth of
July he has us go up on Bob's Hill or somewhere and fight the battle all
over again.
The time I am telling about we built a fire on the hill and rang the
church bells and fired off firecrackers until we were tired and half
starved; then went home to breakfast. Everybody promised to meet again
at my house about nine o'clock.
Soon after nine we all were sitting on our side steps, talking over
where we should go for our battle, when Skinny happened to stand up and
look down the street.
We heard him make a noise like a snake and he dropped off the steps to
the ground so quickly that we thought at first he had a fit or
something, until he made a motion for us to follow him and began to
crawl toward the fence.
We didn't know what the matter was, but knew that it was something
important, so we crawled along after him as fast as we could. When we
reached the pickets he pointed and we peeped over the top, careful not
to let more than our eyes be seen.
What we saw was three members of the Gingham Ground Gang coming up the
street, walking in the middle of the road and looking on both sides as
they came, as if they were expecting trouble and wanted to be ready for
it.
Two of them had red shirts, and that made Skinny mad because it made him
think of his ancestor who was killed at Bunker Hill.
"The Redcoats are coming," said he in a hoarse whisper, so that they
wouldn't hear, but fierce-like, just the same. "Wait until you can see
the whites of their eyes; then, 'charge, the ground's your own, my
braves. Will ye give it up to slaves? Hope ye mercy, still?'"
It was a part of his last day piece at school and sounded fine.
"Charge nothin'!" said Bill. "The Americans didn't do any charging at
Bunker Hill, I guess. The Britishers did the charging. The Americans
waited behind a fence until they got near enough and then let 'em have
it, until their ammunition gave out. Then they ran. That's what they
did."
That was true, too, but, just the same, it was a victory to hold the
hill as long as their powder lasted, and Bill knew it, but he liked to
get Skinny mad.
"Bill Wilson," said Skinny, "you are a nice patriot! You are a Scout and
a half; that's what you are--not! So are we going to run but, bet your
life, we're going to run toward the enemy. If you want to stay here
behind the fence you can do it. The rest of us are going to charge."
Bill gave me a thump in the ribs and grinned, but didn't say anything. I
saw Benny whisper something, his eyes shining with excitement; then
Skinny motioned to us what to do.
Each of us lighted a firecracker and held it with the fuse sputtering
and sizzling, until they were almost opposite. Then we threw the
crackers under their feet. They went off like a volley of musketry. At
the same time we gave a great caw and jumped the fence.
"Give it to 'em, fellers," yelled Skinny. "These are the guys that
wanted to duck Benny in the mill pond."
Say, it was great. The firecrackers surprised them, for they hadn't
seen us, and we were over the fence and upon them before they could run.
Things were lively in Park Street for a few minutes. Then, all of a
sudden, we heard a man's voice say:
"Scouts, attention!"
And there was Mr. Norton, looking surprised and sorry!
We all stood up with a jerk and saluted, and the Gingham Ground boys
started to run. They only went a few steps, however, and then waited to
see what was going to happen.
"Scouts," said Mr. Norton, sternly, "what sort of brawl is this, on the
Fourth of July?"
He was looking at Skinny, he being Scout leader.
"'Tain't a brawl," said Skinny. "It's the battle of Bunker Hill; that's
what it is."
"Oh, it is, is it? On which side are you Scouts fighting?"
"We are Americans, of course."
"Well, if I remember my history right, in that battle a little handful
of Americans faced the British soldiers and held them back until their
powder gave out. And here the American army seems to be attacking a
handful of British."
"That's what I told him," said Bill.
"Anyhow," said Skinny, "those guys tried to duck Benny that time when he
was coming home from his long hike. So we thought that we would duck
them in the race. Didn't they try to duck you, Benny?"
Benny nodded.
"How about Scout law?" asked Mr. Norton.
"Scout law doesn't say we mustn't duck our enemies."
"It does, too," Bill told him. "It says that we must be kind to
animals."
That was a hot one and it made us all laugh.
"How much more should we be kind one to another," said Mr. Norton.
"Well, it wasn't very kind to duck Benny," insisted Skinny.
"No, and they didn't do it. If I have been correctly informed, they let
Benny go because John here was kind to a dumb animal."
That was true and I said so.
"Even if they had ducked him, don't you think that it would be better to
heap 'coals of fire' upon their heads?"
It surprised Benny to hear Mr. Norton talk like that.
"We wouldn't do such a thing," said he. "Besides, we haven't got any hot
coals."
"Yes, you have," laughed Mr. Norton. "The 'hot coals' I mean are kind
words and kind actions. What I meant to say was that you should return
good for evil and then your kind words would make those boys feel as if
you were putting coals of fire on their heads."
"I don't believe we ought to do it," Skinny told him, "if it is going to
hurt that bad."
"Suppose we try it and see. I think perhaps it will not be quite so
painful."
"Boys," said he, turning to the Gingham Ground bunch just as they were
starting away. "I have organized these eight village lads into a patrol
of the Boy Scouts of America and we have planned to have a campfire this
evening on Bob's Hill. These Scouts of mine mean all right. They are
simply working off a little misdirected patriotism. Now, what we want,
is for you to meet with us, you and the rest of the Gang. Will you do
it?"
They didn't want to at first.
"There are Boy Scouts," he went on, "in all parts of the civilized
world; in England, too, Gabriel, as well as in this country, and the Law
says that all Scouts are brothers to every other Scout. There are a half
million in the United States alone. I have been appointed Scoutmaster
for this district and I want to organize one or two more patrols so that
I can have a troop. I have had you boys in mind ever since you so nobly
turned out to help find William, the time he was hurt on Greylock. It
will be much the same as the Gang, only better. You can keep the same
leader if you wish, and I know a man who will buy uniforms for you all.
Will you come to-night so that we can talk it over? What do you say?"
The uniform business settled it.
"We'll come, if the rest of the Gang will," they told him.
"Good! Shake hands on it."
"Attention, Scouts!" shouted Mr. Norton, after he had shaken hands.
"Salute enemy!"
We gave the Scout salute to the Gingham Ground boys, while they stood
there grinning and not knowing what to do.
Then, after whispering together, they gave us the Gang yell. It was
great.
"We'll be there," they called, as they started up the street.
They were, too, ten of them, with Jim Donavan at their head. They came
across lots from the Quaker Meeting House, soon after we had gathered
around the big stone where we have our fires, just as they had come two
years before, the time we had our big fight and came to know Jim.
Mr. Norton saw them coming and went to meet them.
"This is fine," said he, after we all had sat down on the grass around
the fire. "You are a pretty husky bunch of fellows, and Raven Patrol
will have to go some to keep up, after you get started. Skinny--I mean
Gabriel--suppose you tell our visitors something about the Scouts."
"It's great," began Skinny. "We've been bandits and we've been Injuns,
but Scouts beat 'em all. The woods are full of 'em all over the country,
and they go about with uniforms on, doing good and having fun. They are
like an army. We are one company, you will be another. I'm the same as
captain, only they call me patrol leader. Mr. Norton is Scoutmaster, and
there are officers above him, only we never saw them. We learn all about
woodcraft and signs and signaling and how to do a lot of things, and we
rescue people and do all kinds of stunts and get badges. The Ravens are
going across the mountain on an exploring trip. I am going to look for a
cave and maybe there is treasure in it. Our patrol animal is the crow,
and it 'most ought to be yours because you live so near the Raven
Rocks."
Skinny had run down by this time, although Bill was winding him up like
a clock behind his back and making a clicking noise with his tongue.
"G'wan!" said he, turning around and catching him at it, "or I'll biff
you one."
"Perhaps I'd better add a little to that explanation," said Mr. Norton.
Then he told all about it, much as he had told us that first time, and
about Scout law; what it meant to be a Scout; how it made boys manly,
and how much fun they would have.
"What I want is a troop," said he, when he had finished the story.
"Several patrols together are called a troop. I would be in charge as
Scoutmaster. Raven Patrol is now in pretty good shape. We are going on a
camping expedition in a few weeks and we'll have a good chance to
practise up on signaling, swimming, following trails through the woods,
and things like that. Next year I should like to take a whole troop
along. What do you say? Suppose you go over by that other stone and talk
about it among yourselves."
"I know what I'll say, right now," said Jim, "but perhaps we'd better
talk it over just the same."
We saw them whispering together for about five minutes. Then they came
back.
"We'll do it," said Jim. "And we'll do the best we can, only we may make
mistakes at first. We are going to take the American eagle for our
patrol animal on account of this being the Fourth of July."
"Everybody makes mistakes," Mr. Norton told him, "but the boy or the man
who has the right stuff in him never makes the same mistake twice.
Suppose that you elect a patrol leader to-night before we separate,
because we shall want to consult together a great deal in the next few
days and I shall be too busy to see you all."
"Jim," they began to yell, all keeping time. "Jim! Jim! Jim!"
"Jim, you seem to be elected," said Mr. Norton, reaching out and shaking
hands with him.
"Speech!" yelled Hank.
"Ladies and gentlemen," said Benny, getting up on his feet and bowing
right and left, "the Honorable James Donavan will now say a few words,
if he dast."
Jim looked as if he wanted to run, but in a minute he braced up.
"I never made a speech in my life," said he, "and I ain't going to make
one now, but you will find the Gang true blue. We ain't much on clothes,
and our folks haven't got much money, but we'll do the best we can, if
you will tell us how. And we are much obliged for taking us in."
"Three cheers for Captain Donavan and Eagle Patrol," shouted Mr. Norton,
waving his hat. "Now!"
I'll bet they heard us down in the village. After it was quiet again I
saw Skinny whispering something to Bill. Bill nodded his head and passed
it on to Hank, and finally it came to Benny and me, who sat at the end
of the line. We nodded and began to creep nearer the fire while waiting
for the signal.
"Caw!" yelled Skinny, all of a sudden, like you sometimes hear a big
crow in the Bellows Pipe.
As he yelled, he grabbed a burning brand out of the fire, and the rest
of us did the same. Then we formed a circle and danced a war dance
around the Gang, whirling our brands in the air until the sparks flew
in the growing darkness and there seemed to be a ring of fire.
"Shall we eat 'em alive, my braves?" chanted Skinny.
"No," we shouted. "They are brothers."
"Shall we mop the earth with 'em?"
"No," we yelled. "They are Scouts."
"What shall we do?" asked Skinny, stopping in front of Jim, who was too
surprised to say anything.
"Give them the glad hand," we answered.
"'Tis well," said he, grabbing Jim by the hand, while we did the same to
the others.
"I'll tell you what," said Mr. Norton, a little later. "I feel so good
over this that I'll buy. Lead me to a soda fountain."
CHAPTER XIV
PLANNING A CAMPING TRIP
WE boys often think of what a fisherman told us one summer day, out on
Illinois River, at the foot of Buffalo Rock.
[Illustration: "IT GIVES ME PAIN," SHE SAID, "TO INFORM YOU THAT THE
WOODBOX IS EMPTY."]
"Play," said he, "is work that you want to do and don't have to do," or
something like that.
Ma often says, when she sees us playing, that if she should make me work
that hard I would think I was abused.
I guess, maybe, that is so. It surely is some work to chase uphill and
around, play ball, and do all kinds of stunts, and sometimes when night
comes we feel tired.
I went home to supper one day, all fagged out, so tired I hardly could
drag one foot after the other, and flopped down in the nearest chair.
Ma heard me and put her head in at the door.
"It gives me pain," she said, "to inform you that the woodbox is
empty and I need a hotter fire to bake those biscuits that you like so
well."
"Oh, Ma!" I exclaimed. "Can't you get along until morning. I'm all in."
"Why, you haven't done a thing to-day!" she told me.
I had climbed up and down Bob's Hill six times; been up to Peck's Falls
and the cave once; followed the brook over rocks and fallen trees to
where it tumbles out of a sunshiny pasture into the shade of the woods
in a great watery sheet; been swimming in the Basin, on the other side
of the valley; played a match game of baseball at the Eagle ground;
played Indian in Plunkett's woods, tracking the enemy through the
forest; played foot-and-a-half, until I thought my back would break, and
wrestled with Skinny, until he fell on me like a thousand of brick. But
I hadn't done anything all day! Oh, no!
"You don't want me to do it, do you?" she said.
Of course, I didn't want that; so, tired as I was, I dragged out to the
shed and brought in an armful of wood.
Just then I heard a whistle, followed by the caw of a crow from in front
of the house, and I chased out to see what was doing.
It was Benny. He had come over to tell me that there would be a Scout
meeting at his house that night.
"John's too tired," Ma told him. "He hardly was able to bring in four
sticks of wood."
"I feel better now," I hurried to say. "The exercise did me good. After
I have had some of your delicious biscuits and some honey, I'll be all
right again. Besides, I'd hate to miss a Scout meeting; I learn so much
there. Will the wood I brought in last until morning?"
"I thought Mr. Norton was away?" she said.
"He is; but they are going to have a meeting, anyhow."
"Oh, please let him go, Mrs. Smith," put in Benny. "Pedro is our
secretary. We can't have the meeting without him."
Ma likes Benny so well I just knew she would have to give in. She knew
it, too, I guess, for she looked at us a minute, sort of smiling to
herself; then she said:
"Well, if he will come home at nine o'clock and promise to take a nap
to-morrow afternoon, I'll let him go. He has been losing too much sleep
lately."
I didn't think much of that nap business. Daytime wasn't made to sleep
in, except, maybe, the early morning hours when you first wake up.
"I'll promise to lie down and shut my eyes," I told her, "but I can't
promise to take a nap, can I? The sleep may not come."
That is true. I've laid awake a lot of times fifteen or twenty minutes
and maybe more, at night, trying hard to go to sleep and not feeling a
bit sleepy.
That is why I was in bed when Skinny came around the next afternoon. He
knew that I would be, and instead of coming into the back yard and up on
the stoop, as he usually does, he went up the drive between our house
and Phillips' and whistled softly under my window.
With one bound I was out of bed and looking down at him. He had on his
Scout uniform, and his rope was wound around his shoulders.
I was just going to tell him to wait until I could come downstairs, when
he put one finger to his lips, then looked up and down the drive to see
who was watching. There was nobody in sight. Ma was taking a nap in her
room and I guess Mrs. Phillips was, too, across the way.
"S-s-t!" he hissed. "Are you alone?"
I nodded. It didn't seem safe to say anything.
"You ain't chained to the bed, or nothin', are you?"
"Nary a chain," I told him. "We are all out of chains."
"'Tis well!" said he, coiling up the rope in one hand and getting ready
to throw. "Quick, now, and mum's the word!"
I caught the rope as it came in through the window and fastened one end
to the bed. Then I threw out the other end, climbed out myself, and
shinned down.
"What's the matter?" I asked, as soon as I had reached the ground.
"Let's go around and untie the rope; then I'll tell you."
A few minutes later he was showing me a letter which he had from Mr.
Norton, who was away on business. This is what the letter said:
"DEAR FELLOWS:--I shall be at home in a few days
and should like to have a meeting of Raven Patrol
to talk up our camping trip. Are you thinking
about it and planning where to go? The pasture
above Peck's Falls would make an ideal camp. There
is water and sunshine and shade and old Greylock.
That would suit me pretty well, but it is so near
home it might not suit you. If not, I have a
regular trip over the mountain in mind, one that
will take a hike of several days to get us there.
Talk it over among yourselves and ask your folks
about it. Then meet at my house next Saturday
night. We'll decide the matter and begin to get
ready. Yours sincerely,
"CHARLES NORTON, Scoutmaster."
"Ain't he a brick?" said Skinny, when he had finished reading. "What do
you say, old Scout?"
"I say hike," I told him. "That pasture above Peck's Falls is where Tom
Chapin tried to paralyze a bull by the power of the human eye, like the
school reader says, and got thrown over the stone wall by the critter.
No more of that for muh!"
"We'd have a rope along, you know."
"Yes, and who'd tie it and what would the bull be doing all that time?"
"I'd rather go over the mountain on a hike, myself," he said. "Come on,
let's ask the other boys."
"Wait a minute while I fill the woodbox," I told him.
Skinny helped me do that and we were soon on our way.
The other boys felt just as we did about it. Of course, it is always fun
to be near our cave and it is a fine place to get into when it rains,
but we could go there any old time.
The folks seemed to think near home would be better, until we told them
about the bull and how near we all came to getting killed. They had
forgotten about that and so had we, almost.
Finally Pa settled it for me.
"I am willing to leave it to Mr. Norton," he said. "As long as he goes
with you I don't care much where you go, for I know that he will take as
good care of you as I could myself. His hold on you boys is remarkable
and I am willing to back him in anything that he wants to do. I'll say
this much, however. He is going to have his hands full when he
undertakes to look after you boys for a week or two at a time."
We hardly could wait until Saturday night to hear Mr. Norton's plan and
decide what to do.
He seemed glad to see us when the time came, only he wouldn't hurry the
meeting or leave anything out. Skinny, being patrol leader, always acted
as chairman and pounded the table, when he could find one to pound.
"The meetin' will come to order," said he, looking around for something
to thump and not finding anything but Bill Wilson, who dodged out of the
way.
"The secretary will call the roll."
I called the names of the boys, and each one in turn arose and gave the
Scout salute, first to Mr. Norton, then to Skinny.
"Is there any business to come before this 'ere meetin'?" he asked.
"Mr. President," I said, jumping up.
"The gentleman from Park Street," said Skinny, as big as life, just as
Pa had taught us to do at meetings in our barn.
"We have with us this evening our Scoutmaster, who, I think, has
something to say."
"'Tis well," said Skinny. "We'll harken unto his words of wisdom."
"Before I speak the words of wisdom which our patrol leader has so
kindly mentioned," laughed Mr. Norton, "I will ask Mrs. Norton to
refresh and fortify us with some lemonade."
Benny reached the door almost as soon as she did.
"Let me do it, Mrs. Norton," he said.
He grabbed the pitcher and tray and poured out a glass for her; then
went around the circle. It tasted fine on a warm night.
"Mr. Chairman," said Mr. Norton, after we had emptied the pitcher. "I
want to call up the question of our camping trip. Have you boys thought
about the matter?"
"We haven't thought of much else," Hank told him.
"Well, how about it? Shall we camp out above Peck's Falls? What do you
say, William?"
"It's too near home," said Bill. "Ma would get scared the first night
and call me back."
"That certainly would be serious. What do you say, Mr. Secretary?"
"I say so, too," I told him. "It's fine up there and wild and all that,
but let's go where we never have been before."
"How about it, Mr. President?"
"It's me for the hike," said Skinny.
The other boys all said the same.
"It seems to be unanimous," said Mr. Norton. "I thought that probably
you would feel that way. Well, this is what I have in mind, in case you
decide to take the trip, instead of remaining near home. What do you say
to hiking straight east over Florida Mountain, as far as Deerfield and
the Connecticut River? We can get a horse and carry our camping outfit
and supplies in a wagon. We can take turns driving. It will rest us, and
if anybody should give out the wagon will come in handy. We can take as
long a time as we want on the way, camping out each night."
Mr. Norton stopped and looked at us to see how we liked the plan. Say,
it didn't take him long to find out. Every boy jumped to his feet and
shouted. Skinny forgot that he was chairman and started to march around
the room, shooting and striking at the enemy, and we all fell in line
after him except Bill. He stood on his hands, kicked his feet in the
air, and whistled through his teeth.
Mr. Norton looked pleased.
"Mr. Chairman," he said, as soon as we had taken our places again. "I
hardly think it necessary to put that to a vote except, perhaps, as a
matter of form. The next question is, will your folks let you go?
Sometimes fathers and mothers have very decided notions about what they
want their boys to do and more especially what they don't want them to
do."
I told him what Pa had said about being willing to have us go anywhere
with him, and the other boys said that their folks felt the same way.
"Good! We'll consider that settled and get down to details as quickly as
possible. I should like to get started in about two weeks, which will be
early in August. We'll call another meeting in a few days and I'll have
a list of the articles needed and their cost ready to submit to you. I
know where we can get tents, but there are a whole lot of things we
shall need in the woods, besides things to eat. Is there any more
business to come before the meeting, Mr. Chairman?"
"There is," said Skinny, who had been scribbling something on a piece of
paper. He handed it to me to read, and this is what it said:
"Resolved, that Mr. Norton is great stuff."
"All that are in favor of the motion salute the Scoutmaster."
That ended the meeting. We had to have several more like it before we
could get everything ready for the trip.
"It is early yet," said Mr. Norton. "If you would like to have me, I'll
tell you a story about what I think was one of the greatest scouting
trips ever undertaken."
CHAPTER XV
SCOUTING IN THE GREAT NORTHWEST
"SOME of you boys went out to Illinois, last summer," he began. "Did you
go as far as the Mississippi River?"
"No, but we camped out on the Illinois River," I told him, "and that
flows into the Mississippi."
"We explored," explained Benny, "just like LaSalle and Tonty and the
other guys did. Skinny was LaSalle and I was Tonty."
"LaSalle and Tonty were great scouts. Do you remember when they made
those early explorations?"
"I think it was somewhere around 1680 or 1681," said Skinny, who was
always good in history. "Mr. Baxter told us all about it while we were
sitting on top of Starved Rock, where LaSalle once had a fort."
"There was a great country west of the Mississippi, about which LaSalle
knew very little, although when he explored the river he took possession
of the land in the name of his king, and he called the country
Louisiana.
"At that time, with the exception of a few fur traders and missionaries,
all the people who came to America from the Old World settled along the
Atlantic coast and the Great Lakes, in various colonies. Some of these
afterward became the thirteen original states of the United States of
America.
"After Thomas Jefferson became president, he had a chance to buy
Louisiana of Napoleon, who was then at the head of the French
government, and he did so."
"Huh! Napoleon!" said Skinny. "George Washington could lick----"
"Aw, ferget it, can't you?" said Bill. "You are stopping the story."
"That gave us a vast territory, reaching from the Atlantic Ocean to the
Rocky Mountains. Nobody knew very much about it, or about the country
west of the Rockies. Jefferson may have been looking far into the future
when he made the Louisiana purchase, but probably his more immediate
purpose was to secure undisputed possession of the wonderful Mississippi
River.
"That was in 1804, only a little more than a lifetime ago and nearly a
century and a half after LaSalle explored the river and took possession
of the country. Little, if anything, was known about the country at the
time of its purchase by the United States more than was known in
LaSalle's time. A few hardy traders went up and down the river, buying
furs of friendly Indians, and that was all.
"Naturally, after Jefferson had bought it, he wanted to know something
about his purchase. So he appointed two men to explore the new country.
I want you to remember their names, because they did a great work. One
was Meriwether Lewis and the other William Clark, and you will find
their trip described in your school history as 'the Lewis and Clark
expedition.' I can't see why their exploration was not attended by as
much danger and hardship as LaSalle's, which had been undertaken so many
years before. The dense forests and great rivers of the West were all
unknown and there were many hostile Indians.
"What did you boys do, when you made up your minds to explore the rivers
in Illinois last summer?"
"We built a boat," Hank told him.
"Exactly. And that was what Lewis and Clark did, or, rather, it was done
for them at Government expense. A keel boat, fifty-five feet long and
drawing not more than three feet of water, was made for them at
Pittsburgh, where, if you remember, two rivers unite to form the Ohio.
This boat had places for twenty-two oarsmen and carried a large, square
sail. Steamboats were not known in those days, although a few years
afterward Robert Fulton ran one on Hudson River. The Government also
provided two smaller boats and loaded them with coffee, sugar, crackers,
dried meats, carpenter's tools, presents for the Indians, and things
like that. A few horses also were taken along in the large boat.
"The leaders selected a crew of twenty-five men, and one fine day the
whole outfit started down the Ohio River. When they reached the
Mississippi they turned north and soon made their way up the great river
to St. Louis. St. Louis was a French trading station then. Now it is a
large city. A few years ago the hundredth anniversary of the Louisiana
purchase was celebrated by holding a world's fair in St. Louis.
"There more men joined the expedition and considerable information that
President Jefferson wanted was picked up about the Indian tribes who
lived up and down the river.
"Finally, May 14, 1804, the explorers started on the real trip. It
wasn't easy work any longer, for they had to row against the mighty
current of the Mississippi. After they had gone a few miles they came to
another great river, which was pouring a dirty looking, yellow flood
into the Mississippi. Who can tell me what that river was?"
"The Missouri," said Benny, who had been studying about it in school.
"The Mississippi River, with its principal tributary, the Missouri, is
the longest river in the world."
"Right you are. If you will look on some map you will see how it is
possible to go in a boat from Pittsburgh almost across the continent.
Lewis and Clark turned into the Missouri and started for the then
unknown Northwest. They made their way along very slowly, for the river
was swollen with heavy rains and the current was very strong.
"After much labor and hardship they managed to reach the mouth of the
Osage River. There they went into camp and sent out an armed party to
explore the interior. When the party returned they brought back ten deer
and all had a great feast on the river bank.
"Once more they breasted the fierce current, narrowly escaping shipwreck
several times. Once the wind was so strong that they were obliged to
anchor and go ashore. Again they had to pull their boats along with
ropes through some rapids."
"Betcher life they didn't go without a rope," said Skinny. "Why----"
Somebody threw a sofa pillow just then and it struck exactly where his
face happened to be. Before he could find out who did it Mr. Norton went
on.
"At last they reached the mouth of the Kansas River. A large city stands
there now. Does anybody know the name of it?"
"That is too far from home," said Benny. "I know what city is at the
mouth of Hoosac River. There ain't any."
"Kansas City now stands where they went into camp. They divided into two
parties. One went out after game, so that there should be plenty to eat,
and the other explored the country."
"It's fun to explore," said Bill.
"Probably these men found a certain pleasure in it, notwithstanding the
hardships. They were seeing something new every day. After a time they
started once more and late in July reached the mouth of the Platte
River. They had heard that a tribe of Indians were living near there, so
Lewis and Clark went out with a party to find them and tell them that
the country now belonged to the Great Father at Washington. Under some
bluffs, opposite the present city of Omaha, they sat in council with the
Indians, made them gifts, and smoked the peace pipe. The Indians didn't
seem to care who owned the country so long as they received presents and
had room enough to hunt. A city now stands on those bluffs where the
Indian council was held. I guess you can tell me the name of that one."
"Council Bluffs," said two or three of us at the same time.
"Then on went the explorers up the river, through a wonderful country.
Vast prairies, covered with grass and without any trees, stretched away
in every direction, as far as they could see, and great herds of buffalo
roamed up and down. On they went, through what is now Nebraska; then
through South Dakota; then, North Dakota, where some fierce Indians
dwelt. Another council was held and more presents were given. When the
boat was about to put off after this council, the Indians grabbed hold
of the cable and held it. They wouldn't let go."
"Great snakes!" said Bill. "I'll bet they didn't do a thing to those
Injuns. I'll bet they paralyzed them. They had guns, didn't they?"
"Yes, and they did sort of paralyze the savages, I guess.
"'Take aim but don't fire,' Lewis told his men.
"The next second those Indians were looking into the muzzles of about
twenty-five guns."
"That's the stuff!" shouted Skinny, swinging his arms and then
pretending to shoot. "Did they kill them all?"
"I am afraid that you boys are a little bloodthirsty," said Mr. Norton.
"They didn't shoot at all. When the Indians saw the pointed guns they
dropped the cable and pretended that all they wanted was to do some more
trading. The white men were glad enough to let it go at that and get
away as quickly as possible.
"It soon became necessary to go into camp for the winter. An island in
the river was chosen for the purpose and they spent the winter there.
The Indians in the vicinity proved to be friendly. They never had seen
white men before, possibly that was the reason. Some of the things which
are very common to us seemed wonderful to them. Do you remember how I
lighted the fire one day, when we wanted to cook dinner on Bob's Hill
and had forgotten the matches?"
"With a sunglass," I told him.
"Well, that didn't seem very astonishing to us because we were used to
it, but the Indians had never seen a sunglass. They started their fires
by rubbing two sticks together. Even the whites had to use a flint and
steel, for the art of making matches hadn't been discovered. Captain
Clark carried a sunglass in his pocket. One day he went to an Indian
village, intending to smoke a peace pipe with the chief. As he was
entering the village, he saw some wild geese flying over and shot one.
The Indians heard what seemed to be thunder and saw the goose fall, and
it scared them. They ran into their wigwams and closed the skin doors.
Soon after Captain Clark came up to the wigwam of the chief, without
thinking he was doing anything out of the ordinary, he pulled out his
sunglass and lighted his pipe with it.
"The frightened Indians were peeking out of their wigwams, and when they
saw the white man start a blaze in his pipe by holding up one hand,
they felt sure that he was a spirit. The Redskins gave one yell and ran
into the woods. It was a long time before they could be made to
understand.
"Spring came at last and the impatient party started up the river again.
The way grew more and more difficult. They were now a long distance from
the mouth of the river, and the water was shallow in places and filled
with dangerous rocks. Often they had to get out and wade, pulling the
boats along by the cables.
"May 26 they passed the mouth of the Yellowstone River and for the first
time saw the Rocky Mountains in the distance, covered with snow and
looking very grand. They were then in Montana, or what we now call
Montana.
"In June they heard the roaring of a cataract, and Lewis started out
afoot to find it. After he had traveled for hours he climbed a cliff and
at last looked down upon the cataract. So far as we know he was the
first white man who had ever seen it, although thousands see it every
year now. The cascades of the Missouri stretch for thirteen miles, with
foaming rapids between. It is a great sight."
"Gee, Peck's Falls ain't in it," said Skinny. "Did he find a cave?"
"History fails to mention a cave. Lewis went back and ordered the boats
to proceed up the river as far as the first rapids. The question was,
how to get around those cascades. They couldn't go up the river, so they
had to get the boats around in some way. Their horses had died during
the winter. There was nothing to do but drag the boats around eighteen
miles. The men went to work and made rough carts, felled trees, cleared
away bushes, dug out rocks, leveled off the ground, and pulled, pushed,
and struggled on, until at last the work was accomplished and the boats
were launched again in the river above the rapids.
"But soon the river became too shallow for the large boat and they had
to stop again. Then they cut down trees and made 'dugouts.' They paddled
on until finally they came to a most wonderful place. We think that the
ravine below Peck's Falls and that at the Basin are grand and beautiful,
and so they are, but they found a great canyon, whose walls in places
were a thousand feet high.
"Beyond this canyon they could not go in their boats, for they were at
the foot of the first range of the Rockies. They had to leave their
boats there and climb. But, first, Lewis started out alone to find some
Indians for guides.
"The brave man made his way to the top of the ridge and looked down into
the valley beyond. In that valley flowed a river, and far up the stream
he could see an Indian village. It was the home of the Shoshones. He
managed to reach the village, and by offering presents induced some of
the Indians to go back with him, bringing horses, and to guide his men
across the mountains.
"The trip was a very perilous one, even with guides, and it took them a
whole month to cross. Up, up they climbed, so high that they could not
find any game to shoot. One by one, the horses died from exhaustion, and
the starving men ate the flesh to keep themselves alive.
"After terrible hardships, they finally left the mountains behind and
came upon streams which flowed toward the west. Here they rested,
secured a new supply of food, built new boats, and then, when all was
ready, paddled down the Lewis and Clark rivers into the broad Columbia,
which, as you know, pours its waters into the Pacific Ocean. They had
crossed the entire country from Pittsburgh to the Pacific, and made the
whole trip by water except that terrible journey across the Rocky
Mountains.
"It was now November and they were forced to go into camp once more to
spend the winter months. In the spring they started on the long journey
home again and at last reached Washington, where they told the President
about the vast Northwest and what a great country he had purchased from
France."
"I'll tell you what let's do," said Benny, after Mr. Norton had
finished. "When we start on our trip let's play we are Lewis and Clark
'sploring the country."
CHAPTER XVI
CLOUDBURST ON GREYLOCK
SKINNY says that if they would let him run the weather he wouldn't have
it rain daytimes during vacation. All of us Boy Scouts feel that way,
too, because, what's the use? The days are made for boys to have fun in
and the nights are made to sleep. So, why not have it rain nights when
folks are sleeping?
Anyhow, it rained that August as we never had seen it rain before and
never want to see it again. It began in the night, all right, just like
rain ought to do, but it didn't stop. When day came it seemed to take a
fresh start and kept going. It rained all day long and we couldn't have
any fun at all. When it came time to go to bed it quit for a spell, but
it started up again before morning. It wasn't any drizzle, either. It
came down in bucketfuls, until I thought the village would be washed
away and that even Bob's Hill would float off.
Along about ten o'clock in the morning it let up, and pretty soon, who
should come along but Skinny and Bill, barefooted and with old clothes
on. They were worried about the cave, and so was I. While it was raining
so hard I thought about it a lot.
You see, our cave is a little below Peck's Falls, on the bank of the
brook. There are two entrances. One goes in from the top on the upper
side. You first go down into a hole and then wriggle through an opening,
until you come out into the real cave. We don't use that one except when
we want to escape from the enemy, or something like that.
The one we use is below, right at the edge of the water, and leads
straight into the real cave. The floor of the cave is even with the
water at the entrance and then <DW72>s back a little out of the wet.
Once a flood filled the cave and nearly drowned us. We should have been
drowned, if Tom Chapin hadn't been with us. He dove down through the
hole into the upper cave and then pulled us through after him. After
that we built a dam so that it would not happen again. I told all about
that once in the doings of the Band. What we were worrying about was the
dam's giving way.
Almost always in summer the brook is fine. It pours a clear stream down
over the rocks and kind of talks to us and sings, so that we like to be
in the cave and listen to it. But sometimes in the spring of the year,
when the snow on the mountain is melting and old winter is running away
into the valley, and sometimes after very hard rains, the water roars
over the falls and then dashes down through the gulch and over the rocks
below, like some wild beast. At those times, it is a good place to keep
away from, unless you have a dam or a cave that needs looking after.
"Get your hat, Pedro, and come on," said Skinny. "We want to see about
the dam. If it washes out the water will fill our cave."
"And bring a shovel," added Bill. "We'd brought one, only your house is
so much nearer."
"All right," I told them. "Whistle for Benny, while I'm getting it."
The four of us went up through the orchard and took the road around the
hill to the top because the rain had made it too slippery to climb
straight up. We knew by the roaring of the water, long before we came in
sight, that Peck's Falls were going it for all they were worth.
When we finally, one after another, crept out on the ledge of Pulpit
Rock, in front of the falls, the sight almost scared us. It was great,
the way the water came down, fairly jumping from rock to rock, until
with a final leap and roar, it plunged, all white and foaming, into an
angry pool below; then dashed off, with a snarl, through the ravine.
"Gee-whillikens!" said Skinny. "Those are some falls, all right. How'd
you like to go in swimming?"
"It would just about use a fellow up to go through there," I told him.
"Boost me up so that I can look down at the cave."
"We'll boost Benny," he said. "He isn't so heavy."
The pulpit part reaches up several feet above the narrow ledge like a
wall, and back of it there is a straight drop, a hundred feet or more
down.
"The cave is all right, I guess," Benny told us, when we had held him up
so that he could see over without getting dizzy. "I can see where the
upper entrance is, but, say, the brook is fierce."
We crept off from the rock and made our way carefully down the side of
the ravine to the cave.
It was as Benny had said. The dam had held and was keeping the water
from flooding the cave. The upper entrance was all right, although it
was too muddy to use. The water had backed up around the lower entrance
and part way into the cave, but beyond it was dry.
The little mountain brook had turned into a torrent, raging along like
some wild beast, and foaming over the rocks below, almost like Peck's
Falls. Just above these smaller falls, a tree, which had been carried
down into the ravine, stretched across the stream from rock to rock,
with its slippery trunk about two feet above the water.
"I guess everything is all right," said Skinny, "but maybe we'd better
fix the dam a little. Gee, but it's getting dark in here."
We worked a few minutes, throwing rocks and dirt against the dam. I had
just stood off to say that I thought it would hold now, when Skinny gave
an awful yell and slipped off from a rock, on which he had been
standing, into the flood.
I made a grab for him and missed, and in a second he was whirled down
the stream.
It is queer how much thinking one can do in a second. I thought of the
rocks and of the falls below and of how nobody could go through without
being pounded against the stones.
I was afraid to look, until I heard another yell. Then we yelled, too,
for there was Skinny clinging to the tree which stretched across the
stream, just above the lower falls, and yelling to beat the band.
The water pulled and tore at his legs, dragging them under the tree and
to the very edge of the rock which formed the falls. On his face was
such a look, when we came near, that I knew he could not hang on much
longer.
"Hold on tight, Skinny," I called. "We are coming."
It did not take us long to get there, but when we came opposite to where
he was hanging we could not reach him, and the log was too slippery to
walk on.
"Can't you work yourself along the tree?" I asked. "We can't reach, and
even if we could walk out I don't see how we'd ever get back."
He shook his head in despair.
"I can hardly hold on at all," he told us. "I'll have to let go in a
minute, if you don't do something. Get the rope. You always want a
rope."
I hadn't thought of the rope which we have kept in the cave since the
time I told about, when the flood came near drowning us.
Then Bill, being corporal, pulled himself together.
"Run to the cave for the rope," said he, "while I hold him."
Before we could say a word or stop him, he straddled the tree and began
to work his way out, hitching himself along with his hands.
"Run," he yelled again, when he saw us looking with pale faces. "Skinny
saved me and I'll save him, if it takes a leg."
We were halfway to the cave before he had finished speaking. I helped
Benny in through the water, holding him to make sure that he wouldn't
slip, and in two or three seconds he was out again with the rope.
We found Bill clinging to the slippery tree with both legs and holding
Skinny by the collar with both hands. Skinny had a fresh grip and was
hanging on for all he was worth.
We tied a slip noose in one end of the rope and threw it to Bill.
"You'll have to let go with one hand at a time, Skinny," I heard him
say. "Wait until I get a better grip. Now!"
I saw Skinny let go for a second with his left hand. Bill hung to his
collar with one hand and with the other put the loop over his head and
under his arm. Then Skinny grabbed hold again and did the same with the
other hand.
"Pull her tight, boys. Easy now."
We pulled until the noose tightened under Skinny's shoulders. Then we
waded into the water as far as we dared and pulled steadily on the rope.
Skinny scrambled along through the water, digging his finger nails into
the bark, with Bill holding on to his collar as long as he could reach.
By the time we had him out it had grown so dark that we hardly could see
Bill, but we knew he was out there because we heard him say "great
snakes."
"Throw me the rope," he called.
He put the noose around his own shoulders, and with our help was soon
standing on the ground.
"I swam her all right," said Skinny, "but I hadn't ought to have done
it. Ma told me not to go swimming to-day."
Just as he said that something seemed to shut us in. The light was
blotted out and we stood there in the dark, scared and wet, wondering
what was going to happen.
We groped our way along until we reached the cave and crawled in through
the water. I didn't like to do it because I knew that if the dam should
give way the cave would be flooded. But we had made it stronger and we
had the rope to climb out by at the upper hole, if the worst should
come.
The water didn't reach far into the cave, and soon we had a light, for
we always keep candles and matches there.
It didn't seem so scary when we could see, sitting down together on a
piece of old carpet which the folks had given us, where we had sat many
times before.
What happened next, they say, was a cloudburst. Something burst, anyhow.
Skinny had just grinned and said that he thought maybe it was going to
rain, when it started.
And rain! Say, we never had seen it rain before. It came down in chunks
and pailfuls. Pretty soon the water began to creep farther into the
cave, and we got out the rope and made ready to crawl through into the
other part, if it should come much farther.
But the dam held, and there we were, snug and safe, with our candle
throwing dancing shadows, and up against one side of the cave, where we
had hung it long before, our motto:
"Resolved, that the Boys of Bob's Hill are going to make good."
Then we heard a distant roar, different from anything we ever had heard
before and different from any other noise the storm was making. It
scared us because we couldn't think what it was.
"Gee!" said Skinny. "What's broke loose, now?"
"Great snakes!" I heard Bill say. "I wish I hadn't come."
Benny didn't say anything, but he grabbed my hand and by the way he hung
on I knew he was doing a lot of thinking.
That roar seemed to be the end of the storm, for the rain stopped as
quickly as it had come. It began to grow light again and somewhere in
the woods we heard a bird singing.
We were glad enough to get out into daylight once more and make our way
back to the road.
"Let's see what it was that roared so," I said. "It isn't going to rain
any more and Skinny is nearly dry."
We could see great patches of blue sky and knew that the storm was over.
The roaring had seemed to come from the mountain, so we climbed up the
road and went into a field beyond the woods, from which we usually can
see old Greylock looming up, only looking different, it is so near.
This time we couldn't see him at all. The sky was clear overhead, but
clouds still hung about the mountain, shutting him from sight.
Then, as we stood there, the noise came again, only worse this time, and
right in front of us. The ground seemed to tremble under our feet and
from somewhere, back of the cloud which covered the mountainside, came a
mighty roaring and grinding that was awful.
We stood there, clinging to each other and wondering if the end of the
world had come, when suddenly the cloud lifted and Skinny yelled:
"Look! Look!"
Down the face of Greylock, where before trees had been growing, water
was pouring over a great, white scar, which reached from top to bottom,
nearly to where we stood, and over to the south was a smaller scar.
"Guess what," said Benny. "Greylock is crying. What do you know about
that?"
There had been two landslides, the only ones we ever had known to happen
on the mountain.
And to this day, as far as you can see Greylock, you will see those
white scars of bare rock, stretching down his face, as if some monstrous
giant had clawed him, but, of course, no water after that first time.
CHAPTER XVII
ON THE WAY AT LAST
FOLKS in our town think that white streaks down the face of Greylock do
not improve his looks any, but to us boys they seem like scars won in
battle. We feel like cheering some mornings, when we see him fighting to
break away from storm clouds which wrap him around.
At first we can see nothing but clouds from where we stand on Bob's
Hill. Then, the clouds begin to lift a little and Peck's Falls woods
gradually come into view. A little later the very tiptop of the mountain
begins to show, floating like an island in an ocean of mist. While we
look, the clouds fall away still more, making the island larger and
larger, and the bottom mists roll up the wooded sides of the hill.
In a few minutes old Greylock throws them off altogether and stands
there, with his scars showing, except that across his face a narrow
cloud sometimes hangs like a billowy screen, giving him, Ma says, a look
of majesty as if God was living there.
Anyhow, we boys can't help cheering when the mountain shakes off his
bonds and stands forth like a giant Scout, telling us to be cheerful and
brave and reverent and all that.
The great rains did more than scar the face of Greylock. They kept us
from starting on our trip at the time we had planned to go.
"Wait until the woods dry out," Mr. Norton told us. "The roads are too
muddy now to think of starting, and you couldn't have any fun if the
woods were wet. A week of sunshine will fix things all right."
We hated to wait, but there was plenty to do getting ready, so that the
time did not seem long.
"We'll carry no firearms," he went on. "Guns seemed necessary when this
was a wilderness, but we are going over a fairly well traveled road.
Scouts do not believe in wanton killing, anyway."
"How about bears?" asked Skinny, anxiously.
"I have made careful inquiries and have not found anybody who has seen a
bear along that road in years. I know you found one near the Savoy road,
or he found you, but that cub was as badly frightened as you were.
Should any of us see a bear, which is not at all likely, I don't believe
there is anything in Scout law to keep us from running one way while the
bear is running another."
"I don't know about a Scout's running," Skinny told him. "Of course I
ran, but I didn't run far, only to the nearest tree, so that I could
lasso him better."
"Well, that's all right. Run to the nearest tree and then give the Scout
signal. Some of the noises which you boys make, especially William,
would scare a whole drove of bears."
"Anyhow, I'm going to carry my rope."
"I'll tell you what we can do. We'll put in the week making bows and
arrows. Every boy should carry with him a good bow, made of hickory,
hemlock, or mountain ash, and a quiver full of arrows. You never will
have a better chance to become experts in archery."
We thought that we would make them of hemlock, because there are plenty
of hemlock trees up above Peck's Falls and in Plunkett's woods, but Mr.
Norton told us that we ought to make them of seasoned wood. The next day
he sent some seasoned hickory over to our barn and we made the bows and
arrows of that.
We took a lot of pains with them, and a carpenter that Hank knew helped
us some. Before the week was over we had some weapons which Skinny said
he knew we could scare a bear with, anyhow. Each Scout's bow was about
as long as himself and an inch thick in the center. The ends were shaved
down until they bent evenly. For string, we used strong, unbleached
linen threads, twisted together. Benny made his bow so stiff at first
that he couldn't bend it, but Hank showed him how to shave it down,
until he could draw the string back twenty-three inches, like the book
says.
The arrows gave us the most trouble because they had to be so straight
and round. We made them twenty-five inches long and about three-eighths
of an inch thick, and we glued turkey feathers on near the notched end.
The other end we fitted into a brass ferrule, to keep the wood from
splitting. The arrows looked fine, when we had them made and painted.
Each boy painted his a different way, so that we could tell which one
killed the bear.
Mr. Norton showed us how to make guards for the left wrist, to keep the
bow cord from striking it. To protect the fingers of the right hand, we
used an old leather glove, with the thumb and little finger cut away.
I'll never forget the morning we started. After breakfast the boys, all
in uniform, came over to my house. Pretty soon Mr. Norton drove up in a
light wagon, loaded with tents, camp outfit, and things to eat.
We greeted him with cheers, and when he had come close gave him the
Scout salute.
"Come on, boys. Let's get started, if you are ready," he said. "We have
a long walk ahead of us, if we expect to camp on Florida Mountain
to-night."
"Great snakes!" said Bill. "That listens good to little Willie!" And he
gave a yell that brought people out of their houses, all up and down
Park Street.
"Boys," said Pa, just as we were starting, "remember that your folks are
trusting you and, as we understand it, a Scout's honor is to be trusted.
Remember, too, that it is a Scout's duty to obey orders and that the one
to give you orders while you are away is Mr. Norton. And let me add that
he has my full sympathy. If he isn't worn to a frazzle before he gets
back, I'll miss my guess."
In another minute we were off, the folks calling good-bys after us and
shouting for us to remember this and not to forget that and not to do
something else.
Mr. Norton drove the horse at the start because he knew that we would
want to march through town, and away we went, with our bows and arrows
on our backs, and Skinny, with his rope and hatchet, which he called his
tomahawk.
At the Gingham Ground we found the boys of Eagle Patrol drawn up by the
side of the road. They saluted and cheered as we passed.
"If we have good luck this time, we'll take you next year," called Mr.
Norton. "I'm new at the business, myself, and eight youngsters are all I
want to tackle the first time."
"Skinny! Oh, Skinny-y-y!" yelled Jim, when we were almost out of
hearing.
We stopped and waited to find out what was wanted.
"Don't kill all the game-e-e. Save some for seed."
Skinny's only answer was to wave his tomahawk. Then we marched on toward
North Adams, and at nearly every house we passed people came to the door
to see what was going on. It made us feel proud.
We took turns riding, two or three boys in the wagon at a time, because
Mr. Norton said that he didn't want us to get all tired out before we
started and that we shouldn't be really started until we came to the
mountain.
The day was fine and the roads were getting dusty again. We were so
happy that almost before we knew it we came to the foot of a hill, which
led up into the mountain, and there we stopped to eat lunch.
Before leaving home, I asked Pa why they called it Florida Mountain and
why they called a little town on top Florida, and he said because that
was its name. Anyhow, they call 'em that. Before Hoosac Tunnel was built
under the mountain, a stage coach made regular trips over it, along the
road we were going to take. That was the only way people had to get to
Greenfield and the other towns on the east side, without going south to
Pittsfield and from there going over Mount Washington on the Boston &
Albany Railroad. Now, there is a big hole under the mountain, more than
four miles long, and trains go through in a few minutes.
After we had eaten and had a good rest, we started up a road, which we
could see winding up the mountainside, far above us.
"Now, boys," said Mr. Norton, "we don't have to make this trip all in
one day. We are out for fun and to learn something about scouting; if we
climb too far in this hot sun it will get to be work instead of play. I
propose that we climb slowly, taking plenty of time to enjoy the
wonderful views that will unfold before us with every turn of the road.
You boys can stop whenever you feel like it, to rest, or explore, or
shoot. Before we get to the top, we'll pitch our tents near some spring,
in full view of the valley and setting sun. We'll plan it so as to have
several hours of daylight left after we go into camp for the night. What
do you say?"
That suited us all right and away we went, with Benny driving, and the
old horse moving along in good shape.
Say, no tunnels for us, after this! Tunnels are all right when you are
in a hurry. But were we in a hurry? I guess not!
It was just as Mr. Norton had told us. At every turn of the road, and
mountain roads wind around with a lot of turns instead of going straight
up, we stopped to look back over the valley. And every time we stopped
it looked different. It was great. And the higher we climbed, the better
it looked and the farther we could see, until the whole valley lay
before us, all the way to Pittsfield and west toward the Hudson. To the
north, the Green Mountains of Vermont looked blue in the distance.
Across the valley, on the south, old Greylock put his head up above the
other peaks and watched us, wondering, we thought, why we were going up
Florida Mountain instead of climbing over him.
"Hurray!" yelled Skinny. "I'm Captain Clark, exploring the great
Northwest."
"I'm Captain Lewis," shouted Benny, strutting around and waving his bow.
"Me Injun chief," said Bill. "Ugh! Heap pale face get lost. No find
trail. Injun show um way."
Then he gave such a yell that it scared the horse and we hardly could
keep up.
About four o'clock in the afternoon we came to a spring near the top of
the mountain, and a little beyond, through the trees, we could see a
grassy <DW72>, just the place for our camp.
"This looks good to me," said Mr. Norton, driving up to the side of the
road and blocking the wheels of the wagon. "We'll give the horse a
drink after he cools off a little and unload the things which we shall
need to-night."
It looked like an Indian village there, when we had finished setting the
tents up. For beds we went into the woods and cut branches of hemlock,
which we wove into mattresses and covered with blankets.
"Let's play 'Hunt the Deer,'" said Skinny, when all was ready for the
night and Mr. Norton had sat down to rest on a rock, overlooking the
valley.
"All right, boys," he told us. "I want you to have the time of your
lives on this trip and I know that even a view like this will not long
satisfy a boy. But don't go far and remember your Scout training. You
will usually find moss on the north side of tree trunks."
"We know that," said Skinny. "We tried it once on Greylock, when we were
lost, and it worked all right."
"You can't get lost. I believe I could hear William call anywhere on the
mountain. The sun is shining and your shadows will point east. Come
back in time for supper. I'll be cook to-night, but after this you boys
will have to take turns."
"We'll get back in time, never fear," Skinny told him. "We are hungry
enough now to gnaw the bark off the trees."
Then he grabbed a bag which was stuffed with hay, put an ear of corn in
his pocket, and started.
"Give me ten minutes," he said.
It was a game which we had read about in the book. The stuffed bag was
the deer and the corn was for the trail. The game was for Skinny to
scatter corn along, making a crooked trail for us to follow, and then to
hide the deer somewhere for us to find.
After Skinny had made a good start, we scattered, looking for the
trail--corn, footprints, and other signs.
It was great fun and not easy for beginners like we were. Sometimes we
lost the trail altogether. Then one of us would pick it up again, where
Skinny maybe had doubled back toward the camp.
Finally Bill caught sight of the bag in some bushes and yelled:
"Deer!"
Hank hurried up and called, "Second!" I saw it third and all the boys
soon after except Benny. He had lost the trail and was beating around in
the woods somewhere, out of sight and hearing.
It was Bill's first shot and he had to stand where he was when he first
saw the deer. He took out an arrow, aimed carefully, and fired. The
arrow went so fast that I believe it almost would have killed a real
deer if it had hit him, but he aimed too high and it went over.
Then Hank stepped five paces toward the deer and shot. He missed. I
stepped up five paces more and I missed. Harry went five paces closer
and was the first to hit it. After that we all shot from where he had
stood, until we all had hit it.
Skinny had come up and I was just asking him if he had seen Benny, when
we heard a great crashing through the bushes and in a minute he came in
sight, running like sixty.
He was almost tuckered out when he reached us and had only breath enough
left to say:
"Run! It's a bear!"
We ran, all right, but after a little I looked back and could see that
there was nothing following.
"Hold up--a minute," I panted. "It--ain't a-comin'."
"Where was it, Benny?" I asked, when they had come back. "Where did you
see it?"
"I didn't see it. I only heard it. It was stepping around in the bushes
and I heard it grunt. I didn't wait to see it."
"I wish I had my rope," said Skinny. "I left it in the wagon. Come on,
anyhow. We'll surround the critter and shoot him."
Skinny scared us when he said that. I could feel cold chills chasing up
and down my back bone, when I thought of surrounding a live bear.
"Great snakes!" said Bill. "I hope it's a big one, so Skinny can hit it.
He couldn't hit a little one."
"I couldn't, couldn't I?" said he. "I'll show you whether I can hit it
or not. Come on. I'll dare you to."
That settled it. We weren't going to take a dare, but I was hoping all
the time that the bear had run away. So, with Benny keeping close to me
and pointing the way, we crept through the woods, not making any noise,
and each boy held his bow and arrow ready to shoot.
It was scary but it was fun. Finally, with an excited pinch of my arm,
Benny stopped and pointed.
My heart throbbed like a trip-hammer, and I hardly could hold my arrow
on the cord, for, looking through some bushes, I caught sight of
something black and heard the bear tramping around.
I heard Skinny muttering something about a rope; then he whispered:
"Get ready, and run as soon as you shoot."
"Aim."
We stood there, trembling, wanting to run first and shoot afterward, but
too proud to. Each boy pointed his arrow toward where we could see the
bear standing still behind some bushes and only a part, of him showing.
[Illustration: AS WE RAN, WE HEARD A YELL OF PAIN, OR FRIGHT, AND IT WAS
NOT A BEAR'S VOICE AT ALL.]
"Fire!"
I don't know when I fired. I only knew that my arrow was gone and I was
running for the camp like the wind, with the other Scouts chasing after
me.
As we ran, we heard a yell of pain, or fright, and it was not a bear's
voice at all. It was a woman's! Then we heard the voice say:
"For the love of Mike! The woods is full of Injuns and I've got an arrow
in the pit of my stummick."
CHAPTER XVIII
SCOUTING THROUGH A WILDERNESS
"FELLERS," said Skinny, panting and wetting his lips with his tongue.
"We've done it this time. We've killed somebody."
"Killed nothin'!" Bill told him. "Didn't you hear her holler?"
"She's running, too," said Benny. "Killed folks don't run, especially
girls."
We could hear a crashing through the bushes beyond, and knew that what
Benny said was true.
"Let's sneak back and get our arrows, anyhow," said Skinny, when the
noise had stopped.
So we crept back again, ready to run if any one should come, but there
was nobody in sight. One arrow was lying on the ground where the girl
had been standing when we took her for a bear. It was Skinny's; we could
tell by the way it was painted.
It made him real chesty, after he had found out that we had not killed
anybody.
"Didn't I tell you, Bill," said he, "that I'd show you whether I could
hit a bear or not? It must have struck a button or something, or whoever
it was would have bit the dust, and don't you forget it."
While we were standing there talking about it, a man burst through the
bushes, followed by a girl, about eighteen years old, I guess.
"Are these your Injuns?" he asked, before we had time to run. Then he
burst out laughing in such a way that we were not afraid to stay.
In a minute we had found out all about it. They were fern gatherers and
Benny had taken them for bears. A lot of people go up on the mountain in
August, picking what they call Boston ferns to sell to florists. They
put them in cold storage and keep them a long time. There is a crazy
little railroad at the foot of the mountain, on the east side, that
carries whole train loads of those ferns to Hoosac Tunnel station, and
afterward they are shipped all over the country to be put in bouquets.
Skinny's arrow had struck the girl and hurt her a little, but not much.
She was scared half to death.
Mr. Norton had a fine supper ready when we reached the camp again, and
we ate until we couldn't eat any longer.
"You boys ought to know what you are doing every minute you are in the
woods," he told us, after he had heard about the scare. "Suppose that
Gabriel had been carrying a gun, as he wanted to, instead of a bow and
arrows. Just think what would have happened. Hundreds of people have
been killed in exactly that way. Careless hunters have mistaken them for
bear or deer or some other game. You ought to have known what you were
shooting at. It was a foolish thing to do, anyway. I don't believe there
can be any bears around where so many people are looking for ferns and
berries. We'll see dozens of pickers on the other side of the mountain,
probably. If there ever were any bears they have been frightened away
long before this. But suppose that had been a bear. For a bunch of boys
to attack a bear with bows and arrows isn't bravery. It is foolishness.
I am ashamed of you."
We didn't feel quite so chesty when Mr. Norton had finished talking to
us.
"Well, I am not going to spoil the day by scolding," he went on, after
we'd had time to think it over a little. "You can see the folly of it as
well as I. Let us sit here and watch the sun go down behind the west
mountains. Did you ever see such glory? Then, when it grows dark, we'll
build a campfire and I'll tell you about a great scout and a trip he
once made through a wilderness."
It was fine sitting there, watching the sun sink into a golden sea
behind the mountains, while the valley below was already in the shadow
and the dark was creeping up the hillsides.
We sat there a long time without speaking, until finally the golden sea
faded into a streak of gray, and up and down the valley we could see the
twinkling lights of a half dozen towns and the farmhouses between.
Then Mr. Norton threw an armful of brush on the coals, and in the light
of the blaze, which made the shadows dance like ghosts of Indian braves,
he began his story.
"Some of you boys went out to Illinois, last summer," said he, "and I
know from what you have told me that you learned much about the great
French scout, LaSalle; how he explored the Ohio River and went up and
down the Mississippi, taking possession of the country in the name of
the king of France. We already have had one story which grew out of
those early explorations. The Lewis and Clark Expedition through the
Northwest, which I told you about, can be traced back to those scouting
trips of LaSalle and the others, on account of which France claimed the
country.
"This story is of another scouting trip, long after LaSalle's time and
before Lewis and Clark were born, probably. It took place even before
the United States was born, but, in a way, it grew out of those same
trips of LaSalle and Tonty, Marquette and Joliet, the French explorers
of the seventeenth century."
"Was this scout a Frenchman, then?" asked Benny.
"No, he was of English parentage, one of the finest English country
gentlemen who ever lived, but born in America, and one of the greatest
American scouts.
"He was a friend of yours, too, Skinny," he added, laughing to himself.
"Not me," Skinny told him, shaking his head. "I think a lot more of
England than I did, on account of General Baden-Powell and the Boy Scout
business, but I don't know this feller."
"That is strange. It seems to me that I have heard you remark something
about his being able to lick Napoleon Bonaparte with one hand tied
behind his back."
"George Washington!" shouted Skinny. "The Father of his Country. First
in----"
"Say, who's tellin' this story, anyhow?" said Bill, pulling Skinny over
and sitting on him.
"Yes, George Washington, who, it seems to me, would have made the finest
kind of a Boy Scout in his younger days--a scout worthy of membership
in Raven Patrol. He seems to have had all of the Scout virtues. He was
trustworthy, loyal to his home and his native land; he was thrifty; he
was brave; he was reverent."
"I'll bet he couldn't bandage a broken leg like we can," Benny told him.
"Maybe not, but he could find his way through the forest and he didn't
go around shooting at girls, thinking that they were bears. He liked
girls too well for that. I believe he liked the girls better, even, than
our patrol leader does."
We set up a yell at that.
"Aw, I ain't stuck on no girls," said Skinny. "I just rescue 'em, that's
all."
"It's all right," Mr. Norton told him. "A girl is the greatest thing in
the world, unless it is a boy. Anyhow, George Washington was a splendid
type of American boyhood and he surely liked the girls; used to write
poetry about them when he was your age."
I don't know why, but somehow we seemed to think more of Washington
after we had heard that. It seemed to bring him closer to us and make
him a real person, instead of a picture on the wall, praying at Valley
Forge or crossing the Delaware. Most always Washington is crossing the
Delaware when you see him.
"He was a big fellow in the first place, while Napoleon was small. Size
of body doesn't always count. Some of the greatest men the world has
produced have been small of stature. But George Washington was a big
fellow. Like Lincoln, he could outwrestle, outthrow, and outjump any of
his mates. They still show a spot down in Fredericksburg where he stood
and threw a stone across the Rappahannock River. He didn't seem to know
the meaning of fear. From his early youth he was a fine horseman, taming
and riding horses that nobody else could manage."
"Did his mother call him Georgie?" asked Benny, before we could stop
him.
"Perhaps she did, although I hardly can imagine it. At the age of
fourteen George wanted to enter the English navy and he came pretty near
doing it. If he had, perhaps he would have become a great admiral
instead of the father of his country. Who knows?
"A midshipman's warrant was obtained for him, so the story goes, and his
clothes actually had been sent aboard a man-of-war. Then, at the last
minute, his mother found that she could not give up her oldest boy and
she withdrew her consent. It was a great disappointment to the boy, but
like the good Scout that he was he obeyed his mother and went back to
school. He learned to be a surveyor.
"Boys matured earlier in those days when the country was new. When
Washington was only sixteen he set out on horseback through the Blue
Ridge Mountains on a surveying trip. A year afterward he was given
command of the militia in a Virginia district, with the rank of major."
"I don't see what LaSalle had to do with all that," said Harry.
"He didn't have anything to do with it, but he had something to do with
the scouting trip which came later. You see, France and England each had
obtained a strong foothold in this country; France, along the Great
Lakes and Mississippi River; England, along the Atlantic Coast. Between
the Mississippi and the coast stretched a beautiful and fertile country,
the valley of the Ohio. When LaSalle made his explorations he took
possession of the Mississippi in the name of the king of France. On that
account France claimed to own all the land along the Mississippi and
along all the rivers which flowed into the Mississippi. That took in a
great part of the continent."
"I don't see how because LaSalle stood on a rock and hollered out some
words," Hank told him, "that made the whole country belong to France."
"England couldn't see it. Still, the English claim was not much better.
Commissioners from Pennsylvania, Maryland, and Virginia made a treaty
with the Iroquois Indians in 1741. By the terms of that treaty, for
something like $2,000, the Indians gave up all right and title to all
the land west of the Alleghany Mountains, clear to the Mississippi
River. There were all kinds of Indians living in the Ohio Valley but,
according to the traditions of the Iroquois Indians, their forefathers
once upon a time had conquered it."
"It looks like six of one and half a dozen of the other," I said.
"There wasn't a white settlement in the whole territory. Some hardy fur
traders from Pennsylvania had made trips into the valley and this led to
the formation of the Ohio Company of Virginia, with the object of
getting ahead of the French and colonizing the lands. Then the French
began to get busy. France owned Canada at that time, you know. In 1749
the French Governor of Canada sent three hundred men to the banks of the
Ohio River with presents for the Indians. They ordered the English
traders out of the country and nailed lead plates to trees, telling
everybody that the land belonged to France. The Indians liked the
presents well enough, but the lead plates made them mad, when they found
out their meaning. One old chief exclaimed:
"'The French claim all the land on one side of the Ohio; the English
claim all the land on the other. Now, where does the Indian land lie?'
"I have gone into this explanation in order to make it clear to you why
Washington was sent on his scouting trip. Governor Dinwiddie of Virginia
wanted to send some one whom he could trust to the French commander, to
protest against the French coming into the country. At the same time, he
thought the messenger would be able to find out how strong the French
were, how many canoes they had, and all that. It was a perilous mission
to undertake through an unknown wilderness, with winter coming on. Young
Washington was only twenty-two years old, but he was selected as the one
to make the dangerous trip.
"Major Washington started from Williamsburg, October 31, 1753. On the
frontier he procured horses, tents, etc. Later he was joined by a famous
woodsman, named Christopher Gist. They took along a white man to act as
interpreter and some Indian guides. Chief White Thunder was one. Another
was known as the Half King. His friendship was very important to the
English.
"I imagine that the mountains which they went through were much like
these, except that rains and snow had made them almost impassable. The
party pushed on, however, and early in December arrived at the first
French outpost. The French captain gave a feast in their honor, in the
course of which he drank so much wine that it made him talkative. He
began to brag of what the French were going to do. He said that they
were going to take possession of the entire Ohio Valley. The young
American scout kept his head clear and afterward wrote down in a book
all that he had heard.
"Then Washington set out again, and after four more days of weary travel
they came to the French fort on the west fork of French Creek, about
fifteen miles south of Lake Erie. There he delivered his message, and
after a great deal of delay received a sealed reply.
"While pretending to be friendly, the French did their best to win the
Indian guides away from Washington. They plied them with liquor and with
presents, so much so that the young scout had a hard time in starting
them toward home. He succeeded finally in getting away. They first went
up the creek in boats as far as an Indian village, called Venango; then
set out by land. Soon their pack horses became so jaded that Washington
used his saddle horse for a pack horse and walked. After three days of
that, he and Gist took their packs on their shoulders, their guns in
their hands, and started out alone, on a short cut to the Ohio River.
"You will find the story in any history. At one time a treacherous
Indian guide wheeled suddenly and shot at Washington, but did not hit
him. The two men quickly overpowered the savage, and Gist was for
killing him. Young Washington would not permit that, so they did the
next best thing. They took his gun away and sent him home, making him
think that they would follow in the morning. Instead of that, they left
their campfire burning and traveled all night and all the next day, to
get as far away from the spot as possible. At last they reached the
Alleghany River, which they hoped to find frozen. There was open water,
however, and they were forced to build a raft. All they had to work with
was one hatchet, like Skinny's, I mean Gabriel's. On the way across, a
cake of ice struck the raft and threw Washington into the river."
"Gee, I'll bet that it was cold," said Skinny.
"It was, but Washington clung to the raft and finally, in a half-frozen
condition, drifted against an island, where the two men camped that
night. In the morning they found ice cakes so wedged in that they were
able to walk ashore.
"January 16, in the dead of winter, Washington succeeded in reaching
Williamsburg, and delivered the French commander's letter to Governor
Dinwiddie. Soon after that came the French and Indian war, which I am
sure you know all about, in which France lost all her American
possessions except the great tract west of the Mississippi, which
Napoleon later sold to President Jefferson.
"You see, being a scout in those days wasn't all play. It brought many
hardships that we know little about, but, after all, it called for the
same kind of boy. Washington was brave and true, helpful, kind, and
clean, and he was prepared. When the time came, his preparedness put him
in command of the American forces and afterward made him the first
President of the United States."
"Washington was great stuff, all right," said Skinny, shaking his head
sadly, "but everything has been discovered now, and explored, and Injuns
ain't much good outside a show. There ain't anything for a feller to do
any more."
CHAPTER XIX
ON HISTORIC GROUND
WE were one more night on the road before reaching the Connecticut
River.
"This trip is going to be a great part of the fun," Mr. Norton had told
us, "and the best part of it is that we can go as slowly or as fast as
we please. We'll cross over the mountain to-day, stopping whenever we
feel like it, and go into camp somewhere on the other side. I want to
have you do some of our Scout stunts on the way."
I don't know which was the most fun, walking along the mountain road,
which wound through green woods and across laughing brooks, or pitching
our camp at night and, after a good supper of our own cooking, listening
to Mr. Norton's stories, around the campfire.
We started bright and early in the morning, carrying only our bows and
arrows and Skinny's hatchet. The other things were on the wagon. Mr.
Norton drove because we boys wanted to play.
Skinny was George Washington, making his way through the wilderness. He
carried the hatchet because he might have to build a raft to get across
Deerfield River. Benny was bound to be Christopher Gist. Bill had a
right to first choice, on account of being corporal, but Benny wanted to
be Gist and Bill didn't care. He said he'd rather be White Thunder,
anyhow; it sounded so nice and noisy. Hank said that he'd be the Half
King, whatever that was.
"His name was Tanacharisson," said Mr. Norton. "He was a Seneca chief of
great note in those days. He was called 'Half King' because he wasn't a
whole king. He was under the chief of the Six Nations."
I don't know what the rest of us were, but I do know that we had a fine
time, scouting through the forest and along the road. When we came to
the town of Florida, on top of the mountain, Skinny told us that it was
the Indian village of Venango, where we'd find the French outposts. He
wanted to surround it, but White Thunder was for pushing on because he
was getting hungry, although it was still quite early in the forenoon.
So we trudged along, and down the mountain road on the other side, until
we came to Deerfield River.
We found a bridge across the river and didn't have to make a raft. There
wasn't water enough to float one over the rocks, anyhow, although there
was more than usual on account of the big rain.
By night we had left the Florida Mountain far behind. Along in the
afternoon of the next day we marched into Deerfield, which is on the
Connecticut River. Say, the people came out of their houses to see us
pass, with our uniforms on and Skinny in front, swinging his rope and
hatchet.
"This is historic ground," Mr. Norton told us. "At the campfire to-night
we'll have a story of some fights with Indians which were the real
thing. They ought to make your hair stand on end. That stream over there
got its name 'Bloody Brook' from one of those fights."
We camped that night on the bank of Connecticut River, and it seemed a
long way from home.
"This river was discovered by the Dutch," said Mr. Norton, after we had
eaten a big supper and were lying on the river bank in the twilight of
the evening, tired and happy. "The permanent settlements, however, were
made by the English. The river was explored by a Hollander six years
before Gabriel's English ancestors came over in the _Mayflower_. The
first English settlements, you know, were made along the Atlantic coast.
Some years later a few of those settlers hiked over to the Connecticut
Valley, or came up the river, and started a number of towns. One of them
was Deerfield.
"It is hard for us to imagine this fertile and cultivated valley in a
wild state, with a few white settlers here and there surrounded by
Indians. The whites considered themselves a superior race and probably
showed it by their actions. Gradually the savages, who at first had been
kind, grew more sullen and dangerous. This growing hatred on the part of
the Indians made it very difficult for the settlers, but there was
another thing which made it harder. In Europe, two great nations,
England and France, were in almost constant warfare, and each was
striving to get the better of the other in the settlement and possession
of America.
"There were some early Indian wars, with which the French did not have
anything to do, but they had much to do with the later wars and attacks
by Indians. One of those early struggles is known as King Philip's war,
named after a wily Indian chief. It occurred just one hundred years
before the Revolution, where our patrol leader lost his ancestor. Even
at that early day there were one hundred and twenty-five people in
Deerfield. In that war the Indians attacked the town twice."
"Was that what made the brook bloody?" asked Benny.
"No. The bloody event which gave the brook its name happened during the
same war but not during an attack on the town itself. September 18,
1675, I believe, was the date. A company of young men, commanded by
Captain Lothrup, marched out of the town and along a road leading toward
the brook. They were acting as guard and teamsters for a number of
loaded carts, which were being taken to some settler's home. It was a
beautiful day and everything seemed as peaceful as it does now. All were
happy and there was no thought of danger. Some had even placed their
guns in the carts and were walking unarmed.
"At the brook a band of Indian warriors lay in ambush, waiting. On came
the young men, laughing and whistling and chatting with one another.
They stopped occasionally to gather some wild grapes, which grew along
the way. Concealed in the long grass, on each side of the road, lay the
painted savages, motionless and unseen. Their eyes gleamed with hatred
and exultation as they watched their victims approach. Their eager hands
tightly grasped their weapons. Impatient for the slaughter to begin,
they awaited the signal."
"Great snakes!" whispered Bill.
"Snakes is the word. Like snakes in the grass they lay, as silent as the
grave. At last the signal was given. With fierce cries they sprang upon
the surprised whites, and the little brook ran red with blood.
Sixty-four men in all, from the various settlements, were killed that
day. Of seventeen young men, who went out from Deerfield that morning,
not one returned.
"Too late, another company of men came to the rescue. They found nobody
left to rescue. The Indians then were plundering the wagons. The savages
outnumbered the rescuing party ten to one, but the little band did not
hesitate. They fought desperately for five or six hours. They were
unable to drive the savages away, however, and were just going to
retreat, when some soldiers from Northampton, down the river, appeared
and put the Indians to flight. There was sadness in Deerfield that day."
"I don't believe I want to play Indian any more," said Benny, drawing
closer to the fire and looking around as if he might see some savages
hiding in the grass. It made us all feel scary.
"We hardly can imagine it now," Mr. Norton went on, "after more than two
hundred years. Later there were other wars and many attacks by Indians.
The Deerfield people built a stockaded fort, into which all would run at
the first alarm. These later attacks by the savages were a part of the
fight between England and France for the possession of America. The
French induced the Indians to help them drive the English out, but
Englishmen do not drive worth a cent, and at last, as you know, France
was obliged to give up Canada to England, in whose possession it has
remained ever since.
"First came King William's war, in which Deerfield was attacked several
times; then Queen Anne's war, and during that the town was captured and
a great part of it burned."
"Tell us about that," I said.
"War is always a terrible thing, but in those days it seems to have been
more than usually savage and cruel. Take the capture of Deerfield, for
example. The French commander in Canada sent three hundred soldiers to
butcher the people in this little town, in order to make himself solid
with some Indians. The attack occurred a little before daybreak, and
some terrible scenes were enacted. I'll show you an old door up in
Memorial Hall to-morrow, which went through that fight. It was so solid
that they could not break it down. You will see where a hole was cut
through it with axes and bullets.
"That massacre occurred February 29, 1704, about two hundred years ago.
Then came other French and Indian conflicts, until finally England
triumphed. Later the United States Nation was born, and President
Jefferson bought all of the American territory that France had left.
"Everything is peaceful here now, but think how you would feel, to know
that you might be surrounded by savages, fierce and bloodthirsty,
creeping toward you in the darkness, without a sound, until near enough
to strike, and then----"
All of a sudden there came some awful yells and whoops that made our
blood run cold, and a crashing in the bushes that sounded as if all
kinds of Indians were after us.
We jumped to our feet and looked, even Mr. Norton. Benny grabbed tight
hold of my hand, and I could see Skinny feeling around in the grass for
his hatchet.
Then it came again, nearer than before, only worse and over to one
side. It was awful. I don't know about Mr. Norton, but the rest of us
were just going to run, when the yell ended with three caws, like a crow
in the Bellows Pipe at home.
"Shucks!" said Skinny, in disgust. "It's only Bill Wilson!"
We camped there on the river bank nearly a week and never had more fun
in our lives, boating, fishing, swimming, doing Scout stunts and playing
Scout games, and, with it all, eating our heads off, almost.
I can't remember every little thing that we did there, and the boys say
that it will be all right to skip that part in writing this history.
There didn't anything much happen, anyhow, although Mrs. Wade was sure
some of us would get drowned and even Ma told us that she would not feel
real easy in her mind until we were at home again.
"We'll go a little earlier than we intended," said Mr. Norton, when it
was getting near the time for going back. "I want to see some more of
that beautiful Deerfield valley, before the river leaves the mountains.
Perhaps we might do a little exploring on our own account."
We came in sight of Florida Mountain on our homeward trip, not far from
Hoosac Tunnel. The longest part was behind us, but the hardest part, the
climb over the mountain, was ahead.
Wild? Say, if you want to see a wild country, follow Deerfield River as
it fights its way down from Vermont, until finally it breaks through the
mountains and runs off to join the Connecticut. When you get in among
those mountains you will think that you are Christopher Columbus
discovering America.
"The Rockies are higher," said Skinny, when we had stopped to rest and
look around a little. "I read it in a book. Besides, Mr. Norton told us
about Lewis and Clark climbing over them. But these are some mountains
all right; believe me."
That was what we all thought. They were all tumbled and jumbled together
in a topsy-turvy way, with the river winding around in every direction,
trying to get through, and the railroad following the river.
Mr. Norton pointed it out to us and stood there with his hat in his
hand, looking. His eyes were shining, and red was coming into his
cheeks, as if he was seeing something which we boys couldn't see at all.
And maybe he was, for I have noticed that grown folks sometimes can't
see and hear the things which we boys see and hear; at any rate, not in
the same way.
"What does it make you think of?" he asked each of us.
Benny's answer was the best of all.
"There was once a baseball nine made up of real giants," said he. "They
were so big that their heads reached clear up into the sky. One day when
they were practising they lost the ball and so they picked up these 'ere
mountains and began to throw them to each other, playing catch. Every
once in a while some guy would muff the ball, I mean the mountain. Then
he would let it lie where it had fallen and pick up another. That is why
they are all tumbled together every which way."
"That's so," I said. "You can see where the dirt jarred off when they
fell, leaving the bare rocks sticking out in a lot of places."
"It's alive, boys," said Bill, who had been feeling of Benny's head and
looking anxious. "It feels like a nut, but it ain't cracked."
"Benny has given us a good description and something to think about,"
said Mr. Norton. "I don't believe that I should like to live here all
the time, but I should enjoy staying a week and drinking in all this
beauty. Talk about music! Hear the mountain breeze in the treetops. What
does it remind you of, Gabriel?"
"It sounds to me exactly like beefsteak frying," Skinny told him, "and
it makes me hungry. Let's have some eats."
"All right," said Mr. Norton, laughing to himself. "Now that you mention
it, I believe that I can detect a faint resemblance. We can't give you
beefsteak, but there is some bacon left and that ought to make much the
same kind of noise. Whose turn is it to cook?"
"It's mine," Hank told him.
"Well, get busy, and for fear that we might disturb you, we'll go off
somewhere and sit in the shade."
We were all as hungry as wolves when Hank at last called us to dinner
and it tasted fine, although my piece was burnt a little.
"I don't know how you boys feel about it," said Mr. Norton, after the
dishes had been washed and put away, "but I should like to camp here for
a couple of days. We'll do just as you say, however. Perhaps you have
had enough."
We all had been thinking the same thing and told him so.
"All right. We'll find a good place for our tents and go into camp. It
will give us a chance to wash out some clothes in the river and to
explore this delightful wilderness."
We had all kinds of fun practising our Scout stunts, exploring, playing
Indian, and things like that. One of the prettiest places that we found
was a ravine, where two cascades, twins, tumbled over rocky ledges; then
came together and raced down the mountain. I don't mean that they were
as pretty as Peck's Falls, above our cave. They don't make any finer
places than that, only, of course, Niagara Falls are bigger. But they
were worth looking at, just the same.
I am going to put down just how to get there, in case somebody should
want to see them. You probably wouldn't walk over the mountain, as we
did, because it takes so much time, but would go through Hoosac Tunnel.
After you have gone through from the North Adams side and the train
stops to take off the electric engine and put a steam one on, get off
and walk back to the mouth of the tunnel. Then, when you have come to
the mountain, climb up a sort of path, following the brook, and after a
little you will come to the twin cascades. We thought of camping there
at first, but couldn't find any good place for our tents.
Except for the train passing and the engineer leaning out of the cab
window, we seemed out of the world, although we were not more than ten
miles from home, in a straight line. The train was like company, and
when we were around near we always watched it out of sight.
That is a queer little railroad which comes down from Wilmington and
Readsboro, Vermont, as far as Hoosac Tunnel station. Mr. Norton told us
all about it. It is what they call a narrow gauge railroad. That means
that the rails are closer together than on most railroads, and on that
account regular cars cannot run on it. Its rails are three and a half
feet apart, while on a regular railroad they are four feet, eight and
one-half inches apart. It runs along one bank of Deerfield River, a few
feet above the water. The river is mostly stones in summer, with water
in between.
The day after we camped there Skinny, Bill, Benny, Hank, and I sat on a
big stone, opposite our camp, waiting to see the train go by. The other
boys had gone with Mr. Norton part way up the mountain, looking for
berries for our supper.
Pretty soon the train came in sight from toward Readsboro, fifteen miles
north, and it was swinging along at good speed, for it was downhill.
We cheered and waved our hats as it went by. I noticed a girl, who was
sitting at one of the windows in the passenger car, give a look of
surprise when she saw us; then she leaned far out and waved her
handkerchief. It wasn't anybody that I knew, but when Skinny saw her he
jumped to his feet and let out a yell. And what he said was:
"Mary!"
It surprised us some. You may not believe it, but the girl was Mary
Richmond, the one Skinny walked down the mountain with, that time he
lassoed the bear, when he was doing his hike to Savoy and back. She had
been up to Readsboro with her mother, visiting.
"Come on," said he, starting on a run. "She'll have to change cars at
Hoosac Tunnel station."
"Aw, what's the use?" said Bill. "We don't know her."
At that instant, while we stood there watching, we saw the engine give a
sudden lurch and then go bumping over the ties. In another moment it
struck a rock or something and, with an awful crash, the whole train
went off the embankment into the river below.
CHAPTER XX
SCOUTS TO THE RESCUE
YOU may have heard of that wreck, for the papers printed a lot about it
at the time.
After the first crash, there was not a sound. I don't know how long we
stood there, paralyzed with horror, staring at the place where the train
had been. Then we heard a shriek of fear, or pain, we couldn't tell
which, and it was a girl's voice.
That shriek brought us to our senses.
"Scouts to the rescue!"
Skinny shouted at the top of his voice, hoping that Mr. Norton and the
others would hear, and we started on a run.
Before we had gone halfway Skinny turned to Benny.
"Run back to the camp," said he. "Get the bandages and other first-aid
things."
"And bring my rope and hatchet," he called, over his shoulder.
The awful stillness after that first shriek sent us on faster than ever,
while something seemed to clutch at our throats so that we hardly could
breathe.
Bill got there first, but we were not far behind. When we had come close
we could see the train, lying on the stones in the river bed. The engine
had turned bottom side up and lay there on its back with its wheels in
air. The passenger car was on its side and was so badly smashed that it
didn't look like a car at all.
"We've got to have help and have it quick," said Skinny, looking almost
pale. "Who'll go to Hoosac Tunnel station for help? Hank, you go, and
run like Sam Hill."
Hank was off like a deer before the words were out of his mouth, running
toward the station, nearly two miles away.
"Mary!" called Skinny. "Mary! Where are you?"
"Here," we heard a faint voice say. And, climbing down, we found her,
wedged in between some timbers so that she could not move.
"Are you hurt?" we asked, as we commenced to pry her loose.
"A little," she told us, beginning to cry. "I don't know how much, but
I'm all right for now. Find mamma. I don't know where she is."
After a little search we found her, nearly covered with timbers and
bleeding from a cut in her head.
"She's dead," I whispered, while an awful feeling came over me. Her eyes
were closed and she didn't move, even after we had lifted the timbers
away.
We dragged her out as gently as we could and laid her on a couple of car
seats which we took from the train. I sprinkled some water in her face
and pretty soon she opened her eyes.
She stared around for a second or two, trying to understand where she
was. Then she saw Skinny and seemed to remember.
"Mary!" said she. "Have you seen Mary? Oh, save my little girl!"
"Mary's all right," Skinny told her. "We haven't got her out yet, but
we know just where she is. She sent us to find you."
"Thank God!" she whispered, and then she fainted again.
We left her there, lying among the stones on the river bottom, with her
dress floating in the water.
"I wish Mr. Norton was here," groaned Skinny. "I don't know what to do.
Here comes Benny with the things."
There wasn't any time to talk. We hurried back to where we could see
Mary's head sticking out of the wreck. She had her eyes closed, and I
thought she had fainted, but she heard us come up and opened them.
"We've got your mother out," Skinny said. "Now we'll get you out."
Her eyes asked the question which her lips couldn't seem to do.
"Yes, she's alive," we told her. "She's got an ugly cut on her head, but
she seems all right except that."
It was all we could do to get her out, the timbers were so heavy and so
wedged in. They had fallen across each other and made sort of a roof
over her. If it hadn't been for that she would have been killed. By all
pulling on the rope and cutting some with the hatchet, we finally
managed to get her loose.
When we started to lift her out she screamed with pain. We kept on
lifting. There was no other way.
"It's my foot," she moaned. "It feels as if it was all broken to
pieces."
Two of us made a chair with our hands and carried her carefully up on
the river bank; then hurried back to the wreck.
"There is a man groaning somewhere," said Bill. "I think it must be the
conductor."
We found him lying under some wreckage and in great pain.
"Where are you hurt?" we asked, when we had lifted the wreck off from
him.
"My leg!" he groaned. "It's broken. I'm all in."
I took out my knife and ripped his trouser leg and underclothes to above
the spot that hurt him, a little above the knee. Then, by putting one
hand above the break and the other below it, just as Mr. Norton had made
us practise doing a lot of times, and lifting very gently I could see
the broken bone move. He ground his teeth together and great drops of
sweat came out on his forehead, it hurt him so much, although I was
trying to be careful.
"It's broken, all right," I told him. "We've sent for help. The only
thing to do is to lie still and wait."
We straightened him out and piled some coats and things, which we found
in the wreck, around his leg, to make him as comfortable as we could.
"How many are there?" I asked.
"I only had two passengers, a woman and a little girl. They got on at
Readsboro. Then there was the engineer, fireman, and brakeman, besides
myself. We run only a small crew on this train."
The brakeman came up while he was speaking. He had been stunned at first
and when he came to had managed to crawl out.
"Have you seen Jim or George?" he asked.
The conductor shook his head.
"Do you boys know anything about the engineer and fireman?"
We hadn't thought of them before. We had been too busy.
"Then they are under the engine," said he.
He ran through the river to the head of the train, we after him, almost
crazy with the thought of those men at the bottom of that awful heap of
iron and steel. We pulled and lifted at the great pieces, but we might
just as well have tried to move the mountain.
"We can't do it, boys," the brakeman said, at last. "We'll have to wait
for help. There isn't one chance in a hundred that they are alive, but
they may be. Somebody will have to run to the station and make sure that
they bring some jacks. I am 'most done up and don't feel equal to it.
Which one of you will go? Only one, now; the others will be needed
here."
"I'll go," said Benny. "I'm the littlest one in the bunch and can be
spared the easiest. What was that you said you wanted?"
"Jacks; to jack up the engine frame with. There are several in the
baggage room. I saw them there."
Benny hated to leave, when there was so much going on, but before the
brakeman had finished speaking he was climbing up on the river bank. In
another second he had started down the track on a run.
"Now, fellers," Skinny told us, trying to keep his teeth from
chattering, he was so excited, "our Scout book says for us to keep cool
and we've got to do it. While we are waiting for help the thing for us
to do is to be Scouts and to get busy with our bandages."
"And make some stretchers," added Bill. "We can't use our coats and hike
sticks, like the book says, because we didn't bring 'em."
"That's easy. We can use car seats."
The "first-aid kits," which Benny had brought from camp, had everything
that we needed. That was what they were put up for, only we didn't think
we should need them. There were shears and tweezers, carbolized
vaseline, sterilized dressings for wounds, to keep the germs out, all
kinds of bandages and things like that. Say, we looked like a drug
store when we had fairly started.
Skinny cut away the shoe from Mary's foot and Bill brought cold water
from a nearby spring, to bathe it in. The foot was bruised and the ankle
sprained, but no bones were broken. Soon they had her feeling better.
I went to help Mrs. Richmond, but all the time I was thinking of the men
under the engine. She was sitting up on the car seat, trying to keep her
feet out of the water.
"Are you hurt anywhere else, except your head?" I asked.
"No," she said. "I have had a bad shock and my head is cut, but I can
move all my limbs; so I guess there are no broken bones."
Her head looked worse than it was, with a gash cut in it and her hair
matted down with blood.
"I don't dare bathe the cut," I told her, "because the water may be full
of germs, and besides I haven't anything to bathe it with. The book says
to be careful about that."
"What does the book say about my washing my face?" said she, and she
didn't wait for an answer.
It didn't take long to put on a sterilized dressing and bandage her up
in good shape. Then, with Skinny on one side and I on the other, she
managed to walk to a low place on the river bank, where Mary was
waiting, and climb up.
Mrs. Richmond said so much about how we had saved her and her little
girl, it made us feel foolish.
"That ain't anything," Skinny told her. "That's what Scouts are for."
"It may be a long time before a doctor gets here," I said, after a
little. "He will have to come from North Adams or Readsboro. And that
conductor is getting worse every minute. If you will help me, Skinny,
I'll try to put splints on his leg."
You see, I had practised with the splints more than some of the boys
had. They were all for saving folks from drowning.
We first found two pieces of board. There were plenty of them scattered
around, on account of the wreck. We put one piece, which was long
enough to reach from his armpit to below his foot, on the outside of
the leg. The other we put on the inside. It didn't have to be so long,
but reached well below the knee. Then, making sure the broken bones were
in place, we tied the splints on with strips from Skinny's shirt, first
putting a cushion of leaves between the boards and the leg. After that
we tore up Bill's shirt and tied the broken leg to the good one with
three or four strips of that.
"Do you suppose that we can get him up on the river bank?" asked Skinny,
when we had him all fixed.
"We must," a quiet voice answered.
Turning, we saw Mr. Norton, who had come up so still that we had not
heard him.
"Oh, Mr. Norton!" cried Skinny. "We are so glad you have come. It is an
awful wreck and nobody to do anything at first but us, and we didn't
know what to do. I think the engineer and fireman were killed. The
brakeman is over there, trying to get them out."
"You seem to have done remarkably well for boys who didn't know what to
do. I want two poles from the woods, Gabriel. Quick! William, you go
with him. John will help me here."
Skinny grabbed his hatchet, and before we had time to miss them the boys
were back again with two long poles. While they were away Mr. Norton and
I pulled two car seats out of the wreck and were ready to make a
stretcher. By laying the seats end to end on the poles and tying them
fast with Skinny's rope, we had a good one and not bad to ride on,
because of the springs.
Then Mr. Norton and the brakeman, with us boys helping all we could,
lifted the conductor very carefully and laid him on the stretcher. To
lift it by the ends of the poles and carry it up to the river bank was
the easiest part of all.
By that time, Hank and Benny had come back with two or three men from
Hoosac Tunnel station, and they went to work with jacks to get the
engineer and fireman out.
"A special train is coming from Readsboro," Hank told us. "It's bringing
some doctors and the wrecker."
"Do you feel able to continue your journey, Mrs. Richmond?" Mr. Norton
asked. "We could manage to carry the little girl as far as the station
and there is a train due from North Adams in about an hour. Or would you
rather wait for the special and go back?"
"I think we'd better go back to Readsboro," she said. "We have friends
there and I don't feel much like walking."
We didn't have long to wait, for the train soon came puffing down the
valley. Two doctors jumped off before it had time to stop and hurried
over to where we were standing. They were surprised some, when they saw
the people all bandaged up.
"Who did this?" asked one of them, standing over the conductor. "I
thought there were no surgeons here. Did you succeed in getting somebody
from North Adams?"
"These boys," Mr. Norton told him. "They are Boy Scouts and have been in
training some time for this very job."
The doctor gave a little whistle.
"Good thing for him," he said, "that they were around. I couldn't have
done it much better, myself."
We felt proud when he said that, and I could tell by the way Mr. Norton
smiled at us that he was feeling pretty good over it.
All the same, the doctor bandaged him over again, to make sure that
everything was all right. When he had finished, the hurt ones were put
on board the train and made as comfortable as possible. We heard some
cheering over by the wreck and hurried back to find out what had
happened.
"They are alive," a man explained. "We've jacked her up a little, and
the engineer just spoke to us. He says that the fireman is alive, too."
It made us feel better to know that they were alive, and the men worked
like sixty to get them out. By that time the wrecking crew had the big
crane ready. After that it was easy. It didn't take long to swing the
heavy frame clear of the ground and to one side.
The two men were found somewhere in the mass, badly hurt but alive,
which was more than we could understand.
They were lifted out as carefully as possible and carried to the car.
"Good-by, boys!" called Mary out of the window.
"Good-by! God bless you, dear children!" said Mrs. Richmond.
"Good-by,--good-by," yelled the brakeman.
The doctors were too busy to say good-by to anybody. We watched the
train steam up through the valley; then Mr. Norton took each one of us
by the hand, and he squeezed hard.
We heard afterward that both men got well, although many weeks passed
before they were able to work again.
We started for home, bright and early the next morning, taking all day
for the climb over the mountain and camping that night among the
foothills on the west side. It was only six or seven miles from there
home, and we were so tough and hard that it didn't seem far.
"We can do it in two hours, easy," said Skinny.
We were beginning to be in a hurry to see our folks and the cave, after
being away so long.
"Let's get home in time for breakfast," I said. "What do you say?"
"And go without eatin' until we get there? Not much!"
"We can have an early breakfast," Mr. Norton told us, "and start as soon
as we can see; say, about four o'clock. We ought to be able to make it
by seven, easily, and I feel sure that we shall be able to eat again,
after our walk. I'd like to get home early, myself. It is time that I
was going back to work after my vacation."
That is what we did, and we surprised everybody. They had not been
expecting us before afternoon.
After that we didn't see anything of Mr. Norton for several days. Then
he asked us to meet him at a campfire on Bob's Hill, Saturday evening.
"I have spoken to your parents," he told us, "and they have arranged for
a picnic in Plunkett's woods, Saturday afternoon. We will eat supper
together on the grass, at the edge of the woods, and afterward have a
campfire at the old stone. I think that we owe it to your people to make
a sort of official report of what we did on our trip; that will be a
good time to do it."
That was some picnic, all right, and it was great fun, sitting there,
talking and eating; then playing Indian in the woods, surrounding the
palefaces, and all that. But, best of all, was the campfire, after the
sun had gone down and the moon lighted up the hills and made old
Greylock loom up big and shadowy. Of course, we had told our folks all
about everything but they wanted to hear more, and we had to tell it all
over again.
Finally Pa spoke up. "We have heard a great deal from the Scouts," he
said, "and we have enjoyed it all. Now, we'd like to hear from the
Scoutmaster, how the boys behaved. But first I want to tell him how
grateful we all feel for what he is doing for these youngsters."
"I am enjoying it as much as they are," said Mr. Norton, looking fine as
he stood there, with the moonlight on his face. "In fact, I think that I
am getting more out of it than they are. I asked you fathers and mothers
to meet me here to-night because I wanted to tell you how proud I am of
these Bob's Hill boys, the Boy Scouts of Raven Patrol. I understand that
in their cave at Peck's Falls they have a motto hanging, which says that
'The Boys of Bob's Hill are going to make good.' They have made good,
Mr. Smith, every one of them."
He hesitated a moment; then went on:
"I have made official application for Honor Medals for the part they
took in saving human life at that unfortunate train wreck, and I hope
the National Court of Honor will award them. But I, myself, have wanted
to do something personally to show the boys how much I have enjoyed
their companionship and what I think of their conduct--all of them, not
only those who happened to be on hand at the time of the wreck. So I
have had this banner made to hang under the other one, in the cave, or
wherever their place of meeting may be."
He pulled out a fine silk banner from his pocket, as he spoke, and shook
it out until it hung full length in the moonlight, and, looking, we saw
in one corner a black raven and "Patrol 1, Troop 3 Mass."; then, in
large, gold letters, the Scout motto:
"BE PREPARED."
How we did cheer! And our folks cheered louder than anybody.
"Guess what!" said Benny, after all was still again. "When we grow up,
we are going to try and be like Mr. Norton, our Scoutmaster."
"Bet your life we are!" shouted Skinny, springing to his feet and waving
the banner.
Then he stopped and stood there, looking at us, with his arms folded.
"I have spoken," said he. "Let be what is."
THE END
EVERY BOY'S LIBRARY
BOY SCOUT EDITION
SIMILAR TO THIS VOLUME
THE Boy Scouts of America in making up this Library, selected only such
books as had been proven by a nation-wide canvass to be most universally
in demand among the boys themselves. Originally published in more
expensive editions only, they are now, under the direction of the
Scout's National Council, re-issued at a lower price so that all boys
may have the advantage of reading and owning them. It is the only series
of books published under the control of this great organization, whose
sole object is the welfare and happiness of the boy himself. For the
first time in history a _guaranteed_ library is available, and at a
price so low as to be within the reach of all.
=Along the Mohawk Trail=
_Percy K. Fitzhugh_
=Animal Heroes=
_Ernest Thompson Seton_
=Baby Elton, Quarter-Back=
_Leslie W. Quirk_
=Bartley, Freshman Pitcher=
_William Heyliger_
=Be Prepared,= The Boy Scouts in Florida
_A. W. Dimock_
=Boat-Building and Boating=
_Dan. Beard_
=The Boy Scouts of Bob's Hill=
_Charles Pierce Burton_
=The Boys' Book of New Inventions=
_Harry E. Maule_
=Buccaneers and Pirates of Our Coasts=
_Frank R. Stockton_
=The Call of the Wild=
_Jack London_
=Cattle Ranch to College=
_Russell Doubleday_
=Crooked Trails=
_Frederic Remington_
=The Cruise of the Cachalot=
_Frank T. Bullen_
=Danny Fists=
_Walter Camp_
=For the Honor of the School=
_Ralph Henry Barbour_
=Handbook for Boys,= Revised Edition
_Boy Scouts of America_
=Handicraft for Outdoor Boys=
_Dan. Beard_
=The Horsemen of the Plains=
_Joseph A. Altsheler_
=Indian Boyhood=
_Charles A. Eastman_
=Jeb Hutton;= The story of a Georgia Boy
_James B. Connolly_
=The Jester of St. Timothy's=
_Arthur Stanwood Pier_
=Jim Davis=
_John Masefield_
=Last of the Chiefs=
_Joseph A. Altsheler_
=Last of the Plainsmen=
_Zane Grey_
=A Midshipman in the Pacific=
_Cyrus Townsend Brady_
=Pitching in a Pinch=
_Christy Mathewson_
=Ranche on the Oxhide=
_Henry Inman_
=Redney McGaw;= A Circus Story for Boys.
_Arthur E. McFarlane_
=The School Days of Elliott Gray, Jr.=
_Colton Maynard_
=Three Years Behind the Guns=
_Lieu Tisdale_
=Tommy Remington's Battle=
_Burton E. Stevenson_
=Tecumseh's Young Braves=
_Everett T. Tomlinson_
=Tom Strong, Washington's Scout=
_Alfred Bishop Mason_
=To the Land of the Caribou=
_Paul Greene Tomlinson_
=Treasure Island=
_Robert Louis Stevenson_
=Ungava Bob;= A Tale of the Fur Trappers.
_Dillon Wallace_
=Wells Brothers;= The Young Cattle Kings.
_Andy Adams_
=The Wireless Man;= His work and adventures.
_Francis A. Collins_
=The Wolf Hunters=
_George Bird Grinnell_
=The Wrecking Master=
_Ralph D. Paine_
=Yankee Ships and Yankee Sailors=
_James Barnes_
GROSSET & DUNLAP, Publishers, NEW YORK
* * * * *
Transcriber's Notes:
Obvious punctuation errors repaired.
Letter to the Public, "Frenk" changed to "Frank" (Pratt and Frank
Presbrey, with)
End of the Project Gutenberg EBook of The Boy Scouts of Bob's Hill, by
Charles Pierce Burton
*** | {
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{"url":"https:\/\/yo-dave.com\/2011\/01\/29\/a-closure-in-clojure\/","text":"# A Closure in Clojure\n\nBack when closures were first explained to me, a long time ago, I thought \u201csounds like a language with pass-by-reference semantics like Pascal.\u201d Of course, it isn\u2019t quite that simple.\n\nClojure has a lot of nice features that work naturally to give you a \u201cbetter Java than Java\u201d. Here\u2019s an example of using a closure that is not at all easy in Java.\n\n(ns net.dneclark.JFrameAndTimerDemo\n(:import (javax.swing JLabel JButton JPanel JFrame Timer))\n(:gen-class))\n\n(defn timer-action [label counter]\n(proxy [java.awt.event.ActionListener] []\n(actionPerformed\n[e]\n(.setText label (str \"Counter: \" (swap! counter inc))))))\n\n(defn timer-fn []\n(let [counter (atom 0)\nlabel (JLabel. \"Counter: 0\")\ntimer (Timer. 1000 (timer-action label counter))\npanel (doto (JPanel.)\n(.start timer)\n(doto (JFrame. \"Timer App\")\n(.setContentPane panel)\n(.setDefaultCloseOperation JFrame\/EXIT_ON_CLOSE)\n(.setLocation 300 300)\n(.setSize 200 200)\n(.setVisible true)))\n(println \"exit timer-fn\"))\n\n(defn -main []\n(timer-fn))\n\n\nIf you compile and run the program, you will see a small window with a counter than increments every second. You will also see a message displayed that the function timer-fn has exited. Big deal, huh?\n\nBut look at the declaration of the counter. It\u2019s in the definition of the timer-fn. It isn\u2019t a global. But the action listener, timer-action, still has access to the variable and continues to increment it \u2013 even though the function that declared the variable has completed execution. The lexical context of that variable is maintained and the action listener has access to it. Very neat.\n\nI don\u2019t know if Java will ever get closures, but it sure is simple to use them in Clojure.","date":"2018-08-17 23:24:17","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\": 0.5778330564498901, \"perplexity\": 7331.281861759019}, \"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-2018-34\/segments\/1534221213158.51\/warc\/CC-MAIN-20180817221817-20180818001817-00019.warc.gz\"}"} | null | null |
{"url":"http:\/\/fury.gl\/latest\/reference\/fury.shaders.html","text":"# shaders\u00b6\n\n dirname(p) Returns the directory component of a pathname load(filename) pjoin(a,\u00a0*p) Join two or more pathname components, inserting \u2018\/\u2019 as needed.\n\n## dirname\u00b6\n\nfury.shaders.dirname(p)[source]\n\nReturns the directory component of a pathname\n\nfury.shaders.load(filename)[source]\n\n## pjoin\u00b6\n\nfury.shaders.pjoin(a, *p)\n\nJoin two or more pathname components, inserting \u2018\/\u2019 as needed. If any component is an absolute path, all previous path components will be discarded. An empty last part will result in a path that ends with a separator.","date":"2019-03-22 00:53:21","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\": 0.6499428153038025, \"perplexity\": 6789.246975793159}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 5, \"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-2019-13\/segments\/1552912202588.97\/warc\/CC-MAIN-20190321234128-20190322020128-00378.warc.gz\"}"} | null | null |
Although the commercial unmanned aviation industry is still in its infant days, the market already presented a diverse range of fantastic hardware, software and operational products. These prerequisites connected, especially in the last two years', many accelerator, early stage and also later stage investment companies with the market. The investors strongly raised their portfolio by UAV companies, which led to the record-breaking funding year 2015 for the commercial UAV market.
The combination of an increasing investment trend and high technology lead to a successful market entry of the strong and new commercial UAV business.
To enable an outlook of deal value and deal volume Droneii.com provides five key trends for commercial UAV investments.
High risk or early stage venture deals are still in high demand in the new commercial UAV market. 68% of all deals since 2006 were located in Angel, Seed, Accelerator/Incubator, Product Crowdfunding and Early VC investments stages. In 2015 the share increased to 70%, which means the race for the most profitable exits still has just begun.
After the regulatory process is set-up, high acceleration of Mergers & Acquisition activities is expected, which also means that the M&A strategy will change: Enterprises acquisition activities with the goal of integration increase their market investments and slowly supersede the Exit-oriented capital firms.
The Economic climate can be described as careful, but strong enough for an investment increase, the financial policy fosters non-monetary investments as never before, so that the specific UAV regulation situation is the strongest restraint for investments.
In this PDF-file you'll find detailed insights into todays trends in investments in the drone industry. This 30 page report analyzes the history of investments in the UAV sector and provides numbers based trends for the future of investments. Next to a broad wrap up you receive Company Profiles of (10) Drone Companies and (10) Investor Companies (Appendix I) as well as all (184) Drone Investment Deals from 2015 to Q1 2016 (Appendix II).
While the big consumer oriented Hardware companies have already received the biggest lump of deals, we can expect an orientation change to the services and software Sector. Services of and by UAVs will have a big influence on the industrial market and is going to supersede traditional ways in the next five to ten years. Cost-saving and occupational health aspects accelerate this industrial revolution. Speed and Quality of data transfer and analysis are the crucial factor for the Services sector and challenge the software Sector to provide the fastest and most reliable solutions. Software investments still focus on Mapping & Navigation (Airmap, Dronedeploy, PreNAv) and Flight Management (Airware). The trend will move to BVLOS and delivery software.
The regional investments swap slowly from Silicon Valley and China to Europe. France and UK have a very good basis of a regulatory framework and benefit from operation experience and regions high technology, which makes investments very promising.
The Cross-border deal flow is expected to be an important theme in the coming years, as major economies strike agreements and alliances. TPP – USA and Pacific States and TIPP – USA and EU can be agreed soon.
The recent investment deals are application-independent. The products of the Hardware driven investment market is mainly adaptive for a wide range of applications.
Since the sector orientation just began to change the direction to services and software companies, the application itself became more and more important. While the aerial photography and filming application was on his investment peak (DJI, Yuneec, Ehang), the run for 3D imagery/photogrammetry (Spotscale and others) and Agriculture (PrecisionHawk and many others) investments started through last year. The recent investment deals and market data show the trend to the industrial application market: Aerial wind, oil, gas and solar inspection (Sky-Futures, Sharper shape) will become an industrial standard soon. Transportation and Delivery (Medical, Food), the most expected and media-loved application, still needs to overcome some technical and regulative obstacles before a high demand for investments will be developed.
Does Droneii make recommendations on public companies? Alta Vista Ventures is a Canadian drone company specializing in the mining sector. We believe we are the only company in this area but we also do "traditional " drone business. We have rapidly growing revenues and and growing backlog . Please visit our web site or call me for further information. | {
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Q: Test whenever a value x is greater by exactly 1 compared to a value y modulo 3 How do I check that x is 1 greater (ahead) of y in a modulo sense (mod 3)?
This is when x = 0 and y = 2, when x = 1, y = 0 and when x = 2, y = 1.
I tried testing like this:
php -a
php > $x = 0;
php > $y = 2;
php > echo ($x - $y) % 3;
-2
php > $x = 1;
php > $y = 0;
php > echo ($x - $y) % 3;
1
php > $x = 2;
php > $y = 1;
php > echo ($x - $y) % 3;
1
It is not working for the case where x = 0 and y = 2. How can I calculate this so that $x is 'ahead' of $y by 1 in a modulo sense?
A: I will first explain my understanding of your following sentence:
How do I check that x is 1 greater (ahead) of y in a modulo sense (mod 3)?
Following the examples provided, I assume you mean that, if 1 is added to $y, and we take the mod 3 of $y, we would get the mod 3 of $x.
With that in mind, we could write the following code, witch would return true if $x is "ahead" of $y by 1. (I hope you can abstract that example to whatever situation you are facing):
function check($x, $y, $mod) {
return $x % $mod == ($y + 1) % $mod;
}
//$x = 0 and $y = 2
echo check(0,2,3); //returns true
//$x = 1 and $y = 0
echo check(1,0,3); //returns true
//$x = 2 and $y = 1
echo check(2,1,3); //returns true
//$x = 0 and $y = 1
echo check(0,1,3); //returns false because $x is 2 "ahead" of $y
If you want a more generalized version of the function, with an arbitrary difference, you can use this (it should work with positive and negative differences):
function check($x, $y, $mod, $diff) {
return $x % $mod == ($y + $diff) % $mod;
}
A:
Why is it that you used 0 for x and 2 for y? (Sorry, but I'm not much of an algebraist.)
Although your situations are pretty much clarified, it doesn't make sense (at least for me) that x would be 1 bit ahead of y when it is clearly 0 < 2.
As we know, the statement (0 - 2) % 3 would be -2 because the mod is 3, and 0 - 2 is -2, so the result is -2.
Mathematics can have some illogical sense (in my opinion) sometimes, so its worth noting that 0 isn't ahead of 2 (in the sense of programming variables) at all.
And to actually answer your question on calculation, you could have x = 0 and y = -1 as your variables instead, as per se the logic of your statement, then just increment by 1 for both variables, and the result is still the same.
Proof:
<?php
$x = 0
$y = -1
echo ($x - $y) % 3 // outputs 1
echo (++$x - ++$y) % 3 // still outputs 1
echo (++$x - ++$y) % 3 // also outputs 1
echo (++$x - ++$y) % 3 // yep, still the same
echo (++$x - ++$y) % 3 // alright, you get the deal
?>
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,498 |
Cast members from Orange is the New Black attended a photocall at the Soho hotel in London this week as they continue to do press for the new season of the hit Netflix series. Danielle Brooks, Jason Biggs, Taylor Schilling, and Laura Prepon were all present, and have been chatting with press about what to expect – as well as what has changed with season 2.
"The fact that everyone is so amped for a second season is why we went into production a little earlier." Biggs explained seeming eager for fans to see the new season.
"I came into (the show) with some insecurities and needing to exercise some muscles that I hadn't worked in a long time. But the more positive feedback comes through, the more positive I become. It's great to have something to live up to." Biggs continued to say to Independent.ie.
Biggs is definitely right – the second season has A LOT to live up to if it will match the hype from last July's premiere, but we're pretty sure they know exactly what they're doing!
Take a look at our gallery from the photocall below.
Orange is the New Black returns with 13 brand new episodes on June 6th! Will you be watching them all at once like we will be? | {
"redpajama_set_name": "RedPajamaC4"
} | 6,481 |
Q: Setting two sessions, a session and cookie I'm trying to set a session and a cookie for when user logs in.
When the user visits the login page, a session is set and started, with session_start() which is working quite alright, but when the user now fills in the login form (with username and password) and the proper check is done for correct login details, I set the cookie:
$one_week = 60*60*24*7;
setcookie("cookiejarcookie", "cookiejar_value", time()+$one_week, '/', 'localhost');
It's not working, the cookie is not being set. I've tried calling it from the top of the script, but it's not working.
How do I set the cookie after setting the session?
A: Trying to set a cookie on localhost does not work in most browsers. You need to set the domain value to null, empty string or false. Most recommendations I've seen are to set the domain value to false. With that said, I've never understood writing code like that, as it is not something you're going to deploy to a production environment.
See the recommendation by @David. I personally use virtualization to run a server environment and map fake dns using the hosts file.
One tip I can offer is that you have to open your editor (I use notepad++ or wordpad) as administrator on most recent versions of windows that have UAE in order to edit the relevant hosts file.
A: From my comment
You cannot set cookies to localhost, but if you add a my.fake.local in your hosts file ( /etc/hosts or c:\Windows\System32\drivers\etc\hosts ) that should work.
add
127.0.0.1 my.fake.local
in the appropriate hosts file.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,060 |
require 'ostruct'
require 'path'
Before do
@system = Finitio::TEST_SYSTEM
end
Given(/^the System is(, within '(.*?)')?$/) do |path,source|
if path
begin
target = Path.dir.parent/path
target.write(source)
@system = Finitio.system(target)
rescue => ex
@system = ex
end
else
@system = Finitio::TEST_SYSTEM.parse(source)
end
end
Given(/^the type under test is (.*?)$/) do |typename|
@type_under_test = @system[typename]
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 5,234 |
Two concordats were signed in 1817:
Concordat of 24 October 1817, with Bavaria
Concordat of 11 June 1817, with France. | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,535 |
The Falcon in great shape with machine packed powdery corduroy.
Fence to fence again on the Snowy Owl Chair, great surface conditions and a good base on the steeper pitches so will stand up to traffic well.
Fresh tracks on the Buzzard looking over to the main side.
The Lecht has received around a foot of fresh snow since Thursday evening, all runs are now in good shape with a full width cover between the fences and grooming up beautifully.
More photos from Sunday at the Lecht are available in the Pix from the Slopes. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,619 |
National's whiteness problem
Todd Muller's first act on become party leader was to appoint an all-white front bench. This understandably raised eyebrows, so he is now apparently reconsidering caucus and list rankings. Which is obviously necessary, but there's still the underlying problem: National doesn't have a lot of Māori MP's. And the reason for that is that it has consciously chosen not to pursue the Māori vote. Thanks to Bill English's retreat into racism over the foreshore and seabed, National decided not to run in the Māori seats in the 2002 election. Don Brash doubled down on that decision, and because they're a conservative party opposed to change, National has stuck with it. And the result is that they haven't run in the Māori seats since 1999 - over twenty years ago. Officially, they pursue Māori voters on the general roll and for the party vote. But by not standing in these seats, the message they are shouting at the top of their lungs is "we are not interested in Māori votes and we are not interested in Māori".
This is bad for National, leading directly to the diversity problems they have today. Seven fewer Māori candidates means seven fewer Māori on the party list means a caucus which doesn't look like New Zealand. But its also bad for our democracy. One of our two major parties is telling 15% of our population "we don't want your votes", and implicitly, "you don't and never will matter". And that is an appallingly racist message for a party which purports to be for all New Zealanders and which leads the government half the time to be sending. We voters can and should judge them on that.
So, if Muller actually wants to fix his party's whiteness problem, he has an obvious solution available. But that would require reversing 18 years of racist policy, not to mention potentially upset incumbent MPs over list placings. And to be honest, I just don't think they have it in them.
Labels: Maori, Maori Seats, National, Racism
China is committing genocide in Xinjiang
China is sterilising Uighur women in Xinjiang in order to suppress their population:
Chinese authorities are carrying out forced sterilisations of women in an apparent campaign to curb the growth of ethnic minority populations in the western Xinjiang region, according to research published on Monday.
The report, based on a combination of official regional data, policy documents and interviews with ethnic minority women, has prompted an international group of lawmakers to call for a United Nations investigation into China's policies in the region.
Genocide is defined in international law in both the 1948 Convention on the Prevention and Punishment of the Crime of Genocide and the 1998 Rome Statute of the International Criminal Court
any of the following acts committed with intent to destroy, in whole or in part, a national, ethnical, racial or religious group, as such:
(a) Killing members of the group;
(b) Causing serious bodily or mental harm to members of the group;
(c) Deliberately inflicting on the group conditions of life calculated to bring about its physical destruction in whole or in part;
(d) Imposing measures intended to prevent births within the group;
(e) Forcibly transferring children of the group to another group.
[Emphasis added]
China is not a party to the Rome Statute - it fears international law. But it is a party to the 1948 Convention, having assumed the signature of the former Republic of China. It - and members of its regime - are thus subject to international jurisdiction for this crime. And they need to be investigated and prosecuted for it.
Labels: China, Genocide, Human Rights, International Law
One benefit of the pandemic: the 2021 APEC meeting will now be held online, rather than in New Zealand. As for why this is good news, the cost of hosting that summit would have been to turn New Zealand into a police state for the duration of the meeting. The APEC enabling act would have allowed the government to put the army on the streets as "police" with full arrest and search powers, let foreign security staff bring pistols and automatic weapons into New Zealand and use them, arbitrarily close buildings, roads and public places, require proof of identification ("papers, please") from anyone wanting to enter, jam WiFi and any other radio communications, and all with no right of challenge or appeal, and an explicit over-ride of the BORA. But now, none of that is necessary. The bill is still on the Order Paper waiting for its second reading, but hopefully now it can quietly be dumped.
As for the future, I've said it before, and I'll say it again: if the price of hosting such meetings is this sort of erosion of human rights, that price is too high and we should not host them. Letting the PM hob-nob and big-note and get a silly shirt photo with foreign leaders is not worth sacrificing any of our human rights. If she wants to do that, she should do it in some foreign tyranny, not transform New Zealand into a tyranny to accommodate them.
Labels: Foreign Policy, Human Rights, Tyranny
An environmental crime
Meridian energy has been caught manipulating the electricity market for fun and profit:
Meridian Energy pushed up power prices by unnecessary spilling water from its south dams in December that could have been used for generation, the Electricity Authority has ruled in a preliminary decision.
The authority said the "undesirable trading situation" (UTS) could have resulted in other electricity retailers having to pay an extra $80 million for power on the wholesale market, though the final cost could not yet be determined.
"Meridian's activities led to more expensive generation running in the North Island at a time when there was excess fuel in the south," Electricity Authority chief executive James Stevenson-Wallace said.
...and that "more expensive generation" was gas and coal. Running the maths, an estimated 41 GWh of electricity means somewhere between 8 and 13 kilotonnes of extra carbon dioxide spewed into the atmosphere, in just one month. This isn't just a financial crime - its an environmental one as well.
New Zealand electricity companies have a logn history of pulling these sorts of scams. Back in 2011, it was Genesis doing it, and Meridian was the victim. They did it back in the 90's too. And the reason they do it is because in our electricity market, the marginal generator sets the price, so even if Genesis was providing the electricity, Meridian was benefiting from the higher prices. As for how to stop it, sticking sociopathic business managers in jail seems like a reasonable if short-term solution. In the long-term, we need market reform, or better yet, to eliminate the market entirely. A state-run electricity system with the primary goal of providing power rather than profit will simply not be incentivised to pull such scams. And we'll all be better off as a result.
Labels: Coal, Energy, Environment
Removing a hateful symbol
The US state of Mississippi has finally voted to remove the Confederate emblem from its state flag. Good. Its appalling that it was put there in the first place, but even more appalling that it took so long to remove it. Its basicly as if a German Länder still had the Nazi swastika on its flag. and that's just unacceptable.
But while we're talking about removing hateful symbols, its long past time New Zealand got rid of the union jack. Its the flag of the empire which built and profited from the slave trade, which colonised, exploited, and committed horrific crimes across half the world, which invented concentration camps to keep its uncoperative colonial subjects under control, and which is entirely unapologetic about any of that. We had a chance to get rid of that symbol in 2015, and (thanks to shit design choices by the government) we blew it. Its time we had another go.
Labels: Republicanism, USA
The Greens' opening bid is transformational change
Over the weekend the Greens made their first election policy announcement, and promised transformational change: a more progressive tax system, with new rates applying to incomes of $100,000 and $150,000; a new wealth tax on net wealth over $1 million; and a guaranteed minimum income to eliminate poverty. Its bold, its progressive, it would make us a better, more equal society. So naturally, National is against it. As for Labour, the party that promised change at the last election, they don't want to talk about it - they're simply missing in action. Which is I guess what you'd expect from a bunch of people who own that many investment properties in Auckland.
A lot of people are quibbling the details of the wealth tax: is the $1 million threshold to low? Is the effective exemption of family homes by individualisation too complex? Will the rich find ways of giving themselves paper debt or paper partners to evade it? But these are implementation details. You can shift around thresholds and exemptions and this and that, but the core principle - that the rich should be paying tax on their hoarded wealth - is something we should all support (though really, its probably easier to implement the property part of it via a land tax, since its basicly impossible to evade). And it puts Labour in an unpleasant position: for all their left-wing talk, they're essentially a status quo party, who now oppose the more equal society they once stood for, or at least, oppose doing anything serious to get us there. But if they oppose this, they're essentially taking on their own base - and especially the young activists who provide the bulk of their electoral workforce. So hopefully, they'll be forced to respond. And while their response will no doubt be an insipid, watered-down half-measure (remember the "bright-line test"?), by accepting the principle, it will open the door to something better in future.
Posted by Idiot/Savant at 6/29/2020 11:26:00 AM
Labels: 2020 Election, Equality, Greens, Labour, Left
Climate Change: Repealed
One of the big problems with climate change policy is that the government has refused to let local authorities use our major piece of environmental regulation to reduce emissions. Since 2003, local bodies have been explicitly forbidden from considering the impacts of greenhouse gas emissions in their planning documents, and in consent decisions. But now, those restrictions have been repealed:
Coal mines and fossil fuel power plants could be a thing of the past in New Zealand after the Government passed a law which allows environmentally-damaging projects to be refused.
The amendment to the Resource Management Act closes a loophole which allowed consent for new builds without consideration for the environment.
Sadly, this part of the law won't come into effect until 2022, so there's far too much time for dirty polluting infrastructure to be consented and its emissions locked in before then. Also, the parallel provision in the EEZ Act is still in place, so we'll still have the ludicrous situation of the EPA being forbidden to consider climate change impacts when deciding whether to consent new gas wells. But hopefully that will now become a priority for repeal.
The government will supposedly be developing a National Policy Statement on climate change to guide local authorities and their plans. They've been promising that since at least the mid-1990's and the Stratford Power Station decision, and again in 2003 when they passed the ban on considering climate change in the first place. But if they don't, then the courts will effectively do it for them.
Labels: Climate Change, Climate Change Policy, RMA
No future for Marsden Point
RNZ reports that Refining NZ is considering shutting down Marsden Point and effectively turning itself into a fuel distribution service. The problem they're immediately trying to address is that running a refinery is expensive, and it is cheaper just to import processed fuel. But there's a longer-term problem too: climate change. And that problem means there is basicly no future for Marsden Point.
Its simple: if we are to avoid making the Earth uninhabitable, we need to radically decarbonise our economy. And that means running cars, boats and planes on something other than dirty oil. And no oil means no need for dirty great refineries to process it. Which means that if you're running a refinery, you want to look at ways of getting out of that, or face being a giant, valueless stranded asset. For Marsden Point, shutting down the dirty part, and shifting to distributing imported fuel, is a way to manage that risk. If they don't do it now, they'll be doing it in five or ten years time, and the earlier they do it, the lower the risk to them.
(Refining NZ is also finally entering the ETS, after a 20-year exemption, so that's probably helped drive this. But even if they weren't facing finally having to pay for the pollution they cause, an expectation of declining demand for their product should be enough to push this).
The good news: if they shut down the refinery, that's a few hundred thousand tons of carbon we're no longer emitting. Yes, those emissions will now be happening offshore, on someone else's books (probably in a more modern, cleaner refinery). But it means it will be much easier to reduce emissions in future, because we won't have this hulking facility committed to spewing out carbon to maintain minimum throughput.
As for Northland, its in the same situation as Tarankai or the West Coast: they built their economy on an industry with no future, so they need to find something else to do. There's obvious scope for the government to step in here and try and find an alternative - can Northland do wind, or solar, to help power the electric cars and factories of the future? But fundamentally, the world has changed on them, and all they can do is try and cope with it.
Labels: Climate Change, Energy, Oil
More progress on secrecy
Last year, I highlighted a secrecy problem with the government's Infrastructure Funding and Financing Bill. The Bill would establish a new class of public entity, "special purpose vehicles", to hide debt from local government balance sheets fund infrastructure such as roads and sewers in "high-growth" councils. The SPVs would collect and spend public money and enjoy statutory powers, but as originally envisioned, would not have been subject to the Official Information Act or LGOIMA. Not due to any real analysis, but because the Department of Internal Affairs had never really thought about it.
Yesterday the Transport and Infrastructure Committee reported back on the bill, and in response to submissions, have decided to make it subject to the Ombudsmen Act (and therefore the OIA):
We believe it is important for levy payers to have transparency about an SPV's use ofstatutory levy powers. As SPVs would be performing public functions and spending public money, we believe there should be a high level of scrutiny over their operations. We note that the Chief Ombudsman has also expressed his belief that SPVs should be subject to the OIA and the Ombudsmen Act.
We note that other private entities are subject to the OIA to the extent they perform public functions. Scrutiny of public functions should not be avoided because those functions are performed by a private firm.
We therefore recommend inserting a provision in Schedule 2 of the bill to make SPVs subject to the Ombudsmen Act in relation to their administration of, and compliance with, the levy order and this legislation. This amendment would automatically make SPVs subject to the OIA.
So we get not just transparency, but also a right of review by the Ombudsman for unreasonable behaviour. Which could come in handy.
Meanwhile, I'm still shocked that departments aren't bothering to analyse whether new agencies should be subject to the OIA regime. The Law Commission told us the criteria for inclusion back in their 2012 review. Departments should be more than capable of applying these guidelines, rather than leaving it to random members of the public to do so.
Labels: Freedom of Information, Local Government, OIA
Erasing a stain on our democracy
National's prisoner-voting ban was a stain on our democracy. Passed in defiance of the Bill of Rights Act, with an absence of thought, it brought our parliament into disrepute. But now, that stain has been at least partially erased, with the passage of the Electoral (Registration of Sentenced Prisoners) Amendment Bill tonight. But the bullshit the ban symbolised isn't over:
The Green Party have won a last-minute change to the prisoner voting bill that may technically allow all prisoners to register to vote - with the help of the National Party.
But Justice Minister Andrew Little says this change to the bill will be corrected.
"We saw mindless politics from the National Party tonight. The Green Party SOP had two separate amendments. The one that extended the right to vote to all prisoners was voted down. The other one taking away the Electoral Commission's power to remove disqualified voters from enrolling was inexplicably supported by the National Party. We will be correcting this in the House next week," Little said in a statement.
So it seems that Labour is just as keen on performative cruelty and human rights abuse as National. But if they want to "correct" it next week, they're going to need urgency, and they're going to find that difficult without Green votes. Unless they want to club together with the opposition to fuck over their confidence and supply partner - in which case the latter should simply tell them to rely on National to pass the rest of their legislative program for the term.
Meanwhile, in the long-term, this debacle has shown us that we can not trust Parliament to guard our human rights, as venal politicians will always abuse human rights to suck up to arseholes. But it has also shown us the solution: the courts seized the power to declare legislation inconsistent with the BORA, and effectively shamed them into this repeal. The government is currently trying to tame this power with a bullshit half-measures bill, which would channel that public shame into a parliamentary report - another fine example of Labour's status quo "reform". If you think that that's not enough, and you'd like a BORA with real teeth, then please submit on the bill here.
Labels: Democracy, Human Rights, Parliament
Have NZ cops been trained to be killers?
Over the past few weeks, there's been a lot of attention on the question of why American police are so violent, and so often murder people. There are a pile of reasons for this, including the racist roots of their entire policing system, but part of the answer seems to come down to a guy called Dave Grossman. The author of a book called On Killing, he now runs police training courses on "killology" and the "warrior cop", basicly encouraging police to be murderers and trigger-happy psychopaths. According to Grossman, the only good cop is a killer.
So far, so horrifyingly American. But there's an unpleasant local angle: according to his marketing material (example; there are plenty of others), Grossman has trained people in New Zealand:
Col. Grossman is one of the nation's leading law enforcement trainers. He is the author of the Pulitzer-nominated book, On Killing. He has served as a trainer and keynote speaker for all major national and international law enforcement training organizations and has taught the representatives of literally thousands of federal and regional agencies in the U.S., Canada, New Zealand, and Australia.
Which raises the obvious question: has he trained police here? Has he been training New Zealand police officers to be psychopaths and murderers? I think the police - and other government agencies - owe us some answers here.
Labels: Human Rights, Police, USA
Overseeing the COVID-19 law
The COVID-19 Public Health Response Act 2020 is currently being reviewed by a select committee, ahead of the first vote (in early August) on whether to extend it. The law includes clauses requiring orders to be approved by Parliament, and allowing them to be revoked at any time by the House. Today's Order Paper includes a sessional order establishing rules around that. Firstly, no motion to continue or revoke the Act will be allowed before the select committee has reported. Secondly, all new orders will be sent to the Regulations Review Committee, which must report back within 12 working (not sitting) days, and approval motions will not be permitted until the committee has reported back. Finally, disallowance motions moved by a member of the Regulations Review Committee will be given priority over other business.
All up, this looks like a good set of measures to ensure proper Parliamentary oversight of the law, and ensure the House can properly perform its functions.
Labels: Disasters, Parliament
Status quo "reform"
Last year, the government promised that it would strengthen whistleblower protections. Today, they finally released the bill. There's one major change: "[a] discloser is entitled to protection for a protected disclosure made to an appropriate authority at any time" - an "appropriate authority" being public sector heads, officers of Parliament, or various oversight bodies as listed in the schedule to the bill. But note who it specifically excludes: Ministers and MPs. Whistleblowers will only be able to go to the Minister where their organisation has failed to investigate. And if they just drag their feet or stage an internal coverup, you're on very shaky ground trying to do the right thing.
And other than that, its basicly a re-enactment of existing law, with updated language. There's no criminalisation of retaliation against whistleblowers - that remains a strictly civil affair, to be dealt with by the victim at their own expense. And there's no protection of disclosures to the media to ensure a full and proper investigation (permitted in some Australian states). Instead, disclosures to media are explicitly labelled as "bad faith". The law will do nothing to protect whistleblowers in the sort of case which supposedly inspired it.
If this is "reform", its the "reform" from people who think that everything is working as it should, that nothing much needs to change, and that whistleblowers are the problem rather than a solution. As for Chris Hipkins' goal of ensuring that employees feel safe to report cases of serious misconduct - explicitly seen as a failure of the current law - it will change nothing.
Labels: Labour, Whistleblowers
Farmers are poisoning Canterbury
Last year we learned that relatively low levels of nitrates in drinking water were linked to increased incidence of colorectal cancer. New Zealand has a lot of nitrates, thanks to all those cows pissing everywhere. So how bad is the problem here? Bad:
2020 data obtained by Forest & Bird through the Official Information Act shows multiple Canterbury councils have reported nitrate-nitrogen levels this year well above 0.88 mg/l, with Selwyn, Ashburton, Timaru, and Waitaki presenting especially concerning levels.
The supply for the Rangitata Huts reached a recent maximum of 11.43 mg/l in June last year, having since dropped back to 8.85 mg/l in April.
Ashburton's Tinwald treatment plants recorded a nitrate-nitrogen level of 7.01 mg/l, well above the increased risk levels for colorectal, colon, and rectal cancer, and above the Ministry of Health level of 5.65 mg/l that prompts ongoing monitoring of the supply. Numerous other supplies across the wider Ashburton district recorded similar levels.
Forest & Bird says the results are a wakeup call for the government, which last month decided not to put a Dissolved Inorganic Nitrogen limit of 1.0 mg/l (similar to nitrate-nitrogen) in the new freshwater reforms, ignoring the advice of its Science and Technical Advisory Group, as well as submissions from the New Zealand College of Public Health Medicine (that called for a level "considerably lower than 1 mg/L of DIN"), and the Hawke's Bay District Health Board.
The dairy industry is slowly poisoning people in those towns. And by refusing to regulate them properly, the government is effectively colluding in it. And at some stage, we need to decide what is more important: people's lives, or farmers' profits.
Labels: Dirty Dairying, Environment, Water
Getting what they paid for again
It looks like the fishing industry is still getting what they pay for:
The Government has again bowed to fishing industry pressure and refused to extend a marine reserve around Campbell Island, a subantarctic sanctuary recognised for its value in conserving and maintaining unique creatures.
Campbell Island/Motu Ihupuku is uninhabited and is New Zealand's southern-most island. It is accepted as one of the most pristine places on earth and an important breeding ground for seabirds and marine mammals.
The move has exposed a split between the Green Party and its Government partner Labour over protection of the oceans.
Conservation Minister Eugenie Sage says she backed enlarging the sanctuary. But she was overruled by Fisheries Minister Stuart Nash, who sided with industrial fishing interests.
There's no actual fishing down there, but the fishing industry opposes sanctuaries for ideological reasons: they hate the idea that there might be anywhere they are not allowed to pillage. They're not interested in sustainable management, they're not "stewards" ensuring a viable future for their industry, they're purely about environmental destruction. As for how to stop it, a first step is to vote out the politicians they've bought, and elect ones they can't buy.
Labels: Conservation, Corruption, Fishing
One country at a time
Something I missed in the lockdown chaos: Chad has abolished the death penalty:
The European Union has hailed Chad for abolishing the death penalty for terrorism crimes nearly five years since its last execution.
In a statement on Saturday, the EU Spokesperson for Foreign Affairs and Security Policy, Virginie Battu-Henriksson, said the move by Chad should be emulated by other countries that still allow death penalties.
"With this law, Chad has become the 22nd African state to abolish the death penalty for all crimes," said Battu-Henriksson.
80% of African nations are already abolitionist in law or in practice, and hopefully we'll see more countries following suit.
Labels: Chad, Death Penalty
Climate Change: Undermining their own policy
The most effective way New Zealand has of fighting climate change and preventing the earth from becoming uninhabitable in the short term is planting trees to soak up carbon. And, thanks to the carbon price being hard up against the cap for a few years, this has finally been happening. But now, the government is threatening to stop it:
The much criticised conversion of farm land into forestry could be checked by the government if it goes too far, politicians have been told.
Agriculture Minister Damien O'Connor, who is also Minister for Rural Communities, yesterday told Parliament's Primary Production Select Committee that land conversions might have to be reviewed if they reach 40,000 hectares a year.
The conversion of farmland into forestry has been repeatedly accused of undermining thriving rural communities and replacing them with a green desert.
Which is typical for how climate change policy works in this country: when something looks like being effective, someone whines and the government backs off and stops it. Which is why we have 25 years of failed policy and why we expect to completely miss our targets. Which is great for established interests who want to keep profiting from wrecking the planet, but very bad for everyone else.
We can't afford to do this anymore. We are at 100 seconds to midnight. It was over 30 Celsius in Siberia last week. Parts of Antactica are turning green. Australia burned down last summer, and it'll probably do it again this year. We actually need to act. And if its a choice between futureless rural communities and the planet, I'm voting for the planet.
Labels: Agricultural Emissions, Climate Change, Forest Sinks
A ballot for two Member's bills was held today, and the following bills were drawn:
District Court (Protection of Judgment Debtors with Disabilities) Amendment Bill (Anahila Kanongata'a-Suisuiki)
Local Government (Customer Focus) Amendment Bill (Jacqui Dean)
These are both pretty boring bills, which make tweaks rather than big, contentious changes. I don't expect any of them to receive a first reading until after the election.
Labels: Members Bills, Parliament
Climate Change: The final squeals of a dying industry
The Climate Change Response (Emissions Trading Reform) Amendment Bill, which finally make the ETS sortof function, over a decade after it was first established. Farmers are naturally unhappy, claiming that the law will see polluters buying farms to plant trees, effectively driving them off the land. To which the only response should be "good" - because we are already better off without those farms.
I did the maths on this last year, when farmers were squealing about rural communities being killed off by tree planting. Their example then was forestry conversions in Tararua, which would lead to the loss of 47,500 sheep, 20,500 cattle, and ~$1.7 million from the local economy. But at $25 / ton, the carbon cost of the avoided emissions from those animals was already catching up with that economic value. And carbon prices are now at $30/ton, making the value of avoided emissions $1.65 million. Last week, they were over $32, and forward units were over $35. So we're basicly in a situation now where the economic benefit of those farms is less than the climate damage they do. Farmers are protected from that reality because they are effectively subsidised by not being included in the ETS. But we are better off as a society if those farms on marginal land shut down. And that's without considering the value of any carbon stored in trees.
Obviously, some farms will be more efficient. But for those that aren't, the best thing that can happen is for the market to take its course. And the sooner it does, the better.
Labels: Agricultural Emissions, Climate Change, Emissions Trading, Forest Sinks
The Privileges Committee has called for submissions on the New Zealand Bill of Rights (Declarations of Inconsistency) Amendment Bill. The bill recognises that the courts can declare legislation to be inconsistent with the BORA, and requires the Attorney-General to report such declarations to Parliament. While there's no requirement in law for the government to respond in any way, the committee will apparently also be considering changes to the House's standing orders which may include such a requirement.
This is basicly the absolute minimum that the government could do, and its scandalous that it has taken them so long to drag them to even this point. But its better than nothing. And apparently that's all we can expect from this "transformational" government: "better than nothing".
If you're interested in submitting on the bill, you can do so here.
Labels: Human Rights, NZ Constitution, Parliament, Participation
Some progress on secrecy clauses
Over the past year I've noticed a rise in the number of secrecy clauses in legislation: specific clauses requiring that certain information be kept confidential, effectively forbidding its release under the OIA. The most shocking bit is that these are usually applied to information collected from other government agencies, and driven by public sector dysfunction and internal distrust, but a desire to protect "commercially sensitive" information (which is already protected by the OIA) is also a driver. I've investigated these clauses usig the OIA, and submitted on a few bills to argue against them. And in some cases, the message seems to be getting through.
The first case is the Mental Health and Wellbeing Commission Bill, which is currently waiting for its third reading. As drafted, this had a secrecy clause with the usual ambigious "required by law" language. Past Ombudsman's rulings have suggested that this might not be enough to allow disclosure, as the generalised duty of confidentiality would prevail. The good news is that the Ministry of Health didn't intend the section to override the OIA, and the Ombudsman suggested phrasing which explicitly recognises the right of access under the OIA (so, its basicly a "don't randomly share this stuff" clause). Hopefully this will become the standard in future.
The second is the new Forests (Regulation of Log Traders and Forestry Advisers) Amendment Bill, which is currently waiting for its second reading. In this case, the new Forestry Authority would have statutory information-gathering powers, which led to concerns from some submitters about commercially sensitive information. But rather than impose a secrecy clause, the committee instead included a clause saying the OIA applied as usual:
See section 9 of the Official Information Act 1982 for reasons for the Forestry Authority to withhold any official information it holds (including where making the information available would disclose a trade secret or be likely unreasonably to prejudice a person's commercial position).
Looking at the summary of submissions, the driver here was MPI recognising that commercially sensitive information is already protected by the OIA. Unfortunately, other agencies seem to lack this basic understanding of the law. But when they suggest a traditional secrecy clause in future, we will now have a good example to point to.
Labels: Freedom of Information, OIA
Equality wins in the US
The US Supreme Court has ruled that existing anti-discrimination law covers sexual orientation and identity:
The top court in the US has ruled that employers who fire workers for being gay or transgender are breaking the country's civil rights laws.
In a 6-3 decision, the Supreme Court said federal law, which prohibits discrimination based on sex, should be understood to include sexual orientation and gender identity.
The ruling is a major win for LGBT workers and their allies.
This is despite Trump trying to stack the court with Conservatives - it appears that even they can read a statute, and understand that firing someone for being gay or trans necessarily involves consideration of sex, and punishing them for behaviour that they would not question if they had been of a different sex. A lot of people on the American right are probably wondering what the point of court-stacking is if these people turn out to have limits to their hackery (not everyone is as limitless as Samuel Alito or Clarence Thomas). Because for them, its not actually about the law, but the "social order" they think it upholds: an order which has people like them at the top, and everyone else subject to their whim. Hopefully this case will be another rock on their grave.
Labels: Discrimination, Homophobia, Human Rights, USA
Labour chickenshits out on gun control
In the wake of the Christchurch Mosque shootings, Labour moved urgently on gun control. Semi-automatic firearms were banned immediately. But their second tranche of legislation, designed to establish a more permanent framework, got bogged down by NZ First. And now it looks like Labour has chickened out, and will let them gut it, in an effort to get it passed before the election:
Plans to set up a gun register could be delayed for up to three years as the Government aims to establish an independent entity to take over firearms licensing and administration from the police.
The move is part of a suite of changes to the Arms Legislation Bill, which saw the Labour Party give in to most of NZ First's demands.
However, the Bill may not pass before the election, with NZ First MP Ron Mark saying the party was still "disappointed" with the outcome, which did not make exemptions for sporting shooters.
They're also moving to let farmers keep illegal semi-automatic weapons for "pest control", rather than limiting them to a tight group of restricted users. Which basicly means weapons like the ones used to murder people in Christchurch will be in the hands of every farmer in the country, so they can shoot "rabbits".
But this backdown wasn't necessary. Gun control is a hugely popular issue in urban New Zealand. Labour could have simply publicly blamed NZ First for the delay, then campaigned on passing the law without their amendments. Instead, they chose to chicken out, and grovel to a coalition partner who is actively sabotaging a core part of their agenda, when they had no real need to. Its as if having your backbone - or values - surgically removed is a requirement for party membership or something.
Shit like this is why, outside of crises, Labour has been a government of disappointment. And its why their promises mean nothing - whatever they offer, they'll just back down the moment someone disagrees with them. Voters can not and should not trust such cowards.
Labels: 2020 Election, Gun control, Labour, MMP, NZ First
The police fail again
Ever since he killed 51 people in Christchurch, people have been asking how the hell the Christch mosque shooter managed to get a firearms licence. And now we know the answer: the police just didn't bother checking properly:
The March 15 terrorist was wrongly granted a firearms licence due to a string of police failures, sources have told Stuff.
The terrorist, who pleaded guilty to New Zealand's worst mass shooting in March, was not properly inspected by police vetting staff when he applied for a firearms licence in 2017.
Stuff has been told that, among other errors, police failed to interview a family member as required, instead relying on two men who met the terrorist through an internet chatroom.
The error was overlooked when police granted him the firearms licence, allowing the Australian citizen to stockpile the semi-automatic guns later used to murder 51 people.
This wasn't just a single failure - multiple police officers fucked this up. As for why, its hard to escape the conclusion that they just didn't think it was important - and certainly less important than, say, harassing activists for their political views, or kicking in people's doors over cannabis. But ensuring that those who hold firearms licences is a core preventative measure, a key way of ensuring public safety. It speaks volumes that the police took their eye of that ball, and it really makes you wonder what else they're not paying attention to while they chase protesters and harmless drug users.
Labels: Gun control, Police
Ending the DHB scam?
The government has announced the results of its review of the health system, and it looks like a thorough shake-up:
The Government is backing a proposal for the biggest reforms to healthcare in a generation, which includes culling the number of District Health Boards around the country and dropping elections for their members.
Two years in the making, a review team of experts - led by health economist and former Helen Clark confidant Heather Simpson - has proposed a complete overhaul of the health system.
Among the review's main recommendations are:
A new health authority, Health NZ, to take control of the health system.
A reduction in the number of district health boards, from 20 to between 8 and 12, in the next five years.
Ending elections for DHB members and making them all Government-appointed.
A Māori health authority to sit alongside Health NZ and the Ministry of Health.
Good. Because the current system - where DHBs are elected but have no power or independence - is just a scam designed to prevent accountability. When people complain about health funding, the Minister - who makes all the decisions and decides what gets funded where - gets to point the finger at the DHB and say "their fault", while the DHB can do the same right back. The result of this mutual blame-shift is left as an exercise for your next visit to your local hospital. A return to appointed membership will make it clear that they are servants and functionaries and that the real decisions are made by the Minister who appoints them - and that that Minister is the one who should be held accountable for failures of resourcing. So I expect the way this will go is that the government will adopt all the plans - greater centralisation, fewer DHBs - but keep the pointless elections, because Ministers love having a permanent blame sink.
Labels: Health, Public Sector
Steamrolling democracy
The government will be introducing its bill to gut the RMA this week. The bill will replace the current RMA process with a Muldoonist one for selected, projects, removing public submission rights and effectively gagging us and preventing us from speaking up for our interests or presenting evidence to inform the decision. It will also remove appeal rights, meaning that when a bad decision is inevitably made, there will be no legal way of correcting it. And by setting up the Minister as a gatekeeper to this fast-track process, it makes them a nexus for lobbying and corruption.
As for the process, the bill will be introduced to Parliament this week, sent to select committee, and reported back by June 29th - meaning a select committee process of just two weeks. Such a short submissions period means that a) people won't have time to frame a proper submission; and b) committee members won't have time to read or engage with it. The result is to turn the select committee process into a democratic fraud, a box-ticking exercise designed purely to give a semblance of "consultation" to a pre-determined decision. And as with other examples, I do not think people should waste their time on it, except to register their opposition. And again, its appalling that the Greens are supporting this - both the bill and the process. What a pack of fucking footstools.
As for the projects they will be legislatively fast-tracking, at lest they're not all roads. But each and every one of them deserves a full consenting process, so that the community's views can be heard and a good decision made. Our democracy is more important than their economic growth.
Labels: Environment, Greens, Labour, RMA
The Labour list
Labour released its party list today, and it goes all the way to 84 - enough candidates for 70% of the vote (or a lot of mid-term replacements). There are a couple of high profile demotions, though the list spots are still more than winnable, so that's really just stabbing those MPs in their vanity. There's also a lot of people being downranked in the midlist, a decision forced by Labour's Māori MP's suddenly deciding they have competition again. but the lowest-ranked incumbent is at spot 49, so it looks like they're in no danger unless Labour's polling collapses to less than 40%. Meanwhile, one name is conspicuous by its absence: Ōhāriu MP and former Police Association spokesperson Greg O'Connor. I'm not sure if this is because he has chosen to go electorate only, because he is planning a shock retirement-announcement, or whether the Labour Party realised that in the current climate, sticking an advocate of an armed police force on the list was toxic. But whatever, his absence is welcome. And if he is competing in Ōhāriu, I hope the voters throw him out on his arse so his toxic voice for violence is gone forever.
I've done the usual table showing who's up and who's down, though I stopped at 60, because what a) nobody knows who those people are; and b) no-one wins 50% of the vote anyway.
2020 Rank Name 2017 Rank Difference
1 Jacinda Ardern 1 --
2 Kelvin Davis 2 --
3 Grant Robertson 4 +1
4 Phil Twyford 5 +1
5 Megan Woods 6 +1
6 Chris Hipkins 7 +1
7 Andrew Little 3 -4
8 Carmel Sepuloni 8 --
9 David Parker 9 +1
10 Nanaia Mahuta -- --
11 Trevor Mallard 33 +22
12 Stuart Nash 11 -1
13 Iain Lees-Galloway 14 -1
14 Jenny Salesa 19 +5
15 Damien O'Connor 18 +3
16 Kris Faafoi 20 +4
17 David Clark 9 -8
18 Ayesha Verrall -- --
19 Peeni Henare -- --
20 Willie Jackson 22 +2
21 Aupito William Sio 16 -5
22 Poto Williams 25 +3
23 Vanushi Walters -- --
24 Michael Wood 27 +3
25 Adrian Rurawhe -- --
26 Raymond Huo 13 -13
27 Kiri Allan 21 -6
28 Kieran McAnulty 38 +10
29 Louisa Wall 26 -3
30 Meka Whaitiri -- --
31 Rino Tirikatene -- --
32 Camilla Belich -- --
33 Priyanca Radhakrishnan 12 -21
34 Jan Tinetti 15 -19
35 Deborah Russell 30 -5
36 Marja Lubeck 32 -4
37 Angie Warren-Clark 39 +2
38 Willow-Jean Prime 17 -21
39 Tamati Coffey 35 -4
40 Naisi Chen 50 +10
41 Jo Luxton 29 -12
42 Jamie Strange 36 -6
43 Liz Craig 31 -12
44 Ibrahim Omer -- --
45 Duncan Webb 43 -2
46 Anahila Kanongata'a-Suisuiki 37 -9
47 Ginny Andersen 28 -19
48 Rachel Brooking -- --
49 Paul Eagle 34 -15
50 Helen White 40 -10
51 Barbara Edmonds -- --
52 Angela Roberts -- --
53 Shanan Halbert 51 -2
54 Neru Leavasa -- --
55 Tracey McLellan -- --
56 Lemauga Lydia Sosene 44 -12
57 Steph Lewis -- --
58 Dan Rosewarne -- --
59 Rachel Boyack 48 -11
60 Arena Williams -- --
Labels: 2020 Election, Labour, MMP
Even NZ First gets it
NZ First is a party normally associated with racist "law and order" dogwhistling. But even they think our police are over-armed:
New Zealand First is calling for an independent review into the arming of frontline police officers with military-grade assault rifles.
Semi-automatics are routinely stored in a lock box in patrol cars.
The party says scrapping the Police Armed Response Teams was the right move, but examining police use of guns shouldn't stop there.
Now is the time to also rethink the firearms that officers have access to, their training to use them, and when they used, MP Ron Mark says
"In light of global events, and the fact that we have just confiscated all the semi-automatics across New Zealand, let's have a rethink. We strongly believe police are inappropriately armed against a reduced threat of a disarmed public."
On the one hand, this is welcome. On the other, if Mark really thinks that, then the answer isn't an inquiry, but legislation to restrict ordinary use of firearms to designated specialist units and place safeguards and oversight around their use. Because the level of force police are allowed to use and how is absolutely a political question, and one our politicians can and should be held responsible for.
Labels: Disarmament, NZ First, Police, Ron Mark
Gone by lunchtime
Earlier in the week, inspired by events in the UK and USA, New Zealand finally started talking about our toxic legacy of racist, colonial monuments. And that conversation has already paid off, with Hamilton City Council removing a statue of Captain John Fane Charles Hamilton. They were told it would be taken down by local iwi this weekend, and so it is literally gone by lunchtime.
Who was Hamilton? A murderer and a thief. He commanded a British ship during the Second Opium War - when Britain used military force to keep China addicted to drugs - and then died at the Battle of Gate Pā in Tauranga, when the colonial government tried to expand its campaign of land theft from the Waikato to the Bay of Plenty. Why would any modern New Zealander want to celebrate and glorify someone like that?
Now it has been taken down, the statue can be put in a museum, with an appropriate historical context. Alternatively, it could just be returned to the local racists who funded it.
Labels: History, Racism
We should not celebrate racists, thieves and slavers
Following the removal of a number of racist monuments overseas by Black Lives matter protesters, the Māori Party has called for a review of similar monuments and symbols in New Zealand:
Māori Party Co-leader and Te Tai Hauāuru candidate Debbie Ngarewa-Packer is calling on the Government to establish an inquiry that is focused on identifying and getting rid of racist monuments, statues and names from our colonial era.
"We still honour some of the most racist and oppressive figures from our colonial history with monuments, statues and place names in towns and cities across the country.
"I am calling on Prime Minister Jacinda Ardern and her government to work alongside hapū and iwi Māori and other communities of colour in Aotearoa to undertake a comprehensive inquiry into colonial monuments and statues, place names, and street names.
As for why, all over the country Māori have to walk down streets named after people who literally murdered their ancestors. We have place names and public monuments celebrating land thieves, slave-owners, and torturers. The town I live in, Palmerston North, is named after the fucker who started the opium wars and who evicted and starved his tenants to death during the Great Hunger.
Place names and monuments are not about history. They are about what we glorify and celebrate. And who we (or rather, our ancestors) have chosen to glorify and celebrate as a society is sick and wrong. We can, we should, change that. As for how, given the scale of the problem, a national inquiry seems more than justified. The alternative is to leave it to people to chip away locally. And in the case of monuments, that will probably be literal.
Labels: History, Maori, Racism
Armed Response Teams were racist
How racist were the Police's "Armed Response Teams"? Half of all the people they arrested or used force against were Māori:
Māori made up more than 50 percent of arrests and uses of force by members of the now-scrapped Armed Response Teams (ARTs), police data shows.
During the six-month trial, the armed teams went to more than 200 mental health and suicide threat incidents, more than a dozen cannabis offences, and even checked how 17 businesses were adhering to pandemic rules.
Of the 1651 offence or weapon incidents where a person's ethnicity was known, 51 percent were Māori, 36 percent Pākehā, 10 percent Pacific, less than 3 percent Asian and less than 1 percent Middle Eastern, Latin American or African.
There were 49 uses of force towards people; of which 53 percent were used against Māori people, 41 percent against Pākehā, and 4 percent against Pasifika.
16.5% of the population are Māori, so that's a pretty big disparity.
But its not just racism which is the problem, but also the sheer inappropriateness of using armed police for the jobs they were doing. Not just the traffic stops which were the majority of their duties - in addition to the mental health incidents and suicide threats mentioned above, they also turned up, with guns, to 746 family harm incidents. And then there's this:
One suicidal man was pepper sprayed. Others who were suicidal or in mental distress had Tasers presented towards them.
I'm appalled that people with health problems are threatened with chemical or electric torture. It smacks of C19th barbarism.
It is not appropriate for police to be responding to these calls with guns, tasers, pepper-spray and rubber bullets. Instead, we need specialist mental health teams to deal with them. As for how to pay for them, given that police are already doing it, we should take the money straight out of their budget. Defund the police, and fund something better.
Labels: Disarmament, Police, Racism
Why we need cameras on fishing boats
Via Newsroom, the argument for why we need cameras on fishing boats in one graph:
It is pretty obvious what is going on here: a pervasively criminal industry under-reporting to avoid having to clean up its act. Rather than coddle them and drag their feet with endless delays, the government should enforce the law and police them properly.
Labels: Crime, Fishing
Our tone-deaf police
Yesterday, the Police announced that Armed Response Teams would not be part of policing in New Zealand. This morning, Police Commissioner Andrew Coster was on RNZ talking about the "trial". And the first thing he talked about was giving police more weapons:
Coster: The example that we've seen through the course of the ART trial was sponge rounds. They're a less-lethal option that allow incapacitation of someone from a safe distance, and that can avoid the need to actually use a firearm.
Espiner: So like a rubber bullet type of thing?
Coster: Its a sponge round. If you can imagine a 40mm sponge projectile that can wind people or incapacitate them sufficiently to get close and arrest them without needing to use a firearm.
What he's talking about is this. The NZ police bought 850 of them back in 2013 for the armed offenders squad. They're used primarily by oppressive regimes - Israel, Hong Kong, the US - for "crowd control": to suppress protests. They're being used in the US right this minute against Black Lives Matter protesters and the journalists covering them. As for what they do, while Coster talks of "wind[ing] and incapacitat[ing]", in reality these weapons break bones, blind, and kill. A quick google will show you distressing pictures of what they do to people and turn up news reports of kids being killed by them. For the police to be talking about deploying them now, when this is on all our feeds, shows how completely tone-deaf they are on the issue of militarisation. We don't want them to have these weapons. We don't want that sort of policing - the American, beat people and shoot them style - in our country. We can see, right now, what it means and where it goes. And the police just are not getting the message.
The police's enthusiasm for more and more weapons with which to hurt us and their refusal to listen to the public shows that they need to be bought under control. Parliament needs to legislate immediately to do so.
Labels: Disarmament, Police
Undemocratic and unconstitutional
Years ago we learned that the SIS spied on Green MP Keith Locke, both before and after he was elected to Parliament. Now it turns out that he wasn't alone:
A senior Labour politician was spied on by the SIS while an MP in the 1980s and 1990s - even though for part of that time he had an oversight role of the intelligence agency as chair of the justice select committee.
Richard Northey, a Labour MP between 1984 and 1990 and again between 1993 and 1996, said it was "outrageous" that the SIS had kept a file on him while he was a sitting MP with a democratic mandate.
After a request from RNZ, the SIS declassified documents held by Archives New Zealand, including letters sent and received by Northey while he was chair of the justice select committee overseeing the SIS Amendment Bill in 1989.
The documents included correspondence between Northey and then-prime minister David Lange, then-deputy prime minister Geoffrey Palmer and also advice Northey received from the commissioner of security warrants.
[Note: The position of Commissioner of Security Warrants was not established until 1999. So I'm wondering if this was correspondence with the Commissioner of Security Appeals - the forerunner of the modern Inspector-General of Intelligence and Security. If so, it would be even more troubling for the SIS to have their hands on that advice]
The "justification" for the surveillance was Northey's campaigning for racial equality and nuclear disarmament - something which seems contrary to the political neutrality clause the SIS was subject to at the time. But as noted above, he wasn't just an MP, but the chair of the committee responsible for overseeing them, and for considering SIS-related legislation. Browsing the Hansard (pages 94 - 105) on the New Zealand Security Intelligence Service Amendment Bill 1987 (which would have moved the power to issue warrants from the Prime Minister to the Chief Justice; it was sent to select committee and apparently died there), it appears that this was the first time that the SIS had been subjected to any form of Parliamentary oversight. It is bad enough that they were spying on people for their political views, who pursued change peacefully and democraticly. That is grossly undemocratic. But spying on one of the people who was supposed to be keeping an eye on them is downright unconstitutional, a direct attack on our system of democracy. And it really does make you wonder who they thought they were working for back then.
Officially, that sort of thing doesn't happen any more. The SIS officially recognises that it is not appropriate to spy on sitting MPs except in exceptional circumstances, and has a formal Memorandum of Understanding with the Speaker on the subject. Whether that actually means anything in practice is something we probably won't know for another 30 years. But give the change in policy, the least the SIS can do is apologise, to Northey and the people of New Zealand, for their past misdeeds. Until they do, we're perfectly entitled to believe that they haven't really changed a bit.
Labels: Democracy, Parliament, SIS, SMERSH
A victory for public safety
The Police have today announced that "Armed Response Teams" - gangs of heavily-armed police cruising Māori-Pacifica neighbourhoods in juiced-up gun-trucks looking for people to shoot - will not be part of policing in New Zealand. Good riddance. Their "trial" was a bad joke, and really just an excuse to intimidate the public with gun-toting cops (who mostly did traffic stops). But the change in Commissioner has led to a change in policy, with Andrew Coster recognising that policing in New Zealand happens by consent, and waving guns around erodes that consent.
But while this is good news, the fact that this "trial" was even run was appalling, as is the fact that the politicians meant to be supervising the police let them pass it off as an "operational matter". There needs to be accountability for that. And the police still have pistols and assault rifles in every car, with few limits on their use. That needs to change. As for how, Parliament can and should legislate to restrict the ordinary use of firearms to designated specialist units, and implement safeguards and oversight of their use by other officers, including Ministerial signoff and Parliamentary veto for any temporary general arming of the sort we had in the aftermath (rather than immediate response to) the Christchurch Mosque attack (thanks to Graeme Edgeler for the details beyond the first bit of this). And they should implement better oversight of "non-lethal" weapons such as tasers and pepper-spray, to require regular pro-active publication of information on how they are used, and mandatory use-of-force reviews of officers who use them (or any other form of force) too often, with an eye to sacking or desking officers who cannot be trusted. Because as we're seeing from the regular IPCA reports, police are increasingly abusing force. As they are incapable of holding one another to account, Parliament needs to.
The SIS's illegal burglary
RNZ has a story today about the SIS burgling the Czech embassy in 1986 to try and steal a code-book, in violation of the Vienna convention. At the Prime Minister's press conference this afternoon, she was asked if they had broken the law, and wibbled about it. But I think its a good question. Answering it requires delving into the history of SIS intelligence warrant provisions.
When the SIS was put on a legal footing in 1969, its governing legislation included no warrant provisions whatsoever. If they were bugging embassies and conducting black-bag jobs against people Muldoon didn't like, it was all illegal, done on an "above the law" basis. In 1977, probably in response to some scandal, this was finally bought under control by the New Zealand Security Intelligence Service Amendment Act 1977, which gave them a formal warrant provision. This let them, with Ministerial approval, intercept or seize "any communication not otherwise lawfully obtainable by the person making the interception or seizure". This situation prevailed until 1999, when (in response to the Court of Appeal ruling that the SIS had exceeded its powers in burgling the home of GATT Watchdog activist Aziz Choudry) Parliament legislated to allow the SIS to seize documents or things under warrant, to give them formal powers of entry to execute warrants, and to say that the latter - but not the former - applied retrospectively (because, as we all know, when the spies are caught breaking the law, their actions are declared legal, rather than them being held accountable).
What does this mean for the embassy burglary? Well, the break-in - the illegal entry - was illegal at the time, but retrospectively legalised in 1999. But if they actually took any document or thing (rather than a "communication", which from context is something transitory like a phone call, or else is a letter; a code book probably doesn't count), then they committed a crime, and that crime was not retrospectively legalised in 1999. And the people who took it or conspired to do so - and what is a planned operation, if not a conspiracy in law - can be prosecuted for theft (not burglary, because the trespass was retrospectively legal), and jailed for up to 7 years. And, if the rule of law means anything, they should be.
Meanwhile, Andrew Little is refusing to say whether he authorises embassy burglaries. He also says that the spies operate within the law, including international law which has been incorporated into domestic law. If the latter is true, then he needs to look at s5 of the Diplomatic Privileges and Immunities Act 1968, which incorporates Articles 1, 22 to 24, and 27 to 40 of the Vienna Convention into NZ law. This includes provisions such as "the premises of the mission shall be inviolable" and "the archives and documents of the mission shall be inviolable", and "the official correspondence of the mission shall be inviolable". Then he needs to look at s54 of the Intelligence and Security Act 2017 (and its long series of predecessors), which specifically establishes a lower standard of scrutiny for intelligence warrants which target only foreign citizens, and ask why we should believe him when the law itself calls him a liar. If he wants to be taken seriously, he should legislate a specific exemption, just so there is no confusion. If he doesn't want to do that, then he should be more careful about which lies he lets the SIS put in his mouth.
Labels: Crime, SIS, SMERSH
New Zealand now has zero active cases of Covid-19. None. After 75 days, and 22 deaths, the epidemic is over... for now. But its still wildly out of control in the rest of the world. Which is why those people demanding that we re-open the border to countries which still have the disease - invariably so they or their friends can make money - need to be told to fuck off. We're currently in a little bubble of normality. But it takes only one case to get through undetected, one person to break quarantine, and we're back where we started. Except next time, it will be harder, because people are tired of the restrictions and will be less willing to comply. Its obviously better to not end up in that situation, which means keeping the border closed, and only opening it to countries which have eradicated the virus and which have similarly strong border restrictions to keep it out.
Meanwhile, I'm wondering: will the epidemic notice - an important policy instrument which gives the government all sorts of special powers - be revoked, or allowed allowed to expire on its current date of June 24? There are big policy implications from this - the entire order system under the COVID-19 Public Health Response Act depends on it, as do various special powers under the Health Act, plus things like visa extensions for people trapped here by the border closure - so it might take a little time to work through. But its a question the government should be prepared to front up and answer.
Labels: Disasters, Health
On political capital
Over the weekend, Green co-leader James Shaw gave an interview where he expressed disappointment that Labour hadn't spent more political capital than they had. Stuff's Andrea Vance has followed that up with an opinion piece asking "what good is popularity if you fail to do anything with it?" arguing that Labour basicly wasted its honeymoon. I'd go stronger than that. Because if you can remember back to before (gestures vaguely) all this happened, Labour's re-election was looking uncertain. Unless something big changed, they were on track to be a one-term government. And they were headed in that direction precisely because they had repeatedly refused to spend their political capital, repeatedly refused to provide any real policy payoff that would remind their voters that they were worth voting for. They had become a government of disappointment, who had left their supports disillusioned.
We've since been reminded that government matters, that who is in charge matters, and Ardern's stellar response in a crisis has left her and her government one of the most popular in New Zealand history. It seems almost certain they will be re-elected in September, probably in a landslide. Which invites the question: will Ardern actually do anything with that popularity this time? Or will she waste it - again - and simply collect a fat salary for presiding over an unjust and unequal status quo? And if its going to be the latter, why should anyone vote for her?
Labels: 2020 Election, Labour
Its not an experiment if you don't collect data
When the New Zealand Police started using Armed Response Teams - gangs of heavily-armed police cruising Māori-Pacifica neighbourhoods in juiced-up gun-trucks - they told us it was a "trial". But it turns out that those armed police - who were mainly doing traffic stops and bail checks, rather than the "high-risk incidents" we were told they'd be handling - weren't even bothering to collect data on their callouts:
Police in the Armed Response Teams failed to record their callouts properly on almost every occasion during the trial's first two months.
The six-month experiment ended in April. The trial involved a group of officers in three regions - Counties Manukau, Waikato and Canterbury - equipped with guns on their hips at all times.
Officers were expected to record and submit data on every single call-out. In the first two months, data from five out of every six callouts was missing.
Police did not provide the total rate of responses for the remainder of the trial when asked by RNZ. Instead a spokesperson said the evaluation of the trial would "only be one of the factors taken into consideration as part of our decision making".
Which is just heaping bullshit on bullshit. It's not an experiment if you don't collect the data. But we already knew that this "trial" was a lie from start to finish. Its primary purpose was to accustom the public to a US-style, permanently-armed, militarised police-force. And what's happening in the US at the moment ought to be a warning against ever going down that path.
Meanwhile, the Police Minister is still pretending that the question of whether police are armed is an operational decision for police, rather than a political one for politicians. Bullshit. To point out the obvious: politicians pass the laws which set the parameters for police use of force. They make the financial decisions around what sorts of weapons police can buy and how many. And they make the decisions about how they can deploy them. The decision to "trial" Armed Response Teams was signed off by the Minister, and he announced it on the beehive website - hardly an "operational" decision then. Which means that if we want to keep having an unarmed, rather than a US-style militarised police force, then we need to vote for politicians who will insist on it and refuse to fund militarisation. Rather than chickenshits who will sign off on whatever the police say they want. Sadly, at the moment, we have the latter. But we will have a chance to fix that in September.
The pandemic law, deadlines, and the election
One of the safeguards built into the COVID-19 Public Health Response Act 2020 was automatic expiry. The law must be renewed by Parliament every 90 days, or it is automatically repealed. Orders made under the law are automatically revoked unless confirmed by Parliament within 10 sitting days, or 60 calendar days, whichever is shorter. So what does this look like in practice?
The first order - the Alert level 2 Order - made under the law was issued on 14 May. If not confirmed, it will lapse automatically on 13 July. The 10 sitting day limit is harder to calculate, because there has been urgency, but if there is no further urgency it will need to be confirmed by 24 June. Of course, by that stage the government will have made a decision on Level 1, and it is likely that it will have been revoked anyway and that a whole new Order will be in place. If that Order come into force by 15 June, then it will need to be confirmed by 21 July, assuming no urgency. As for the law itself, its first 90-day period expires on August 11, which is after Parliament is expected to be dissolved for the election. So obviously, they'll do it beforehand - say in the last week, by August 5. Which would then set a deadline of confirmation by November 3. Which could lead to a tight timeline, since Parliament usually doesn't sit for 4-6 weeks after an election (the previous Parliament first sat on October 20, 2014, the current one on November 7, 2017, both following mid-September elections), and then wastes its first week on ceremony. There's a serious risk that if government formation is at the long end of the range (AKA if Winston is involved), then the law will expire before it can be renewed.
And then there's another problem: if the law is confirmed as quickly as possible after the election, then it runs the risk of expiring over the holiday period, as the House traditionally doesn't return until after Waitangi Day.
Parliament can solve this problem - the House can, when confirming the law, set an alternative period (say, 120 days) for renewal under s3(2)(b). But it looks like they're going to need to do it at least once, and maybe twice. The alternative is to confirm again just before the House rises for the holidays (which would give them until March. And I think that would be far more in the spirit of keeping these powers under regular and strong scrutiny than giving themselves extra time for a second time.
Labels: 2020 Election, Disasters, Parliament
Still getting what they paid for
Last year, in the face of public pressure to better regulate New Zealand's pervasively criminal fishing industry, the government finally anounced a bullshit trial scheme for cameras on boats. The trial would apply only to a limited number of boats fishing in Māui's dolphin habitat. But all fishing boats would eventually be required to have cameras, from 1 July this year.
The government has just quietly delayed that, until 1 October 2021 - a 15 month extension. There's no press release, and no explanation, so they were clearly hoping no-one would notice. And in the absence of any proper explanation, the only one we have is the history of large donations from the fishing industry to a key government Minister. I guess his donors are still getting what they paid for.
Labels: Corruption, Fishing
Time to double sick leave
The CTU is planning to mount a campaign to double sick leave, from 5 to 10 days a year. Good. We've all just had an extremely strong reminder of the need to stay at home when sick, and of the effectiveness of doing so (the lockdown, social distancing, and responsible attitude to disease ATM has absolutely crushed seasonal flu as well as protecting us from the pandemic). But the current sick leave entitlement of a mere 5 days does not enable this. Nor does the intrusive provision allowing employers to demand proof of illness. And then there's the problems caused by "lean staffing" - AKA employer cheapness - which lets employers guilt workers into coming into work when sick, because if they don't their co-workers will suffer (because the employer hasn't ensured there are sufficient staff to cope with people being away). The result is that disease spreads. We can all think of a time when someone came into work when sick, because they were out of leave, or felt they had to, or just stupid, and then everybody got it. And that needs to stop.
But in order for that to happen, we need to actually enable people to do the right thing. And that means ensuring they have plenty of no-questions-asked sick leave, and that they can actually take it. The first is the easy bit: it requires changing one number in one section of law (and two more numbers if you want to increase carry-over to match the increased entitlement). Ensuring that people can actually use their entitlement will require a shift in management styles. As for how to do that, public health is already recognised as a workplace health and safety issue, included in an employer's obligations under the Health and Safety at Work Act 2015. Policies, practices, or a workplace culture which endanger public health by encouraging people to infect others violate an employer's primary duty of care under that Act. And that's punishable by a $500,000 fine - or $1.5 million if it is deemed to expose people to a risk of serious illness (like, say, COVID-19). So my solution to changing workplace culture is for Worksafe to actually enforce the law, and start prosecuting and fining employers over this. On the other end, workers should refuse to work when sick if they are able to, and complain to their unions (join a union!), or complain directly to Worksafe if their employer's practices are unsafe. Hopefully a few investigations and prosecutions will sharpen employer's minds, and force them to eliminate their present unsafe practices.
You can sign the CTU's petition in support of their campaign here.
Labels: Health, Left, Worker's Rights
This is not rehabilitation
When miners pillage conservation land, they are typically required to "rehabilitate" it afterwards. Its not much compared to the damage they do, but its something. But apparently even that is now too much to expect. NZG Limited, a company owned by Oravida directors James Blackwell, Julia Jiyan Xu, Stone Shi, and David Wong-Tung, has been mining gold in the Mikonui Valley near Hokitika. As a condition of access to conservation land, they were required to restore it afterwards, including recontouring it, replacing topsoil, and replanting it in native forest. But none of that has been done. Instead, all of these requirements have been quietly waived by DoC, and the area has been left looking like a moonscape:
A mine on conservation land on the West Coast conditionally signed off as rehabilitated has been described as a desolate wasteland.
The remaining condition for sign-off is one year of weeding.
There's no topsoil, a gaping hole remains, and replanting, written as a concession condition, wasn't done. The Department of Conservation (DoC), which signed off the rehabilitation, said conditions were changed after a discussion with the mining company. If rehabilitation is not completed, the company can lose the bond it paid when it gained the concession.
So rather than forcing them to meet their commitments (or take their money to pay for any failure), the conditions were weakened to allow them to walk away with their pillage free and clear. Its absolutely disgusting, and I am boggled as to why DoC or the Minister would allow it. But the article also notes that "no mining companies have left sites that have not been rehabilitated to the satisfaction of the Department of Conservation on public conservation land in the West Coast area". If that's the case, maybe someone should start looking at those sites to check that its not a similar story there.
Meanwhile, if this is an example of how a Green voice in government leads to better conservation outcomes, we're better off with them in opposition.
Labels: Conservation, Environment, Mining
America burns
Like everyone else, I've spent too much of the weekend watching the protests in America, and the increasingly brutal response to them. The overwhelming impression is of a nationwide police riot, as people speaking out against a murder and demanding change are beaten, gassed and shot by racist, militarised thugs outraged that people would question their unaccountability. And today it has got worse, with President Trump demanding that state governors "dominate" protesters, and now threatening to send in the military.
Let's be clear: using the military against protesters is what tyrants do. Its what China did in Tiananmen Square. Its what Uzbekistan did in Andijan. Its what Britain did at Peterloo. And when the President of the United States behaves like a tyrant, he deserves to be treated like one.
Labels: Democracy, Murder, Protests, Tyranny, USA
Wales makes it 16
Wales has lowered the voting age to 16:
16 and 17 year olds can now officially vote in Wales for Senedd elections.
Votes at 16 & 17 come into force on Monday, as part of the Senedd and Elections (Wales) Act 2020.
Next year's Senedd elections will be the first in which 16 and 17 year olds and legally resident foreign nationals are allowed to vote in Wales, in a major expansion of the franchise. Around 65,000 16/17s are expected to benefit.
The move is a victory for young people, following campaigning from ERS Cymru and a coalition of youth and civil society campaigners.
Congratulations to the Welsh for recognising that teenagers are equal citizens! But isn't it time New Zealand followed suit?
Labels: Democracy, Voting Age, Wales
Climate Change: Banking failure
The government announced more changes to the ETS today, including to the emissions budget for 2021 - 2025. The overall budget for that period will stay at 354 million tons of CO2-equivalent. But the ETS component of that - stockpile reduction, free allocation, and credits to be auctioned - budget will increase to 160 million tons:
[Graph by the Ministry of the Environment from here]
This is an increase of 10 million tons from what was proposed in the consultation document. Comparing the two, 10 million tons has been moved from "emissions outside the ETS" to auction volume. Which would be fine if it was because farmers are suddenly going to have to pay for those emissions. But they're not - agriculture isn't coming into the scheme until 2025. So what's going on? Well, if you go and look at the government's latest net position report, agricultural emissions over 2021 - 25 are expected to decrease due to land-use changes. The scale of that decrease? Just over 10 million tons. But rather than banking that success - or just going "agriculture will do what its going to do, and its outside the ETS so it doesn't matter for this calculation" - they've instead given that projected saving straight to industrial polluters, who were not expected to increase their emissions to compensate (and were in fact projected to have a net reduction as well, but now can be expected not to because they will have the credits to pollute). They've taken success, and turned it into failure. Meanwhile, if agricultural trends reverse, and their projected emissions increase again, we've already locked in higher emissions through the allocation budget, and we will blow our first carbon budget, and make it that much harder to meet our 2050 "net-zero-but-not-really" target.
Heckuva job they're doing there. Aren't you glad the Green voice in government is ratcheting down emissions?
Labels: Agricultural Emissions, Climate Change, Climate Change Policy, Emissions Trading | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,128 |
← Film Study: Adam Wingard's "Death Note" remake is NOW to become a Netflix Movie..
thanks! :] I'd been meaning to get to talking about Alice in Wonderland for awhile but couldn't find something "unique" enough [til now!] THANKS FOR THE COMMENT!
It is certainly a unique article! Genuinely refreshing to read something that focuses on an area that people know less about. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,183 |
CALA to follow legislative session to protect small business owners
By S. Laney Griffo | Apr 10, 2017
CALA has found several bills to support and oppose. | By Griffin5 at English Wikipedia, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=10086152
SACRAMENTO — With the California legislative session in full swing, California Citizens Against Lawsuit Abuse has found a series of bills it is supporting and opposing in order to prevent courts from being clogged and to protect small business owners.
"We have some good bills we'll be supporting," Maryann Marino, Southern California regional director, California Citizens Against Lawsuit Abuse (CALA) told Northern California Record.
The group supports Assembly Bill 913, a bill sponsored by Assemblymember Adam Gray, which would prevent frequent filers.
The bill would allow a judge to block people who have filed more than 15 Americans with Disability Act cases.
"This would provide relief to small business owners," Marino said.
California CALA is also in support of Assembly Bills 1429 and 1430, both sponsored by Assemblymember Vince Fong.
"These bills would curb wage and hour lawsuits that are exploding in California," Marino said.
Aggrieved employees must file a complaint with the labor board first rather than immediately filling complaints with the courts.
The group has offered support of AB 281, sponsored by Assemblymember Rudy Salas.
"This is a bill that would amend the private attorney general act and tighten up what employees can sue employers for," Julie Griffiths, Northern and Central California regional director for California CALA.
When there is an issue, such as an incorrect wage statement, employers have a corrective action period to address the issue before a lawsuit can we filed.
The final bill the group will be supporting is AB 1583, sponsored by Assemblymember Ed Chau.
This bill would require a plaintiff receive a certification of merit on a complaint before being allowed to file a Proposition 65, warning labels suit.
There are a couple of bills the group will be opposing this legislative session, including AB 5.
This bill would mandate that employers offer their part-time employees more hours before a hiring a new employee.
Griffith said the bill is too ambiguous and could "open a can of worms" if the language is not made clearer.
Finally, the group is opposed to AB 814, sponsored by Assemblymember Richard Bloom. This bill would allow city attorneys to go on what Griffith refers to as "fishing expeditions," by allowing the attorneys to subpoena documents before a case is filled.
"This could be really dangerous," Griffith said.
On April 4, CALA joined with Civil Justice Association of California to host a rally in front of the Capitol building to show support and opposition to these bills. They will continue their support and opposition as the session moves forward.
Griffith said that have focused on these "to keep frivolous lawsuits from the courts," which would hurt small business owners and to prevent clogging in the courts.
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California Citizens Against Lawsuit Abuse
California State Legislature
Civil Justice Association of California | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,727 |
By Wildy Zumwalt
CRAFTMAN'S
The craftsman's corner highlights artisans, craftspeople and inventors who make the world of the saxophone better.
True craftsmanship seldom comes from a singular focus but rather from one's overall attitude towards life. There is no trade school that can teach passion, and technical manuals typically omit chapters on developing one's intuition. Passion is at the heart of everything Kim Bock does. It was what drew the Dane to study saxophone in the United States, and it was his intuition that brought him to open one of the most recognized saxophone shops on the East Coast. But it is his dedication to the saxophone that prompted him to craft some of the most sought-after saxophone necks on the market. So it is fitting that the first person featured in our new segment, The Craftsman's Corner, be saxophonist, repairman, creator, Kim Bock.
Kim Bock began performing professionally in the late 1980s in his native Denmark and moved to Boston in 1994 to study at the famed Berklee School of Music. He studied with some of the finest names in jazz, including Jerry Bergonzi, George Garzone, Billy Pierce, and Rich Perry. He would later obtain his master's degree from the University of South Florida and move to New York in 2000. Along the way, he would perform with some of the most recognized names in music, including a stint as the lead tenor player with Maynard Ferguson. Bock was very active as a player, most notably with his organ trio FLOW (with Soren Moller, organ and Peter Retzlaff, drums) until 2013 when the success of his shop and the coming neck line demanded his full attention. It is his background as a player that would serve as a base of trust when he opened his saxophone shop, KB Sax, in 2010.
KB Sax Shop
Bock has always been interested in mechanics. He developed a curiosity with repair at an early age and over time has done in-depth repair study, most notably with Randy Jones at Tenor Madness in Waterloo, Iowa. Beyond his work with the saxophone, he is an avid motorcyclist who specializes in riding and building classic bikes. It is his perfectionism as a technician that has allowed him to develop a world-renowned clientele with such saxophonists as Ravi Coltrane, Seamus Blake, Ben Wendel, Donny McCaslin, John Ellis, Steve Coleman, Wayne Shorter, and Joe Lovano, all who entrust their prized instruments with the Dane. As the shop grew, it moved beyond repair to become one of the centers for the sale of vintage saxophones and equipment.
KB Saxophone Necks
It was the depth of his clientele that served as the testing ground for the creation of his KB Saxophone Necks. Started in 2014, it would take two years of research, development, and extensive testing before they would be released to the public in 2016. It was through his extensive work with vintage saxophones served as both the base and impetus for his creations. Today he offers two models of tenor saxophone necks in a variety of metals to help the saxophonist craft their tonal concept. His latest model, the hand hammered bronze Vanguard, is already being met with tremendous enthusiasm.
John Lennon wrote, "Life is what happens to you while you're busy making other plans." Such is the case with Kim Bock. In 2014 he welcomed his daughter into the world, and with this addition to his family comes an awareness of priority. Nevertheless, Bock is not resting on his accomplishments. 2019 will bring the development of a new line of alto saxophone necks and with it a whole new level of bespoke offerings for the saxophonist.
Read the KB Neck Review | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,700 |
Hu L, Liu P, Ma L, Xin X, Chen J, Xie Q, Luo F, Xie X, Huang J.
Individuals with a 47,XXY karyotype usually present with a male phenotype due to the additional Y chromosome. In this paper, we describe a 47,XXY female who was pregnant with a fetus of the same karyotype based on chromosome analysis of amniotic fluid cells. Further analysis of her Y chromosome indicated that the additional Y chromosome contains no SRY gene on the short arm but carries the azoospermia factor region on the long arm, including azoospermia factor a, b and c (AZFa, AZFb, AZFc). This region may influence her female phenotype. Fertile females with a 47,XXY karyotype and loss of SRY are extremely rare. This paper is the first report of a 47,XXY pregnant woman with a normal phenotype and may enrich our knowledge on 47,XXY individuals. | {
"redpajama_set_name": "RedPajamaC4"
} | 6,226 |
There is now a large body of evidence to support the promotion of the natural environment for physical activity and mental wellbeing and the provision of programmes of activity in natural surroundings as effective in helping to increase people's access.
champs has partnered with Mersey Forest to support the commissioning and delivery of services that improve people's health and wellbeing by increasing their access and use of local green space. A group of providers will work together to develop a consortia business model for commissioning. This will include a clear offer of the most effective interventions and their impact on people's health and wellbeing and reducing inequalities. Consultants from the Confederation of Social Entrepreneurs and the Hope Street Centre will lead this work with Mersey Forest and the Investment Forum of partners. NHS Liverpool, Western Cheshire and Wirral are the lead ChaMPs members supporting this work.
More information on green infrastructure work can be found here. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,891 |
{"url":"http:\/\/pol346.com\/precepts\/precept02_handout.html","text":"# Handout for Precept 2\n\nCalculating pooled sample standard deviations.\n\nTeam pol346 pol346.com (Princeton University, Department of Politics)princeton.edu\/politics\n2020-03-06\n\n## Chapter 2, Problem 21.\n\nBumpus Natural Selection Data. In 1899, biologist Hermon Bumpus presented as evidence of natural selection a comparison of numerical characteristics of moribund house sparrows that were collected after an uncommonly severe winter storm and which had either perished or survived as a result of their injuries. Display 2.15 [see data below] shows the length of the humerus (arm bone) in inches for 59 of these sparrows, grouped according to whether they survived or perished. Analyze these data to summarize the evidence that the distribution of humerus lengths differs in the two populations. Write a brief paragraph of statistical conclusion, using the ones in Section 2.1 as a guide, including a graphical display, a conclusion about the degree of evidence of a difference, and a conclusion about the size of the difference in distributions.\n\n\nlibrary(Sleuth3)\nsuppressMessages(library(dplyr))\nsuppressMessages(library(janitor))\nbumpus <- ex0221 %>% clean_names()\nbumpus %>% head(2) \n\nhumerus status\n1 0.687 Survived\n2 0.703 Survived\n\n## Chapter 2, Problem 13.\n\nFish Oil and Blood Pressure. Researchers used 7 red and 7 black playing cards to randomly assign 14 volunteer males with high blood pressure to one of two diets for four weeks: a fish oil diet and a standard oil diet. The reductions in diastolic blood pressure are shown in Display 1.14 [not shown, see data below]. (Based on a study by H. R. Knapp and G. A. FitzGerald, \u201cThe Antihypertensive Effects of Fish Oil,\u201d 320 (1989): 1037-43.) Do the following steps to compare the treatments.\n\n\nstudy <- ex0112 %>% clean_names()\nstudy %>% head(2)\n\nbp diet\n1 8 FishOil\n2 12 FishOil\n1. Compute the averages and the sample standard deviations for each group separately.\n\n2. Compute the pooled estimate of standard deviation using the formula in Section 2.3.2. [See below]\n\nFormula: Pooled estimate of standard deviation for two independent samples\n\n\\begin{aligned} \\textrm{We assume:} & \\\\ \\\\ \\sigma_1 & = \\sigma_2 = \\sigma \\\\ s_1, s_2 & = \\textrm{ independent estimates of }\\sigma\\\\\\\\ \\textrm{Therefore:} \\\\ \\\\ s_p & = \\sqrt{\\frac{(n_1 -1)s^2_1 + (n_2-1)s^2_2}{(n_1 + n_2 -2)} }\\\\ \\textrm{d.f.} & = (n_1 + n_2 - 2) \\end{aligned}\n\n1. Compute SE( $$\\bar{Y_2} - \\bar{Y_1}$$ ) using the formula in Section 2.3.2. [See below]\n\nFormula: Standard error for the difference\n\n$\\textrm{SE}(\\bar{Y}_2 - \\bar{Y}_1) = s_p \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2} }$ d.\u00a0What are the degrees of freedom associated with the pooled estimate of standard deviation? What is the 97.5th percentile of the $$t$$-distribution with this many degrees of freedom?\n\n1. Construct a 95% confidence interval for $$\\mu_2 - \\mu_1$$ using the formula in Section 2.3.3. [Set $$\\alpha$$ to $$0.05$$] $$100(1- \\alpha)$$ % :$$(\\bar{Y_2} - \\bar{Y}_1) \\pm t_{df}(1-\\alpha\/2)\\textrm{SE}(\\bar{Y_2} - \\bar{Y}_1).$$\n\n2. Compute the $$t$$-statistic for testing equality as shown in Section 2.3.5. [Under null hypothesis, assume hypothesized value for $$(\\mu_2 - \\mu_1) = 0$$.]\n\n\\begin{aligned} \\textit{t-statistic} = \\frac{(\\bar{Y}_2 - \\bar{Y}_1) - [\\textit{Hypothesized value for } (\\mu_2 - \\mu_1)]}{\\textrm{SE}(\\bar{Y}_2 - \\bar{Y}_1)} \\end{aligned}\n\n1. Find the one-sided $$p$$-value (as evidence that the fish oil diet resulted in greater reduction of blood pressure) by comparing the $$t$$-statistic in (f) to the percentiles of the appropriate $$t$$-distribution (by reading the appropriate percentile from a computer program or calculator). [Sample code below.]\n\n# Example: if t-statistic is 2.7 and degrees of freedom are 13\npt(2.7, 13)\n\n[1] 0.990903\n\n# for a t-distribution with 13 degrees of freedom, 99 percent of the area under\n# the curve is to the left of 2.7\n\n# to calculate 1-sided p-value for probability of values equal to or\n# greater than 2.7 (with df = 13)\n1 - pt(2.7, 13) \n\n[1] 0.009096983","date":"2020-07-07 11:55:54","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\": 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\": 0.9524844288825989, \"perplexity\": 3500.680501799494}, \"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-2020-29\/segments\/1593655892516.24\/warc\/CC-MAIN-20200707111607-20200707141607-00056.warc.gz\"}"} | null | null |
// Licensed to the Apache Software Foundation (ASF) under one
// or more contributor license agreements. See the NOTICE file
// distributed with this work for additional information
// regarding copyright ownership. The ASF licenses this file
// to you under the Apache License, Version 2.0 (the
// "License"); you may not use this file except in compliance
// with the License. You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing,
// software distributed under the License is distributed on an
// "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
// KIND, either express or implied. See the License for the
// specific language governing permissions and limitations
// under the License.
using System;
using Org.Apache.REEF.Common.Context;
using Org.Apache.REEF.Driver;
using Org.Apache.REEF.Driver.Context;
using Org.Apache.REEF.Driver.Evaluator;
using Org.Apache.REEF.Tang.Annotations;
using Org.Apache.REEF.Tang.Util;
using Org.Apache.REEF.Tests.Functional.Bridge.Exceptions;
using Org.Apache.REEF.Utilities;
using Org.Apache.REEF.Utilities.Logging;
using Xunit;
namespace Org.Apache.REEF.Tests.Functional.Failure.User
{
/// <summary>
/// This class contains a test that tests the behavior upon throwing an Exception when
/// sending a Context Message from the Evaluator's IContextMessageSource.
/// </summary>
[Collection("FunctionalTests")]
public sealed class SendContextMessageExceptionTest : ReefFunctionalTest
{
private static readonly Logger Logger = Logger.GetLogger(typeof(SendContextMessageExceptionTest));
private static readonly string ContextId = "ContextId";
private static readonly string ExpectedExceptionMessage = "ExpectedExceptionMessage";
private static readonly string ReceivedFailedEvaluator = "ReceivedFailedEvaluator";
/// <summary>
/// This test validates that a failure in the IContextMessageSource results in a FailedEvaluator.
/// </summary>
[Fact]
public void TestSendContextMessageException()
{
string testFolder = DefaultRuntimeFolder + TestId;
TestRun(DriverConfiguration.ConfigurationModule
.Set(DriverConfiguration.OnDriverStarted, GenericType<TestSendContextMessageExceptionDriver>.Class)
.Set(DriverConfiguration.OnEvaluatorAllocated, GenericType<TestSendContextMessageExceptionDriver>.Class)
.Set(DriverConfiguration.OnEvaluatorFailed, GenericType<TestSendContextMessageExceptionDriver>.Class)
.Set(DriverConfiguration.OnContextFailed, GenericType<TestSendContextMessageExceptionDriver>.Class)
.Build(),
typeof(TestSendContextMessageExceptionDriver), 1, "SendContextMessageExceptionTest", "local", testFolder);
ValidateSuccessForLocalRuntime(0, 0, 1, testFolder);
ValidateMessageSuccessfullyLoggedForDriver(ReceivedFailedEvaluator, testFolder);
CleanUp(testFolder);
}
private sealed class TestSendContextMessageExceptionDriver :
IObserver<IDriverStarted>,
IObserver<IAllocatedEvaluator>,
IObserver<IFailedContext>,
IObserver<IFailedEvaluator>
{
private readonly IEvaluatorRequestor _requestor;
[Inject]
private TestSendContextMessageExceptionDriver(IEvaluatorRequestor requestor)
{
_requestor = requestor;
}
public void OnNext(IDriverStarted value)
{
_requestor.Submit(_requestor.NewBuilder().Build());
}
public void OnNext(IAllocatedEvaluator value)
{
var contextConfig = ContextConfiguration.ConfigurationModule
.Set(ContextConfiguration.Identifier, ContextId)
.Set(ContextConfiguration.OnSendMessage, GenericType<SendContextMessageExceptionHandler>.Class)
.Build();
value.SubmitContext(contextConfig);
}
/// <summary>
/// Throwing an Exception in a Context message handler will result in a Failed Evaluator.
/// </summary>
public void OnNext(IFailedEvaluator value)
{
// We will not be expecting any failed contexts here, this is because the Exception
// is thrown on the heartbeat to the Driver, and will thus fail before the initial heartbeat
// to the Driver is sent.
Assert.Equal(0, value.FailedContexts.Count);
Assert.NotNull(value.EvaluatorException.InnerException);
Assert.True(value.EvaluatorException.InnerException is TestSerializableException);
Assert.Equal(ExpectedExceptionMessage, value.EvaluatorException.InnerException.Message);
Logger.Log(Level.Info, ReceivedFailedEvaluator);
}
public void OnNext(IFailedContext value)
{
throw new Exception("The Driver does not expect a Failed Context message.");
}
public void OnError(Exception error)
{
throw new NotImplementedException();
}
public void OnCompleted()
{
throw new NotImplementedException();
}
}
/// <summary>
/// A Context message source that throws an Exception.
/// </summary>
private sealed class SendContextMessageExceptionHandler : IContextMessageSource
{
[Inject]
private SendContextMessageExceptionHandler()
{
}
public Optional<ContextMessage> Message
{
get { throw new TestSerializableException(ExpectedExceptionMessage); }
}
}
}
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 6,009 |
Q: h2o cross validation standard deviation is different from the manually computed standard deviation, why is that I tried to use h2o's cross-validation. metrics, but I found the standard deviation is different from what I have calculated
mean sd cv_1_valid cv_2_valid cv_3_valid cv_4_valid cv_5_valid model
auc 0.9512599 0.0058884337 0.96505374 0.9521365 0.9406109 0.9451134 0.95338506 drf
sd(c(0.96505374,0.9521365,0.9406109,0.9451134,0.95338506) )
[1] 0.009310416
mean(c(0.96505374,0.9521365,0.9406109,0.9451134,0.95338506) )
[1] 0.9512599
Mean is the same but sd is different, is there other ways of calculating this?
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,708 |
{"url":"https:\/\/www.shaalaa.com\/textbook-solutions\/c\/selina-solutions-concise-mathematics-class-10-icse-chapter-13-section-mid-point-formula_355","text":"Share\n\n# Selina solutions for Concise Mathematics Class 10 ICSE chapter 13 - Section and Mid-Point Formula [Latest edition]\n\nCourse\n\n## Chapter 13: Section and Mid-Point Formula\n\nExercise 13(A)Exercise 13(B)Exercise 13(C)\n\n### Selina solutions for Concise Mathematics Class 10 ICSE Chapter 13 Section and Mid-Point Formula Exercise 13(A) [Pages 177 - 178]\n\nExercise 13(A) | Q 1.1 | Page 177\n\nCalculate the co-ordinates of the point P which divides the line segment joining: A (1, 3) and B (5, 9) in the ratio 1 : 2\n\nExercise 13(A) | Q 1.2 | Page 177\n\nCalculate the co-ordinates of the point P which divides the line segment joining:\u00a0\u00a0A (-4, 6) and B(3, -5) in the ratio 3 : 2\n\nExercise 13(A) | Q 2 | Page 177\n\nIn what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis.\n\nExercise 13(A) | Q 3 | Page 177\n\nIn what ratio is the line joining (2, -4) and (-3, 6) divided by the y \u2013 axis.\n\nExercise 13(A) | Q 4 | Page 177\n\nIn what ratio does the point (1, a) divide the join of (-1, 4) and (4,-1)?\u00a0Also, find the value of a.\n\nExercise 13(A) | Q 5 | Page 177\n\nIn what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8)?\u00a0Also, find the value of a.\n\nExercise 13(A) | Q 6 | Page 177\n\nIn what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of\u00a0the point of intersection.\n\nExercise 13(A) | Q 7 | Page 177\n\nFind the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the coordinates\u00a0of the point of intersection.\n\nExercise 13(A) | Q 8 | Page 177\n\nPoints A, B, C and D divide the line segment joining the point (5, -10) and the origin in five\u00a0equal parts. Find the co-ordinates of B and D.\n\nExercise 13(A) | Q 9 | Page 177\n\nThe line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that (PB)\/(AB)=1\/5\u00a0Find the co-ordinates of P.\n\nExercise 13(A) | Q 10 | Page 177\n\nP is a point on the line joining A(4, 3) and B(-2, 6) such that 5AP = 2BP. Find the co-ordinates\u00a0of P.\n\nExercise 13(A) | Q 11 | Page 177\n\nCalculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x\u00a0= 2. Also, find the co-ordinates of the point of intersection.\n\nExercise 13(A) | Q 12 | Page 177\n\nCalculate the ratio in which the line joining A(6, 5) and B(4, -3) is divided by the line y = 2.\n\nExercise 13(A) | Q 13 | Page 177\n\nThe point P (5, -4) divides the line segment AB, as shown in the figure, in the ratio 2 : 5. Find\u00a0the co-ordinates of points A and B.\n\nExercise 13(A) | Q 14 | Page 177\n\nFind the co-ordinates of the points of tri-section of the line joining the points (-3, 0) and (6, 6)\n\nExercise 13(A) | Q 15 | Page 177\n\nShow that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate\u00a0axes.\n\nExercise 13(A) | Q 16 | Page 177\n\nShow that A (3, -2) is a point of trisection of the line segment joining the points (2, 1) and (5, -8).\nAlso, find the co-ordinates of the other point of trisection.\n\nExercise 13(A) | Q 17 | Page 177\n\nIf A = (-4, 3) and B = (8, -6)\n(i) Find the length of AB\n(ii) In what ratio is the line joining A and B, divided by the x-axis?\n\nExercise 13(A) | Q 18 | Page 177\n\nThe line segment joining the points M(5, 7) and N(-3, 2) is intersected by the y-axis at point L.\u00a0Write down the abscissa of L. Hence, find the ratio in which L divides MN.\u00a0Also, find the co-ordinates of L.\n\nExercise 13(A) | Q 19 | Page 177\n\nA (2, 5), B (-1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P\nand Q lie on AB and AC respectively,\nSuch that: AP : PB = AQ : QC = 1 : 2\n(i) Calculate the co-ordinates of P and Q.\n(ii) Show that PQ =1\/3BC\n\nExercise 13(A) | Q 20 | Page 177\n\nA (-3, 4), B (3, -1) and C (-2, 4) are the vertices of a triangle ABC. Find the length of line\u00a0segment AP, where point P lies inside BC, such that BP : PC = 2 : 3\n\nExercise 13(A) | Q 21 | Page 177\n\nThe line segment joining A (2, 3) and B (6, -5) is intercepted by x-axis at the point K. Write\u00a0down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also find the coordinates\u00a0of the point K.\n\nExercise 13(A) | Q 22 | Page 177\n\nThe line segment joining A (4, 7) and B (-6, -2) is intercepted by the y \u2013 axis at the point K.\u00a0write down the abscissa of the point K. hence, find the ratio in which K divides AB. Also, find\u00a0the co-ordinates of the point K.\n\nExercise 13(A) | Q 23 | Page 177\n\nThe line joining P(-4, 5) and Q(3, 2) intersects the y-axis at point R. PM and QN are\u00a0perpendicular from P and Q on the x-axis Find:\n\n(i) the ratio PR : RQ\n\n(ii) the coordinates of R.\n\n(iii) the area of the quadrilateral PMNQ.\n\nExercise 13(A) | Q 24 | Page 177\n\nIn the given figure line APB meets the x-axis at point A and y-axis at point B. P is the point (-4,2) and AP : PB = 1 : 2. Find the co-ordinates of A and B.\n\nExercise 13(A) | Q 25 | Page 177\n\nGiven a line segment AB joining the points A (-4, 6) and B (8, -3). Find:\n(i) the ratio in which AB is divided by the y-axis\n(ii) find the coordinates of the point of intersection\n(iii) the length of AB.\n\nExercise 13(A) | Q 26 | Page 178\n\nIf P(-b, 9a - 2) divides the line segment joining the points A(-3, 3a + 1) and B(5, 8a) in the ratio 3: 1, find the values of a and b.\n\n### Selina solutions for Concise Mathematics Class 10 ICSE Chapter 13 Section and Mid-Point Formula Exercise 13(B) [Page 182]\n\nExercise 13(B) | Q 1.1 | Page 182\n\nFind the mid \u2013 point of the line segment joining the point:\u00a0(-6, 7) and (3, 5)\n\nExercise 13(B) | Q 1.2 | Page 182\n\nFind the mid \u2013 point of the line segment joining the point: (5, -3) and (-1, 7)\n\nExercise 13(B) | Q 2 | Page 182\n\nPoints A and B have co-ordinates (3, 5) and (x, y) respectively. The mid point of AB is (2, 3).\u00a0Find the values of x and y.\n\nExercise 13(B) | Q 3 | Page 182\n\nA (5, 3), B(-1, 1) and C(7, -3) are the vertices of triangle ABC. If L is the mid-point of AB and\u00a0M is the mid-point of AC, show that LM =1\/2BC\n\nExercise 13(B) | Q 4.1 | Page 182\n\nGiven M is the mid point of AB, find the co-ordinates of: A; if M = (1, 7) and B = (-5, 10)\n\nExercise 13(B) | Q 4.2 | Page 182\n\nGiven M is the mid point of AB, find the co-ordinates of: B; if A = (3, -1) and M = (-1, 3)\n\nExercise 13(B) | Q 5 | Page 182\n\nP (-3, 2) is the mid-point of line segment AB as shown in the given figure. Find the co-ordinates\u00a0of points A and B.\n\nExercise 13(B) | Q 6 | Page 182\n\nIn the given figure, P (4, 2) is mid-point of line segment AB. Find the co-ordinates of A and B.\n\nExercise 13(B) | Q 7 | Page 182\n\n(-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the length of its median through the\u00a0vertex (3, -6)\n\nExercise 13(B) | Q 8 | Page 182\n\nGiven a line ABCD in which AB = BC = CD, B= (0, 3) and C = (1, 8)\nFind the co-ordinates of A and D.\n\nExercise 13(B) | Q 9 | Page 182\n\nOne end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, of the\u00a0centre of the circle is (2, -1)\n\nExercise 13(B) | Q 10 | Page 182\n\nA (2, 5), B (1, 0), C (-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the coordinates\u00a0of the mid-points of AC and BD. Give a special name to the quadrilateral.\n\nExercise 13(B) | Q 11 | Page 182\n\nP (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of\u00a0the point of intersection of its diagonals. Find co-ordinates of R and S.\n\nExercise 13(B) | Q 12 | Page 182\n\nA (-1, 0), B (1, 3) and D (3, 5) are the vertices of a parallelogram ABCD. Find the co-ordinates\u00a0of vertex C.\n\nExercise 13(B) | Q 13 | Page 182\n\nThe points (2, -1), (-1, 4) and (-2, 2) are mid points of the sides of a triangle. Find its vertices\n\nExercise 13(B) | Q 14 | Page 182\n\nPoints A (-5, x), B (y, 7) and C (1, -3) are collinear (i.e. lie on the same straight line) such that\u00a0AB = BC. Calculate the values of x and y.\n\nExercise 13(B) | Q 15 | Page 182\n\nPoints P (a, \u22124), Q (\u22122, b) and R (0, 2) are collinear. If Q lies between P and R, such that PR =\u00a02QR, calculate the values of a and b.\n\nExercise 13(B) | Q 16 | Page 182\n\nCalculate the co-ordinates of the centroid of the triangle ABC, if A = (7, -2), B = (0, 1) and C =(-1, 4).\n\nExercise 13(B) | Q 17 | Page 182\n\nThe co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8);\u00a0calculate the co-ordinates of vertex P.\n\nExercise 13(B) | Q 18 | Page 182\n\nA (5, x), B (-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin.\u00a0Calculate the values of x and y.\n\n### Selina solutions for Concise Mathematics Class 10 ICSE Chapter 13 Section and Mid-Point Formula Exercise 13(C) [Pages 182 - 183]\n\nExercise 13(C) | Q 1 | Page 182\n\nGiven a triangle ABC in which A = (4, -4), B = (0, 5) and C = (5, 10). A point P lies on BC such\u00a0that BP : PC = 3 : 2, Find the length of line segment AP.\n\nExercise 13(C) | Q 2 | Page 182\n\nA(20, 0) and B(10, -20) are two fixed points Find the co-ordinates of the point P in AB such that\u00a0: 3PB = AB, Also, find the co-ordinates of some other point Q in AB such that AB = 6 AQ.\n\nExercise 13(C) | Q 3 | Page 183\n\nA(-8, 0), B(0, 16) and C(0, 0) are the verticals of a triangle ABC. Point P lies on AB and Q lies\u00a0on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5\n\nShow that : PQ =3\/8 BC\n\nExercise 13(C) | Q 4 | Page 183\n\nFind the co-ordinates of points of trisection of the line segment joining the point (6, -9) and the\u00a0origin.\n\nExercise 13(C) | Q 5 | Page 183\n\nA line segment joining A(-1,5\/3) and B (a, 5) is divided in the ratio 1 : 3 at P, the point where the\u00a0line segment AB intersects the y-axis.\n\n(i) calculate the value of \u2018a\u2019\n(ii) Calculate the co-ordinates of \u2018P\u2019.\n\nExercise 13(C) | Q 6 | Page 183\n\nIn what ratio is the line joining A(0, 3) and B (4, -1) divided by the x-axis? Write the co-ordinates\u00a0of the point where AB intersects the x-axis\n\nExercise 13(C) | Q 7 | Page 183\n\nThe mid-point of the segment AB, as shown in diagram, is C(4, -3). Write down the co-ordinates\u00a0of A and B.\n\nExercise 13(C) | Q 8 | Page 183\n\nAB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7), find\n(i) the length of radius AC\n(ii) the coordinates of B.\n\nExercise 13(C) | Q 9 | Page 183\n\nFind the co-ordinates of the centroid of a triangle ABC whose vertices are:\u00a0A(-1, 3), B(1, -1) and C(5, 1)\n\nExercise 13(C) | Q 10 | Page 183\n\nThe mid point of the line segment joining (4a, 2b -3) and (-4, 3b) is (2, -2a). Find the values of\u00a0a and b.\n\nExercise 13(C) | Q 11 | Page 183\n\nThe mid point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the values of a\u00a0and b.\n\nExercise 13(C) | Q 12 | Page 183\n\n(i) write down the co-ordinates of the point P that divides the line joining A(-4, 1) and B(17, 10)\u00a0in the ratio 1 : 2.\n(ii) Calculate the distance OP, where O is the origin.\n(iii) In what ratio does the Y-axis divide the line AB?\n\nExercise 13(C) | Q 13 | Page 183\n\nProve that the points A(-5, 4); B(-1, -2) and C(5, 2) are the vertices of an isosceles right angled\u00a0triangle. Find the co-ordinates of D so that ABCD is a square.\n\nExercise 13(C) | Q 14 | Page 183\n\nM is the mid-point of the line segment joining the points A(-3, 7) and B(9, -1). Find the coordinates\u00a0of point M. Further, if R(2, 2) divides the line segment joining M and the origin in the\u00a0ratio p : q, find the ratio p : q\n\nExercise 13(C) | Q 15 | Page 183\n\nCalculate the ratio in which the line joining A(-4,2) and B(3,6) is divided by point p(x,3). Also, find x\n\nExercise 13(C) | Q 16 | Page 183\n\nFind the ratio in which the line 2x+y=4 divides th line segment joining the point p(2,-2) and Q (3,7).\n\nExercise 13(C) | Q 17 | Page 183\n\nIf the abscissa of point P is 2. find the ratio in which this point divides the line segment joining the point (-4,3) and (6,3). Also find the co-ordinates of point P.\n\nExercise 13(C) | Q 18 | Page 183\n\nThe line joining the points (2,1) and (5,-8) is trisected at the point P and Q, point P lies on the line 2x-y+k=0, find the value of k Also, find the co-ordinates of point Q.\n\nExercise 13(C) | Q 19 | Page 183\n\nFind the image of the point A(5,3) under reflection in the point P(-1,3).\n\nExercise 13(C) | Q 20 | Page 183\n\nM is the mid-point of the line segment joining the points A(0,4) and B(6,0). M also divides the line segment OP in the ratio 1:3 find :\n\n(a) co-ordinates of M\n\n(b) Co-ordinates of P\n\n(C) Length Of BP\n\nExercise 13(C) | Q 21 | Page 183\n\nA (-4,2)2, B(0,2) and C (-2,-4) are the vertices of a triangle ABC. A,Q and R are mid-Points of sides BC,CA and AB respectively.Show that the centroid of \u0394 PQR is the same as the centroid of \u0394 ABC.\n\nExercise 13(C) | Q 22 | Page 183\n\nA(3, 1), B(y, 4) and C(1, x) are vertices of a triangle ABC. P, Q and R are mid - points of sides BC, CA and AB respectively. Show that the centroid of\u00a0\u0394PQR is the same as the centroid\u00a0\u0394ABC.\n\n## Chapter 13: Section and Mid-Point Formula\n\nExercise 13(A)Exercise 13(B)Exercise 13(C)\n\n## Selina solutions for Concise Mathematics Class 10 ICSE chapter 13 - Section and Mid-Point Formula\n\nSelina solutions for Concise Mathematics Class 10 ICSE chapter 13 (Section and Mid-Point Formula) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Concise Mathematics Class 10 ICSE solutions in a manner that help students grasp basic concepts better and faster.\n\nFurther, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Selina textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.\n\nConcepts covered in Concise Mathematics Class 10 ICSE chapter 13 Section and Mid-Point Formula are Co-ordinates Expressed as (x,y), Distance Formula, Section Formula, Mid-point Formula.\n\nUsing Selina Class 10 solutions Section and Mid-Point Formula exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Selina Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 10 prefer Selina Textbook Solutions to score more in exam.\n\nGet the free view of chapter 13 Section and Mid-Point Formula Class 10 extra questions for Concise Mathematics Class 10 ICSE and can use Shaalaa.com to keep it handy for your exam preparation","date":"2020-09-21 12:24:41","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\": 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\": 0.5730870962142944, \"perplexity\": 1350.2224992448444}, \"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-2020-40\/segments\/1600400201699.38\/warc\/CC-MAIN-20200921112601-20200921142601-00169.warc.gz\"}"} | null | null |
A six-month trial of live CCTV on a London bus route has begun.
London Mayor Boris Johnson announced that the technology had been installed on 21 double-decker buses on a north London route.
The real-time images will be beamed to a control room manned by Transport for London (TfL) and police.
The trial will be monitored to determine whether live images can help transport staff deal with disorder more effectively.
Mr Johnson said: "I am determined to banish the sad minority of hoodlums and trouble makers that have blighted our buses.
"Having the facility to access live pictures from buses travelling around the capital will mean our bus controllers can play a far more effective role in sending police officers to sort out troublemakers.
"If this trial is successful then we will consider rolling out the system on other routes as part of our campaign to stamp out the casual disorder that led to a culture of fear on public transport."
Andy Thompson, of London Buses, said: "While crime is low on London buses, we know there is more to do, which is why we are always looking at ways to improve safety, and our passengers' perception of safety, while they travel on our network."
Currently there are 60,000 recordable CCTV cameras operating on the 8,000 London buses.
The images can be downloaded on request. | {
"redpajama_set_name": "RedPajamaC4"
} | 9,006 |
Exalarius huachucensis är en stekelart som beskrevs av Lasalle 1994. Exalarius huachucensis ingår i släktet Exalarius och familjen finglanssteklar. Inga underarter finns listade i Catalogue of Life.
Källor
Finglanssteklar
huachucensis | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,173 |
Q: Ef Core 3.1 and Identity framework combining DB Context I am trying to merge my application context with my db context I have tried various solutions but none of them are working for ef core 3.1.
public class AppManagerDBContext : IdentityDbContext
{
public AppManagerDBContext(DbContextOptions options) : base(options) { }
protected override void OnConfiguring(DbContextOptionsBuilder optionsBuilder)
{
if (!optionsBuilder.IsConfigured)
{
optionsBuilder.UseSqlServer("name=DefaultConnection");
}
}
public DbSet<BmiInformation> BmiInformation { get; set; }
public DbSet<WorkOuts> WorkOuts { get; set; }
protected override void OnModelCreating(ModelBuilder modelBuilder)
{
modelBuilder.Entity<BmiInformation>().ToTable("BmInformation");
}
}
This is the full error that I am getting I thought all I had to do is inherit from IdentityDbContext is there something else I have to do in .NET Core 3.1 and EF Core?
Add-Migration MyFirstMigration -Context AppManagerDBContext
Build started...
Build succeeded.
Error
The AppManagerDBContext constructor must be a
DbContextOptions. When registering multiple
DbContext types make sure that the constructor for each context type
has a DbContextOptions parameter rather than a non-generic
DbContextOptions parameter.
Edit 1
Ok I solved the issue with the following however I am now getting a new error.
My User Info class
public class UserInfo : IdentityUser
{
// These two new fields are added here
[PersonalData]
public string Name { get; set; }
[PersonalData]
public DateTime DOB { get; set; }
}
Error 2
The entity type 'IdentityUserLogin' requires a primary key to
be defined. If you intended to use a keyless entity type call
'HasNoKey()'.
A: You must inject a DbContextOptions object into your DbContext, which type paremeter must be the object type you are injecting into.
In your case
public class AppManagerDBContext : IdentityDbContext
{
public AppManagerDBContext(DbContextOptions<AppManagerDBContext> options) : base(options) { }
protected override void OnConfiguring(DbContextOptionsBuilder optionsBuilder)
{
// ...
}
// ...
}
Both errors are quite explanatory.
You must mark one or more properties of your entity to be a primary key.
If you want so, try
public class UserInfo : IdentityUser
{
// DatabaseGenerated is optional.
// Only in case you want to delegate the key generation
[Key]
[DatabaseGenerated(DatabaseGeneratedOption.Identity)]
public int Id { get; set; }
[PersonalData]
public string Name { get; set; }
[PersonalData]
public DateTime DOB { get; set; }
}
Or, if you want none, configure the entity as unkeyed.
protected override void OnModelCreating(ModelBuilder modelBuilder)
{
modelBuilder.Entity<BmiInformation>().ToTable("BmInformation");
modelBuilder.Entity<UserInfo>(u =>
{
u.HasNoKey();
});
}
Hope it helps.
| {
"redpajama_set_name": "RedPajamaStackExchange"
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This remastered collection is faithful to the original 1960 performances, which were recorded on 35mm film in order to accurately capture the orchestra. The music here is big, bright, and beautiful, with full instrumentation and precise direction. It's a perfect introduction to Beethoven's symphony cycle, from the passionate highs and sweeping lows of "Eroica" to Symphony No. 9, fragments of which haunt the creations of so many subsequent composers as well as the scores of countless film and television works.
Beethoven: The Complete Symphony Collection London Symphony Orchestra
Symphony No. 1 in C Major, Op. 21
I. Adagio molto - Allegro con brio
II. Andante cantabile con moto
III. Menuetto. Allegro molto e vivace
IV. Finale. Adagio - Allegro molto e vivace
Symphony No. 2 in D Major, Op. 36
II. Larghetto
III. Scherzo. Allegro
IV. Allegro molto
Symphony No. 3 in E-Flat Major, Op. 55 "Eroica"
II. Marcia Funebre. Adagio assai
III. Scherzo. Allegro vivace
IV. Finale. Allegro molto
Symphony No. 4 in B-Flat Major, Op. 60
I. Adagio - Allegro vivace
III. Allegro vivace
IV. Allegro ma non troppo
II. Andante con moto
III. Allegro (attaca) - IV. Allegro
Symphony No. 6 in F Major, Op. 68 "Pastoral"
I. Allegro ma non troppo
II. Andante molto mosso
III. Allegro (attaca) - IV. Allegro (attaca) - V. Allegretto
Symphony No. 7 in A Major, Op. 92
I. Poco sostenuto. Vivace
II. Allegretto
III. Presto
IV. Allegro con brio
Symphony No. 8 in F Major, Op. 93
I. Allegro vivace e con brio
II. Allegretto scherzando
III. Tempo di menuetto
IV. Allegro vivace
Symphony No. 9 in D Minor, Op. 125 "Choral"
I. Allegro ma non troppo, un poco maestoso
II. Molto vivace
Josef Krips & London Symphony Orchestra
III. Adagio molto e cantabile
IV. Presto - Allegro assai - Choral Finale (Ode to Joy)
London Symphony Orchestra, Jennifer Vyvyan, Rudolf Petrak, Donald Bell, Shirley Carter, Josef Krips & BBC Symphony Chorus
℗ 1960 Everest Records
oooble , 17/04/2009
great music - poor conversion
I love Beethoven. I justs a shame the QC is so poor in places.
Glide & Swerve , 05/09/2009
Very poor quality
spoilt by the ridiculous hissing!!! find another recording to buy!!!
xmb53 , 31/05/2009
9th isn't worth it
Very poor quality recorded at the start of the last movement of teh 9th symphony - don't buy, if this is of any importance to you.
More By London Symphony Orchestra
Holst: The Planets, Op. 32
Handel: Messiah
Beethoven: Symphony No. 9 "Choral"
Tchaikovsky: Swan Lake
Beethoven: Symphonies Nos. 1-9
Rachmaninov: Piano Concerto No. 2 & Rhapsody on a Theme of Paganini
Immortal Beloved (Original Motion Picture Soundtrack)
London Symphony Orchestra & Sir Georg Solti
Beethoven's 7th
Antal Doráti & London Symphony Orchestra
Classics for the Heart
The 50 Greatest Pieces of Classical Music
London Philharmonic Orchestra & David Parry
Vivaldi: The Four Seasons
Takako Nishizaki, Capella Istropolitana & Stephen Gunzenhauser | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 765 |
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<p class="styleSans9.0000<enum PANGO_WEIGHT_NORMAL of type PangoWeight><enum PANGO_STYLE_NORMAL of type PangoStyle>">Triangle USA Petroleum Corporation <br/>Wisness 150-100-23-14-7H <br/>SWSE, 300' FSL and 1981' FEL, Sec. 23-T150N-R100W (Surface) NENE, 250' FNL and 1005' FEL, Sec. 14-T150N-R100W (Btm Hole) <br/>DRILLING PROGRAM McKenzie Countl, ND <br/> </p>
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<p class="styleSans9.0000<enum PANGO_WEIGHT_NORMAL of type PangoWeight><enum PANGO_STYLE_NORMAL of type PangoStyle>">Maximum anticipated bottom hole pressure equals approximately 6,054 psi* and maximum anticipated surface pressure equals approximately 3,614 psi** (bottom hole pressure minus the pressure of a partially evacuated hole calculated at 0.22 psi/foot). <br/>*Max Mud Wt x 0.052 x TVD = A (bottom hole pressure) **Maximum surface pressure = A — (0.22 x TVD) </p>
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<p class="styleSans9.0000<enum PANGO_WEIGHT_NORMAL of type PangoWeight><enum PANGO_STYLE_NORMAL of type PangoStyle>">Hydraulic Fracturing Stimulation <br/>None of the diesel compounds listed below will be used to hydraulically fracture the referenced well. <br/>68334-30-5 (Primary Name: Fuels, diesel) 68476-34-6 (Primary Name: Fuels, diesel, No. 2) 68476-30-2 (Primary Name: Fuel oil No. 2) 68476-31-3 (Primary Name: Fuel oil, No. 4) 8008-20-6 (Primary Name: Kerosene) <br/>Contact Information <br/>Permit Matters Drilling and Completion Matters Triangle Petroleum Corporation Triangle Petroleum Corporation 1200 17th Street 1200 17'h Street <br/>Suite 2600 Suite 2600 <br/>Denver, CO 80202 Denver, CO 80202 <br/>Brittany Vannoy Tony Hale <br/>303-260-1801 (office) 303-260-1684 (office) 303-260-5080 (fax) 303-981-0005 (cell) <br/>bvanno trian le etroleum.com thale@trianglepetroleum.corn </p>
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} | 7,847 |
Q: How to send broadcast to all connected client in node js I'm a newbie working with an application with MEAN stack. It is an IoT based application and using nodejs as a backend.
I have a scenario in which I have to send a broadcast to each connected clients which can only open the Socket and can wait for any incoming data. unless like a web-browser they can not perform any event and till now I have already gone through the Socket.IO and Express.IO but couldn't find anything which can be helpful to achieve what I want send raw data to open socket connections'
Is there any other Node module to achieve this. ?
Here is the code using WebSocketServer,
const express = require('express');
const http = require('http');
const url = require('url');
const WebSocket = require('ws');
const app = express();
app.use(function (req, res) {
res.send({ msg: "hello" });
});
const server = http.createServer(app);
const wss = new WebSocket.Server({ server });
wss.on('connection', function connection(ws) {
ws.on('message', function(message) {
wss.broadcast(message);
}
}
wss.broadcast = function broadcast(msg) {
console.log(msg);
wss.clients.forEach(function each(client) {
client.send(msg);
});
};
server.listen(8080, function listening() {
console.log('Listening on %d', server.address().port);
});
Now, my query is when this code will be executed,
wss.on('connection', function connection(ws) {
ws.on('message', function(message) {
wss.broadcast(message);
}
}
A: var WebSocketServer = require("ws").Server;
var wss = new WebSocketServer({port:8100});
wss.on('connection', function connection(ws) {
ws.on('message', function(message) {
wss.broadcast(message);
}
}
wss.broadcast = function broadcast(msg) {
console.log(msg);
wss.clients.forEach(function each(client) {
client.send(msg);
});
};
A: Try the following code to broadcast message from server to every client.
wss.clients.forEach(function(client) {
client.send(data.toString());
});
Demo server code,
const WebSocket = require('ws')
const wss = new WebSocket.Server({ port: 2055 },()=>{
console.log('server started')
})
wss.on('connection', (ws) => {
ws.on('message', (data) => {
console.log('data received \n '+ data)
wss.clients.forEach(function(client) {
client.send(data.toString());
});
})
})
wss.on('listening',()=>{
console.log('listening on 2055')
})
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,129 |
{"url":"https:\/\/www.physicsforums.com\/threads\/fourier-integral-representation-of-function.672244\/","text":"# Fourier integral representation of function\n\n1. Feb 16, 2013\n\n### izen\n\n1. The problem statement, all variables and given\/known data\n\nHow can I get to that answer?\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nI'm stuck dont know how to get to the answer that I got from wolframalpha. there is no solution there unfortunately.\n\nI know how to get to that denominator w^2-1 but I dont know how to get to sin(pi*w)\n\nThanks\n\nLast edited: Feb 16, 2013","date":"2018-01-18 00:41:56","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\": 0.8290624022483826, \"perplexity\": 745.9321291439476}, \"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-2018-05\/segments\/1516084887024.1\/warc\/CC-MAIN-20180117232418-20180118012418-00386.warc.gz\"}"} | null | null |
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On the beach, sea views, first line: Beach House for 4 people plus baby. Very well maintained home in small village fishermen. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,828 |
\part*{Abstract}
\addcontentsline{toc}{section}{\protect\numberline{}{Abstract}}
\vspace{10ex}
Let $\,G\,$ be a simply connected simple algebraic
group over an algebraically closed field $\,K\,$ of
characteristic $\,p>0\,$ with root system $\,R,\,$
and let $\,{\mathfrak g}\,=\,{\cal L}(G)\,$ be its restricted Lie algebra.
Let $\,V\,$ be a finite dimensional $\,{\mathfrak g}$-module over $\,K.\,$
For any point $\,v\,\in\,V$,
the {\it isotropy subalgebra} of $\,v\,$ in $\,\mathfrak g\,$ is
$\,{\mathfrak g}_v\,=\,\{\,x\,\in\,{\mathfrak g}\,/\,x\cdot
v\,=\,0\,\}\,$
A restricted $\,{\mathfrak g}$-module $\,V\,$ is called {\bf exceptional} if
for each $\,v\in V\,$ the isotropy subalgebra $\,{\mathfrak g}_v\,$ contains
a non-central element (that is, $\,{\mathfrak g}_v\,\not\subseteq\,
{\mathfrak z(\mathfrak g)}\,$).
This work is devoted to classifying irreducible exceptional
$\,\mathfrak g$-modules.
A necessary condition for a $\,\mathfrak g$-module to be exceptional
is found and a complete classification of modules over
groups of exceptional type is obtained.
For modules over groups of classical type, the general problem is
reduced to a short list of unclassified modules.
The classification of exceptional modules is expected to have applications
in modular invariant theory and in classifying modular simple Lie
superalgebras.
\newpage
\part*{Declaration}
\vspace{10ex}
\begin{center}
No portion of the work referred to in this thesis has been \\
submitted in support of an application for another degree\\
or qualification of this or any other institution of learning.
\end{center}
\part*{Copyright}
\vspace{10ex}
Copyright in text of this thesis rests with the Authoress. Copies (by
any process) either in full, or of extracts, may be made {\bf only} in
accordance with instructions given by the Authoress and lodged in the
John Rylands University Library of Manchester. Details may be obtained from
the Librarian. This page must form part of any such copies made.
Further copies (by any process) of copies made in accordance with such
instructions may not be made without the permission (in writing) of
the Authoress.
\newpage
\part*{The Authoress}
\addcontentsline{toc}{section}{\protect\numberline{}{The Authoress}}
\vspace{5ex}
I graduated from Universidade Federal de Santa Maria, Santa Maria -
RS, Brazil in 1988 with a Teaching degree in
Mathematics (Licenciatura em Matem\'atica).
I was awarded an M. Sc. in Mathematics (Mestre em Matem\'atica)
from Universidade
de Bras\'{\i}lia, Bras\'{\i}lia - DF, Brazil in 1991. My
M. Sc. Dissertation was ``On finite groups admitting automorphisms of
prime order with centralizers of controlled order''. Since
August 1991 I have been a Lecturer in the Department of Mathematics of the
Universidade Federal de Vi\c cosa, Vi\c cosa - MG, Brazil.
I commenced the studies for this degree at the University of
Manchester in October 1993, under Professor Brian Hartley's
supervision. This thesis is the result of research work done
since January 1995 under the supervision of Dr. Alexander Premet.
\begin{center}
\vspace*{12ex}
{\bf\Large\it Dedicated to \vspace{2ex}\\
Professor Brian Hartley\vspace{2ex}\\
(In Memoriam)\vspace{12ex}}
{\it I thank God for the opportunity
of working with a great Master,
even for such a short time.}
\end{center}
\newpage
\part*{Acknowledgements}
\addcontentsline{toc}{section}{\protect\numberline{}{Acknowledgements}}
I thank God for more this achievement in my life.
I thank my supervisor, Dr. Alexander Premet, for teaching me
the Representation Theory of Algebraic
Groups and Lie Algebras, and for his guidance throughout the
development of this work.
I also thank Dr. Grant Walker for his support
during the most difficult time of my course, for his constant encouragement,
and for very useful hints on solving combinatorial problems.
I would like to thank my office mates Ersan Nal\c cacio\u{g}lu,
Mohammad Ali Asadi and Suo Xiao for the
good chats that helped to keep our spirits up.
I have very good friends that have helped me to stay sane
whilst doing my Ph. D.. In particular, I would like to
thank Noreen Pervaiz for being such a good hearted person and an excellent
flatmate and Jackie Furby for her friendship.
A very special thanks to David D. Garside for helping me to cope with
the stress
and mainly for bringing so much happiness into my heart
and life over the last year or so. I also thank his family
for being so caring and generous.
I thank the Universidade Federal de Vi\c cosa (MG - Brazil)
and all my colleagues of the Departamento de Matem\'atica for their
support throughout these years. In particular, I would like to
thank Jos\'e Geraldo Teixeira for always attending promptly my
numerous enquiries.
I am forever in debt to my good friend Maria de Lourdes Carvalho for her
extreme dedication on looking after all my things in Brazil, as if they
were hers.
I also would like to thank Dr. Nora\'{\i} Rocco for his advice during
the moments of important decisions and for always keeping one eye at my
process at CAPES.
Finally, I most gratefully acknowledge over four years of financial support
given to me by CAPES - MEC - Brazil.
\newpage
\vspace{20ex}
\part*{Chapter 1}
\part*{Introduction}
\addcontentsline{toc}{section}{\protect\numberline{1}Introduction}
\vspace*{8ex}
Let $\,G\,$ be a simply connected simple algebraic
group over an algebraically closed field $\,K\,$ of
characteristic $\,p>0,\,$ and let
$\,{\mathfrak g}\,=\,{\cal L}(G)\,$ be its restricted Lie algebra.
Let $\,V\,$ be a finite dimensional vector space over $\,K.\,$
If $\,G\,$ acts on $\,V\,$ via a rational representation
$\,\pi\,:\,G\,\longrightarrow\,GL(V),\,$ then $\,{\mathfrak g}\,$ acts
on $\,V\,$ via the differential
$\,d\pi\,:\,{\mathfrak g}\,\longrightarrow\,{\mathfrak
{gl}}\;(V)$ (see Section~\ref{tlaag}).
For any point $\,v\,\in\,V$, the {\it isotropy subgroup}
of $\,v\,$ in $\,G\,$ is $\,G_v\,=\,\{\,g\,\in\,G\,/\,
g\cdot v\,=\,v\,\}\,$ and the {\it isotropy subalgebra} of
$\,v\,$ in $\,\mathfrak g\,$ is
$\,{\mathfrak g}_v\,=\,\{\,x\,\in\,{\mathfrak g}\,/\,x\cdot
v\,=\,0\,\}\,$ (see Section~\ref{sec3.2}).
This thesis is devoted to classifying irreducible rational
$\,G$-modules $\,V\,$ for which the isotropy subalgebras of all points have at
least one non-central element. We call such modules {\bf exceptional}
(Definition~\ref{excpmod}).
For Lie algebras of exceptional type a complete classification
is obtained. For Lie algebras of classical type partial results are
given.
The classification of such modules is interesting for the following
reasons.
For a rational action of an algebraic group $\,G\,$ on
a vector space $\,V$, we say that a subgroup $\,H\,\subset\,G\,$ is an
{\it isotropy subgroup in general position} (ISGP for short) if $\,V\,$
contains a non-empty Zariski open subset $\,U\,$ whose points have
their isotropy subgroups conjugate to $\,H$. The points of $\,U\,$ are
called {\it points in general position}.
Isotropy subalgebras $\,{\mathfrak h}\,\subset\,{\mathfrak g}\,$ in general
position are defined similarly.
A rational $\,G$-action $\,G\longrightarrow \GL(V)\,$
is said to be {\it locally free} if the ISGP $\,H\,$ exists
and equals $\,\{ e\}$. Similarly, a linear $\,\mathfrak g$-action $\,\mathfrak
g\longrightarrow\mathfrak{gl}(V)\,$ is said to be {\it locally free} if the
isotropy subalgebra in general position $\,\mathfrak h\,$ exists
and equals $\,\{ 0\}$.
In \cite[1967]{ave}, E.M. Andreev, E.B. Vinberg and A.G. \'Elashvili
have given a necessary and sufficient condition for an irreducible
action of a complex simple Lie group $\,G\,$ on a vector space $\,V\,$
to have zero isotropy subalgebra in general position.
Their method is based on the notion of the {\it index of the trace
form} associated to a representation and cannot be used
in positive characteristic, as in this case the trace form may vanish.
Let $\,G\,$ be a simple Lie group and $\,V\,$ be a finite
dimensional vector space over a field of characteristic zero.
Define the {\it trace form} associated to the representation
$\,\psi\,:\,G\longrightarrow\GL(V)\,$ by $\,\kappa_{\psi}(X,\,Y)\,=\,
{\rm tr}\,({\rm d}\psi(X)\circ {\rm d}\psi(Y))\,$ for all $\,X,\,Y\in
G\,$.
Let $\,G\,$ act irreducibly on $\,V.\,$
The {\it index} $\,l_V\,$ of the trace form associated to $\,\psi\,$
is given by
$\,{\rm tr}({\rm d}\psi(X))^2\,=\,l_V\,{\rm tr}({\rm ad}
({\rm d}\psi(X)))^2\,$, for $\,X\in G.\,$
$\,l_V\,$ is a positive
rational number, independent of $\,X\,$ \cite[p. 257]{ave}.
The index of a direct sum of irreducible representations is the sum of
the indices of the irreducible representations \cite[p. 260]{ave}.
The index of a one-dimensional representation of a simple Lie algebra
$\,L\,$ is $\,0\,$, and, for irreducible representations, it takes
value $\,1\,$ only for the adjoint representation
\cite[p. 259]{ave},~\cite[1.4.3]{kac1}.
The sufficient condition obtained by Andreev, Vinberg and
\'Elashvili establishes that if the index of the trace form of
$\,\pi\,$ is greater than 1, then $\,\pi\,$ is locally free
\cite[Theorem]{ave}.
Table~\ref{table1} has been taken from~\cite{ave} and
lists the irreducible representations of the simple Lie algebras
for which $\,0\,<\,l_V\,<\,1\,$ (up to graph automorphism).
\begin{table}[p]
\begin{tabular}{llllll}
\hline\hline
Type & & Rank & Highest & $\dim\,V$ & Index \\
& & & Weight & & \\ \hline\hline \vspace{.5ex}
A & $\mathfrak{sl}_{n}$ & $\,n-1\,$ & $\,\omega_1\,$
& $n$ & $\displaystyle\frac{1}{2n}$ \vspace{.5ex}\\
& $\textstyle\bigwedge^2\mathfrak{sl}_{n}$ & $\,n-1\,$ &
$\,\omega_2\,$ & $\displaystyle\frac{n\,(n-1)}{2}$ &
$\displaystyle\frac{n-2}{2n}$ \vspace{.5ex} \\
& $S^2\mathfrak{sl}_{n}$ & $\,n-1\,$ & $\,2\omega_1\,$
& $\displaystyle\frac{n\,(n+1)}{2}$ &
$\displaystyle\frac{n+2}{2n}$ \vspace{.5ex} \\
& $\textstyle\bigwedge^3\mathfrak{sl}_{n}$ & $\,5,6,7\,$ &
$\,\omega_3\,$ & $20,\,35,\,56$ &
$\displaystyle\frac{1}{2},\,\frac{5}{7},\,\frac{15}{16} $\\
& $n\,=\,6,\,7,\,8$ & & & & \vspace{.5ex}\\ \hline \vspace{.5ex}
B & $\mathfrak{so}_{n}$ & $\,\displaystyle\frac{n-1}{2}\,$ & $\,\omega_1\,$
& $n$ & $\displaystyle\frac{1}{n-2}$ \vspace{.5ex} \\
& ${\rm spin}_{n}$ & $\,3,\,4,\,5,\,6\,$ & $\,\omega_{\ell}\,$
& $8,\,16,\,32,\,64$ &
$\displaystyle\frac{1}{5},\,\frac{2}{7},\,\frac{4}{9},\,\frac{8}{11}$ \\
& $n\,=\,7,\,9,\,11,\,13$ & & & & \vspace{.5ex}\\ \hline \vspace{.5ex}
C & $\mathfrak{sp}_{n}$ & $\,\displaystyle\frac{n}{2}\,$ & $\,\omega_1\,$
& $n$ & $\displaystyle\frac{1}{n+2}$ \vspace{.5ex}\\
& $\textstyle\bigwedge^2_0\mathfrak{sp}_{n}$ &
$\,\displaystyle\frac{n}{2}\,$
& $\,\omega_2\,$ & $\displaystyle\frac{n\,(n-1)}{2}-1$ &
$\displaystyle\frac{n-2}{n+2}$ \vspace{.5ex} \\
& $\textstyle\bigwedge^3_0\mathfrak{sp}_{6}$ & $\,3\,$ & $\,\omega_3\,$
& $\,14\,$ &
$\displaystyle\frac{5}{8}$ \vspace{.5ex} \\ \hline \vspace{.5ex}
D & $\mathfrak{so}_{n}$ & $\,\displaystyle\frac{n}{2}\,$ & $\,\omega_1\,$
& $n$ & $\displaystyle\frac{1}{n-2}$ \vspace{.5ex} \\
& ${\rm spin}_{n}$ & $\,5,\,6,\,7\,$ & $\,\omega_{\ell}\,$
& $\,16,\,32,\,64$ &
$\displaystyle\frac{1}{4},\,\frac{2}{5},\,\frac{2}{3} $\\
& $n\,=\,10,\,12,\,14$ & & & & \vspace{.5ex}\\ \hline \vspace{.5ex}
E & $\,E_6\,$ & $6$ & $\,\omega_1\,$ & $27$ &
$\displaystyle\frac{1}{4} $ \vspace{.5ex}\\
& $\,E_7\,$ & $7$ & $\,\omega_7\,$ & $56$ & $\displaystyle\frac{1}{3} $
\vspace{.5ex}\\ \hline \vspace{.5ex}
F & $\,F_4\,$ & $4$ & $\,\omega_4\,$ & $26$ & $\displaystyle\frac{1}{3} $
\vspace{.5ex}\\ \hline \vspace{.5ex}
G & $\,G_2\,$ & $2$ & $\,\omega_1\,$ & $7$ & $\displaystyle\frac{1}{4} $
\vspace{.5ex}\\ \hline \hline
\end{tabular}
\caption[Irreducible Representations with index
$\,0\,<\,l_V\,<\,1\,$ - Characteristic Zero]{\label{table1}}
\end{table}
In Table~\ref{table1}, $\,\mathfrak{sl}_{n}\,$, $\,\mathfrak{sp}_{n}\,$, and
$\,\mathfrak{so}_{n}\,$ stand for natural representations of these Lie
algebras; $\,S^k\,$ and $\,\textstyle\bigwedge^k\,$ denote
$\,k$th symmetric and
exterior powers, respectively.
One can show that the set of weights of the $\,G$-module
$\,\textstyle\bigwedge^k\,$ contains a unique maximal weight. We
denote by $\,\textstyle\bigwedge^k_0\,$
the corresponding irreducible component of $\,\textstyle\bigwedge^k\,$.
$\,{\rm spin}_{n}\,$ stands for the irreducible
spinor representation of $\,\mathfrak{so}_{n}\,$; $\,G_2,\,F_4,\,
E_6,\, E_7\,$ denote the natural representations of the
corresponding Lie algebras.
This list was extensively used by V.G. Kac to classify
the simple Lie superalgebras in characteristic $\,0\,$~\cite[1977]{kac1}.
In~\cite[1972]{ela1},\cite{ela2}, A.G. \'Elashvili lists the
simple and irreducible semisimple linear Lie groups $\,G\,$ such that
the ISGP's $\,H\,$ for the action of $\,G\,$ on $\,V\,$ have positive
dimension. He also proves the existence of
isotropy subalgebras in general position for actions of simple, and
irreducible semisimple, linear Lie groups, and finds an
explicit form for them.
As proved by Richardson~\cite[1972]{rich}, in characteristic zero,
the ISGP always exists for an arbitrary rational (linear) action
of a reductive group on an algebraic variety.
In~\cite[1986]{pop1}, \cite[1987]{pop2}, A.M. Popov classified the
irreducible linear actions of (semi)simple complex linear Lie groups
with finite (but nontrivial) ISGP's.
\medskip
These results find applications in Invariant Theory.
The determination of ISGP's is interesting for the following reasons.
If $\,m_G\,$ is the maximal dimension of orbits of an algebraic $\,F$-group
$\,G\subset \GL(V)\,$ (acting on a finite dimensional vector space
$\,V\,$ over an algebraically closed field $\,F\,$), then the
maximal number of algebraically independent rational invariants for the
action of $\,G\,$ on $\,V\,\cong\,F^N\,$ equals $\,N-m_G\,$ \cite{rosen}.
On the other hand, if $\,H\,$ is the ISGP, then $\,m_G\,=\,\dim\,G\,
-\,\dim\,H.\,$ In the case of a semisimple group $\,G\,$, the algebra
$\,F(V)^G\,$ of rational invariants of $\,G\,$ is the field of quotients of
the algebra $\,F[V]^G\,$ of polynomial invariants of $\,G.\,$
Therefore, the degree
of transcendence of $\,F(V)^G\,$ equals the Krull dimension of
$\,F[V]^G.\,$
Hence, $\,{\rm tr.deg.}\,F(V)^G\,=\,\dim\,V\,-\,\dim\,G\,+\,\dim\,H\,$.
Thus, to find the Krull dimension of the
algebra $\,F[V]^G\,$ in the case of a semisimple group
$\,G\,$ it suffices to know $\,\dim\,H.\,$
Knowledge of the ISGP $\,H\,$ can be used to investigate the properties of
$\,\mathbb C[V]^G.\,$
Namely,
if for an action of a connected reductive linear group $\,G\,$ on a vector
space $\,V\,$ the ISGP $\,H\,$ is reductive (for example, finite), then
$\,V\,$ contains a Zariski open $\,G$-invariant subset whose points
all have closed orbits \cite{popvl}.
Then, for the action of the normalizer $\,N(H)\,$ of the
subgroup $\,H\,$ in $\,G\,$ on the subspace $\,V^H,\,$ the restriction
homomorphism $\,\mathbb C[V]^G\longrightarrow \mathbb C[V^H]^{N(H)}\,$ is an
isomorphism \cite{luna}. Then $\,W:=N(H)/H\,$ acts on $\,V^H\,$
and $\,\mathbb C[V^H]^{N(H)}\,\cong\,\mathbb C[V^H]^W.\,$
Thus, knowledge of the reductive (in particular,
finite) ISGP's $\,H\,$ enables us to reduce the determination of
$\,\mathbb C[V]^G\,$ to that of $\,\mathbb C[V^H]^W.\,$
(These results are expected to generalise to prime characteristic.)
Knowledge of the ISPG is important in constructing some moduli spaces
in algebraic geometry \cite{dcm}, \cite{mum}.
Also, information on the ISGP's may be very helpful in establishing
the rationality of the field of invariants $\,\mathbb C(V)^G\,$
\cite{boka}.
\bigskip
We now give an overview of this thesis.
Chapter 2 is a background chapter, where we introduce
the notation used throughout this work.
In Section~\ref{centsection} we give
some well-known properties of centralizers that are used in the main
sections of Chapter 3. In Section~\ref{sec3.2} we prove
that, for Lie algebra actions, if the stabilizer of any
point of a Zariski open subset $\,W\,$ of the module $\,V\,$
contains a non-central element then the stabilizer of any point $\,v\in
V\,$ has the same property (see Proposition~\ref{3.1.1}).
We define the exceptional modules in~\ref{excpmod}.
In Section~\ref{necess} we prove the following theorem, which gives
a necessary condition for a module to be exceptional.\\
{\bf Theorem~\ref{???}}~~{\it
Let $\,p\,$ be a non-special prime for $\,G$.
Let $\,\pi\,:\,G\longrightarrow\GL(V)\,$
be a non-trivial faithful rational representation of $\,G\,$ such that
$\,\ker\,{\rm d}\pi\subseteq\mathfrak z(\mathfrak g)\,$.
If $\,V\,$ is an exceptional $\,\mathfrak g$-module, then it satisfies
the inequalities
\[
r_p(V)\,:=\,\sum_{\stackrel{\scriptstyle\mu\;good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,\frac{|W\mu|}{|R_{long}|}\,
|R_{long}^+-R^+_{\mu,p}|\,\leq |R|\,,
\]
and
\begin{equation
s(V)\,:=\,\sum_{\stackrel{\scriptstyle\mu\,good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,|W\mu|\;\leq\;\mbox{\bf limit}\,,\nonumber
\end{equation}
where $\,m_{\mu}\,$ denotes the multiplicity of the weight $\,\mu\in
{\cal X}_{++}(V)\,$ and the {\bf limit}s for the different types of
Lie algebras are given in Table~\ref{table2} (see p.~\pageref{table2}).}
Observe that the necessary condition for a module to be exceptional is
given in terms of sums involving orbit sizes of weights. We start
Chapter 4 with some well-known facts on weights, their orbits and
centralizers with respect to the natural action of the Weyl group.
In Section~\ref{proced} we describe the procedure used
to classify exceptional modules
(it relies heavily on Theorem~\ref{???}). Finally, in Section~\ref{resul} we start proving the main results of
this thesis.
The complete classification of exceptional modules for Lie algebras of
exceptional type is as follows.\vspace{1ex}\\
{\bf Theorem~\ref{frlaet}}~~{\it
Let $\,G\,$ be a simply connected simple algebraic group of exceptional
type, and $\,\mathfrak g\,=\,{\cal L}(G).\,$
Let $\,V\,$ be an infinitesimally irreducible $\,G$-module.
If the highest weight of $\,V\,$ is listed in Table~\ref{table3},
then $\,V\,$ is an exceptional $\,\mathfrak g$-module. If $\,p\,$
is non-special for $\,G\,$, then the modules listed
in Table~\ref{table3} are the only exceptional $\,\mathfrak g$-modules.
\begin{table*}[htb]\begin{center}
\begin{tabular}{llcll}
\hline\hline
Type & Rank & Weights & $\dim\,V$ & Module
\\ \hline\hline \vspace{.5ex}
$\,E_6\,$ & $6$ & $\,\omega_1\,$ & $27$ & natural \\ \cline{3-5}
& & $\,\omega_2\,$ & $78,\;p>3$ & adjoint \\
& & & $77,\;p=3$ & \\ \cline{3-5}
& & $\,\omega_6\,$ & $27$
\vspace{.5ex} & twisted-natural\\ \hline \vspace{.5ex}
$\,E_7\,$ & $7$ & $\,\omega_1\,$ & $133,\;p>2$ & adjoint \\
& & & $132,\;p=2$ &\\ \cline{3-5}
& & $\,\omega_7\,$ & $56$
\vspace{.5ex} & natural \\ \hline \vspace{.5ex}
$\,E_8\,$ & $8$ & $\,\omega_8\,$ & $248$
\vspace{.5ex} & adjoint \\ \hline \vspace{.5ex}
$\,F_4\,$ & $4$ & $\,\omega_1\,$ & $52,\;p>2$ & adjoint \\
& & & $26,\;p=2$ & \\ \cline{3-5}
& & $\,\omega_4\,$ & $26,\;p\neq 3$ & natural\\
& & & $25,\;p=3$
\vspace{.5ex} & \\ \hline \vspace{.5ex}
$\,G_2\,$ & $2$ & $\,\omega_1\,$ & $7,\;p>2$ & natural\\
& & & $6,\;p=2$ & \\ \cline{3-5}
& & $\,\omega_2\,$ & $14,\;p\neq 3$ & adjoint\\
& & & $7,\;p=3$ &
\vspace{.5ex}\\ \hline \hline
\end{tabular}\end{center}\caption[The Exceptional Modules for Groups or Lie
Algebras of Exceptional Type]{Exceptional Modules for Groups of
Exceptional Type{\label{table3}}}
\end{table*}}
Comparing Tables~\ref{table1} and~\ref{table3}, we see that in the case
of exceptional groups the list of exceptional modules is the same
as in characteristic zero. (Observe that Table~\ref{table1} omits the
highest weights corresponding to the adjoint representations.)
For Lie algebras of type $\,A,\,B,\,C,\,D\,$ (classical types),
certain reduction lemmas are proved in Section~\ref{lact}.
For classical groups of low characteristics,
we expect some new highest weights to appear in the list of
exceptional modules (that is, the list of exceptional modules for
classical groups of low characteristics will differ from the list
obtained by Andreev-Vinberg-\'Elashvili).
The results obtained for the groups of classical type are as
follows.\vspace{2.2ex}\\
{\bf Theorem~\ref{anlist}}~~{\it
If $\,V\,$ is an infinitesimally irreducible $\,A_{\ell}(K)$-module
with highest weight listed
in Table~\ref{tablealall}, then $\,V\,$ is an exceptional
$\,\mathfrak g$-module.
If $\,V\,$ has highest weight different from the ones listed in Tables
~\ref{tablealall} or~\ref{leftan}, then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\begin{table*}[htb]\begin{center}
$\begin{array}{c|c|c|c|c|c}\hline\hline
{\rm N.} & {\rm Rank} & {\rm Prime} & {\rm Weights} & {\rm Module} &
\dim\,V \\
\hline \hline
1 &\ell=1 & {\rm any}
& \omega_1 & {\rm natural} & 2 \\ \hline
2 & \ell=1 & p\geq 3 & 2 \omega_1 & {\rm adjoint} &
\ell^2\,+\,2\ell\,-\,\varepsilon \\ \cline{2-4}
& \ell\geq 2 & {\rm any} & \omega_1\,+\,\omega_{\ell} & &
\varepsilon\,\in\,\{ 0,\,1\} \\ \hline
3 & \ell\geq 2 & p\geq 3 & 2\omega_1,\;2\omega_{\ell} & S^2,\,{S^2}^* &
\displaystyle\binom{\ell+2}{2} \\ \hline
4 &\ell\geq 2 & {\rm any} & \omega_1,\;\omega_{\ell} &
\mathfrak{sl},\,\mathfrak{sl}^* & \ell\,+\, 1\\ \hline
5 &\ell\geq 3 & {\rm any} & \omega_2,\;\omega_{\ell-1} &
\bigwedge^2,\,{\bigwedge^2}^* & \displaystyle\binom{\ell +1}{2} \\ \hline
6 & 5\leq\ell\leq 7 & {\rm any} & \omega_3,\;\omega_{\ell-2} &
\bigwedge^3,\,{\bigwedge^3}^* & \displaystyle \binom{\ell +1}{3}\\
\hline\hline
\end{array}$
\caption[Exceptional $\,A_{\ell}$-Modules]
{Exceptional $\,A_{\ell}$-Modules \label{tablealall}}\end{center}
\end{table*}}
\newpage\noindent
{\bf Theorem~\ref{listbn}}~~{\it Suppose $\,p\,>\,2.\,$
If $\,V\,$ is an infinitesimally irreducible
$\,B_{\ell}(K)$-module with highest weight listed in
Table~\ref{tableblall}, then $\,V\,$ is an exceptional $\,\mathfrak g$-module.
If $\,V\,$ has highest weight different from the ones listed in Tables
~\ref{tableblall} or~\ref{leftbn}, then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\begin{table*}[htb]
\[
\begin{array}{c|c|c|c|c|c}\hline\hline
{\rm N.} & {\rm Rank} & {\rm Prime} & {\rm Weights} & {\rm Module} &
\dim\,V \\ \hline\hline
1 & \ell\geq 2
& {\rm any} & \omega_1 & {\rm natural} & 2\ell\,+\,1 \\ \hline
2 & \ell=2 & p\geq 3 & 2\,\omega_2 & {\rm adjoint} &
\ell\,(2\ell\,+\,1) \\ \cline{2-4}
& \ell\geq 3 & {\rm any} & \omega_2 & & \\ \hline
3 & 2\leq\ell\leq 6 & {\rm any} & \omega_{\ell} & {\rm spin}_{2\ell+1}
& 2^{\ell} \\
\hline\hline
\end{array}
\]
\caption[Exceptional $\,B_{\ell}$-Modules]
{Exceptional $\,B_{\ell}$-Modules\label{tableblall}}
\end{table*}
\vspace*{2.5ex}\\
{\bf Theorem~\ref{listcn}}~~{\it Suppose $\,p\,>\,2.\,$
If $\,V\,$ is an infinitesimally irreducible
$\,C_{\ell}(K)$-module with highest weight listed in
Table~\ref{tableclall}, then $\,V\,$ is an exceptional $\,\mathfrak g$-module.
If $\,V\,$ has highest weight different from the ones listed in Tables
\ref{tableclall} or~\ref{leftcn}, then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\begin{table*}[htb]
\[
\begin{array}{c|c|c|c|c|c}\hline\hline
{\rm N.} & {\rm Rank} & {\rm Prime} & {\rm Weight} & {\rm Module} &
\dim\,V \\ \hline\hline
1 & \ell\geq 2 & p\geq 3 & 2\omega_1 & {\rm adjoint} &
\ell\,(2\ell\,+\,1) \\ \hline
2 & \ell\geq 2 & {\rm any} & \omega_1 & {\rm natural} & 2\,\ell \\ \hline
3 & \ell\geq 2 &
{\rm any} & \omega_2 & & \ell\,(2\ell-1)\,-\,\nu \\
& & & & & \nu\,\in\,\{1,\,2\} \\ \hline
4 & \ell=3 & {\rm any} & \omega_3 & & 14 \\ \hline\hline
\end{array}
\]
\caption[Exceptional $\,C_{\ell}$-Modules]
{Exceptional $\,C_{\ell}$-Modules\label{tableclall}}
\end{table*}
\begin{table*}[htb]
\[
\begin{array}{c|c|c|c|c|c}\hline\hline
{\rm N.} & {\rm Rank} & {\rm Prime} & {\rm Weights} & {\rm Module} & \dim\,V
\\ \hline\hline
1 & \ell=2 & p\neq 3 & \omega_1\,+\,\omega_2 & & \\ \hline
2 & \ell\geq 4 & {\rm any} & \omega_3 & & \\ \hline
3 & 7\leq \ell\leq 11 & {\rm any} & \omega_{\ell} & & \\ \hline
4 & \ell=3,\,4 & {\rm any} & 2\omega_{\ell} & & \\ \hline
5 & \ell=3 & {\rm any} & \omega_1\,+\,\omega_3 & & \\ \hline\hline
\end{array}
\]
\caption[Unclassified $\,B_{\ell}$-Modules]
{Unclassified $\,B_{\ell}$-Modules\label{leftbn}}
\end{table*}}
\begin{table*}[htb]
\[
\begin{array}{c|c|c|c|c|c}\hline\hline
{\rm N.} & {\rm Rank} & {\rm Prime} & {\rm Weight} & {\rm Module} &
\dim\,V \\ \hline\hline
1 & \ell= 2 & p\neq 3 & \omega_1\,+\,\omega_2 & & \\ \hline
2 & \ell\geq 4 & {\rm any} & \omega_3 & & \\ \hline
3 &\ell= 5 & {\rm any} & \omega_{4} & & \\ \hline
4 &\ell=4,\,5 & {\rm any} & \omega_{\ell} & & \\ \hline \hline
\end{array}
\]
\caption[Unclassified $\,C_{\ell}$-Modules]
{Unclassified $\,C_{\ell}$-Modules\label{leftcn}}
\end{table*}}
\vspace*{2ex}\noindent
{\bf Theorem~\ref{dnlist}}~~{\it
If $\,V\,$ is an infinitesimally irreducible
$\,D_{\ell}(K)$-module with highest weight listed
in Table~\ref{tabledlall}, then $\,V\,$ is an exceptional
$\,\mathfrak g$-module.
If $\,V\,$ has highest weight different from the ones listed in Tables
~\ref{tabledlall} or~\ref{leftdn}, then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\begin{table*}[tbp]
\[
\begin{array}{c|c|c|c|c|c}\hline\hline
{\rm N.} & {\rm Rank} & {\rm Prime} & {\rm Weights} & {\rm Module} &
\dim\,V \\ \hline\hline
1 & \ell\geq 4 & {\rm any} & \omega_1 & {\rm natural} & 2\,\ell \\ \hline
2 & \ell\geq 4 & {\rm any} & \omega_2 & {\rm adjoint} &
2\,\ell^2\,-\,\ell\,-\,\nu \\
& & & & & \nu\,\in\,\{1,\,2\} \\ \hline
3 & \ell= 4 & {\rm any} & \omega_3,\;\omega_4 & {\rm twisted-natural}
& 8 \\ \hline
4 & 5\leq\ell\leq 7 & {\rm any} & \omega_{\ell-1},\; \omega_{\ell}
& {\rm semi-spinor} & 2^{\ell -1} \\ \hline\hline
\end{array}
\]
\caption[Exceptional $\,D_{\ell}$-Modules]{Exceptional
$\,D_{\ell}$-Modules \label{tabledlall}}
\end{table*}
\begin{table*}[tbp]
\[
\begin{array}{c|c|c|c|c|c}\hline\hline
{\rm N.} &{\rm Rank} & {\rm Prime} & {\rm Weights} & {\rm Module} &
\dim\,V \\ \hline\hline
1 & \ell=5 & p=2, 5 & \omega_1\,+\,\omega_{4} & & \\
& & & \omega_1\,+\,\omega_5 & & \\ \hline
2 & \ell=5 & p=2 & \omega_{4}\,+\,\omega_{5} & & \\ \hline
3 & \ell=4 & {\rm any} & \omega_1\,+\,\omega_{3} & & \\
& & & \omega_1\,+\,\omega_4 & & \\
& & & \omega_{3}\,+\,\omega_{4} & & \\ \hline
4 & \ell=5 & p\geq 3 & 2\omega_{4},\;2\omega_{5} &
& \\ \hline
5 & \ell\geq 5 & {\rm any} & \omega_{3} & & \\ \hline
6 & 8\leq\ell\leq 10 & {\rm any} & \omega_{\ell-1},\; \omega_{\ell}
& & \\ \hline\hline
\end{array}
\]
\caption[Unclassified $\,D_{\ell}$-Modules]{Unclassified
$\,D_{\ell}$-Modules \label{leftdn}}
\end{table*}}
Thus, with a few exceptions, the problem of classifying
exceptional modules for the classical types is reduced to the groups
of small rank and a finite list of highest weights in each rank.
Computer calculations can be used to sort out these remaining cases
by producing central stabilizers of some
generic vectors (see e.g., \cite{cohwa}).
Due to the length and technicality of the proofs of the main theorems, we have
opted to give in Chapter 5 the complete proof of the classification of
exceptional modules just for the exceptional types. The
proof of Theorem~\ref{frlaet} is given in a series of
lemmas, and for each type separately in Section~\ref{proofexcp}.
The proofs of Theorems~\ref{anlist}, \ref{listbn}, \ref{listcn}, \ref{dnlist}
are given in the Appendix, where we also deal with
some combinatorial inequalities.
\begin{table*}[tbp]\begin{center}
$\begin{array}{c|c|c|c|c|c}\hline\hline
{\rm N.} & {\rm Rank} & {\rm Prime} & {\rm Weights} & {\rm Module} &
\dim\,V \\ \hline \hline
1 & \ell=2 & p\geq 3 & 2\omega_1\,+\,\omega_2 & & \\
& & & \omega_1\,+\,2\omega_2 & & \\ \hline
2 & \ell=3 & p=5 & 2\omega_1\,+\,\omega_3 & & \\
& & & \omega_1\,+\,2\omega_3 & & \\ \hline
3 & \ell=5 & p= 2 & \omega_1\,+\,\omega_3 & & \\
& & & \omega_{4}\,+\,\omega_{5} & & \\ \hline
4 & \ell=\,4 & p=2,\,3 & \omega_2\,+\,\omega_3 & & \\ \hline
5 & \ell\geq 3 & p\neq 3 & \omega_1\,+\,\omega_2 & & \\
& & & \omega_{\ell-1}\,+\,\omega_{\ell} & & \\ \hline
6 & 4\leq\ell\leq 6 & p\geq 3 & \omega_1\,+\,\omega_{\ell-1} & &
\\ \cline{2-3}
& \ell=7 & p=7 & & & \\ \cline{2-3}
& 4\leq\ell\leq 8 & p=2 & \omega_2\,+\,\omega_{\ell} & & \\ \hline
7 & \ell=3,\,4 & p\geq 3 & 2\omega_2,\;2\omega_{\ell-1} & & \\ \hline
8 & \ell = 9 & {\rm any} & \omega_5 & & \\ \hline
9 & 7\leq\ell\leq 11 & {\rm any} & \omega_4,\;\omega_{\ell-3} & &
\\ \hline
10 & \ell\geq 8 & {\rm any} & \omega_3,\;\omega_{\ell-2} & & \\
\hline \hline
\end{array}$\caption[Unclassified $\,A_{\ell}$-Modules]
{Unclassified $\,A_{\ell}$-Modules \label{leftan}}\end{center}
\end{table*}
\newpage
\vspace{50ex}
\part*{Chapter 2}
\part*{Preliminaries}
\addcontentsline{toc}{section}{\protect\numberline{2}Preliminaries}
\label{background}
\setcounter{section}{2}
\setcounter{subsection}{0}
As this work is about classifying certain representations of algebraic
groups in prime characteristic and the corresponding representations
of the associated restricted Lie algebras,
I will give in this Chapter a brief description of the structure of
restricted Lie algebras as well as an introduction to the results I will
need from the representation theory of algebraic groups.
The main objective is to establish notation, while more detailed
information on these structures appears in the references.
I will assume familiarity with the basic concepts of algebraic groups
and Lie algebras, while referring to the usual literature for proofs of the
results.
\subsection{Restricted Lie Algebras}\label{rla}
In this section I define restricted Lie algebra
and discuss its main properties, as well as some results that are
used in this work. The main references to this section are
\cite{jac}, \cite{stfa}, \cite{wint}.
Throughout this section, $\,F\,$ is a field of characteristic
$\,p > 0\,$ ($\,p\,$ being a prime). Lie algebras and Lie modules are
finite dimensional over $\,F$.
\begin{definition}\label{restliealg}
A {\bf Lie $\,p$-algebra} ({\bf restricted Lie algebra of
characteristic $\,p\,$}) $\,L\,$ is a Lie algebra over a field $\,F\,$
of characteristic $\,p > 0\,$ in which there is defined a mapping
$\,x\,\longmapsto\,x^{[p]}$, called {\bf $\,p$-mapping} such that
(i) $\;\,(t\,x)^{[p]}\,=\,t^p\, x^{[p]},\;(\,t\,\in\,F,\,x\,\in\,L\,)$;
(ii) $\;\,(x\,+\,y)^{[p]}\,=\,x^{[p]}\,+\,y^{[p]}\,+\,\displaystyle
\sum_{i=1}^{p-1}\,i^{-1}s_i(x,\,y)$;
(iii) $\;\,{\rm ad}(x^{[p]})\,=\,({\rm ad}\,x)^p,\;(\,x\,\in\,L\,)$;\\
where $\,s_i(x,\,y)\,$ is the coefficient of $\,t^i\,$ in
$\,{\rm ad}(t x\,+\,y)^{p-1}(x),\;(\,x,\,y\in\,L\,)\,$.
\end{definition}
Formula {\it (ii)} is called {\bf Jacobson's Identity}.
As usual, we write $\,{\rm ad}(x)(y)\,=\,[\,x,\,y\,]$.
Note that when $\,[\,x,\,y\,]\,=\,0\,$ then {\it (ii)} becomes
{\it (ii')} $\;\,(x\,+\,y)^{[p]}\,=\,x^{[p]}\,+\,y^{[p]}$.\\
\notation $\,(L,\,[p])\,$ denotes a restricted Lie algebra of
characteristic $\,p$.
\begin{example} \label{examrest}
{\rm (1) A typical example of a restricted Lie algebra is a Lie subalgebra
$\,L\,$ of an associative algebra $\,\cal A\,$ stable under the $\,p$th
power map in $\,\cal A\,$ (in this case the $\,p$th power map defines the
$\,[p]$-structure in $\,L\,$).
For an associative algebra $\,A\,$ consider a new operation defined
by $\,[x,\,y]\,=\,x\,y\,-\,y\,x\,$, for all $\,x,\,y\in A\,$. This
gives $\,A\,$ a Lie algebra structure. Denote this Lie algebra
by $\,A^{(-)}\,$.
If the base field of $\,A\,$ has characteristic $\,p,\,$ then
$\,A^{(-)}\,$ carries a canonical restricted Lie algebra structure,
given by $\,x\longmapsto x^p\,$ (the $\,p$th power map).
In particular, $\,\mathfrak{gl}(V)\,:=\,({\rm End}(V))^{(-)},\,$
where $\,V\,$ a finite dimensional vector space over a field $\,F\,$
of characteristic $\,p,\,$ is a restricted Lie algebra.}
{\rm (2) Let $\,{\mathfrak U}\,$ be an arbitrary (non-associative) algebra.
A {\it derivation} of $\,{\mathfrak U}\,$ is a linear mapping
$\,D\,:\,{\mathfrak U}\longrightarrow {\mathfrak U}\,$ satisfying the
rule for the derivative of a product, namely,
$\,D(a\,b)\,=\,D(a)\,b\,+\,a\,D(b)\,$, for all $\,a,\,b\,\in\mathfrak U\,$. Let
$\,{\mathfrak D}({\mathfrak U})\,$ be the set of all derivations of
$\,{\mathfrak U}$. Then
$\,{\mathfrak D}({\mathfrak U})\,$ is a subalgebra of the Lie
algebra $\,(\End({\mathfrak U}))^{(-)}$.
One has the Leibniz formula
\begin{equation} \label{leibniz}
D^k(ab)\,=\,\sum_{i=1}^k\,\binom{k}{i}\,D^i(a)\,D^{k-i}(b)
\end{equation}
for any $\,D\,\in\,{\mathfrak D}({\mathfrak U})\,$
(this can be established by induction on
$\,k$). Now, assuming that the base field
is of characteristic $\,p\,$ and taking
$\,k\,=\,p\,$ in~(\ref{leibniz}), we have that the binomial
coefficients $\,\displaystyle\binom{p}{i}\,$ vanish for $\,1\leq i\leq p-1$.
Hence~(\ref{leibniz}) reduces to
\begin{equation}
D^p(ab)\,=\,D^p(a)\,b\,+\,a\,D^p(b),
\end{equation}
which implies $\,D^p\,\in\,{\mathfrak D}({\mathfrak U})\,$.
Thus $\,{\mathfrak D}({\mathfrak U})\,$
is closed under the mapping $\,D\,\longmapsto\,D^p\,$.
Hence $\,{\mathfrak D}({\mathfrak U})\,$ is a
restricted Lie algebra.}
\end{example}
\medskip
Ideals, subalgebras and modules of restricted Lie algebras are defined in the
obvious way.
\begin{definition}
Let $\,(L,\,[p])\,$ be a restricted Lie algebra over $\,F$.
A {\bf restricted Lie subalgebra or $\,p$-subalgebra} (respectively {\bf
$\,p$-ideal)} of $\,L\,$ is a subalgebra (respectively ideal) which is
stable under the $\,p$-mapping $\,[p].\,$
Such a subalgebra
(respectively, ideal) is regarded as a Lie $\,p$-algebra by taking its
$\,p$-mapping to be the restriction of that of $\,L$.
\end{definition}
\begin{definition}
Let $\,(L_1,\,[p]_1)\,$ and $\,(L_2,\,[p]_2)\,$ be restricted Lie
algebras over $\,F$. A homomorphism $\,f\,:\,L_1\,\rightarrow\,L_2\,$ is
called {\bf restricted} (or {\bf $\,p$-homomorphism}) if
$\,f(x^{[p]_1})\,=\,f(x)^{[p]_2},\;\forall\,x\,\in\,L_1$.
\end{definition}
\begin{definition}
A {\bf $\,p$-representation} (or {\bf restricted representation})
of a restricted Lie algebra $\,L\,$ in a vector space $\,V\,$ is a
restricted homomorphism from $\,L\,$ into $\,{\mathfrak{gl}\,}(V).\,$
A {\bf Lie $\,p$-module} for a Lie $\,p$-algebra $\,L\,$ is a
$\,L$-module $\,V\,$ such that
\[
(x^{[p]})\cdot v \,=\,x\cdot (x\cdot(\cdots (x\cdot v)\cdots
))\;({p}\;\mbox{times}),\;\forall\,x\,\in\,L,\;v\in V.
\]
\end{definition}
\begin{example}{\rm
If $\,L\,$ is a Lie $\,p$-algebra, then by \ref{restliealg}(iii)
$\,{\rm ad}\,:\,L\longrightarrow \mathfrak{gl}(L)\,$ is a
$\,p$-rep\-re\-sent\-a\-tion of $\,L.\,$
Consequently, the centre $\,Z\,$ of $\,L\,$ is a $\,p$-ideal,
for $\,Z\,=\,\ker\,{\rm ad}\,$ (see \cite[p.\ 69]{stfa})}.
\end{example}
\subsubsection{Nilpotent, Semisimple and Toral Elements}\label{nste}
Let $\,(L,\,[p])\,$ be a restricted Lie algebra over $\,F$.
Given $\,i\in\mathbb Z^+\,$ denote by
$\,x^{[p]^i}\,$ the image of $\,x\,$ under
the $\,i$th iterate of $\,x\longmapsto x^{[p]}\,$ (with
$\,x^{[p]^0}\,=\,x\,$).
\begin{definition}
A $\,p$-ideal $\,I\,$ of $\,L\,$ is called {\bf $\,p$-nilpotent}
if there is $\,n\,\in\,\mathbb N\,$ such that $\,I^{[p]^n}\,=\,0$. An
element $\,x\,\in\,L\,$ is called {\bf $\,p$-nilpotent} if there is
$\,n\,\in\,\mathbb N\,$ such that $\,x^{[p]^n}\,=\,0$. The $\,p$-ideal
$\,I\,$ is called {\bf $\,p$-nil} if every element $\,x\,\in\,I\,$ is
$\,p$-nilpotent. The set $\,{\cal N}(L)\,:=\,\{\,x\in L
\,/\,x^{[p]^e}\,=\,0\;\,\mbox{for}\;\,e>>0\,\}\,$ is called the {\bf
nilpotent cone} of $\,L.\,$
\end{definition}
Let $\,V\,$ be a finite dimensional vector space over a field $\,F$.
An endomorphism
$\,\sigma\,:\,V\,\rightarrow\,V\,$ is called {\bf semisimple} if its
minimal polynomial has distinct roots in some field extension of
$\,F\,$ (so that $\,\sigma\,$ is diagonalizable after some base field
extension). From general algebra we have the following characterization:
$\,\sigma\,$ is semisimple if the ideal, in $\,F[X]$, generated by
the minimum polynomial of $\,\sigma\,$ and its derivative, contains $\,1$.
Let $\,<x>\,$ denote the smallest restricted
subalgebra of $\,L\,$ containing $\,x$, i.e., the linear span
of $\,\{ \ x^{[p]^i}\,/\,i\in\mathbb Z^+ \ \}\,$.
\begin{definition}
Let $\,(L,\,[p])\,$ be a restricted Lie algebra over $\,F$. An element
$\,x\,\in\,L\,$ is called {\bf $\,p$-semisimple} (or {\bf semisimple}
for short) if
$\,x\,\in\,<x^{[p]}>\,$; {\bf toral} if $\,x^{[p]}\,=\,x$.
\end{definition}
\begin{proposition}{\rm\cite[p. 80]{stfa},~\cite[V.7]{seli}} \label{properties}
Let $\,(L,\,[p])\,$ be a restricted Lie algebra. Then the following
statements hold:
(1) Each toral element is $\,p$-semisimple.
(2) If $\,x\in L\,$ is $\,p$-semisimple, then the endomorphism
$\,\psi (x)\,$ is semisimple for
every finite dimensional restricted representation
$\,\psi\,:\,L\,\rightarrow\,\mathfrak{gl}(V)$.
(3) If $\,x,\,y\,\in\,L$ are $\,p$-semisimple and $\,[x,y]\,=\,0$, then
$\,x\,+\,y\,$ is $\,p$-semisimple.
(4) If $\,x\in L\,$ is $\,p$-semisimple, then $\,<x>\,$ consists of
$\,p$-semisimple elements.
(5) If $\,K\,$ is perfect and $\,L\,$ contains no nontrivial
$\,p$-nilpotent elements, then each
element in $\,L\,$ is $\,p$-semisimple.
(6) An endomorphism $\,\sigma\,\in\,End_F(V)\,$ is semisimple if and
only if it is semisimple as an element of the restricted Lie algebra
$\,\mathfrak{gl}(V)\,:=\,{\rm End}(V)^{(-)}\,$.
\end{proposition}
The significance of semisimple elements rests on the following result.
\begin{theorem}{\rm\cite[p. 80]{stfa}}\label{sigsemi}
Let $\,(L,\,[p])\,$ be a restricted Lie algebra over $\,F$. For each
$\,x\,\in\,L\,$ there exists $\,k\,\in\,\mathbb N\,$ such that
$\,x^{[p]^k}\,$ is $\,p$-semisimple.
\end{theorem}
If $\,F\,$ is perfect, one can prove a much stronger result
closely related to the well-known Jordan-Chevalley decomposition of an
endomorphism.
\begin{theorem}{\rm\cite[p. 81]{stfa}}\label{jcd}
Let $\,F\,$ be a perfect field and let $\,(L,\,[p])\,$ be a finite
dimensional restricted Lie algebra over $\,F$. Then for any
$\,x\,\in\,L\,$ there are uniquely determined elements $\,x_n,\,x_s\,\in\,L\,$
such that
(1) $\,x_n\,$ is $\,p$-nilpotent, $\,x_s\,$ is $\,p$-semisimple.
(2) $\,x\,=\,x_s\,+\,x_n,\;\,[x_s,\,x_n]\,=\,0$.
\end{theorem}
The decomposition obtained can be refined in the case of an
algebraically closed field $\,K$.
\begin{definition}
A $\,p$-mapping $\,[p]\,$ on $\,L\,$ is called {\bf nonsingular}
if~$\,x^{[p]}\,\neq\,0$ for all $\,x\,\in\,L\setminus\,\{ 0\}$.
\end{definition}
The following useful result (due to N. Jacobson) can be found in
\cite[p. 81]{stfa}.
\begin{theorem}{\rm\cite[p. 82]{stfa}} \label{3.6}
Let $\,(L,\,[p])\,$ be a finite dimensional restricted Lie algebra
over an algebraically closed field $\,K$. Then the following statements
hold:
(1) If $\,L\,$ is abelian and the $\,p$-mapping is nonsingular, then $\,L\,$
possesses a basis consisting of toral elements.
(2) For any $\,x\,\in\,L\,$ there exist toral elements
$\,x_1,\,\ldots,\,x_r\,\in\,L,\,$ scalars
$\,c_1,\,\ldots,\,c_r\,\in\,K,\,$ and a $\,p$-nilpotent element
$\,y\in L\,$ such that
\[
x\,=\,y\,+\,\sum_{i=1}^r\,c_i\,x_i\,,\;\;\;[y,\,x_i]\,=\,[x_i,\,x_j]\,
=\,0\;\;\;\forall\, i,\,j.
\]
\end{theorem}
\begin{definition} \label{resttorus}
Let $\,(L,\,[p])\,$ be a restricted Lie algebra over $\,F$. A subalgebra
$\,T\subset L\,$ is called a {\bf torus} or a {\bf toral subalgebra} if
$\,T\,$ is an abelian $\,p$-subalgebra, consisting of
$\,p$-semisimple elements.
\end{definition}
It follows from the definition and Proposition~\ref{properties}(2),
that if $\,T\,$ is a torus of $\,L\,$ and $\,\varphi\,$ is a
$\,p$-representation of $\,L,\,$ then $\,\varphi(T)\,$ is
diagonalizable (see \cite[4.5.5]{wint}).
\begin{lemma} \label{ppp}
Let $\,(L,\,[p])\,$ be a finite dimensional restricted Lie algebra
over an algebraically closed field $\,K$. If $\,(L,\,[p])\,$
contains no nonzero $\,p$-nilpotent elements, then $\,(L,\,[p])\,$ is toral.
\end{lemma}\noindent
\begin{proof} By Theorem~\ref{jcd}(2), each element of $\,(L,\,[p])\,$
is semisimple, so it remains to prove
that $\,(L,\,[p])\,$ is abelian, i.e.,
$\,{\rm ad}\;x\,=\,0$, for each $\,x\,\in\,L$. As $\,{\rm ad}\;x\,$ is
diagonalizable ($\,{\rm ad}\;x\,$ being semisimple and $\,K\,$ algebraically
closed), we have to show that $\,{\rm ad}\;x\,$ has no nonzero eigenvalues.
Suppose, on the contrary, that $\,[x,\,y]\,=\,a\,y\;(a\neq 0)\,$ for some
nonzero $\,y\in L$. Then
\[
({\rm ad}\;y)^2\,(x)\,=\,[y,\,[y,\,x]]\,=\,-a\,[y,\,y]\,=\,0,\,\qquad\mbox{(*)}
\]
i.e., $\,[y,\,x]\,$ is an eigenvector of $\,{\rm ad}\;y\,$ of
eigenvalue $\,0$.
Now write $\,x\,$ as a linear combination of eigenvectors of
$\,{\rm ad}\;y\,$ ($\,y\,$ is also semisimple).
Clearly, $\,[y,\,x]\,$ is a combination of $\,{\rm ad}\;y$-eigenvectors which
belong to nonzero eigenvalues, if any. This, however, contradicts (*).
\end{proof}
\subsubsection{Restricted Universal Enveloping Algebras}
For restricted Lie algebras there is an analogue of the universal
enveloping algebra.
This structure plays an important role in the
representation theory of algebraic groups over a field of positive
characteristic.
\begin{definition} \label{ruel}
Let $\,(L,\,[p])\,$ be a restricted Lie algebra. A pair $\,({\mathfrak
u}(L),\,i)\,$ consisting of an associative $\,K$-algebra with unity
and a restricted homomorphism $\,i\,:\,L\,\longrightarrow\,
{\mathfrak u}(L)^{(-)}\,$
is called a {\bf restricted universal enveloping algebra} if given any
associative $\,K$-algebra $\,A\,$ with unity and any restricted
homomorphism $\,f\,:\,L\,\longrightarrow\,A^{(-)}$, there is a unique
homomorphism $\,F\,:\,{\mathfrak u}(L)\,\longrightarrow\,A\,$ of
associative $\,K$-algebras such that $\,F\circ i\,=\,f$.
\end{definition}
The restricted universal enveloping algebra $\,{\mathfrak u}(L)\,$ is
the quotient of the ordinary universal enveloping
algebra $\,{\mathfrak U}(L)\,$ by the two-sided ideal generated by all
$\,x^p - x^{[p]}$, where $\,x\in L\,$ (see \cite[V,Theorem 12]{jac}).
The universal property of $\,{\mathfrak u}(L)\,$ shows the
uniqueness of the restricted universal enveloping algebra of $\,L\,$ up to
isomorphism (see \cite[I.8.1]{stfa}).
\begin{theorem}{\rm\cite[p. 91]{stfa}} \label{ruelbasis}
Let $\,(L,\,[p])\,$ be a restricted Lie algebra. Then the following
statements hold:
(1) The restricted universal enveloping algebra of $\,L\,$ exists.
(2) If $\,({\mathfrak u}(L),\,i)\,$ is a restricted universal enveloping
algebra of $\,L\,$ and $\,(e_j)_{j\in J}\,$ is an ordered basis of
$\,L\,$ over $\,K$, then the elements
$\,i(e_{j_1})^{s_1}\,\cdots\,i(e_{j_n})^{s_n}\,$, where
$\,j_1\,<\,\cdots\,<\,j_n,\;n\,\geq\,1$, $\,0\,\leq\,s_k\,\leq\,p-1$,
form a basis of $\,{\mathfrak u}(L)\,$ over $\,K$. In particular,
$\,i\,:\,L\,\longrightarrow\,{\mathfrak u}(L)\,$ is injective and
$\,\dim_F\,{\mathfrak u}(L)\,=\,p^n\,$ if $\,\dim_F\,L\,=\,n$.
\end{theorem}
We identify $\,L\,$ with its image $\,i(L).\,$
Note that the
finiteness of the dimension is preserved when passing from restricted
Lie algebras to their restricted enveloping algebras, in contrast with
the ordinary Lie algebras and their envelopes. \vspace{1.5ex}\\%However,
\noindent
\begin{remark}\label{extrest}
It follows from Definition~\ref{ruel} that a $\,p$-representation of
the restricted Lie algebra $\,L\,$ extends uniquely to a
representation of $\,\mathfrak u(L)\,$ and, conversely,
any representation of $\,\mathfrak u(L)\,$ restricts to a
$\,p$-representation of $\,L.\,$
\end{remark}
\subsection{The Lie Algebra of an Algebraic Group}\label{tlaag}
Throughout this section let $\,K\,$ be an algebraically closed field
and $\,G\,$ be a connected algebraic group over $\,K$.
In this section I define the Lie algebra $\,{\cal L}(G)\,$ of the
group $\,G\,$ and show that if $\,{\rm char}\,K\,>\,0,\,$ then $\,{\cal
L}(G)\,$ is a restricted Lie algebra. I also recall some
results related to the general structure of algebraic groups
and their Lie algebras. Standard notions of
algebraic geometry used here can be found in the first chapter of
\cite{bor2} or \cite{hum2}.
Denote by $\,K[G]\,$
the coordinate ring of (the irreducible affine variety) $\,G$.
Let $\,T_e(G)\,$ denote the tangent space of $\,G\,$ at the
identity element $\,e\in G$. $\,T_e(G)\,$ can be identified with the space
$\,{\rm Der}(K[G],\,K_e)\,$ of the point-derivations of $\,G\,$ at $\,e\,$
(for definition and notation see~\cite[1.3]{car1} or~\cite[I.3.3]{bor2}).
If $\,f\in K[G]\,$ and $\,g\in G,\,$ the map
$\,f^g\,:\,G\longrightarrow K\,$ defined by $\,f^g(t)\,=\,f(tg)\,$
lies in $\,K[G]\,$ (since right multiplication by $\,g\,$ is a morphism
of $\,G\,$ \cite[I.1.9]{bor2}).
Let $\,\alpha_g\,:\,K[G]\longrightarrow K[G]\,$ be defined by
$\,\alpha_g(f)\,=\,f^g$. Then $\,\alpha_g\,$ is a $\,K$-algebra
automorphism of $\,K[G].\,$ Furthermore,
$\,\alpha_{gh}\,=\,\alpha_g\alpha_h\,$ and so we have a homomorphism
from $\,G\,$ into the group of $\,K$-algebra automorphisms of $\,K[G]$.
Recall that the set of all derivations $\,{\rm Der}_K(K[G])\,$ of the
algebra $\,K[G]\,$ forms a Lie algebra under
the Lie multiplication $\,[D_1,D_2]\,=\,D_1D_2\,-\,D_2D_1\,$.
(cf.~Example~\ref{examrest}(2)).
A derivation $\,D\in{\rm Der}_K(K[G])\,$ is said to be {\bf invariant}
if $\,D(f^g)\,=\,(Df)^g\,$ for all $\,f\in K[G]\,$ and $\,g\in G$.
The set $\,{\rm Der}_K(K[G])^G\,$ of all invariant derivations forms a Lie
subalgebra of $\,{\rm Der}_K(K[G])\,$. If $\,K\,$ has characteristic
$\,p\,$ and $\,D\,$ is an invariant derivation, then so is $\,D^p\,$.
This gives $\,{\rm Der}_K(K[G])^G\,$ a restricted Lie algebra structure
(see~\cite[1.3]{car1}).
Given $\,D\in {\rm Der}_K(K[G])^G$, the map $\,f\longmapsto Df(e)\,$
is a point-derivation of $\,G\,$ at the identity element. This gives
rise to a map $\,\varphi : {\rm Der}_K(K[G])^G\longrightarrow
{\rm Der}(K[G],\,K_e)\,$ from the Lie algebra of invariant derivations
of $\,K[G]\,$ to the space of point-derivations of $\,G\,$ at $\,e.\,$
By~\cite[I.3.4, I.3.5]{bor2},
$\,\varphi\,$ is an isomorphism of vector spaces. Since
$\,{\rm Der}_K(K[G])^G\,$ has a Lie algebra structure, $\,\varphi\,$ gives a
restricted Lie algebra structure to $\,{\rm Der}(K[G],\,K_e).\,$
Thus, the tangent space $\,T_e(G)\,$ to $\,G\,$ at the
identity element has a canonical restricted Lie algebra structure.
This restricted Lie algebra is denoted by $\,\mathfrak g\,=\,{\cal L}(G)\,$ and
called the {\bf Lie algebra of the algebraic group $\,G$}.
The Lie algebra of an algebraic group $\,G,\,H,\,M,\ldots\,$ is often denoted
by the corresponding German character $\,\mathfrak{g,\,h,\,m,}\ldots\,$.
We have $\,\dim\,{\cal L}(G)\,=\,\dim\,G\,$ since $\,G\,$ is a smooth
variety~\cite[1.3]{car1}. One can also define the Lie algebra of a
disconnected algebraic group, but then $\,{\cal L}(G)\,=
\,{\cal L}(G^{\circ})\,$, where $\,G^{\circ}\,$
is the connected component of $\,G\,$ containing the identity element
$\,e\in G\,$ \cite[I.3.6]{bor2}.
If $\,H\,$ is a closed subgroup of $\,G\,$ then
$\,\mathfrak h\,$ is a restricted subalgebra of $\,\mathfrak g\,$
\cite[I.3.8]{bor2}.
If $\,U\,$ is a unipotent subgroup of $\,G$, then $\,{\cal L}(U)\,$
consists of $\,p$-nilpotent elements of the restricted Lie algebra
$\,{\cal L}(G)\,$ \cite[I.4.8]{bor2}.
\medskip
Let $\,\phi : G \rightarrow G'\,$ be a homomorphism of algebraic groups.
There is an associated Lie algebra homomorphism
$\,{\rm d}\phi : {\cal L}(G) \rightarrow {\cal L}(G')$, given by
$\,{\rm d}\phi (D) = D \circ
\phi^*$. This homomorphism is called the {\bf differential of} $\,\phi\,$
\cite[Theorem~9.1]{hum2}. Here $\,\phi^* : K[G']\rightarrow K[G]\,$
denotes the {\it comorphism} associated with the morphism
$\,\phi : G\rightarrow G'\,$ \cite[1.5]{hum2}.
If $\,\phi_1 : G_1\rightarrow G_2\,$ and
$\,\phi_2 : G_2\rightarrow G_3\,$ are homomorphisms of algebraic
groups, then their differentials satisfy
$\,{\rm d}(\phi_2\circ \phi_1)\,=\,{\rm d}(\phi_2)\circ{\rm d}
(\phi_1) \,$.
Let $\,V\,$ be a finite dimensional vector space over the field $\,K.\,$
A representation $\,\pi\,:\,G\longrightarrow \GL(V)\,$ which is also a
morphism of algebraic varieties is called a
{\bf rational representation} of $\,G$.
The differential $\,{\rm d}\pi\,:\,\mathfrak g\longrightarrow
\mathfrak{gl}(V)\,$ defines a $\,p$-representation of the restricted
Lie algebra $\,\mathfrak g\,=\,{\cal L}(G)\,$ \cite[I.3.10]{bor2}.
The {\bf adjoint representation} of an algebraic group $\,G\,$ is defined
as follows.
For each $\,x\in G$, the inner automorphism $\,{\rm Int}\,x\,(y)\,=
\,x\,y\,x^{-1}\;(x,\,y\,\in G\,)\,$ of $\,G\,$ preserves the identity
element. This induces a linear action of $\,G\,$ on the tangent space
$\,{\cal L}(G)\,=\,T_e(G),\,$ denoted by
$\,\Ad\,:\,G\longrightarrow\GL({\cal L}(G))\,$ and called the
{\it adjoint representation} of $\,G\,$.
The differential of $\,\Ad\,$ is $\,{\rm ad}\,:\,
{\cal L}(G)\longrightarrow \mathfrak{gl}({\cal L}(G)),\,$ where
$\,({\rm ad}\,a)(b)\,=\,[a,\,b]\,$ for all $\,a,\,b\in {\cal L}(G)\,$
\cite[10.3, 10.4]{hum2}.
\subsection{General Notions Related to Algebraic Groups}\label{gennot}
Let $\,G\,$ be a connected reductive algebraic group over $\,K\,$ and
let $\,T\,$ be a maximal torus of $\,G.\,$
Let $\,B\,$ be a Borel subgroup of $\,G\,$ containing $\,T.\,$ Then
$\,B\,$ has a semidirect product decomposition $\,B\,=\,T\,U ,\,$ where
$\,U\,=\,R_{\rm u}(B)\,$ is the unipotent radical of $\,B\,$
\cite[19.5]{hum2}. There is a unique Borel subgroup $\,B^-\subset G\,$
containing $\,T\,$ and such that $\,B\cap B^-\,=\,T.\,$ $\,B\,$ and $\,B^-\,$
are called {\bf opposite} Borel subgroups. We have $\,B^-\,=\,T\,U^-\,$,
where $\,U^-\,=\,R_{\rm u}(B^-)\,$ \cite[Cor. 26.2]{hum2}.
$\,U\,$ and $\,U^-\,$ are connected subgroups normalized by
$\,T\,$ and satisfying $\,U\cap U^-\,=\,\{ e\}.\,$ They
are maximal unipotent subgroups of $\,G.\,$
\subsubsection{The Root System}\label{therootsys}
Consider the minimal subgroups of positive dimension in $\,U\,$ and $\,U^-\,$
which are normalized by $\,T.\,$ These are connected unipotent groups,
isomorphic to the additive group $\,{\mathbf G}_a\,$ \cite[20.5]{hum2}.
$\,T\,$ acts on each of them by conjugation, giving rise to
a homomorphism
from $\,T\,$ to the group of algebraic automorphisms of $\,{\mathbf G}_a.\,$
However, the only algebraic automorphisms of $\,{\mathbf G}_a\,$ are the
maps $\,\lambda\longmapsto\mu\lambda\,$ for some $\,\mu\in K^*.\,$
Thus $\,{\rm Aut}\,{\mathbf G}_a\,$ is isomorphic to $\,{\mathbf G}_m,\,$
the multiplicative group of $\,K$.
Hence, each of our $\,1$-dimensional unipotent groups determines
an element of $\,{\rm Hom}(T,{\mathbf
G}_m)\,=\,X\,$. (Here $\,X\,=\,X(T)\,$ is the so-called {\it character
group} of $\,T\,$.)
The elements of $\,X\,$ arising in this way are called the {\bf roots} of
$\,G$.
Distinct $\,1$-dimensional
unipotent subgroups give rise to distinct roots. The roots form a finite
subset $\,R\,$ of $\,X\,$ (which is independent of the choice of the
Borel subgroup $\,B\,$ containing $\,T\,$)~\cite[1.9]{car1}.
For each root $\,\alpha\in R,\,$ the $\,1$-dimensional unipotent subgroup
giving rise to it is denoted by $\,U_{\alpha}\,$.
The $\,U_{\alpha}\,$ are called the {\bf root subgroups} of $\,G.\,$
The roots arising from $\,U^-\,$ are the negatives
of the roots arising from $\,U.\,$
Let $\,W\,=\,N(T)/T\,$ be the {\bf Weyl group} of $\,G\,$ (see
\cite[24.1]{hum2}).
\begin{theorem}{\rm\cite[27.1]{hum2}}\label{ssars}
Let $\,G\,$ be a semisimple algebraic group, and \linebreak
$\,E\,=\,X\otimes_{\mathbb Z}{\mathbb R}\,$. Then $\,R\,$ is an abstract
root system in $\,E$, whose rank is the rank of $\,G\,$ and whose abstract
Weyl group is isomorphic to $\,W$.
\end{theorem}
Thus, all results on abstract root systems are applicable to
$\,R.\,$ Here we summarize some of them.
A {\bf basis} of $\,R\,$ is a subset
$\,\Delta\,=\,\{\alpha_1,\ldots,\alpha_{\ell}\},\,$ $\,\ell\,=\,{\rm
rank}\,G,\,$ which spans $\,E\,$ (hence is a basis of $\,E\,$) and
relative to which each root $\,\alpha\,$ has a (unique) expression
$\,\alpha\,=\,\sum\,c_i\,\alpha_i\,$, where the $\,c_i$'s are integers
of like sign. The elements of $\,\Delta\,$ are called {\bf simple
roots}. Bases do exist; in fact, $\,W\,$ permutes them simply
transitively, and each root lies in at least one basis. Moreover,
$\,W\,$ is generated by the reflections $\,s_{\alpha}\,$
($\,\alpha\in\Delta\,$), for any basis $\,\Delta\,$
(see~\cite[III]{hum1}). There is an inner product
$\,(\,\cdot\,,\,\cdot\,)\,$ on $\,E\,$ relative to which $\,W\,$ consists
of orthogonal transformations. The formula for
$\,s_{\alpha}\,$ becomes $\,s_{\alpha}(\beta)\,=\,\beta\,-\,
\langle\beta,\,\alpha\rangle\,\alpha\,$, where
$\,\langle\beta,\,\alpha\rangle\,=\,2(\beta,\,\alpha)/(\alpha,\,\alpha).\,$
Bases of $\,R\,$ correspond one-to-one to Borel subgroups
containing $\,T\,$ \cite[27.3]{hum2}. In particular, each choice
of a Borel subgroup $\,B\,$ containing $\,T\,$ amounts to a choice of
$\,\Delta,\,$ or to a choice of {\bf positive} roots $\,R^+\,$ (those
for which all $\,c_i\,$ above are nonnegative). Moreover, if $\,\Delta\,$ is a
basis of $\,R,\,$ then $\,G\,=\,\langle\, T,\,U_{\alpha}\,/\,\pm\alpha\in
\Delta\,\rangle\,$ \cite[Theorem 27.3]{hum2}.
We recall that $\,R\,$ is called {\bf irreducible} if
$\,\Delta\,$ cannot be partitioned into two disjoint
``orthogonal'' subsets \cite[10.4]{hum1}.
Every root system is the disjoint union of
(uniquely determined) irreducible root systems in suitable subspaces
of $\,E\,$ \cite[11.3]{hum1}.
Up to isomorphism, the irreducible root systems correspond
one-to-one to the following {\bf Dynkin diagrams} \cite[11.4]{hum1}:
\begin{center}
\setlength{\unitlength}{1cm}
\begin{picture}(12,15)
\put(0,14.5){$A_{\ell}\,(\ell\geq 1)$:}
\multiput(2.5,14.6)(1.2,0){6}{\circle*{0.2}}
\put(2.6,14.6){\line(1,0){3.6}}
\multiput(6.35,14.5)(0.2,0){4}{$\cdot$}
\put(7.4,14.6){\line(1,0){1.2}}
\put(0,13){$B_{\ell}\,(\ell\geq 2)$:}
\multiput(2.5,13.1)(1.2,0){6}{\circle*{0.2}}
\put(2.6,13.1){\line(1,0){2.4}}
\multiput(5.15,13)(0.2,0){4}{$\cdot$}
\put(6.2,13.1){\line(1,0){1.2}}
\put(7.2,13.15){\line(1,0){1.2}}
\put(7.3,13.05){\line(1,0){1.2}}
\put(7.7,13){$>$}
\put(0,11.5){$C_{\ell}\,(\ell\geq 3)$:}
\multiput(2.5,11.6)(1.2,0){6}{\circle*{0.2}}
\put(2.6,11.6){\line(1,0){2.4}}
\multiput(5.15,11.5)(0.2,0){4}{$\cdot$}
\put(6.2,11.6){\line(1,0){1.2}}
\put(7.3,11.65){\line(1,0){1.2}}
\put(7.3,11.55){\line(1,0){1.2}}
\put(7.8,11.5){$<$}
\put(0,10){$D_{\ell}\,(\ell\geq 4)$:}
\multiput(2.5,10.1)(1.2,0){5}{\circle*{0.2}}
\put(8.5,10.65){\circle*{0.2}}
\put(8.5,9.55){\circle*{0.2}}
\put(2.6,10.1){\line(1,0){2.4}}
\multiput(5.15,10)(0.2,0){4}{$\cdot$}
\put(6.2,10.1){\line(1,0){1.2}}
\put(7.39,10.1){\line(2,1){1.2}}
\put(7.39,10.1){\line(2,-1){1.2}}
\put(0,8.5){$E_{6}$:}
\multiput(1.5,8.6)(1.2,0){5}{\circle*{0.2}}
\put(1.6,8.6){\line(1,0){4.8}}
\put(3.9,7.4){\circle*{0.2}}
\put(3.9,7.4){\line(0,1){1.2}}
\put(0,6.8){$E_{7}$:}
\multiput(1.5,6.9)(1.2,0){6}{\circle*{0.2}}
\put(1.6,6.9){\line(1,0){6}}
\put(3.9,5.7){\circle*{0.2}}
\put(3.9,5.7){\line(0,1){1.2}}
\put(0,5.1){$E_{8}$:}
\multiput(1.5,5.2)(1.2,0){7}{\circle*{0.2}}
\put(1.6,5.2){\line(1,0){7.2}}
\put(3.9,4){\circle*{0.2}}
\put(3.9,4){\line(0,1){1.2}}
\put(0,3.3){$F_{4}$:}
\multiput(1.5,3.4)(1.2,0){4}{\circle*{0.2}}
\put(1.5,3.4){\line(1,0){1.2}}
\put(3.9,3.4){\line(1,0){1.2}}
\put(2.7,3.45){\line(1,0){1.2}}
\put(2.7,3.35){\line(1,0){1.2}}
\put(3.2,3.3){$>$}
\put(0,1.8){$G_{2}$:}
\multiput(1.5,1.9)(1.2,0){2}{\circle*{0.22}}
\multiput(1.5,2)(0,-0.1){3}{\line(1,0){1.2}}
\put(2,1.8){$\langle$}
\end{picture}
\end{center}
\vspace*{-6ex}
The vertices of the Dynkin diagram correspond to the simple roots
$\,\alpha_i\,$ ($\,i=1,\ldots,\ell\,$). The vertices corresponding to
$\,\alpha_i,\,\alpha_j\,$ are joined by
$\,\langle\alpha_i,\,\alpha_j\rangle\,\langle\alpha_j,\,\alpha_i\rangle\,$
edges, with an arrow pointing to the shorter of the two roots if they
are of unequal length.
The order of $\,s_{\alpha_i}\,s_{\alpha_j}\,$ in $\,W\,$ is
$\,2,\,3,\,4,\,$ or $\,6,\,$ according to whether $\,\alpha_i\,$ and
$\,\alpha_j\,$ are joined by $\,0,\,1,\,2,\,$ or $\,3\,$ edges
($\,\alpha_i\,\neq\,\alpha_j\,$). The giving of the Dynkin diagram is
equivalent to the giving of the matrix of {\bf Cartan integers}
$\,\langle\alpha_i,\,\alpha_j\rangle\,$
($\,\alpha_i,\,\alpha_j\,\in\Delta\,$). See \cite[11.2]{hum1}.
By~\cite[27.5]{hum2}, if $\,G\,$ is a semisimple algebraic group,
then it has a decomposition $\,G\,=\,G_1\,\cdots\,G_n\,$, where each
$\,G_i\,$ is a minimal Zariski closed connected normal subgroup of positive
dimension. Moreover, this decomposition corresponds precisely to the
decomposition of $\,R\,$ into its irreducible components.
An algebraic group is {\bf simple} if it is non-commutative and has no
Zariski closed, connected, normal subgroups other than itself and $\,e\,$.
We say that a simple group is of {\bf exceptional type} if its (irreducible)
root system has Dynkin diagram of type $\,E_6,\,E_7,\,E_8,\,F_4,\,$ or
$\,G_2\,$. A simple group is of {\bf classical type} if its (irreducible)
root system has Dynkin diagram of type $\,A_{\ell},\,B_{\ell},\,
C_{\ell},\,$ or $\,D_{\ell}.\,$
\subsubsection{The Structure of the Lie Algebra}\label{structliealg}
Let $\,{\mathfrak g}\,=\,{\cal L}(G)\,$ be the Lie algebra of the
connected reductive algebraic group $\,G.\,$
The various structural properties we have described for $\,G\,$
have analogues for the Lie algebra $\,\mathfrak g.\,$ See
\cite[IV.13.18]{bor2},~\cite[26.2]{hum2}.
Let $\,T\,$ be a maximal torus of $\,G\,$ and $\,{\mathfrak t}\,=\,
{\cal L}(T).\,$ For each root $\,\alpha\in R\,$ let $\,U_{\alpha}\,$ be
the corresponding root subgroup of $\,G\,$ and $\,{\mathfrak g}_{\alpha}\,=
\,{\cal L}(U_{\alpha}).\,$ Then we have a
decomposition
\[
{\mathfrak g}\,=\,{\mathfrak t}\oplus\sum_{\alpha\in R}
{\mathfrak g}_{\alpha}\;.
\]
This is called the {\bf Cartan decomposition} of $\,{\mathfrak g}\,$
\cite[1.13]{car1}.
Let $\,{\mathfrak n}\,=\,{\cal L}(U)\,$ and $\,{\mathfrak n^-}\,=\,{\cal L}
(U^-).\,$ Then $\,{\mathfrak n}\,=\,\displaystyle\sum_{\alpha\in R^+}
{\mathfrak g}_{\alpha}\,$ and $\,{\mathfrak n^-}\,=\,\displaystyle
\sum_{\alpha\in R^-}{\mathfrak g}_{\alpha}\,$, where
$\,R^-\,=\,-R^+\,$. Hence we can write
\[
{\mathfrak g}\,=\,{\mathfrak t}\oplus{\mathfrak n}\oplus{\mathfrak n^-}.
\]
Each of the spaces $\,{\mathfrak g}_{\alpha}\,$ is $\,1$-dimensional and
invariant under the adjoint action of $\,T\,$ on $\,{\mathfrak g}.\,$
Moreover the $\,1$-dimensional representation of $\,T\,$ afforded by the
module $\,{\mathfrak g}_{\alpha}\,$ is $\,\alpha.\,$
(The roots of $\,G\,$ can be defined in the following
way: as $\,T\,$ acts on $\,G\,$ by conjugation, $\,T\,$ acts
on the Lie algebra $\,{\mathfrak g}\,=\,{\cal L}(G)\,$
via the adjoint representation $\,\Ad\,$.
For each $\,\alpha\in X\,$ (that is, for each character of $\,T\,$) let
\[
{\mathfrak g}_{\alpha}\,=\,\{\,x\in{\mathfrak g}\,/\,{\rm Ad}t\cdot x\,=\,
\alpha(t)\,x\;\,\mbox{for all}\;\,t\in T\,\}.
\]
As $\,T\,$ is diagonalizable, $\,{\mathfrak g}\,$ is the direct
sum of the $\,{\mathfrak g}_{\alpha}$'s \cite[III.8.17]{bor2}.
Those $\,\alpha\,$ for
which $\,{\mathfrak g}_{\alpha}\neq 0\,$ are called the {\it weights}
of $\,T\,$ in $\,{\mathfrak g}.\,$ The set $\,R\,$ of the {\it nonzero}
weights of $\,T\,$ in $\,{\mathfrak g}\,$ is called the
set of {\it roots} of $\,G\,$ with respect to $\,T.\,$
Thus the space $\,{\mathfrak g}\,$ is the direct sum of
$\,{\mathfrak t}\,=\,{\cal L}(T)\,=\,{\mathfrak g}_0\,$ and
of the $\,{\mathfrak g}_{\alpha},\;\alpha\in R.\,$ As
$\,{\mathfrak g}_{\alpha}\,$ is $\,1$-dimensional for each $\,\alpha\in
R,\,$ we can write $\,{\mathfrak g}_{\alpha}\,=\,K\,e_{\alpha}\,$.
We call $\,e_{\alpha}\,$ a {\bf root vector} corresponding to $\,\alpha$.)
Let $\,B\,=\,T\,U\,$ be a Borel subgroup of $\,G.\,$
By~\cite[14.25]{bor2},
$\,\mathfrak g\,=\,({\rm Ad}\,G)\cdot {\mathfrak b},\,$ where
$\,{\mathfrak b}\,=\,{\cal L}(B).\,$ Let
$\,{\mathfrak t}\,=\,{\cal L}(T)\,$ and $\,{\mathfrak n}\,=\,{\cal L}(U).\,$
Clearly, $\,\mathfrak b\,$ is spanned by $\,\mathfrak t\,$ and all
$\,e_{\alpha}\,$, where $\,\alpha\in R^+.\,$ Moreover,
$\,\mathfrak b\,=\,\mathfrak t\,+\,\mathfrak n.\,$ We call
$\,\mathfrak b\,$ a {\bf Borel subalgebra} of $\,\mathfrak g\,$.
\begin{lemma}{\rm\cite[4.8]{veld}}\label{veldkamp}
Let $\,B'\,=\,H'U'\,$ with $\,B'\,$ a connected solvable group,
$\,H'\,$ a maximal torus, and $\,U'\,$ the maximal unipotent subgroup
of $\,B'$. Let $\,{\mathfrak b}',\,{\mathfrak h}',\,$ and
$\,{\mathfrak n}'\,$ be the respective Lie algebras, so that
$\,{\mathfrak b}'\,=\,{\mathfrak h}'\,+\,{\mathfrak n}'.\,$ If
$\,t\in{\mathfrak h}'\,$ and $\,n\in{\mathfrak n}'\,$, then there
exists $\,n'\in{\mathfrak n}'\,$ such that $\,t\,+\,n'\,$ is
(${\rm Ad}\,U'$)-conjugate to $\,t\,+\,n\,$, and $\,t\,$ and $\,n'\,$ commute.
\end{lemma}
Recall that $\,x\in{\mathfrak g}\,$ is called toral if
$\,x^{[p]}\,=\,x\,$.
\begin{proposition}\label{anytoral}
Any toral element in $\,{\mathfrak g}\,$ is
(${\rm Ad}\,G$)-conjugate to an element in $\,{\mathfrak t}\,$.
\end{proposition}\noindent
\begin{proof}
Let $\,x\in{\mathfrak g}\,$ be toral.
As $\,\mathfrak g\,=\,({\rm Ad}\,G)\cdot {\mathfrak b}\,$ \cite[14.25]{bor2},
we may assume that $\,x\,=\,t\,+\,n,\,$
where $\,t\in\mathfrak t\,$ and $\,n\in\mathfrak n\,$.
By Lemma~\ref{veldkamp}, $\,x\,$ is (${\rm Ad}\,U'$)-conjugate to
$\,t\,+\,n'\,$ and $\,t\,$ and $\,n'\,$ commute. As
$\,x^{[p]}\,=\,x,\,$ we must have $\,(t\,+\,n')^{[p]}\,=\,t\,+\,n'\,$.
On the other hand,
$\,(t\,+\,n')^{[p]}\,=\,t\,+\,{n'}^{[p]}\,$, by Jacobson's
identity (see Definition~\ref{restliealg}(ii')) and the fact that
$\,[t,\,n']\,=\,0.\,$
Hence, $\,\,{n'}^{[p]}\,=\,n'\,$. As $\,n'\,$ is nilpotent, we get
$\,n'\,=\,0$, finishing the proof.
\end{proof}
\begin{lemma}\label{centretoral}
Let $\,G\,$ be a reductive algebraic group, and
$\,\mathfrak g\,=\,{\cal L}(G)\,$. Then the centre $\,\mathfrak z\,=\,
\mathfrak z(\mathfrak g)\,$ is a toral subalgebra of $\,\mathfrak g\,$.
\end{lemma}\noindent
\begin{proof}
Clearly, $\,\mathfrak z\,$ is ($\Ad\,G)$-stable.
So a maximal torus $\,T\,$ of $\,G\,$ acts diagonalisably on $\,\mathfrak z\,$.
If the root space $\,\mathfrak z_{\alpha}=\{\,x\in{\mathfrak z}\,/\,
{\rm Ad}t\cdot x\,=\,
\alpha(t)\,x\;\,\mbox{for all}\;\,t\in T\,\}\,$ is nonzero for some
$\,\alpha\neq 0$, then $\,e_\alpha\in \mathfrak z\,$
(as $\,\mathfrak g_{\alpha}\,$ is $\,1$-dimensional).
But $\,[e_{\alpha},e_{-\alpha}]\,\neq\, 0$, for every $\,\alpha\in
R\,$ \cite[8.3]{hum1}. Hence $\,\mathfrak z\subset \mathfrak t\,
=\,{\cal L}(T)$. As $\,\mathfrak t\,$ is toral, so is any of its
restricted subalgebras. The result follows.
\end{proof}
At this point we should emphasize some properties of the $\,p$-mapping
in $\,\mathfrak g\,$.
Let $\,{\bf G}_a\,$ denote the additive group $\,K^+$.
Then $\,K[{\bf G}_a]\,=\,K[x]\,$ so that $\,{\cal L}
({\bf G}_a)\,$ is spanned by the invariant derivation
$\,D\,=\,\displaystyle\frac{\rm d}{{\rm d}x}\,$ \cite[9.3]{hum2}.
Since
$\,D^{[p]}\,x^n\,=\,n(n-1)\cdots(n-(p-1))\,x^{n-p}\,$ (or zero if
$\,n<p\,$) it follows that
the $\,p$-mapping is zero in $\,{\cal L}({\bf G}_a)\,$ (because the
product of $\,p\,$ consecutive integers is divisible by $\,p\,$).
Thus,
the $\,p$-mapping of $\,\mathfrak g\,$ vanishes on
$\,{\mathfrak g}_{\alpha}\,\cong\,{\cal L}(U_{\alpha}),\,$ for each
root $\,\alpha\in R,\,$ implying that $\,e_{\alpha}^{[p]}\,=\,0\,$.
Now let $\,{\bf G}_m\,$ denote the multiplicative group
$\,(K,\,\cdot\,).\,$ Then $\,K[{\bf G}_m]\,=\,K[x,x^{-1}]\,$
(the ring of Laurent polynomials). For $\,D\in {\cal L}({\bf G}_m),\,$
one has $\,D\,x\,=\,a\,x\,$, where $\,a\in K\,$ \cite[I.3.9]{bor2}.
It follows that $\,D^{[p]}\,x\,=\,a^p\,x\,$. Thus $\,{\cal L}
({\bf G}_m)\,$ is isomorphic to the $\,1$-dimensional Lie algebra
$\,K\,$ with $\,p$-mapping $\,a\longmapsto a^p\,$.
From this it follows that $\,{\cal L}({\bf G}_m)\,$ is a
$\,1$-dimensional torus.
A maximal torus $\,T\,$ of a connected algebraic group $\,G\,$
is isomorphic to a direct product $\,K^*\times\cdots\times K^*\,$
($\,\ell\,$ times), where $\,\ell\,=\,{\rm
rank}\,G\,$ \cite[III.8.5]{bor2}. Thus
$\,{\cal L}(T)\,\cong\,K\oplus\cdots\oplus K\,$
($\,\ell\,$ times) is a toral subalgebra of $\,{\cal L}(G)\,$
\cite[III.8.2 Corollary]{bor2}.
By Theorem~\ref{3.6}(1), $\,\mathfrak t\,$ has a basis consisting of
toral elements. This implies the following:
\begin{lemma} \label{toralbase}
$\mathfrak t\,=\,{\cal L}(T)\,$ contains $\,p^{\dim T}\,$ toral
elements. If $\,G\,$ is reductive, there are finitely many toral
$\,(\Ad\,G)$-conjugacy classes in $\,\mathfrak g$.
\end{lemma}\noindent
\begin{proof}
Let $\{\,t_1,\ldots ,\,t_{\ell}\,\}$ be a basis of $\,\mathfrak t\,$
consisting of toral elements. Suppose $\,x\,\in\,\mathfrak t\,$ is toral,
that is, $\,x\,=\,\sum_{i=1}^{\ell}\,\lambda_i\,t_i\,$ and $\,x^{[p]}\,=\,x$.
Then Jacobson's identity shows that
$\,x^{[p]}\,=\,\sum_{i=1}^{\ell}\,\lambda_i^p\,t_i\,=\,
\sum_{i=1}^{\ell}\,\lambda_i\,t_i\,=\,x\,$. Therefore,
$\,\lambda_i^p\,=\,\lambda_i$, implying $\,\lambda_i\,\in\,\mathbb
F_p,\,$ for each $\,i\leq \ell\,$. It follows that $\,\mathfrak t\,$ contains
$\,p^{\ell}\,$ toral elements. As $\,\dim\,T\,=\,\ell\,$, the first
part of the lemma is proved. Now let $\,y\,\in\,\mathfrak g\,$ be toral.
By Proposition~\ref{anytoral}, $\,y\,$ is $\,(\Ad\,G)$-conjugate to a
toral element in $\,\mathfrak t\,$. This finishes the proof.
\end{proof}
\subsubsection{Special and Good Primes}
Let $\,R\,$ denote the root system of a reductive algebraic
group $\,G\,$ relative to a maximal torus $\,T\subset G.\,$
\begin{definition}\label{badprime}
A prime $\,p\,$ is said to be {\bf bad} for
$\,G\,$ if $\,p\,$ divides a coefficient of a root $\,\alpha\in R\,$
when expressed as a combination
of simple roots.
\end{definition}
The bad primes for the simple algebraic groups are as follows:
none if $\,G\,$ is of type $\,A_{\ell}\,$
$\,p=2\,$ if $\,G\,$ is of type $\,B_{\ell},\,C_{\ell},\,D_{\ell}\,$
$\,p=2\,$ or $\,3\,$ if $\,G\,$ is of type $\,G_2,\,F_4,\,E_6,\,E_7\,$
$\,p=2,\,3\,$ or $\,5\,$ if $\,G\,$ is of type $\,E_8\,$ (see
\cite[p. 106]{stein2}). \\
Primes which are not bad for $\,G\,$ are called {\bf good}.
\begin{definition}\label{notvgp}
Let $\,G\,$ be a simple algebraic group. A prime $\,p\,$ is said to be
{\bf very good} for $\,G\,$ if either
$\,G\,$ is not of type $\,A_{\ell}\,$ and $\,p\,$ is a good prime for $\,G,\,$
or
$\,G\,$ is of type $\,A_{\ell}\,$ and $\,p\,$ does not divide $\,\ell+1$.
\end{definition}
\begin{definition}\label{pspecial}
Let $\,\alpha_0\,$ be the maximal short root
and $\,\tilde{\alpha}\,$ the longest root in $\,R^+.\,$ Put
$\,d=(\tilde{\alpha},\,\tilde{\alpha})/(\alpha_0,\,\alpha_0).\,$ We say that
$\,p\,$ is {\bf special} for a reductive group $\,G\,$ if $\,p|d.\,$
\end{definition}
By Jantzen~\cite{jant2}, $\,p\,$ is special
for $\,G\,$ if either $\,p=3\,$ and $\,R\,$ has a component of type $\,G_2\,$
or $\,p=2\,$ and $\,R\,$ has a component of type $\,B_{\ell},\,C_{\ell},\,
\ell\geq 2,\,$ or $\,F_4.\,$
\medskip
Let $\,\mathfrak g\,$ be the restricted Lie algebra of a
reductive algebraic group $\,G\,$ defined over an algebraically closed
field of characteristic $\,p > 0\,$.
The classification of nilpotent $\,(\Ad\,G)$-orbits in $\,\mathfrak
g\,$ reduces easily
to the case where $\,G\,$ is simple, and the following important result holds.
\begin{theorem}\label{badhs}
There are finitely many nilpotent $\,(\Ad\,G)$-orbits in $\,\mathfrak g\,$.
In other words, there are finitely many $\,[p]$-nilpotent conjugacy
classes in $\,\mathfrak g\,$.
\end{theorem}\noindent
\begin{proof}
For $\,p\,$ good (or zero), the result is proved by Richardson
by reducing the general case to the case $\,\mathfrak
g\,=\,\mathfrak{gl}(V)\,$ \cite{rich2} (resp.,
\cite{kos},~\cite{dynk}). See also \cite[I.5.6]{spst}.
For $\,p\,$ bad, the result is established by Holt and
Spaltenstein by using computer calculations~\cite{holspa}.
\end{proof}
\medskip
\subsubsection{Chevalley Basis}
Let $\,\mathfrak g_{\mathbb C}\,$ be the complex simple Lie algebra
associated with the root system $\,R.\,$
Denote by $\,\mathfrak h_{\mathbb C}\,$ a maximal toral subalgebra of
$\,\mathfrak g_{\mathbb C}\,$. Let
\[
\mathfrak g_{\mathbb C}\,=\,\mathfrak h_{\mathbb
C}\,\oplus\,\sum_{\alpha\in R}\mathfrak g_{\mathbb C,\alpha}
\]
be a Cartan decomposition of $\,\mathfrak g_{\mathbb C}\,$.
Let $\,h_{\alpha}\,=\,\displaystyle\frac{2\alpha}{(\alpha,\,\alpha)}\,\in
\mathfrak h_{\mathbb C}\,$ be the {\it coroot} corresponding to
$\,\alpha\in R\,$ \cite[3.6.1]{car2},~\cite[8.2]{hum1}.
According to~\cite[25.2]{hum1}, one can choose root
vectors $\,e_{\alpha}\in\mathfrak g_{\mathbb
C, \alpha}\,$ ($\,\alpha\in R\,$) so that
(a) $\,[e_{\alpha},\,e_{-\alpha}]\,=\,h_{\alpha}\,$.
(b) If $\,\alpha,\,\beta,\;\alpha\,+\,\beta\,\in R\,$ and
$\,[e_{\alpha},\,e_{\beta}]\,=\,N_{\alpha\,\beta}\,e_{\alpha+\beta}$,
then $\,N_{\alpha\beta}\,=\,-N_{-\alpha,\,-\beta}$.
(c) $\,N_{\alpha\beta}^2\,=\,q(r+1)\,\displaystyle\frac{(\alpha\,+
\,\beta,\,\alpha\,+\,\beta)}{(\beta,\,\beta)}\,$,
where $\,\beta\,-\,r\alpha,\,\ldots ,\,\beta\,+\,q\alpha\,$ is
the $\,\alpha$-string through $\,\beta\,$.
Let $\,\Delta\,=\,\{\alpha_1,\,\ldots,\,\alpha_{\ell}\,\}\,$ be a basis of
$\,R\,$. By definition a {\bf Chevalley basis of $\,\mathfrak g_{\mathbb C}\,$}
is any basis $\,\{\,h_i=h_{\alpha_i},\,\alpha_i\in\Delta\,\}\cup
\{\,e_{\alpha},\,\alpha\in R\,\}\,$ of $\,\mathfrak g_{\mathbb C}\,$
with the $\,e_{\alpha}$'s satisfying the conditions (a),(b),(c) above.
Now fix a Chevalley basis
$\,\{\,h_i=h_{\alpha_i},\,\alpha_i\in\Delta\,\}\cup\{\,e_{\alpha},\,
\alpha\in R\,\}\,$ of $\,\mathfrak g_{\mathbb C}\,$. Then one has
\begin{align}\label{relations}
[h_i,\,h_j]\, &=\,0,\qquad 1\,\leq\, i,\,j\,\leq\,\ell\nonumber\\
[h_i,\,e_{\alpha}]\, &=\,\langle \alpha,\,\alpha_i\rangle\,e_{\alpha},
\qquad 1\,\leq i \leq\,\ell\,\nonumber\\
[e_{\alpha},\,e_{-\alpha}]\, &=\,h_{\alpha}, \qquad \alpha\,\in\,R\\
[e_{\alpha},\,e_{\beta}]\, &=\,0\,\qquad\mbox{if}\;\,\alpha+\beta\not\in R,
\nonumber\\
[e_{\alpha},\,e_{\beta}]\, &=\,N_{\alpha,\beta}\,e_{\alpha+\beta}
\qquad\mbox{if}\;\,\alpha+\beta\in R,\nonumber
\end{align}
where $\,N_{\alpha,\beta}\,=\,\pm(r+1),\,$ with $\,r\,$ the greatest
integer for which $\,\beta-r\alpha\,$ is a root
\cite[\S 25]{hum1},~\cite[4.2]{car2}.
We denote $\,e_i\,=\,e_{\alpha_i},\; f_i\,=\,e_{-\alpha_i}\,$
and rewrite the relations~\eqref{relations}
accordingly.
Let $\,\mathfrak g_{\mathbb Z}\,$ be the $\,\mathbb Z$-span of the
Chevalley basis. By~\eqref{relations}, $\,\mathfrak g_{\mathbb Z}\,$
is a Lie algebra over $\,\mathbb Z.\,$
Let $\,\mathfrak g\,=\,{\cal L}(G)\,$ denote the (restricted) Lie
algebra of $\,G\,$. By \cite[Sect.2.5]{bor1},
$\,\mathfrak g\,=\,\mathfrak g_{\mathbb Z}\otimes_{\mathbb Z}K,\,$
provided that $\,G\,$ is simply connected.
For convenience, we also denote by $\,e_{\alpha}\,$ and
$\,h_{\alpha}\,$ the corresponding basis elements $\,e_{\alpha}\otimes
1\,$ and $\,h_{\alpha}\otimes 1\,$ of $\,\mathfrak g\,$.
By the discussion at the end of Section~\ref{structliealg}, we have
that
$\,e_{\alpha}^{[p]}\,=\,0,\;h_{\alpha}^{[p]}\,=\,
h_{\alpha}\,$ for all $\,\alpha\in R$.
\subsubsection{Graph Automorphisms}\label{graphaut}
Let $\,R\,$ be a root system and $\,\Delta\,$ a basis of $\,R$.
The automorphism group of
$\,R\,$ is the semidirect product of $\,W\,$
and the group $\,\Gamma\,:=\,\{\sigma\in{\rm Aut}\,R\,/\,\sigma
(\Delta)=\Delta\}\,$ of
{\it graph} (or {\it diagram}) automorphisms of $\,R$.
Each $\,\sigma\in\Gamma\,$ determines an automorphism of the Dynkin
diagram of $\,R.\,$ Conversely, each automorphism of the Dynkin
diagram determines an automorphism of $\,R\,$ \cite[12.2]{hum1}.
If $\,R\,$ is irreducible, then $\,\Gamma\,$ is
trivial, except for types $\,A_{\ell}\,(\ell\geq 2),\;D_{\ell}\,$ and
$\,E_6\,$ \cite[Table 1, p.\ 66]{hum1}.
Now let $\,R\,$ be the (irreducible) root system of a simply connected
simple algebraic group $\,G.\,$
Each automorphism $\,\sigma\in\Gamma\,$
gives rise to a (rational) automorphism of $\,G,\,$ denoted by
$\,\hat{\sigma}\,$ and called a {\bf graph automorphism of $\,G\,$}
\cite[12.2 and 12.3]{car2}.
Each graph automorphism $\,\hat{\sigma}\,$ of $\,G\,$ induces a Lie
algebra automorphism of $\,\mathfrak g\,=\,{\cal L}(G),\,$ also
denoted by $\,\hat{\sigma}.\,$ The latter has the property:
\[
\hat{\sigma}(e_{\alpha})\, =\,e_{\sigma(\alpha)},\quad
\hat{\sigma}(f_{\alpha})\, =\,f_{\sigma(\alpha)}\qquad\quad\mbox{for
all}\;\;\alpha\in\Delta\,.
\]
The compatibility between the graph automorphisms of $\,G\,$ and
$\,\mathfrak g\,$ is as follows. For $\,g\in G,\;x\in\mathfrak g\,$ we
have $\,\hat{\sigma} (({\rm Ad}\,g)\cdot x)\,=\,({\rm
Ad}\,g^{\hat{\sigma}})\cdot x^{\hat{\sigma}}\,$.
By~\cite[Proposition 12.2.3]{car2}, there exist graph automorphisms of
order $\,2\,$ of $\,A_{\ell}(K),\;\ell\geq 2;\;D_{\ell}(K),\;\ell\geq
4; \;E_{6}(K)$. There is also a graph automorphism of order $\,3\,$ of
$\,D_4(K)$.
\subsection{Basic Representation Theory}\label{reptheo}
From now on let $\,G\,$ denote a simply connected simple algebraic
group (universal Chevalley group) over an algebraically closed field $\,K\,$
of characteristic $\,p>0$.
Fix a maximal torus $\,T\,$ of $\,G,\,$ and let
$\,B\,=\,T\,U\,$ be a Borel subgroup
containing $\,T$.
Denote by $\,X\,=\,X(T)\,$ the character group of $\,T.\,$
Let $\,R\,$ be the (irreducible) root system of $\,G\,$ with respect to
$\,T,\,$ $\,R^+\,$ (resp. $\,R^-\,$) the set of positive (resp. negative)
roots relative to $\,B$, and
$\,\Delta\,=\,\{\alpha_1,\ldots,\alpha_{\ell}\}\,$ the corresponding basis
of simple roots, where $\,\ell\,=\,{\rm rank}\,G\,$. Let
$\,W\,=\,N_G(T)/T\,$ be the Weyl group
of $\,G,\,$ $\,s_{\alpha}\,$ the reflection in $\,W\,$ corresponding to
$\,\alpha\in R$. Put $\,s_i\,=\,s_{\alpha_i}\,$, and let $\,w_0\,$
denote the unique element of
$\,W\,$ sending $\,R^+\,$ to $\,R^-\,$.
The cardinality of $\,R\,$ is denoted by $\,|R|\,=\,2\,|R^+|,\,$ and
$\,\dim\,G\,=\,|R|\,+\,\ell\,$ \cite[26.3]{hum2}.
\subsubsection{Weights}\label{introback}
Let $\,(\;\cdot\;,\,\cdot\;)\,$ denote a
nondegenerate $\,W$-invariant symmetric bilinear form on $\,X,\,$
extended to $\,E\,=\,X\otimes_{\mathbb Z}\mathbb R\,$. Set
$\,\langle\alpha,\,\beta\rangle\,=\,\displaystyle\frac{2\,(\alpha,\,\beta)}
{(\beta,\,\beta)}\,$, for $\,\alpha,\,\beta\,\in\,E,\,$ as in
Section~\ref{therootsys}. Since $\,R\,$ is an abstract root system,
$\,\langle\alpha,\,\beta\rangle\,\in\,\mathbb Z\,$ for all
$\,\alpha,\,\beta\,\in\,R\,$ \cite[\S 27]{hum2}.
A vector $\,\lambda\,\in\,E\,$ is called a
{\bf weight} provided that $\,\langle\lambda,\,\alpha\rangle\in\mathbb
Z\,$ for all $\,\alpha\in R$.
The weights form a lattice $\,P\,$ of $\,E,\,$ in which the lattice
$\,\mathbb ZR\,$ spanned by $\,R\,$ is a subgroup of finite index.
$\,P/\mathbb ZR\,$ is called the {\bf fundamental group} of $\,G.\,$
$\,W\,$ acts naturally on $\,P\,$ and $\,\mathbb ZR,\,$ and acts trivially
on the fundamental group $\,P/\mathbb ZR\,$ \cite[14.7]{bor2}.
If $\,\Delta\,=\,\{\alpha_1,\ldots,\alpha_{\ell}\},\,$
then $\,P\,$ is generated by the system of {\bf fundamental weights}
$\,\{\omega_1, \ldots ,\omega_{\ell}\},\,$ defined by
$\,\langle \omega_i,\alpha_j\rangle = \delta_{ij}\,$.
Define $\,\rho\,:=\,\sum_{i=1}^{\ell}\,\omega_i\,$ (it equals half the sum
of the positive roots).
A weight $\,\lambda\,$ is said to be {\bf dominant} if
$\,\langle\lambda, \alpha \rangle \geqslant 0\,$ for all roots
$\,\alpha\in R^+$. Let $\,P_{++}\,$ denote the set of all dominant weights.
Clearly, $\,P_{++}\,=\,\displaystyle \sum_{i=1}^{\ell}\,\mathbb Z^+\omega_i\,$.
\linebreak A weight $\,\lambda\in P\,$ is $\,W$-conjugate to one and only one
dominant weight. There is a natural partial ordering of $\,E\,$:
given $\,\lambda,\,\mu\in E,\,$ we write $\,\mu\,\leq\,\lambda\,$ if and only
if $\,\lambda\,-\,\mu\,$ is a sum of positive roots \cite[Appendix]{hum2}.
As $\,G\,$ is simply connected, the group $\,X\,=\,X(T)\,$
is the full weight lattice $\,P\,$ \cite[\S 31]{hum2}.
\subsubsection{Maximal Vectors and Irreducible Modules}\label{wamv}
Let $\,V\,$ be a finite dimensional vector space over $\,K.\,$
If $\,\pi\,:\,G\,\rightarrow\,GL(V)\,$ is a rational representation
of $\,G\,$, then we call $\,V\,$ a {\bf rational $\,G$-module}.
Denote $\,\pi(g)\,v\,=\,g\cdot v\,$ for all $\,g\in G,\,v\in V$.
Being a diagonalizable group, the maximal torus $\,T\,$ acts
completely reducibly on $\,V,\,$ so that
$\,V\,$ is the direct sum of weight spaces
\[
V_{\mu}\,:=\,\{\,v\in V\,/\,t\cdot v\,=\,\mu(t)\,v\;\;\mbox{for
all}\;t\in T\}\,,
\]
where $\,\mu\in X\,$ \cite[III.8.4]{bor2}.
The dimension $\,\dim\,V_{\mu}\,$ is called the {\bf multiplicity} of $\,\mu$.
Put $\,{\cal X}(V)\,:=\,\{\,\mu\in X\,/\,V_{\mu}\,\neq\,(0)\,\}.\,$
The elements of $\,{\cal X}(V)\,$ are called the {\bf weights of
$\,V\,$}.
We call $\,v\in V_{\mu}\,$ a {\bf weight vector} (of weight $\,\mu\,$).
Put $\,{\cal X}_{++}(V)\,=\,{\cal X}(V)\cap P_{++}\,$.
A (nonzero) vector $\,v\in V\,$ is called a {\bf maximal vector}
if it is fixed by all $\,u\in U.\,$ If $\,V\,$ is nonzero, then
maximal vectors exist \cite[31.2]{hum2}.
If $\,V\,$ is
irreducible, then a maximal vector $\,v_0\,$ of weight $\,\lambda\,$
is unique up to a scalar multiple. Moreover, $\,\lambda\,$ is a dominant weight
of multiplicity $\,1\,$ and is called the {\bf highest weight} of $\,V.\,$
The vectors in $\,V_{\lambda}\,$ are called {\bf highest weight vectors}.
All other weights of $\,V\,$ are of the form
$\,\lambda\,-\,\sum\,c_{\alpha}\,\alpha,\,$ where $\,\alpha\in
R^+\,$ and $\,c_{\alpha}\in \mathbb Z^+\,$. They are permuted by $\,W,\,$ with
$\,W$-conjugate weights having the same multiplicity \cite[31.3]{hum2}.
It follows that $\,{\cal X}(V)\,=\,W\cdot {\cal X}_{++}(V)\,$.
If $\,V'\,$ is an irreducible rational $\,G$-module with highest weight
$\,\lambda'\,$, then $\,V\,\cong\,V'\,$ (as
$\,G$-module) if and only if $\,\lambda\,=\,\lambda'\,$
\cite[31.3]{hum2}. We denote by $\,E(\lambda)\,$ the irreducible
rational $\,G$-module with highest weight $\,\lambda\,$.
In the next section we will review the method of construction of
irreducible rational $\,G$-modules.
\subsubsection{The Construction of Irreducible Modules}
The construction of irreducible rational modules for semisimple
algebraic groups given below
shows a connection between the representation theory of such groups
with the representation theory of {\it hyperalgebras}.
The main results of this section are due to Chevalley
\cite{stein1}, \cite{wong1}.
Recall some notation:
$\,\mathfrak g_{\mathbb C}\,$ is the complex
simple Lie algebra associated with the root system $\,R,\,$ and
$\,\Delta\,$ is a basis of $\,R$. Fix a Chevalley basis
$\,\{ h_{\beta},\,\mbox{$\beta\in\Delta$};$ \linebreak
$\,\mbox{$e_{\alpha}$},\, \mbox{$\alpha\in R$} \}\,$
of $\,\mathfrak g_{\mathbb C}\,$ and
let $\,\mathfrak g_{\mathbb Z}\,$ be the $\,\mathbb Z$-span of the
Chevalley basis. The basis of the restricted Lie algebra
$\,\mathfrak g\,=\,{\cal L}(G)\,=\,\mathfrak g_{\mathbb
Z}\otimes_{\mathbb Z}K\,$ is also denoted by $\,e_{\alpha}\,$ and
$\,h_{\beta}\,$ (that is, we identify $\,e_{\alpha}\,$ and
$\,h_{\beta}\,$ with $\,e_{\alpha}\otimes 1\,$ and
$\,h_{\beta}\otimes 1\,$ in $\,\mathfrak g\,$).
The $\,p$th power map in $\,\mathfrak g\,$ has the property
$\,e_{\alpha}^{[p]}\,=\,0,\;h_{\alpha}^{[p]}\,=\,h_{\alpha}\,$ for all
$\,\alpha\in R$.
Using Kostant's Theorem~\cite{wong1}, which describes a $\,\mathbb
Z$-basis of the $\,\mathbb Z$-form $\,U_{\mathbb Z}\,$ of the
universal enveloping algebra $\,U\,=\,U(\mathfrak g_{\mathbb C})\,$
of $\,\mathfrak g_{\mathbb C}\,$ generated by all
$\,\displaystyle\frac{e^n_{\alpha}}{n!}\,$ ($\,\alpha\in
R,\;n\in\mathbb Z^+$), we can construct
an {\it admissible lattice} (a lattice stable under $\,U_{\mathbb Z}\,$)
in an arbitrary $\,\mathfrak g_{\mathbb C}$-module \cite[\S 27]{hum1}.
In particular, let $\,V\,=\,V_{\lambda}\,$ be an irreducible
$\,\mathfrak g_{\mathbb C}$-module with highest weight
$\,\lambda\in P_{++}\,$,
and let $\,v_0\,$ be a maximal vector of $\,V.\,$ This is
a nonzero weight vector annihilated by all $\,e_{\alpha},\,\alpha\in
R^+\,$. By Theorem~\cite[27.1]{hum1}, $\,V_{\mathbb
Z}\,:=\,U_{\mathbb Z}\,v_0\,=\,U_{\mathbb Z}^-\,v_0\,$ is
the (unique) smallest $\,\mathbb Z$-form of $\,V\,$ containing
$\,v_0\,$ and stable under $\,U_{\mathbb Z}\,$. In particular,
$\,V_{\mathbb Z}\,$ is an admissible lattice.
Now tensoring with $\,K,\,$ one obtains a $\,K$-space
$\,V_K(\lambda)\,:=\,V_{\mathbb Z}\otimes K\,$. By construction,
$\,V_K(\lambda)\,$ carries a canonical module structure over
the {\bf hyperalgebra} $\,U_K\,:=\,U_{\mathbb Z}\otimes_{\mathbb Z}K\,$
\cite[20.2]{hum1},~\cite[1.3]{wong1}.
Note that $\,U_K\,$ is generated over $\,K\,$ by $\,x_{\alpha}^{(m)}\,=\,
\displaystyle\frac{e_{\alpha}^m}{m!} \otimes 1,\,$ where $\,\alpha\in
R\,$ and $\,m\in\mathbb Z^+\,$.
Let $\,e_0\,=\,v_0\otimes 1\,$. Then
\[
U_K\cdot e_0\,=\,(U_{\mathbb Z}\otimes_{\mathbb Z}K)\cdot (v_0\otimes
1)\,=\,U_{\mathbb Z}\cdot v_0\otimes K\,=\,V_{\mathbb Z}\otimes K\,=\,
V_K(\lambda)\,.
\]
One easily sees that $\,V_K(\lambda)\,=\,U_K^-\cdot e_0\,$, where
$\,U_K^-\,=\,U_{\mathbb Z}^-\otimes_{\mathbb Z}K\,$.
Moreover, $\,V_K(\lambda)\,$ is the direct sum
of its weight spaces $\,V_K(\lambda)_{\mu}\,=\,(V_{\mathbb Z}\cap
V_{\lambda,\mu})\otimes_{\mathbb Z}K\,$, where $\,\mu\in {\cal
X}(V)\,$ \cite[Theorem 20.2]{hum1}.
It follows that $\,V_K(\lambda)\,$ is a rational $\,G$-module of
highest weight $\,\lambda,\,$ whose dimension over $\,K\,$ equals
$\,\dim_{\mathbb C}\,V_{\lambda}\,$.
\begin{theorem}{\rm\cite{cps}}\label{connection}
There is a one-to-one correspondence between finite dimensional
$\,U_K$-modules and finite dimensional rational $\,G$-modules.
\end{theorem}
Since $\,\mathfrak g_{\mathbb Z}\,$ embeds in
$\,U_{\mathbb Z}\,$, $\,V_K(\lambda)\,$
carries a canonical $\,\mathfrak g$-module structure.
One can show that the corresponding representation
$\,\mathfrak g\longrightarrow\mathfrak{gl}(V_K(\lambda))\,$ is nothing
but the differential of the rational representation
$\,G\longrightarrow\GL (V_K(\lambda))\,$ induced by the action of
$\,U_K\,$ on $\,V_K(\lambda)\,$ (see Theorem~\ref{connection}).
The rational $\,G$-module $\,V_K(\lambda)\,$ is called the
{\bf Weyl module} with highest weight $\,\lambda$.
The vector $\,e_0\in V_K(\lambda)\,$ is a maximal vector of
weight $\,\lambda\,$ relative to the action of a Borel subgroup
$\,B\,=\,T\,U\,$ of $\,G$.
There is also an intrinsic construction of Weyl modules. Let
$\,\lambda\,$ be a dominant weight of $\,X(T)\,\cong\,P\,$
and $\,\langle v \rangle\,$ a
$\,1$-dimensional module for $\,B\,$ affording $\,\lambda\,$
upon restriction to $\,T$. Now $\,G\times B\,$ acts on $\,K[G]$,
with $\,G\,$ acting on the left and $\,B\,$ on the right.
This yields an action of $\,G\times B\,$ on $\,K[G] \otimes_K\langle
v \rangle $. Set $\,H(\lambda)
= {(K[G]\otimes_K\langle v \rangle )}^B$, where $\,B\,$
denotes the direct factor of $\,G \times B$. One checks that $\,H(\lambda) =
D(\lambda)\otimes_K\langle v \rangle $, where $\,D(\lambda) = \{f
\in K[G]\,/\, f(x) = \lambda(b)f(xb), \; x \in G, \; b \in B\}$.
It turns out that $\,H(\lambda)\,$ has a simple socle isomorphic to
$\,E(\lambda)\,$.
Moreover, $\,H(-\omega_0\lambda)^*\,$ is isomorphic to the Weyl module
$\,V_K(\lambda)\,$ \cite[Part II, 2.12]{jant1}.
\medskip
From now on denote by $\,V(\lambda)\,$ the Weyl module $\,V_K(\lambda)\,$.
Recall some fundamental properties of the Weyl modules.
\begin{theorem}{\rm\cite[2.13-14,5.11]{jant1}}\label{univprop}~Let
$\lambda$ be a dominant weight of \mbox{$\,X(T)\,\cong\,P$}.
\begin{itemize}
\item[(1)] $V(\lambda)\,$ has a unique maximal submodule
$\,\Phi(\lambda)\,$ such that $\,V(\lambda) / \Phi(\lambda)
\cong E(\lambda).\,$
In particular $\,V(\lambda)\,$ is indecomposable.
\item[(2)] ${\rm dim}(V(\lambda))$ is given by the Weyl dimension formula
(see {\rm \cite[24.3]{hum1}}).
\item[(3)] Given a rational $\,G$-module $\,M\,$ generated by a
maximal weight vector of weight $\,\lambda,\,$ there is an epimorphism
$\,V(\lambda)\longrightarrow M\,$. In other words, $\,V(\lambda)\,$ is
a universal object among the rational $\,G$-modules of highest weight
$\,\lambda$.
\end{itemize}
\end{theorem}
In general, the Weyl module $\,V(\lambda)\,$
is reducible. Its composition factors are of the form $\,E(\mu),\,$
where $\,\mu\leq\lambda$, and $\,E(\lambda)\,$ always occurs
in $\,V(\lambda)\,$ with (composition) multiplicity $\,1$.
However, in one important case the Weyl module
$\,V(\lambda)\,$ is irreducible and its dimension is known.
\begin{theorem}[Steinberg]{\rm\cite{stein3}}\label{stmod}
The {\bf Steinberg module} $\,E((p-1)\rho)\,$ is an irreducible
$\,G$-module of dimension $\,p^{|R^+|}\,$.
\end{theorem}
\subsubsection{Infinitesimally Irreducible Modules}\label{construct}
If $\,\pi\,:\,G\,\rightarrow\,GL(V)\,$ is a rational representation,
then the differential
$\,{\rm d}\pi\,:\,\mathfrak g\,\rightarrow\mathfrak{gl}(V)\,$ is a
representation of $\,\mathfrak g\,$ (see Section~\ref{tlaag}).
Let $\,\lambda\,=\,\lambda(\pi)\,$ be the highest weight of
$\,V\,$.
The differential $\,{\rm d}\mu\,: \mathfrak t\longrightarrow K\,$
of a weight $\,\mu\in X\,$ is a linear function on $\mathfrak t$.
The torus $\,\mathfrak t\,$ acts on the weight space $\,V_{\mu}\,$
via the differential $\,{\rm d}\mu\,$. As usual, abusing notation,
we identify $\,\mu\,$ with $\,{\rm d}\mu\,$ and call $\,\mu\,=\,{\rm
d}\mu\,$ a {\it weight of $\,\mathfrak t\,$}.
However, the elements of $\,p\,X\,$ have zero differentials, so that
the weights of $\,\mathfrak t\,$ correspond one-to-one to the elements
of $\,X/pX\,$, a set of cardinality $\,p^{\ell}.\,$
A weight $\,\lambda\in P_{++}\,$ is called {\bf $\,p$-restricted} if
$\,\lambda= \sum_ic_i\omega_i\,$ with $\,0\leq c_i \leq p-1\,$
for all $\,i\,$. Denote by $\,\Lambda_p\,$ the subset
of $\,P_{++}\,$ consisting of all $\,p$-restricted weights.
\begin{definition}\label{infirred}
A rational $\,K$-representation $\,\pi\,$ of $\,G\,$ is called {\bf
infinitesimally irreducible} if its differential $\,{\rm d}\pi\,$
defines an irreducible representation of the Lie algebra $\,\mathfrak
g\,=\,{\cal L}(G)\,$.
\end{definition}
\begin{lemma}{\rm\cite[6.2]{bor1}}\label{bor42}
Let $\,\pi\,:\,G\,\rightarrow\,GL(V)\,$
be an infinitesimally irreducible rational representation
of $\,G\,$ with highest weight $\,\lambda\,=\,\lambda(\pi)\,$. Then
(i) $\,V\,=\, U(\mathfrak n^-)\cdot V_{\lambda}\,$.
(ii) $\,V_{\lambda}\,$ is the only subspace of $\,V\,$ annihilated by
$\,\mathfrak n\,$.
\end{lemma}
Denote by $\,M(G)\,$ the set of all infinitesimally irreducible
representations of $\,G$. The following two theorems (due to Curtis and
Steinberg) are among the fundamental results of the modular
representation theory.
\begin{theorem}{\rm\cite{curt1},\cite{stein3}}\label{curt}
(i) A rational representation $\,\pi\,$ of $\,G\,$ is infinitesimally
irreducible if and only if $\,\lambda(\pi)\,\in\,\Lambda_p\,$ (that
is, $\,\lambda(\pi)\,$ is a $\,p$-restricted weight).
(ii) Each irreducible $\,p$-representation of the restricted Lie
algebra $\,\mathfrak g\,=\,{\cal L}(G)\,$ is equivalent to the
differential of a unique infinitesimally irreducible representation of
$\,G$.
\end{theorem}
\begin{theorem}[Steinberg's Tensor Product Theorem]
{\rm\cite{stein3}}\label{bts}
For any irreducible rational $\,K$-rep\-re\-sent\-ation $\,\psi\,$ of
$\,G\,$ there exist infinitesimally irreducible rep\-re\-sent\-a\-tions
$\,\pi_0,\,\pi_1,\,\ldots,\,\pi_m\,$ such that
\begin{equation}\label{decomp}
\psi\cong \pi_0\,\otimes\,\pi_1^{\rm Fr}\,\otimes
\,\ldots\,\otimes\,\pi_m^{{\rm Fr}^m}\,.
\end{equation}
Here $\,{\rm Fr}\,$ is the Frobenius endomorphism of the field $\,K$.
\end{theorem}
Let $\,{\cal X}(\psi)\,$ denote the set of weights of a rational
$\,G$-representation $\,\psi\,$. If $\,\psi\,$ is irreducible, then
by formula~\eqref{decomp} we have
\[
{\cal X}(\psi)\,=\,{\cal X}(\pi_0)\,+\,p\,{\cal X}(\pi_1)\,+\,
\ldots\,+\,p^m\,{\cal X}(\pi_m)\,.
\]
Steinberg's tensor product theorem can be stated in another way.
If $\,\lambda\,$ is a dominant weight, then it has a unique
$\,p$-adic expansion $\lambda = \lambda_0 + p\lambda_1 +
\cdots + p^m\lambda_m$, where $\lambda_0, \ldots , \lambda_m\,$ are
$\,p$-restricted weights.
\begin{theorem}~{\rm \cite{stein3}}~~Let $\,G\,$ be a simply connected
semisimple algebraic group and let $\,\lambda\,$ be a dominant weight
with $\,p$-adic expansion $\,\lambda = \lambda_0 + p\lambda_1 +
\cdots + p^m\lambda_m\,$. Then
$$
E(\lambda)
\cong E(\lambda_0) \otimes E(\lambda_1)^{Fr} \otimes \cdots \otimes
E(\lambda_m)^{{Fr}^m}.
$$
\end{theorem}
This theorem reduces many questions concerning irreducible representations
of $\,G\,$ to the study of the infinitesimally irreducible ones.
One of such questions is a description of systems of weights of
irreducible rational representations of simple algebraic groups.
\subsubsection{Weight Space Decomposition}
In this section, we explore the relationship between the
weight spaces of the Weyl module $\,V(\lambda)\,$ and the
weight spaces of its irreducible quotient
$\,E(\lambda)\,$. The main result (Theorem~\ref{premprin}) asserts that
although in passing from
the Weyl module to its irreducible top factor the dimensions of weight spaces
may decrease, weight spaces rarely disappear entirely.
Denote by $\,\pi_{\mathbb C}\,:\,\mathfrak g_{\mathbb
C}\longrightarrow \mathfrak{gl}(V_{\lambda})\,$ the irreducible complex
representation of the Lie algebra $\,\mathfrak g_{\mathbb C}\,$ in the
vector space $\,V_{\lambda}\,$. Let $\,{\cal X}(\pi_{\mathbb
C})\,$ denote the set of weights of this representation (that is,
$\,{\cal X}(\pi_{\mathbb C})\,=\,{\cal X}(V_{\lambda})\,$). Let
$\,\pi\,:\,G\longrightarrow \GL(E(\lambda))\,$ be the irreducible
representation of $\,\mathfrak g\,$ in the vector space $\,E(\lambda).\,$
It follows from our discussion in
Section~\ref{construct} that
$\,{\cal X}(\pi)\,\subseteq\,{\cal X}(\pi_{\mathbb C})\,$.
Recall that
$\,{\cal X}(\pi)\,$ and $\,{\cal X}(\pi_{\mathbb C})\,$ are both
$\,W$-invariant, and any weight of $\,P\,$
is $\,W$-conjugate to precisely one dominant weight. Therefore, we can write
$\,{\cal X}(\pi)\,=\,W\cdot {\cal X}_{++}(\pi)\qquad$ and
$\qquad{\cal X}(\pi_{\mathbb C})\,=\,W\cdot {\cal X}_{++}(\pi_{\mathbb
C})\,$.
By \cite[Chap. VIII \S 7]{bourb3}, it is true that
$\,{\cal X}_{++}(\pi_{\mathbb C})\,=\,(\lambda(\pi)\,-\,Q_+)\,\cap\,P_{++},
\,$ where $\,Q_+\,=\,\{ \sum c_{k}\,\alpha_k\,/\,\alpha_k\in\Delta,\,
c_{\alpha_k}\in\mathbb Z^+\,\}\,$ and $\,(\lambda(\pi)\,-\,Q_+)\,=\,
\{\,\lambda(\pi)\,-\,\beta\,/\,\beta\in Q_+\,\}$.
Let $\,T\,$ be a maximal torus of $\,G\,$ with which the root system
$\,R\,$ is associated, and let
\[
V(\lambda)\,=\,\bigoplus_{\mu\in {\cal X}(\pi_{\mathbb
C})}\,V(\lambda)_{\mu}\,\quad\mbox{and}\,\quad
\Phi(\lambda)\,=\,\bigoplus_{\mu\in {\cal X}(\pi_{\mathbb
C})}\,\Phi(\lambda)_{\mu}
\]
be the decompositions of the $\,G$-modules $\,V(\lambda)\,$ and
$\,\Phi(\lambda)\,$ into the direct sum of weight subspaces with
respect to $\,T\,$. One has
$\,V(\lambda)_{\mu}\,=\,(V_{\mathbb Z}\cap V_{\lambda,\mu})
\otimes_{\mathbb Z}K\,$ and
$\,\Phi(\lambda)_{\mu}\,\subseteq\,V(\lambda)_{\mu}\,$.
The quotient module $\,V(\lambda)/\Phi(\lambda)\,=\,E(\lambda)\,$
is irreducible and has highest weight $\,\lambda$. It is clear
that
$\,E(\lambda)_{\mu}\,\cong\,V(\lambda)_{\mu}/\Phi(\lambda)_{\mu}\,$ as
vector spaces. Since $\,\pi\,$ is realized in
$\,E(\lambda)\,$, the following equality holds:
\begin{equation}\label{iiii}
{\cal X}(\pi)\,=\,\{\,\mu\in {\cal X}(\pi_{\mathbb C})\,/\,
V(\lambda)_{\mu}\,\neq\,\Phi(\lambda)_{\mu}\,\}.
\end{equation}
\begin{theorem}{\rm\cite[p. 169]{pre1}}\label{theop1}
Let $\,p\,$ be non-special for $\,G$, and
$\,\lambda\,\in\,\Lambda_p$. For $\,\lambda\,=\,\omega_1\,$, assume
also that $\,p\neq 2\,$ for groups of type $\,G_2\,$.
Then $\,V(\lambda)_{\mu}\,\neq\,\Phi(\lambda)_{\mu}\,$ for any weight
$\,\mu\,$ of the Weyl module $\,V(\lambda)$.
\end{theorem}
In~\cite[p. 169]{pre1}, Premet shows that Theorem~\ref{theop1} is
equivalent to the following:
\begin{theorem}\label{premprin}
Let $\,p\,$ be as in Theorem~\ref{theop1}.
Then for any $\,\pi\in M(G)\,$ the equality
$\,{\cal X}(\pi)\,=\,{\cal X}(\pi_{\mathbb C})\,$ holds. In particular,
\[
{\cal X}_{++}(\pi)\,=\,(\lambda(\pi)\,-\,Q_+)\,\cap\,P_{++}.
\]
\end{theorem}
For special primes the inequality
$\,{\cal X}(\pi)\,\neq\,{\cal X}(\pi_{\mathbb C})\,$ becomes an
ordinary occurrence. The simplest examples are the natural
representation of the group $\,B_{\ell}(K)\,$ for $\,p=2\,$ and
the adjoint representation of the group of type $\,G_2\,$ for $\,p=3$.
\medskip
The following result tells us about the weights of irreducible
restricted \mbox{$\,{\mathfrak g}$-modules} (cf. Theorem~\ref{curt}(ii)).
\begin{corollary}{\rm\cite{pre1}}\label{pppweights}
Let $\,p\,$ be as in Theorem~\ref{theop1}. Let $\,\phi\,$ be an
irreducible \mbox{$\,p$-representation} of
$\,{\mathfrak g}\,$ with highest weight $\,\bar{\lambda}\,\in\,
P\,\otimes_{\mathbb Z}\,\mathbb F_p$, and $\,\lambda\,$ the unique
inverse image of $\,\bar{\lambda}\,$ under the reduction homomorphism
$\,P\,\longrightarrow\,P\,\otimes_{\mathbb Z}\,\mathbb F_p\,$,
lying in $\,\Lambda_p$.
Then the set of $\,\mathfrak t$-weights of $\,\phi\,$ coincides with
the image of $\,W\cdot ((\lambda\,-\,Q_+)\,\cap\,P_{++})\,$
in $\,P\,\otimes_{\mathbb Z}\,\mathbb F_p\,$.
\end{corollary}
\subsubsection{Weight Multiplicities}
The main unsolved problem concerning the modules $\,E(\lambda)\,$ is
the determination of their weight multiplicities (or formal characters
of $\,E(\lambda)\,$).
A lot of work has been done in the direction of solving this problem
in the last 30 years and good progress has been made.
Thanks to Steinberg's tensor product Theorem~\ref{bts}, it is enough
to obtain this kind of information about the collection
$\,\{\,E(\lambda)\,/\,\lambda\in \Lambda_p\,\}\,$. However, we observe
that even if $\,\lambda\in \Lambda_p\,$, some of the dominant weights below
$\,\lambda\,$ in the partial ordering defined in
Section~\ref{introback} may lie outside $\,\Lambda_p\,$
unless $\,R\,$ has type $\,A_1,\,A_2,\,B_2\,$ (cf. \cite{verma1})
For a given $\,\lambda\,$ and a given $\,p,\,$ it is possible (at least in
principle) to compute effectively the weight multiplicities of
$\,E(\lambda)\,$. Burgoyne~\cite{burg} carried out computer calculations
along these lines for small ranks and small primes $\,p$. The
underlying idea is to write down a square matrix over the integers
(of size equal to the dimension
of a weight space of $\,V_{\lambda}\,$). The number of elementary divisors
of the matrix divisible by $\,p\,$ counts the decrease in the
dimension of the weight space when we pass to $\,E(\lambda).\,$
In this work we use information on weights and their
multiplicities given in \cite{buwil} (for Lie algebras of
small rank) and in \cite{gise} (for Lie algebras of exceptional type).
\newpage
\part*{Chapter 3}
\part*{The Exceptional Modules}
\addcontentsline{toc}{section}{\protect\numberline{3}{The
Exceptional Modules}}
\setcounter{section}{3}
\setcounter{subsection}{0}
In this Chapter we give the definition of exceptional modules and
find a necessary condition for a module to be exceptional.
\subsection{Some Results on Centralizers}\label{centsection}
This section contains some results on centralizers of elements in the
algebraic group and its Lie algebra that will be used throughout the
main sections of this work.
If $\,x\,$ is an element of an algebraic group $\,G\,$ then
$\,C_G(x)\,=\,\{ g\in G\,/\,gx\,=\,xg\}\,$ is the {\bf centralizer} of
$\,x\,$. This is a closed subgroup of $\,G\,$ \cite[8.2]{hum2}.
We now discuss the relationship between centralizers in an algebraic group
$\,G\,$ and in its Lie algebra $\,{\mathfrak g}\,=\,{\cal L}(G).\,$
Recall that $\,G\,$ acts on $\,{\mathfrak g}\,$ via the adjoint
representation $\,{\rm Ad}\,:\,G\longrightarrow \GL(\mathfrak{g}).\,$
(cf. Section~\ref{tlaag}). Let $\,x\in G\,$ and $\,a\in{\mathfrak
g}.\,$ Define
\begin{gather*}
C_G(x)\,=\,\{ g\in G\,/\,x^{-1}gx\,=\,g\} \\
\mathfrak c_{\mathfrak g}(x)\,=\,\{ b\in {\mathfrak g}\,/\,{\rm
Ad}x\cdot b\,=\,b\}\\
C_G(a)\,=\,\{ g\in G\,/\,{\rm Ad}g\cdot a\,=\,a\}\\
\mathfrak c_{\mathfrak g}(a)\,=\,\{ b\in {\mathfrak g}\,/\,[b,a]\,=\,0\}.
\end{gather*}
The relationship between these subgroups and subalgebras is as follows.
\begin{gather*}
{\cal L}(C_G(x))\subseteq \mathfrak c_{\mathfrak g}(x)\;\;\mbox{for all}\;\;
x\in G\;\;{\rm\cite[10.6]{hum2}}\,, \\
{\cal L}(C_G(a))\subseteq \mathfrak c_{\mathfrak g}(a)\;\;\mbox{for all}\;\;
a\in {\mathfrak g}\;\;{\rm\cite[9.1]{bor2}}.
\end{gather*}
Equality holds in certain important special cases.
\begin{lemma}{\rm \cite[III.Prop. 9.1]{bor2}}\label{semisim}
If $\,s\in G\,$ is semisimple,
then $\,{\cal L}(C_G(s))\,=\,\mathfrak c_{\mathfrak g}(s).\,$
If $\,a\in {\mathfrak g}\,$
is semisimple, then $\,{\cal L}(C_G(a))\,=\,\mathfrak c_{\mathfrak
g}(a)$.
\end{lemma}
\begin{lemma}{\rm\cite[p.38]{slo}}\label{centra}
Let $\,a\in{\mathfrak g}\,$.
If the characteristic of $\,K\,$ is
either $\,0\,$ or a very good prime for $\,G,\,$ then
$\,{\cal L}(C_G(a))\,=\,{\mathfrak c}_{\mathfrak g}(a)\;$ for any
$\,a\in{\mathfrak g}\,$.
\end{lemma}
Now $\,G\,$ acts on the nilpotent cone $\,{\cal N}\,$ of $\,\mathfrak g\,$
via the adjoint representation and $\,{\cal N}\,$ splits into finitely
many ($\Ad\,G$)-orbits (see \cite{baca1}, \cite{baca2}, \cite{pom1},
\cite{pom2}, \cite{holspa}).
\begin{lemma}\label{help}
Let $\,{\cal Z}\,$ be an irreducible component of $\,{\cal N}\,$
and let $\,\cal T\,$ be a torus of $\,\mathfrak g\,$ of dimension
$\,\ell\,=\,{\rm rank}\,G.\,$ Then
$\,\dim\,{\cal Z}\,\leq\,\dim\,\mathfrak g\,-\,\ell$.
\end{lemma}\noindent
\begin{proof}
Suppose $\,\dim\,{\cal Z}\,>\,\dim\,\mathfrak g\,-\,\ell$. As
$\,{\cal N}\,$ is a cone, so is $\,{\cal Z}.\,$ Hence, $\,0\,\in\,{\cal
Z}\cap{\cal T}\,$ and, as a consequence, $\,{\cal Z}\cap{\cal
T}\,\neq\,\emptyset\,$. Since
$\,{\cal Z}\,$ and $\,\mathfrak t\,$ are irreducible varieties of
$\,\mathfrak g,\,$ by the Affine Dimension Theorem~\cite[Prop. 7.1]{har}
we have that $\,\dim\,{\cal Z}\cap\mathfrak t\,>\,0$. By
contradiction, the result follows.
\end{proof}
\begin{proposition}\label{othercases}
Let $\,x\,$ be a nilpotent element of $\,\mathfrak g\,$. Then
\[
\dim\,C_G(x)\,\geq\,\ell\,.
\]
\end{proposition}\noindent
\begin{proof}
Suppose $\,\dim\,C_G(x)\,<\,\ell\,$. Then $\,\dim\,G\cdot x\,>\,
\dim\,\mathfrak g\,-\,\ell\,$. But since $\,G\cdot x\subset {\cal
N},\,$ this implies
$\,\dim\,{\cal N}\,>\,\dim\,\mathfrak g\,-\,\ell$, contradicting the
previous lemma. This proves the proposition.
\end{proof}
\begin{proposition}\label{semicases}
Let $\,a\,$ be a semisimple element of $\,\mathfrak g\,$. Then
\[
\dim\,C_G(a)\,\geq\,\ell\,.
\]
\end{proposition}\noindent
\begin{proof}
Let $\,Z\,=\,\{ x\in\mathfrak g\,/\,x^{[p]}\,=\,t^{(p-1)}x\;\mbox{for
some}\;t\in K\}\,$ be the so called {\it cone} over the variety of
toral elements $\,\cal T\,$. Let $\,\mathfrak t\,$ be a torus of
dimension $\,\ell\,=\,{\rm rank}\,G\,$. The intersection $\,Z\cap
\mathfrak t\,$ consists of finitely many affine lines (as every toral
subalgebra has finitely many toral elements). Each irreducible
component $\,Z_i\,$ of $\,Z\,$ is homogeneous whence contains $\,0$.
It follows that $\,Z_i\cap \mathfrak t\,$ is non-empty. Hence,
$\,\dim\,Z_i\leq\dim\,\mathfrak g\,-\,\ell\,+\,1$. Each component of
$\,\cal T\,$ is contained in one of the $\,Z_i$'s, but cannot be equal
to it as given a nonzero $\,y\in{\cal T}\cap Z_i\,$, the scalar
multiple $\,\lambda y\,$ is not toral, for $\,\lambda\not\in {\bf
F}_p\,$. However, $\,\lambda y\in Z_i\,$ as $\,Z_i\,$ is homogeneous.
Therefore, each homogeneous component of $\,\cal T\,$ has dimension
$\,\leq\,\dim\,\mathfrak g\,-\,\ell\,$. Now proceed as in the proof
of Proposition~\ref{othercases}.
\end{proof}
\subsection{The Exceptional Modules}\label{sec3.2}
Let $\,V\,$ be a finite dimensional vector space over $\,K$.
If $\,G\,$ acts on $\,V\,$ via a rational representation
$\,\pi\,:\,G\,\longrightarrow\,GL(V),\,$ then $\,{\mathfrak g}\,$ acts
on $\,V$, via the differential
$\,d\pi\,:\,{\mathfrak g}\,\longrightarrow\,{\mathfrak {gl}}\;(V)$.
For $\,v\in V\,$ and $\,x\in \mathfrak g$, put
$\,d\pi(x)\cdot v\,=\,x\cdot v\,$.
Let $\,G\cdot v\,=\,\{\,g\cdot v\,/\,g\,\in\,G\,\}\,$
be the $\,G$-orbit of $\,v\,$ and
$\,{\mathfrak g}\cdot v\,=\,\{\,x\cdot v\,/\,x\,
\in\,{\mathfrak g}\,\}.\,$
\begin{definition}\label{isotropy}
The {\bf isotropy (sub)group} (or
{\bf stabilizer}) of $\,v\in V\,$ in $\,G\,$ is $\,G_v\,=\,\{\,g\,\in\,G\,/\,
g\cdot v\,=\,v\,\}\,$ and the {\bf isotropy (Lie) subalgebra} of
$\,v\in V\,$ in $\,\mathfrak g\,$ is
$\,{\mathfrak g}_v\,=\,\{\,x\,\in\,{\mathfrak g}\,/\,x\cdot v\,=\,0\,\}$.
\end{definition}
\noindent
\begin{remark}\label{rem2}~{\bf (1)}~The stabilizer $\,{\mathfrak
g}_v\,=\,\{\,x\,\in\,{\mathfrak g}\,/\,
x\cdot v\,=\,0\,\}\,$ is a restricted Lie subalgebra of $\,{\mathfrak g}$, as
$\,x^{[p]}\cdot v\,=\,x^p\cdot v$. Hence, by Lemma~\ref{ppp},
either $\,{\mathfrak g}_v\,$ is a torus or $\,{\mathfrak g}_v\,$
contains a nonzero element $\,x\,$ such that $\,x^{[p]}\,=\,0.\,$
\end{remark}
By~\cite[Ex. 10.2]{hum2}, $\,{\cal L}(G_v)\,\subseteq\,
{\mathfrak g}_v$.
For $\,x\,\in\,{\mathfrak g}$, let $\,V^x\,=\,\{\,v\,\in\,V\,/\,x\cdot
v\,=\,0\,\}\,$ and $\,x\cdot V\,=\,\{\,x\cdot v\,/\,v\in V\,\}$.
\begin{definition} \label{isgp}
For a rational action of an algebraic group $\,G\,$ on a vector space
$\,V$, we say that a subgroup $\,H\,\subset\,G\,$ is an {\bf isotropy
subgroup in general position} (or an ISGP for short) if $\,V\,$
contains a non-empty Zariski open subset $\,U\,$ whose points have
their isotropy subgroups conjugate to $\,H$. The points of $\,U\,$ are
called {\bf points in general position}.
\end{definition}
Isotropy subalgebras $\,{\mathfrak h}\,\subset\,{\mathfrak g}\,$ in general
position are defined similarly.
\begin{definition} \label{locfree}
A rational $\,G$-action $\,G\longrightarrow\GL(V)\,$ is said to be
{\bf locally free} if the ISGP $\,H\,$ exists and equals $\,\{ e\}$.
\end{definition}
\begin{definition} \label{locfreega}
A linear $\,\mathfrak g$-action $\,\mathfrak
g\longrightarrow\mathfrak{gl}(V)\,$ is said to be {\bf locally free} if the
isotropy subalgebra in general position $\,\mathfrak h\,$
equals $\,\{ 0\}$.
\end{definition}
We are interested in studying $\,G$- and $\,\mathfrak g$-actions which are
not locally free.
\subsubsection{Lie Algebra Actions in Positive Characteristic}
The following result is well-known and applies to all
characteristics. We include the proof for reader's convenience, as
references are hard to find.
\begin{proposition}\label{rem1}~~If there exists $\,0\neq\,v\in V\,$ such
that $\,{\mathfrak g}_v\,=\,\{ 0\}$, then there exists a Zariski open subset
$\,W\subseteq V\,$ such that $\,{\mathfrak g}_w\,=\,0\,$ for all
$\,w\in W$.
\end{proposition}\noindent
\begin{proof}
Consider the morphism $\,\psi\,:\,{\mathfrak g}\times V\longrightarrow
V\times V\,$ given by $\,(x,\,v)\longmapsto (x\cdot v,\,v)\,$.
Let $\,(v_1,\,v_2)\in V\times V\,$. Then
\begin{align*}
\psi^{-1}(v_1,\,v_2)\,&=\,\{\,(x,\,v)\in {\mathfrak g}\times V\,/\,
(x\cdot v,\,v)\,=\,(v_1,\,v_2)\,\}\\
& =\,\{\,(x,\,v_2)\in {\mathfrak g}\times V\,/\,x\cdot v_2\,=\,v_1\,\}.
\end{align*}
Let $\,d\,$ denote the minimal dimension of the fibres of $\,\psi.\,$
Suppose $\,\psi^{-1}(v_1,\,v_2)\,\neq\,\emptyset\,$ and
let $\,(x,\,v_2),\;(y,\,v_2)\,\in\,\psi^{-1}(v_1,\,v_2)\,$. Then
$\,x\,-\,y\,\in\,{\mathfrak g}_{v_2}\,$. Hence,
$\,\psi^{-1}(v_1,\,v_2)\,=\,(\,{\mathfrak g}_{v_2}\,+\,y,\,v_2),\,$ where
$\,y\cdot v_2\,=\,v_1\,$. Thus, either $\,\dim\,\psi^{-1}(v_1,\,v_2)\,=\,\dim
{\mathfrak g}_{v_2}\,$ or $\,\psi^{-1}(v_1,\,v_2)\,=\,\emptyset\,$.
Let $\,U\,=\,\{\,(x,\,v)\in\mathfrak g\times V\,/\
\,\dim\,\psi^{-1}(\psi(x,\,v))\,=\,d\,\}.\,$ By the theorem on
dimension of fibres~\cite[4.1]{hum2}, $\,U\,$ is a non-empty Zariski
open subset of \mbox{$\,{\mathfrak g}\times V$}. Thus, $\,\dim {\mathfrak
g}_v\,=\,\dim\,\psi^{-1}(\psi(x,\,v))\,$
is the minimal possible for all $\,v\in V.\,$
As the composition $\,{\mathfrak g}\times
V{\stackrel{\psi}{\longrightarrow}}V\times V
{\stackrel{{\rm pr}_2}{\longrightarrow}}V \,$ is surjective, we have
$\,d=0\,$ (by our assumption).
Now the projection $\,\pi_2\,:\,{\mathfrak g}\times V\longrightarrow V\,$
is surjective, whence dominant (see~\cite[4.1]{hum2}).
Therefore, $\,\pi_2(U)\,$ contains a Zariski open
subset $\,W\,$ of $\,V.\,$ Thus, all points of $\,W\,$ have trivial
stabilizer, proving the proposition.
\end{proof}
\begin{proposition} \label{3.1.1}
If there is a non-empty Zariski open subset $\,W\,\subset\,V\,$ such that
$\,{\mathfrak g}_w\,\neq\,0,\;\;\forall\,w\in W$, then
$\,{\mathfrak g}_v\,\neq\,0$, for all $\,v\in V$.
\end{proposition}\noindent
\begin{proof} Consider $\,{\cal C}\,=\,\{\,(x,\,v)\in {\mathfrak
g}\times V\,/\,x\cdot v\,=\,0\,\}$, the commuting variety of
$\,{\mathfrak g}\times V$. Note that $\,{\cal C}\,$ is ``bihomogeneous'',
i.e., $\,\forall\,\lambda,\,\mu\in K-\{ 0\},\;(\lambda\,x,\,\mu\,v)\,
\in\,{\cal C}$, whenever $\,(x,\,v)\,
\in\,{\cal C}$. Let $\,\bar{\cal C}\,$ be the closed subset of
$\,\mathbb P({\mathfrak g})\,\times\,\mathbb P(V)\,$ corresponding to
$\,{\cal C}$. As $\,\mathbb P({\mathfrak g})\,$ is complete, the
projection
$\,{\rm pr}_2\,:\,\mathbb P({\mathfrak g})\,\times\,\mathbb P(V)\,\longrightarrow\,
\mathbb P(V)\,$ is a closed map (see \cite[\S 6]{hum2}) so that
$\,{\rm pr}_2 (\bar{\cal C})\,$ is a closed set.
Hence the image of $\,{\cal C}\,$ under the projection map
$\,{\mathfrak g}\times V\longrightarrow V\,$ is also closed
(as the closed subsets of $\,\mathbb P(V)\,$ are the images of the
Zariski closed subsets of $\,V\,$ corresponding to
collections of homogeneous polynomials in $\,K[V]$).
Let $\,\varepsilon\,:\,{\cal C}\longrightarrow V\,$ denote the
restriction of the projection map $\,{\mathfrak g}\times V\longrightarrow V\,$
to $\,{\cal C}.\,$ For each $\,v\in V$, the fibre
$\,\varepsilon^{-1}(v)\,=\,\{\,(x,\,v)\in {\cal C}\,/\,x\in {\mathfrak g}_v\,\}
\,$ is isomorphic to a vector space and so is irreducible as an algebraic
variety. Moreover, $\,\dim\,\varepsilon^{-1}(v)\,=\,\dim\,{\mathfrak g}_v$.
Now, for each $\,n\in\mathbb N$, consider the set
\[
{\cal C}_n\,=\,\{\,(x,\,v)\in {\cal C}\,/\,\dim\, {\mathfrak g}_v\,
\geq\,n\,\}\,=\,\{\,(x,\,v)\in {\cal C}\,/\,\dim\,
\varepsilon^{-1}(\varepsilon(x,\,v))\,\geq\,n\,\}.
\]
By \cite[AG.10.3]{bor2}, each $\,{\cal C}_n\,$ is a closed set.
Let $\,s\,=\,\min\,\{\,\dim {\mathfrak g}_w\,/\,w\in W\,\}$.
There exists a closed variety $\,\bar{{\cal C}_s}\,$ in $\,\bar{\cal
C}\,$ that corresponds to $\,{\cal C}_s$.
Therefore, $\,{\rm pr}_2(\bar{{\cal C}_s})\,$ is closed in $\,\mathbb P(V)$,
implying that $\,\varepsilon ({\cal C}_s)\,$ is also closed in $\,V$.
Now as $\,\varepsilon\,$ is surjective, $\,\varepsilon ({\cal
C}_s)\,\supseteq\,W$. Finally, as $\,W\,$ is dense in $\,V$, this implies
$\,\varepsilon ({\cal C}_s)\,=\,V$, proving the proposition.
\end{proof}
\begin{definition} \label{nonlocfree}
A $\,{\mathfrak g}$-module $\,V\,$ is said to be {\bf non-free} if
$\,{\mathfrak g}_v\,\neq\,0\,$ for all $\,v\,\in\,V$.
\end{definition}
\noindent
\begin{remark}~{\bf (2)} Comparing the two propositions above, we can
say that if $\,V\,$ is not a non-free $\,{\mathfrak g}$-module, then
it is locally free.
\end{remark}
By Lemma~\ref{centretoral}, the centre $\,{\mathfrak z(\mathfrak g)}\,$ of
the restricted Lie algebra $\,{\mathfrak g}\,=\,{\cal L}(G)\,$ is a
toral subalgebra. Hence, by Proposition~\ref{properties}(2),
$\,{\mathfrak z(\mathfrak g)}\,$ acts diagonalisably on $\,V\,$.
Observe that sometimes $\,{\mathfrak z(\mathfrak g)}\,$
is nonzero but acts trivially on $\,V,\,$ implying that
$\,{\mathfrak g}_v\,\supseteq\,{\mathfrak z
(\mathfrak g)}\,$ for all $\,v\in V$. This happens, for instance, when
$\,\mathfrak g\,=\,\mathfrak{sl}(\ell+1, K),\,$ $\,p|(\ell +1)\,$
and $\,V\,=\,\mathfrak g\,$ is the adjoint $\,\mathfrak g$-module.
The following statement is true.
\begin{proposition} \label{3.1.a}
If there is a non-empty Zariski open subset $\,W\,\subset\,V\,$ such that
$\,{\mathfrak g}_w\,$ has a non-central element for all $\,w\in W$, then
$\,{\mathfrak g}_v\,$ has a non-central element for all $\,v\in V$.
\end{proposition} \noindent
\begin{proof}
The proof repeats almost verbatim the proof of Proposition~\ref{3.1.1}.
\end{proof}
Our next definition is crucial for the rest of this work.
\begin{definition}\label{excpmod}
A $\,{\mathfrak g}$-module $\,V\,$ is called {\bf exceptional} if
for each $\,v\in V\,$ the isotropy subalgebra $\,{\mathfrak g}_v\,$ contains
a non-central element (that is, $\,{\mathfrak g}_v\,\not\subseteq\,
{\mathfrak z(\mathfrak g)}\,$).
\end{definition}
\begin{example}\label{adjoint} {\rm
Clearly, if $\,\mathfrak g\,$ is non-abelian, then the trivial
$\,1$-dimensional $\,\mathfrak g$-module is exceptional.
For any non-abelian Lie algebra $\,{\mathfrak g},\,$ the adjoint
module is always
exceptional, as $\,{\mathfrak z(\mathfrak g)}\,$ is a proper subset of
$\,{\mathfrak g}_x\,$ for all $\,x\in{\mathfrak g}\,$ (note that
$\,[x,\,x]\,=\,0\,$).}
\end{example}
For the centreless Lie algebras the notions of {\it non-free} and
{\it exceptional} modules coincide.
It follows from the definitions that all exceptional modules are
non-free. The converse statement is not true, as the following lemma shows.
(The proof of this lemma will be given in Section~\ref{lact}.1.)
\begin{lemma}\label{nonexcepnonfree}
Let $\,\mathfrak g\,=\,\mathfrak{sl}(np,K),\,$ where
$\,p>2\,$ and $\,np\geq 4.\,$ Then the Steinberg module
$\,V\,=\,E((p-1)\rho)\,$ is not an exceptional module, but it is non-free.
\end{lemma}\noindent
It follows from Proposition~\ref{3.1.a} that if there is a nonzero
$\,v\in V\,$ such that $\,{\mathfrak
g}_v\,\subset\,{\mathfrak z(\mathfrak g)}\,$ then $\,V\,$ is not
an exceptional module. In particular, locally free modules cannot be
exceptional.
\begin{lemma}\label{notbadhw}
If $\,V\,$ is an exceptional $\,\mathfrak g$-module, then so is each of its
composition factors.
\end{lemma}\noindent
\begin{proof}
Let $\,V\,=\,V_0\,\supset\,V_1\,\supset\,V_2\,\supset\,\cdots\,\supset\,
V_k\,\supset \,\{ 0\}\,$
be a composition series of $\,V.\,$ Take $\,0\neq\bar{v}\in\,V_i/V_{i+1}\,$
and let $\,v\in V_i\,$ be its preimage. Then $\,\mathfrak
g_v\,\subset\,\mathfrak g_{\bar{v}}\,$. The result follows.
\end{proof}
\medskip
\subsection{A Necessary Condition}\label{necess}
Let $\,\pi\,:\,G\longrightarrow\GL(V)\,$ be a non-trivial faithful
rational representation of $\,G\,$ and
let $\,d\pi\,:\,{\mathfrak g}\,\longrightarrow\,{\mathfrak{gl}}\;(V)\,$
be the differential of $\,\pi\,$ at $\,e\in G$.
Suppose that $\,\ker d\pi\,\subseteq\,{\mathfrak z}
({\mathfrak g})\,$ and assume that $\,V\,$ is an exceptional
$\,{\mathfrak g}$-module.
Under these hypotheses, we want to study highest weights of $\,V.\,$
(We do not assume that $\,V\,$ is irreducible.)
Note that if $\,V\,$ has no $\,1$-dimensional $\,\mathfrak g$-submodules, then
$\,\dim {\mathfrak g}_v\,<\,\dim {\mathfrak g}\,$ for each nonzero $\,v\in V$.
First we aim to find an upper bound for $\,\dim\,V.\,$
\begin{proposition} \label{prev}
Given $\,x\,\in\,{\mathfrak g},\,$ define $\,\varphi_x\,:\,G\,\times\,V^x\,
\longrightarrow\,V\,$ by setting $\,\varphi_x(g,v)\,=\,g\cdot v.\,$
Then for any $\,v\in \Im\varphi_x,\;\dim \varphi_x^{-1}(v) \,\geq\, \dim
C_G(x)$, where $\,C_G(x)\,=\,\{\,g\in G\,/\,(\Ad\,g)(x)\,=\,x\,\}$.
\end{proposition}\noindent
\begin{proof}
First note that given $\,v\in \Im\varphi_x$, there exist
$\,h\,\in\,G\,$ and $\,v_1\,\in\,V^x\,$ such that $\,v\,=\,h\cdot
v_1\,$ (we set $\,h=e\,$ for $\,v\in V^x)$.
By definition,
\[
\varphi_x^{-1}(v) \,=\,\{\,(g,\,w)\,\in\,G\times V^x\,/\,g\cdot
w\,=\,v\,\}\,.
\]
Let $\,{\rm pr}_2\,:\,\varphi_x^{-1}(v)\,\longrightarrow\,V^x\,$ be the
projection map. Then its image is $\,G\cdot v\,\cap\,V^x$.
For $\,w\in\, {\rm pr}_2(\varphi_x^{-1}(v))$, if
$\,(g,\,w),\;(h,\,w)\,\in\,{\rm pr}_2^{-1}(w)\,\subset\,\varphi_x^{-1}(v)$,
then $\,g\cdot w\,=\,v\,=\,h\cdot w\,$ which implies $\,h\,\in\,G_v\,g$.
Hence, for each $\,w\in\; {\rm pr}_2(\varphi_x^{-1}(v))$, there exists
$\,g\,\in\,G\,$ such that $\,g\cdot w\,=\,v\,$ and
$\,(\,G_v\,g,\,w)\,\subset\,\varphi_x^{-1}(v)$, so that
$\,\dim\, {\rm pr}_2^{-1}(w)\,=\,\dim\,
\{(x\,g,\,w)\,/\,x\in G_v\}\,=\,\dim\, G_v$. Hence all fibres of
$\,{\rm pr}_2|_{\varphi_x^{-1}(v)}\,$ have the same dimension
equal to $\,\dim\, G_v$. Therefore,
\begin{equation} \label{var1}
\dim\, \varphi_x^{-1}(v)\,=\,\dim\, G_v\,+\,\dim\, (G\cdot v\,\cap\,V^x).
\end{equation}
Now $\,(G\cdot v\,\cap\,V^x)\,\supseteq\,C_G(x)\cdot v_1\,$ yielding
\begin{equation}\label{var2}
\begin{split}
\dim\, (G\cdot v\,\cap\,V^x)\, & \geq\,\dim\, (C_G(x)\cdot v_1) \\
& =\, \dim\,
C_G(x)\,-\,\dim\, (G_{v_1}\,\cap\,C_G(x)).
\end{split}
\end{equation}
Combining~(\ref{var1}) and~(\ref{var2}) we get
\begin{eqnarray}
\dim\, \varphi_x^{-1}(v) & \geq & \dim\, G_v\,+\,\dim\, C_G(x)\,-\,
\dim\, (G_{v_1}\,\cap\,C_G(x)) \nonumber\\
& \geq & \dim\, G_v\,-\,\dim\, G_{v_1}\,+\,\dim\, C_G(x)\,=\,\dim\,
C_G(x), \nonumber
\end{eqnarray}
as $\,G_v\,\cong\,G_{v_1}$. This proves the proposition.
\end{proof}
\medskip
\begin{proposition} \label{dimxvxg}
Given an exceptional $\,{\mathfrak g}$-module $\,V$,
there exists a non-central $\,x\,\in\,{\mathfrak g}\,$ such that
either $\,x^{[p]}\,=\,x\,$ or $\,x^{[p]}\,=\,0\,$ and
\begin{equation}\label{eqa}
\dim x\,V\,\leq\,\dim\,G\,-\,\dim\,C_G(x)\,\leq\,|R|\,.
\end{equation}
\end{proposition}\noindent
\begin{proof}
Let $\,v\,\in\,V$. By Remark~\ref{rem2}(1),
the stabilizer $\,{\mathfrak g}_v\,$ is either toral or contains a
non-zero element $\,x\,$ such that $\,x^{[p]}\,=\,0\,$.
Let $\,{\cal T}_1^*\,=\,\{\,x\in\,{\mathfrak g}\setminus\{ 0\}\,/\,x^{[p]}\,
=\,x\,\}\,$ and $\,{\cal N}_1^*\,
=\,\{\,x\in\, {\mathfrak g}\setminus\{ 0\}\,/\,x^{[p]}\,=\,0\,\}$.
By Lemma~\ref{centretoral}, $\,{\mathfrak z}(\mathfrak g)\,$ is toral,
hence consists of semisimple elements.
If $\,{\mathfrak g}_v\,$ contains a nilpotent element $\,x$,
then $\,x\not\in{\mathfrak z}(\mathfrak g)\,$. If $\,{\mathfrak
g}_v\,$ contains no nilpotent elements, then the $\,p$-mapping
$\,[p]\,$ is nonsingular on $\,{\mathfrak g}_v\,$, whence
$\,{\mathfrak g}_v\,$ is toral. Then by Theorem~\ref{3.6}
$\,{\mathfrak g}_v\,$ is spanned by toral elements. Thus, $\,{\mathfrak g}_v\,$
contains at least one non-central toral element, as $\,V\,$ is
exceptional.
Therefore, for each $\,v\in V,\,$ there is a non-central
$\,x\in \,{\cal N}_1^*\,\cup\,{\cal T}_1^*\,$ such that $\,v\in V^x$. Hence
\[
V\,=\,\bigcup_{x\in ({\cal N}_1^*\cup {\cal T}_1^*)\setminus\mathfrak
z(\mathfrak g)}\,V^x.
\]
If $\,x^{[p]}\,=\,x$, then there exists $\,y\in G\,$ such
that $\,(\Ad\,y)(x)\,\in\,{\mathfrak t}\,$ (Proposition~\ref{anytoral}).
By Lemma~\ref{toralbase}(2), $\,{\mathfrak t}\cap
{\cal T}_1^*\,$ is finite. Combining this with Theorem~\ref{badhs}, we
obtain that
there are finitely many nilpotent and toral conjugacy classes in
$\,{\mathfrak g}$. Let $\,n_1,\,n_2,\,\ldots\,,n_s\,$
(resp. $\,t_1,\,t_2,\,\ldots\,,t_{\ell}\,$) be
representatives of the non-central conjugacy classes in $\,{\cal
N}_1^*\,$ (resp.,
$\,{\cal T}_1^*$). As $\,{\mathfrak g}_v\,$ contains a non-central
element from $\,{\cal N}_1^*\cup {\cal T}_1^*$, we have
$\,v\,\in\,\left(\bigcup_{j=1}^{s}\;G\cdot V^{n_j}\right)\,\cup\,
\left(\bigcup_{i=1}^{\ell}\;\,G\cdot V^{t_i}\right)$. Therefore,
\begin{equation} \label{uniongti}
V\,=\,\left(\bigcup_{j=1}^{s}\;G\cdot V^{n_j}\right)\,\cup\,
\left(\bigcup_{i=1}^{\ell}\;\,G\cdot V^{t_i}\right)
\end{equation}
For any $\,x\,\in\,{\mathfrak g}$, the set $\,G\cdot V^{x}\,$ is the image of
the morphism
\[
\varphi_x\,:\,G\,\times\,V^x\,\longrightarrow\,V,\qquad\;(g,v)\,
\longmapsto\,g\cdot v\,.
\]
In particular, each $\,G\cdot V^{x}\,$ is constructible
\cite[4.4]{hum2}. This implies that $\,G\cdot V^{x}\,$
contains an open dense subset of $\,\overline{G\cdot V^{x}}\,$
\cite[Cor.10.2]{bor2}.
Hence at least one of the subsets $\,G\cdot V^{n_j},\;G\cdot V^{t_i}\,$ is
Zariski dense in $\,V\,$ (note that the number of the subsets in the
decomposition~\eqref{uniongti} is finite).
It follows that $\,V\,$ contains a non-empty Zariski dense subset of the form
$\,G\cdot V^{x}.\,$ So $\,\varphi_x\,$ is dominant
for some non-central $\,x\in{\cal N}_1^*\cup {\cal T}_1^*$.
Now, using \cite[3.1]{hum2} and the theorem on dimension of fibres of
a morphism (see \cite[Theorem AG.10(ii)]{bor2}), we have
\begin{equation} \label{first}
\dim {\mathfrak g}\,+\,\dim V^x\,-\,\min_{v\in V}\;\,\dim\,
\varphi^{-1}_x(v) \,= \, \dim V.
\end{equation}
By Proposition~\ref{prev}, for any $\,v\in\, \Im\varphi_x,
\;\dim\, \varphi_x^{-1}(v)\,\geq\,\dim C_G(x)$, forcing
\begin{equation} \label{second}
\dim {\mathfrak g}\,+\,\dim V^x\,-\,\dim C_G(x)\,\geq\, \dim V.
\end{equation}
As $\,\dim{\mathfrak g}=\dim G\,$ and $\,\dim x\cdot V=\dim V\ -\
\dim V^x,\,$ we get $\,\dim x\cdot V\,\leq\,\dim\,G\,-\,\dim\,C_G(x)\,$.
Now if $\,x\,$ is toral, then it is semisimple by
Proposition~\ref{properties}(1). In this case Proposition~\ref{semicases}
applies. If $\,x\,$ is nilpotent we
apply Proposition~\ref{othercases}. Thus, in both cases the result follows.
\end{proof}
From now on we assume that $\,G\,$ is a simply connected simple
algebraic group over $\,K\,$.
Fix a maximal torus $\,T\,$ of $\,G\,$ and
a basis $\,\Delta\,$ of simple roots in the root system $\,R\,$ of
$\,G\,$ (with respect to $\,T\,$). Let $\,R^+\,$ denote the system of positive
roots with respect to $\,\Delta\,$ and
let $\,B\,$ be the Borel subgroup
of $\,G\,$ generated by $\,T\,$ and the $\,1$-parameter unipotent
subgroups $\,U_{\alpha}\,=\,\{\,x_{\alpha}(t)\,/\,t\in K\,\}\,$,
where $\,\alpha\in R^+\,$.
Let $\,\tilde{\alpha}\,$ denote the maximal root in $\,R^+.\,$
Decompose $\,\mathfrak g\,=\,{\cal L}(G)\,$ into
root spaces with respect to $\,T\,$ giving a Cartan decomposition
\[
\mathfrak g\,=\,\mathfrak t\,\bigoplus\,\sum_{\alpha\in R}\,K\,e_{\alpha}
\]
where $\,\mathfrak t\,=\,\Lie (T)\,$ (cf. Section\ref{gennot}).
By Section~\ref{structliealg},
$\,\mathfrak g\,=\,({\rm Ad}\,G)\cdot\mathfrak b,\,$
where $\,\mathfrak b\,=\,{\cal L}(B).\,$
Put $\,{\cal E}\,=\,(\Ad G)\cdot e_{\tilde{\alpha}}\cup\{0\}.\,$
It is well-known (cf.~\cite{kraft},~\cite{pre4}) that $\,{\cal E}\,$
is a Zariski closed, conical subset of $\,\mathfrak g.\,$
The following result is well-known in the characteristic zero
case. It was generalized by A. Premet in~\cite{pre4} to the case
$\,p>0$. Here we reproduce the proof for the reader's convenience.
\begin{lemma}{\rm\cite[Lemma 2.3]{pre4}}\label{varep}
Suppose $\,p\,$ is non-special for $\,G$.
Let $\,Z\,$ be a Zariski closed, conical, (${\rm Ad}\,G$)-invariant
subset of $\,\mathfrak g.\,$ Then either $\,Z\subseteq{\mathfrak z}
(\mathfrak g)\,$ or $\,{\cal E}\subseteq Z\,$.
\end{lemma}\noindent
\begin{proof}
If $\,x\in\mathfrak b\,$ then $\,x\,=\,t\,+\,\sum_{\alpha\in R^+}\,
n_{\alpha}\,e_{\alpha}\,$ where $\,t\in \mathfrak t\,$ and
$\,n_{\alpha}\in K$. Define $\,{\rm Supp}(x)\,=\,\{\alpha\in R^+\,/\,
n_{\alpha}\neq 0\}$.
As $\,\mathfrak g\,=\,({\rm Ad}\,G)\cdot\mathfrak b\,$ and $\,Z\,$ is
$\,({\rm Ad}\,G)$-stable, $\,Z\cap \mathfrak b
\neq\{ 0\}$. If $\,Z\cap \mathfrak b \subseteq \mathfrak t$, then
$\,({\rm Ad}\,x_{\alpha}(t))\cdot z\,=\,z\,$ for all $\,\alpha\in R^+,\,
t\in K,\,$ and $\,z\in Z\cap \mathfrak b $. It follows that
$\,({\rm d}\alpha)(z)\,=\,0\,$ for every $\,\alpha\in R$. In other words,
$\,Z\cap \mathfrak b \subseteq\mathfrak z(\mathfrak g)\,$. As
$\,Z\,=\,({\rm Ad}\,G)\cdot Z\cap \mathfrak b\,$ we get $\,Z\subseteq
\mathfrak z(\mathfrak g)$.
Now suppose that $\,{\rm Supp}(z)\,\neq\, \emptyset\,$ for some
$\,z\in Z\cap \mathfrak b$. Let $\,X_*(T)\,$ denote the lattice of one
parameter subgroups in $\,T\,$. If $\,\lambda(t)\in X_*(T)\,$ and
$\,\gamma\in R$, then $\,\gamma(\lambda(t))\,=\,t^{m(\gamma)}\,$ for some
$\,m(\gamma)\in \mathbb Z.\,$ A standard argument shows that there
exists $\,\lambda(t)\,\in\,X_*(T)\,$ such that $\,m(\gamma)\,>\,0\,$
for every $\,\gamma\in R^+\,$ and the numbers
$\,m(\gamma),\,\gamma\in R^+\,$ are pairwise distinct. Let
$\,r\,=\,\max\,\{m(\alpha)\,/\,\alpha\,\in\,{\rm Supp}(z)\}.\,$ As
$\,Z\,$ is conical,
\[
\{\,t^r\cdot ({\rm Ad}\,\lambda(t^{-1}))\cdot z\,/\,t\in
K^*\,\}\,\subseteq \,Z\,.
\]
By construction, there are $\,z_0,\,z_1,\ldots,\,z_r\,\in\,\mathfrak b\,$
such that $\,z_0\in Ke_{\delta}\setminus\{ 0\}\,$ for some
$\,\delta\in R^+\,$ and
\[
t^r\cdot ({\rm Ad}\,\lambda(t^{-1}))\cdot
z\,=\,z_0\,+\,tz_1\,+\,\cdots\,+\,t^r z_r\,.
\]
As $\,Z\,$ is Zariski closed we must have $\,z_0\in Z.\,$ If
$\,\delta\,$ is long, there is $\,w\in W\,$ such that
$\,w\delta\,=\,\tilde{\alpha}\,$. It follows that $\,({\rm
Ad}\,N_G(T))\cdot z_0\,$ contains $\,e_{\tilde{\alpha}}\,$. Thus in this
case $\,{\cal E}\subseteq Z.\,$ If $\,\delta\,$ is short, there is
$\,w\in W\,$ such that $\,w\delta\,=\,{\alpha}_0\,$. So we may assume
that $\,\delta\,=\,{\alpha}_0\,$. By \cite{hogew},
$ \,\tilde{\alpha}\,\neq \,{\alpha}_0\,$ implies that $\,\mathfrak
g\,$ is simple. Hence, there is $\,\gamma\in R^+\,$ such that
$\,[e_{\gamma},\,e_{\alpha_0}]\,\neq\,0$. Obviously, all roots in
$\,\alpha_0\,+\,\mathbb N\gamma\,$ are long. Let
$\,q\,=\,\max\,\{\,i\in\mathbb N\,/\,\alpha_0\,+\,i\gamma\in R\,\}.\,$
Applying the same argument as above to the subset
\[
\{\,t^q\cdot ({\rm Ad}\,x_{\gamma}(t^{-1}))\cdot z\,/\,t\in
K^*\,\}\,
\]
we get $\,Z\cap Ke_{\alpha_0+q\gamma}\neq\,\{ 0\}.\,$ Since
$\,\alpha_0+q\gamma\,$ is long, we are done.
\end{proof}
\begin{corollary}\label{maxcent}
Let $\,V\,$ be any rational $\,G$-module and let $\,x\in{\mathfrak
g}$. The maximum value of $\,\dim\,V^x,\;x\not\in
{\mathfrak z(\mathfrak g)}$, is $\,\dim\,V^{e_{\tilde{\alpha}}}$.
\end{corollary}\noindent
\begin{proof}
Given $\,d\in\mathbb Z^+,\,$ let
$\,X_d\,=\,\{\,x\in\mathfrak g\,/\,\dim\,x\cdot V\,\leq\,d\}.\,$
It is clear that $\,X_d\,$ is ($\Ad\,G$)-invariant
and a conical Zariski closed subset of $\,\mathfrak g$.
By Lemma~\ref{varep}, either $\,X_d\subseteq {\mathfrak z}({\mathfrak
g})\,$ or $\,{\cal E}\subseteq X_d\,$. Now take $\,d_1\,=\,\min\,\{
d\,/\, X_d\not\subseteq{\mathfrak z}({\mathfrak g})\}\,$.
Then $\,{\cal E}\subseteq X_{d_1}\,$ and the result follows.
\end{proof}
\vspace{2ex}\noindent
\begin{remark} \label{after}
It follows from Proposition~\ref{dimxvxg}(a) and this last Corollary
that
\[
\dim\,e_{\tilde{\alpha}}\cdot V\,\leq\,|R|.
\]
\end{remark}
\subsubsection{An Upper Bound for $\,\dim\,V\,$}
In this section we use results proved before to produce an upper
bound for the dimension of the exceptional modules.
Recall that $\,G\,$ is a simply connected simple algebraic group and
$\,\mathfrak g\,=\,{\cal L}(G)\,$ (cf. Section~\ref{introback}).
Let $\,\tilde{\alpha}\,$ be the highest root for $\,{\mathfrak g},\,$
and $\,E=e_{\tilde{\alpha}}\,\in\,{\mathfrak g}\,$ denote a highest
root vector. Then $\,E^{[p]}= 0.\,$
There exist $\,F=e_{-\tilde{\alpha}}\in{\mathfrak g}_{-\tilde{\alpha}}\,$
and $\,H=h_{\tilde{\alpha}}\in{\mathfrak h}\,$
such that $\,(E,H,F)\,$ form a standard basis of an
$\,{\mathfrak{sl}}(2)$-triple $\,{\mathfrak s}_{\tilde{\alpha}}\,$ in
$\,\mathfrak g$.
Let $\,V\,$ be a finite dimensional exceptional $\,G$-module.
As $\,H\,=\,[E,\,F],\,$ we have $\,H\,V\,\subseteq\,E\,V\,+\,F\,V\,$ implying
\[
\dim\,H\,V\,\leq\,\dim\,E\,V\,+\,\dim\,F\,V\,=\,2\,\dim\,E\,V\,.
\]
Thus $\,\displaystyle
\frac{1}{2}\,\dim\,H\,V\,\leq\,\dim\,E\,V\,\leq\,|R|$, by
Remark~\ref{after}.
Now $\,V\,=\,\displaystyle\bigoplus_{\mu\in{\cal X}(V)}\,V_{\mu}$,
where the $\,V_{\mu}$'s are the weight spaces of $\,V\,$ with respect to $\,T$.
Recall from Section~\ref{wamv}
that the differential of a weight $\,\mu\in X\,$ is a linear function
on $\,\mathfrak t\,$, also denoted by $\,\mu\,$.
If $\,H\in\mathfrak t,\,$ then
$\,H\cdot v\,=\,\mu(H)\,v,\,$ for all $\,v\in V_{\mu}\,$. Therefore,
$\,H|_{V_{\mu}}\,=\,\mu (H)\,Id_{m_{\mu}}$.
Now the action of $\,H\,$ on $\,V\,$ induced by the differential
$\,{\rm d}\pi\,$ (at the identity $\,e\in G\,$) turns $\,V\,$ into a
$\,\mathbb Z/p\mathbb Z$-graded vector space
$\,V\,=\,\displaystyle\sum_{i\in\mathbb Z/p\mathbb Z}\,V_{i}\,$, where
\[
V_i\,=\,\bigoplus_{\mu(H)=i}\,V_{\mu}\,.
\]
Thus, we can calculate $\,\dim\,H\,V\,$ as follows:
\begin{equation} \label{hvhv}
\begin{split}
\dim\,H\,V\,& =\,\sum_{\mu (H)\not\equiv 0\pmod p}
\,\dim\,V_{\mu}\\
& = \sum_{\mu\in{\cal X}_{++}(V)}\,m_{\mu}\,|\{\,\nu\in W\mu\,/\,
\nu(H)\not\equiv 0\pmod p\,\}|\,,
\end{split}
\end{equation}
where $\,m_{\mu}\,=\,\dim\,V_{\mu}\,$.
The second equality in~\eqref{hvhv} comes from the facts that
$\,{\cal X}(V)\,=\,W\cdot {\cal X}_{++}(V)\,$
and that the multiplicity is constant on the $\,W$-orbits of $\,{\cal X}(V)$.
Let $\,(\,\cdot\,,\,\cdot\,)\,$ denote a scalar product on the
$\,\mathbb R$-span of $\,\Delta,\,$ invariant under the action of the
Weyl group $\,W\,$ of $\,R.\,$ We adjust $\,(\,\cdot\,,\,\cdot\,)\,$ in such
a way that $\,(\alpha,\alpha)=2\,$ for every short root $\,\alpha\in R.\,$
As usual, $\,\langle\beta,\,\alpha\rangle\,=\,2\,(\beta,\,\alpha)/
(\alpha,\,\alpha),\,$ for $\,\beta,\,\alpha\in R$.
This adjustment on $\,(\,\cdot\,,\,\cdot\,)\,$ and the assumption that
$\,p\,$ is not special for $\,G\,$ imply $\,p\nmid\frac
{(\tilde{\alpha},\,\tilde{\alpha})}{2}\,$. Hence
\begin{equation}\label{mual}
\mu (H)\,=\,\langle\mu,\,\tilde{\alpha}\rangle\,=\,\frac{2\,(\mu,\,
\tilde{\alpha})}{(\tilde{\alpha},\,\tilde{\alpha})}
\not\equiv\,0\!\pmod{p}
\,\Longleftrightarrow\,(\mu,\,\tilde{\alpha})\not\equiv 0\!\pmod{p}
\end{equation}
Note that $\,w\mu (H)\,=\,\langle w\mu,\,\tilde{\alpha}\rangle\,=\,
\langle\mu,\,w^{-1}\tilde{\alpha}\rangle$. Hence
$\,w\mu (H)\,\not\equiv\,0\pmod{p}
\,\Longleftrightarrow\,(w\mu,\,\tilde{\alpha})\,\not\equiv 0\pmod{p}$,
by~(\ref{mual}). Also if $\,w_1\in C_W(\mu)$, then
$\,w\,w_1\mu (H)\,=\,w\mu (H).\,$ Thus, we obtain that
\begin{multline*}
\sharp\,\{ w\in W\,/\,w\mu(H)\not\equiv 0\!\!\pmod p\}\,= \\
|C_W(\mu)|\cdot |\{\nu\,\in\,W\mu\,/\,\nu(H)\not\equiv 0\!\!\pmod p\}|\,.
\end{multline*}
Observe that $\,(w\mu,\,\tilde{\alpha})\,=
\,(\mu,\,w^{-1}\tilde{\alpha})$, as the scalar product is
$\,W$-invariant. Thus $\;w\mu (H)\,\not\equiv\,0\!\!\pmod{p}
\;\Longleftrightarrow\;(\mu,\,w^{-1}\tilde{\alpha})\,
\not\equiv 0\!\!\pmod{p}$.
Also, $\,\{\,w^{-1}\tilde{\alpha}\,/\,w\in W\,\}\,=\,R_{long}$, the set
of all long roots in $\,R$, forms a subsystem of $\,R\,$
\cite[VI \S 1 Ex.14]{bourb2}.
In particular,
$\,R_{long}\,=\,R_{long}^+\,\cup\,-R_{long}^+\,$. Hence
\begin{multline*}
\sharp\,\{ w\in W\,/\,w\mu(H)\not\equiv 0\!\!\pmod p\}\,= \\
2\,|C_W(\tilde{\alpha})|\cdot
|\{\,\gamma\in R^+_{long}\,/\,(\mu,\,\gamma)\,
\not\equiv\,0\!\!\pmod p\,\}|\,.
\end{multline*}
We summarize the discussion in:
\begin{lemma}\label{lem345}
Given $\,\mu\,\in\,{\cal X}_{++}(V)\,$ one has
\begin{multline*}\label{ineq345}
|C_W(\mu)|\cdot |\{\,w\mu\in W\mu\,/\,(w\mu,\,\tilde{\alpha})
\,\not\equiv\,0\pmod p\,\}|\\
=\,2\,|C_W(\tilde{\alpha})|\cdot
|\{\,\gamma\in R^+_{long}\,/\,(\mu,\,\gamma)\,
\not\equiv\,0\pmod p\,\}|.
\end{multline*}
\end{lemma}
\hfill$\Box$
Set $\,R_{\mu,p}\,=\,\{\,\gamma\in R_{long}\,/\,
(\mu,\,\gamma)\,\equiv\,0\!\pmod p\,\}\,$ and
$\,R_{\mu,p}^+\,=\,R_{\mu,p}\,\cap R_{long}^+\,=\,\{\,\gamma\in R_{long}^+\,/\,
(\mu,\,\gamma)\,\equiv\,0\!\pmod p\,\}\,$.
We have
\begin{equation} \label{split345}
\begin{split}
\frac{1}{2}\,|\{\,\nu\in W\mu\,/\,(\nu,\,\tilde{\alpha})
\,\not\equiv\,0\!\!\pmod p\,\}| & = \frac{|C_W(\tilde{\alpha})|}
{|C_W(\mu)|}\,|R_{long}^+-R^+_{\mu,p}| \vspace{1ex} \\
& = \frac{|W|}{|C_W(\mu)|}\,\frac{|R_{long}^+-R^+_{\mu,p}|}{|R_{long}|}
\vspace{1ex}\\
& = \frac{|W\mu|}{|R_{long}|}\,
|R_{long}^+-R^+_{\mu,p}|
\end{split}
\end{equation}
Now observe that $\,R^+_{\mu,p}\,=\,R_{long}^+\,$ if and only if
$\,(\mu,\,\gamma)\,\equiv\,0\pmod p\,$ for all $\,\gamma\in R_{long}^+$.
In this case the contribution of the weight $\,\mu\,$ to the
sum~(\ref{hvhv}) is $\,0$. This fact gives rise to the following definition:
\begin{definition}\label{badweight}
A weight $\,\mu\in P(R)\,$ is called {\bf bad} if
$\,\mu(h_{\gamma})\equiv 0\pmod p\,$ for all $\,\gamma\in R^+_{long}$,
or equivalently, if $\,(\mu,\,\gamma)\equiv 0\pmod p\,$ for all
$\,\gamma\in R^+_{long}$.
If $\,\mu\,$ is not a bad weight we say that $\,\mu\,$ is {\bf good}.
\end{definition}
Recall that each weight $\,\mu\in P\,$ can be written as
$\,\mu\,=\,\sum\,a_i\,\omega_i$, with $\,a_i\in\mathbb Z_+$.
\begin{lemma}
Suppose $\,p\,$ is not special for $\,G$. Then
$\,\mu\in P\,$ is bad if and only if
$\,\mu\,=\,\displaystyle\sum_{i=1}^{\ell}\,a_i\,\omega_i$, where
$\,a_i\in p\mathbb Z\,$ for all $\,i\,$.
\end{lemma}\noindent
\begin{proof}
If $\,R_{long}\,=\,R\,$, the lemma is obvious.
If $\,R\,$ is of type $\,B_{\ell}$, then $\,R_{long}\cong D_{\ell}$. A basis
of $\,R_{long}\,$ is given by $\,\alpha_1=\varepsilon_1
-\varepsilon_2\,$, $\,\alpha_2=\varepsilon_2 -\varepsilon_3,\ldots,
\,\alpha_{\ell -1}=\varepsilon_{\ell -1} -\varepsilon_{\ell},
\,\alpha_{\ell}=\varepsilon_{\ell -1} +\varepsilon_{\ell}$.
Hence $\,(\mu,\,\alpha_i)=a_i\,$, if $\,1\leq i\leq \ell -1,\,$ and
$\,(\mu,\,\alpha_{\ell})=(\mu,\,\varepsilon_{\ell-1} +\varepsilon_{\ell})=
a_{\ell -1} + a_{\ell}$. Therefore, $\,\mu\,$ is bad for $\,R\,$
if and only if $\,a_i\equiv 0\pmod p\,$ for all $\,i$.
If $\,R\,$ is of type $\,C_{\ell}$, then $\,R_{long}\cong A_1^{\ell}$. A basis
of $\,R_{long}\,$ is given by $\,2\varepsilon_1$, $\,2\varepsilon_2$,
$\,\ldots,\,$$\,2\varepsilon_{\ell -1},\,2\varepsilon_{\ell}$.
One has $\,(\mu,\,2\varepsilon_i)=
2(a_i +\cdots +a_{\ell}),\;1\leq i\leq \ell\,$.
Therefore, $\,\mu\,$ is bad for $\,R\,$
if and only if $\,a_i\equiv 0\pmod p\,$ (recall that in this case
$\,p\neq 2$).
For $\,R\,$ of type $\,G_2$, $\,R_{long}\cong A_2\,$ has basis
$\,-2\varepsilon_1 +\varepsilon_2+\varepsilon_3,\,\varepsilon_1-2\varepsilon_2
+\varepsilon_3$.
Thus $\,(\mu,\,-2\varepsilon_{1} +\varepsilon_{2}+\varepsilon_3)= 3a_2\,$ and
$\,(\mu,\,\varepsilon_1-2\varepsilon_2 +\varepsilon_3)= 3(a_1+a_2)\,$, implying
that $\,\mu\,$ is bad for $\,G_2\,$
if and only if $\,a_i\equiv 0\pmod p\,$ (in this case $\,p\neq 3$).
For $\,R\,$ of type $\,F_4$, $\,R_{long}\cong D_4\,$ has basis
$\,\varepsilon_1 -\varepsilon_2,\,\alpha_1=\varepsilon_2 -\varepsilon_3,
\,\alpha_2=\varepsilon_3 -\varepsilon_4,\,
\varepsilon_3 +\varepsilon_4$. Hence,
$\,(\mu,\,\varepsilon_{1} -\varepsilon_{2})= a_2 +a_3 +a_4$,
$\,(\mu,\,\alpha_1)=a_1,\;(\mu,\,\alpha_2)=a_2\,$ and
$\,(\mu,\,\varepsilon_3 +\varepsilon_4)=a_2 +a_3$.
Therefore, $\,\mu\,$ is bad for $\,F_4\,$
if and only if $\,a_i\equiv 0\pmod p\,$.
\end{proof}
\smallskip
\noindent
\begin{remark} It is easy to check that if $\,\mu\,=\,\sum_{i=1}^{\ell}\,
a_i\,\omega_i\,$ is a bad weight with $\,a_k\neq 0$, then
$\,\mu -\alpha_k\,$ is a good weight, unless $\,p=2\,$ and $\,R\,$ is of type
$\,A_1\,$ or $\,B_{\ell}\,$.
\end{remark}
\smallskip
We further observe that for $\,\mu\,$ good, $\,R_{\mu,p}\,$ is a
proper subsystem of $\,R_{long}\,$, so it is contained in a maximal
subsystem of $\,R_{long}$. We refer to~\cite{borsie} for a list of
maximal subsystems in $\,R_{long}$.
From (\ref{hvhv}) and~(\ref{split345})
we conclude that if $\,V\,$ is an exceptional $\,\mathfrak g$-module
then it satisfies
\[
r_p(V)\,:=\,\sum_{\stackrel{\scriptstyle\mu\;good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,\frac{|W\mu|}{|R_{long}|}\,
|R_{long}^+-R^+_{\mu,p}|\,=\,\frac{1}{2}\,\dim\,H\,V\,\leq\,\dim\,E\,V\,
\leq |R|
\]
But then
\[
\left(\,\sum_{\stackrel{\scriptstyle\mu\;good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,\frac{|W\mu|}{|R_{long}|}\,\right){\displaystyle
\min_{\mu\,good}}\,|R_{long}^+-R^+_{\mu,p}|\,\leq\,|R|
\]
implying
\[
s(V)\,:=\,\sum_{\stackrel{\scriptstyle\mu\;good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,|W\mu|\,\leq\,\frac{|R|\,\;|R_{long}|}
{\displaystyle\min_{\mu\;good}\,|R_{long}^+-R^+_{\mu,p}|}
\]
Now note that $\,|R_{long}^+-R^+_{\mu,p}|\,$ is minimal if and only if
$\,|R_{\mu,p}|\,$ is maximal. (Recall that $\,R_{\mu,p}\subseteq
R_{long}$.)
Put $\,M\,=\,\displaystyle\min_{\mu\,good}\,|R_{long}^+-R^+_{\mu,p}|\,$
and $\,L_G\,=\,\displaystyle\frac{|R|\,\;|R_{long}|}
{\displaystyle\min_{\mu\,good}\,|R_{long}^+-R^+_{\mu,p}|}\,$. We
call $\,\lceil L_G\rceil\,$ the {\bf limit}. Table~\ref{table2} lists all
related subsystems and gives
the values of $\,\lceil L_G\rceil\,$ for different types of $\,G$.
In the column
headed by $\,MLS,\,$ we list the possible maximal subsystems of $\,R_{long}$.
\begin{table}[tb]
\begin{center}
\begin{tabular}{|c|c|c|c|c|}\hline \hline
Type of $\,G$ & $R_{long}$ & $MLS$ &
$\,M\,$ &
$\,\lceil L_G \rceil\,$ \\ \hline \hline
$A_{\ell}$ & $A_{\ell}$ & $A_{\ell -1}$ & $\ell$ &
$\ell^3\,+\,2\ell^2\,+\,\ell$ \\ \hline
$B_{\ell}$ & $D_{\ell}$ & $D_{\ell -1}\,$ if $\,\ell\geq 5$ &
$2(\ell -1)$ & $2\ell^3\,$ if $\,\ell\geq 5$ \\
& & $A_{\ell -1}\,$ if $\,\ell\leq 4$ &
$\frac{(\ell -1)\ell}{2}$ & $8\ell^2\,$ if $\,\ell\leq 4$
\\ \hline
$C_{\ell}$ & $A_1^{\ell}$ & $A_1^{\ell -1}$ & $1$ &
$\,4\ell^3$ \\ \hline
$D_{\ell}$ & $D_{\ell}$ & $D_{\ell -1}\,$ if $\,\ell\geq 5$ &
$2(\ell -1)$ & $2\ell^3\,-\,2\ell^2\,$ if
$\,\ell\geq 5$\\
& & $A_{\ell -1}\,$ if $\,\ell\leq 4$ &
$\frac{(\ell -1)\ell}{2}$ & $\frac{8\ell^2(\ell -1)}{(\ell -2)}$
\\ \hline
$G_2$ & $A_2$ & $A_1$ & $2$ & $36$ \\ \hline
$F_4$ & $D_4$ & $A_3$ & $6=2\cdot 3$ & $192$ \\ \hline
$E_6$ & $E_6$ & $D_5$ & $16=2^4$ & $324$ \\ \hline
$E_7$ & $E_7$ & $E_6$ & $27=3^3$ & $588$ \\ \hline
$E_8$ & $E_8$ & $E_7$ & $57=3\cdot 19$ & $1011$ \\ \hline \hline
\end{tabular}
\caption[The Limits]{The Limits\label{table2}}
\end{center}
\end{table}
We summarize these facts in the following theorem.
\begin{theorem}[The Necessary Condition]\label{???}
Let $\,p\,$ be a non-special prime for $\,G$.
Let $\,\pi\,:\,G\longrightarrow\GL(V)\,$
be a non-trivial, faithful, rational representation of $\,G\,$ such that
$\,\ker\,{\rm d}\pi\subseteq\mathfrak z(\mathfrak g)\,$.
If $\,V\,$ is an exceptional $\,\mathfrak g$-module, then it satisfies
the inequalities
\begin{equation}\label{equata}
r_p(V)\,:=\,\sum_{\stackrel{\scriptstyle\mu\;good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,\frac{|W\mu|}{|R_{long}|}\,
|R_{long}^+-R^+_{\mu,p}|\,\leq |R|\,,
\end{equation}
and
\begin{equation}\label{?ref123}
s(V)\,:=\,\sum_{\stackrel{\scriptstyle\mu\,good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,|W\mu|\;\leq\;\mbox{\bf limit}\,,
\end{equation}
where $\,m_{\mu}\,$ denotes the multiplicity of $\,\mu\in
{\cal X}_{++}(V)\,$ and the {\bf limit}s for the different types of
algebraic groups are given in Table~\ref{table2}.
\end{theorem}
\hfill $\Box$
\newpage
\part*{Chapter 4}
\part*{The Procedure and The Results}
\addcontentsline{toc}{section}{\protect\numberline{4}{The
Procedure and The Results}}
\setcounter{section}{4}
\setcounter{subsection}{0}
The main section of this Chapter describes the techniques used to
classify the exceptional modules. Everywhere below the indexing of
the simple roots in the basis
$\,\Delta\,=\,\{\,\alpha_1,\,\ldots\,,\alpha_{\ell}\,\}\,$ and of the
fundamental weights in the basis
$\,\{\,\omega_1,\,\ldots,\,\omega_{\ell}\,\}\,$
corresponds to Bourbaki's tables~\cite[Chap. VI, Tables I-IX]{bourb2}.
\subsection{Some Facts on Weights}
In this section we formulate some known facts and obtain a number of
preliminary results that will be extensively used to prove lemmas
in the next sections.
Given a weight $\,\mu\,=\,a_1\,\omega_1\,+\,\cdots\,+
\,a_{\ell}\,\omega_{\ell}\,\in\,P\,$,
the size of its orbit under the action of the Weyl group is given by
\[
|W\mu|\,=\,\frac{|\,W\,|}{|\,C_W(\mu)\,|},
\]
where $\,C_W(\mu)\,=\,\langle s_{\alpha_i}\,/\,a_i\,=\,0\rangle\,$ is
the centralizer of $\,\mu\,$ in $\,W\,$ \cite[1.10]{seitz1}.
\begin{lemma}\label{remark}
Let $\,\lambda,\,\mu\,\in\,P_{++}\,$ be such that
$\,\langle\lambda,\,\alpha_i\rangle\,=\,0\,$ if
$\,\langle\mu,\,\alpha_i\rangle\,=\,0.\,$ Then $\,|W\mu|\,\geq\,|W\lambda|$.
If $\,\langle\mu,\,\alpha_i\rangle\,\neq\,0\,$ and
$\,\langle\lambda,\,\alpha_i\rangle\,=\,0\,$ for some $\,i,\,$ then
$\,|W\mu|\,\geq\,2\,|W\lambda|$.
\end{lemma}\noindent
\begin{proof}
From the definition of centralizer it is clear that
$\,C_W(\mu)\,$ is a subgroup of $\,C_W(\lambda)\,$ and a proper one
if $\,\langle\mu,\,\alpha_i\rangle\,\neq\,0\,$ and
$\,\langle\lambda,\,\alpha_i\rangle\,=\,0.\,$ This yields the assertion.
\end{proof}
If $\,Y\subseteq \Delta$, then
we write $\,W(Y)\,=\,\langle\,s_{\alpha}\,/\,\alpha\in Y\,\rangle\,$ for the
subgroup of $\,W\,$ generated by the reflections $\,s_{\alpha}\,$
corresponding to roots $\,\alpha\in Y$.
If $\,Y\,$ is a root (sub)system, then $\,W(Y)\,$ denotes the Weyl
group associated with the system $\,Y\,$.
Moreover, we denote
$\,\Delta_i\,=\,\Delta-\{\alpha_i\},\;\,\Delta_{ij}\,=\,\Delta
-\{\alpha_i,\alpha_j\},\,$
and so on. Note that $\,C_W(\mu)\,=\,W(Y)$, for some $\,Y\subseteq \Delta$.
In the following lemmas, $\,i_0\,$ and $\,i_{-1}\,$ are both $\,0.\,$
Recall that $\,P\,(=\,P(G)\,)\,$ denotes the weight lattice of the
group $\,G\,$.
\begin{lemma}\label{orban}
Let $\,G\,$ be an algebraic group of type $\,A_{\ell}.\,$ Let
$\,\mu\,=\,a_{i_1}\,\omega_{i_1}\,+\,a_{i_2}\,\omega_{i_2}\,+\,\cdots\,+
\,a_{i_m}\,\omega_{i_m}\,\in\,P$,
with $\,i_{k-1}\,<\,i_{k}\,$ and $\,a_{i_k}\neq 0\,$ for $\,1\leq
k\leq m.\,$ Then
\[
|W\mu|\,=\,\binom{i_2}{i_1}\,\cdots\,\binom{i_m}{i_{m-1}}\,
\binom{\ell+1}{i_m}.
\]
\end{lemma}\noindent
\begin{proof} For $\,G\,$ of type $\,A_{\ell},\,$
$\,|W|\,=\,(\ell+1)!\,$ and
$\,C_W(\mu)\,\cong\,S_{n_1+1}\times S_{n_2+1}\times\cdots\times
S_{n_{m+1}+1},\,$
where the $\,n_k$'s are obtained from the Dynkin diagram in the
following way: $\,n_1= i_1 -1,\quad n_{m+1}= \ell - i_m,\;$
$\,n_k =i_k-1-i_{k-1}=$ number of
nodes between the $\,i_{k-1}$th and the $\,i_{k}$th nodes, for $\,1<k\leq m$.
Therefore, $\,|\,C_W(\mu)\,|\,=\,i_1!\,(i_2-i_1)!\,\cdots
\,(i_m-i_{m-1})!\,(\ell-i_m+ 1)!$, yielding the result.
\end{proof}
\begin{lemma}\label{orbbncndn}
Let $\,G\,$ be an algebraic group of type $\,B_{\ell},\,C_{\ell}\,$ or
$\,D_{\ell}$. Let $\,\mu\,=\,a_{i_1}\,\omega_{i_1}\,+\,
a_{i_2}\,\omega_{i_2}\,+\,\cdots\,+\,a_{i_m}\,\omega_{i_m}\,\in\, P$,
with $\,i_{k-1}\,<\,i_{k}\,$ and
$\,a_{i_k}\neq 0\,$ for \mbox{$\,1\leq k\leq m\,$}. Then
\[
|W\mu|\,=\,2^r\,\binom{i_2}{i_1}\,\cdots\,\binom{i_{m-1}}{i_{m-2}}\,
\binom{t}{i_{m-1}}\,\binom{\ell}{t}
\]
where $\,r=t=i_m\,$ if $\,G\,$ is not of type $\,D_{\ell}\,$ or
$\,i_{m}<\ell -1;\,$ if $\,G\,$ is of type $\,D_{\ell},\,$
and $\,i_{m}=\ell-1\,$ or ($\,i_{m}=\ell\,$ and $\,i_{m-1}\leq
\ell-2\,$), then $\,r=\ell-1,\,t=\ell;\,$ for ($\,i_{m}=\ell\,$ and
$\,i_{m-1}=\ell-1\,$), $\,r=\ell-1\,=\,t\,=\,i_{m-1}.\,$
\end{lemma}\noindent
\begin{proof}
For $\,G\,$ of type $\,B_{\ell}\,$ or $\,C_{\ell},\,$
$\,|W|\,=\,2^{\ell}\,\ell!.\,$ For $\,i_m\,<\,\ell-1\,$ we have
$\,C_W(\mu)\,\cong\,S_{n_1+1}\times S_{n_2+1}\times\cdots\times
S_{n_{m}+1}\times W(B_{\ell-i_m}\;\mbox{or}\;C_{\ell-i_m}),\,$
where the $\,n_k$'s ($\,1\leq k\leq m\,$) are obtained from the
Dynkin diagram as in the proof of Lemma~\ref{orban}.
Thus $\,|\,C_W(\mu)\,|\,=\,i_1!\,(i_2-i_1)!\,\cdots
\,(i_m-i_{m-1})!\,2^{\ell-i_m}\,(\ell-i_m)!\,$ and so
\[
|W\mu|\,=\,2^{i_m}\,\binom{i_2}{i_1}\,\cdots\,\binom{i_{m-1}}{i_{m-2}}\,
\binom{i_m}{i_{m-1}}\,\binom{\ell}{i_m}.
\]
Hence, in these cases, $\,r=t=i_m.\,$
For $\,i_m\,=\,\ell-1,\,$
$\,C_W(\mu)\,\cong\,S_{n_1+1}\times S_{n_2+1}\times\cdots\times
S_{n_{m}+1}\times S_2\,$ so
$\,|\,C_W(\mu)\,|\,=\,i_1!\,(i_2-i_1)!\,\cdots
\,(i_{m-1}-i_{m-2})!\,(\ell-1-i_{m-1})!\,2!\,$ and this implies
\[
|W\mu|\,=\,2^{\ell-1}\,\binom{i_2}{i_1}\,\cdots\,\binom{i_{m-1}}{i_{m-2}}\,
\binom{\ell-1}{i_{m-1}}\,\binom{\ell}{\ell-1}.
\]
Here $\,r=\ell-1\,=t=i_m.\,$
For $\,i_m\,=\,\ell,\,$
$\,C_W(\mu)\,\cong\,S_{n_1+1}\times S_{n_2+1}\times\cdots\times
S_{n_{m}+1},\,$ so
\[
|W\mu|\,=\,2^{\ell}\,\binom{i_2}{i_1}\,\cdots\,\binom{i_{m-1}}{i_{m-2}}\,
\binom{\ell}{i_{m-1}}\,\binom{\ell}{\ell}\,,
\]
and $\,r=\ell=t=i_m.\,$
For $\,G\,$ of type $\,D_{\ell},\,$
$\,|W|\,=\,2^{\ell-1}\,\ell!.\,$ For $\,i_m\,<\,\ell-1\,$ we have
$\,C_W(\mu)\,\cong\,S_{n_1+1}\times S_{n_2+1}\times\cdots\times
S_{n_{m}+1}\times W(D_{\ell-i_m}),\,$
where the $\,n_k$'s ($\,1\leq k\leq m\,$) are obtained from the
Dynkin diagram as in the proof of Lemma~\ref{orban}.
Thus $\,|\,C_W(\mu)\,|\,=\,i_1!\,(i_2-i_1)!\,\cdots
\,(i_m-i_{m-1})!\,2^{\ell-i_m-1}\,(\ell-i_m)!\,$ and so
\[
|W\mu|\,=\,2^{i_m}\,\binom{i_2}{i_1}\,\cdots\,\binom{i_{m-1}}{i_{m-2}}\,
\binom{i_m}{i_{m-1}}\,\binom{\ell}{i_m}.
\]
Here $\,r=t=i_m.\,$
For $\,i_m\,=\,\ell-1\,$ (or $\,i_m\,=\,\ell\,$ and
$\,i_{m-1}\,\leq\,\ell-2\,$),
$\,C_W(\mu)\,\cong\,S_{n_1+1}\times S_{n_2+1}\times\cdots\times
S_{n_{m-1}+1}\times S_{\ell-1-i_{m-1}+1}\,$ so
\[
|W\mu|\,=\,2^{\ell-1}\,\binom{i_2}{i_1}\,\cdots\,\binom{i_{m-1}}{i_{m-2}}\,
\binom{\ell}{i_{m-1}}\,\binom{\ell}{\ell}.
\]
Here $\,r=\ell-1,\,t=\ell.\,$
For $\,i_m\,=\,\ell\,$ and $\,i_{m-1}\,=\,\ell-1\,$
$\,C_W(\mu)\,\cong\,S_{n_1+1}\times S_{n_2+1}\times\cdots\times
S_{n_{m-1}+1},\,$ so
\[
|W\mu|\,=\,2^{\ell-1}\,\binom{i_2}{i_1}\,\cdots\,\binom{i_{m-2}}{i_{m-3}}\,
\binom{\ell-1}{i_{m-2}}\,\binom{\ell}{\ell-1}\,,
\]
whence $\,r=\ell-1\,=\,t=i_{m-1}.\,$ This proves the Lemma.
\end{proof}
\begin{lemma}\label{orbexcp}
Some formulae for exceptional groups.
(1) Let $\,G\,$ be of type $\,E_6.\,$ Then $\,|W|\,=\,2^7\cdot 3^4\cdot 5\,$,
\begin{gather*}
C_W(\omega_1)\,\cong\,C_W(\omega_6)\,\cong\,W(D_5),\quad
C_W(\omega_2)\,\cong\,S_6,\\
C_W(\omega_3)\,\cong\,C_W(\omega_5)\,\cong\,S_2\times S_5,\quad
C_W(\omega_4)\,\cong\,S_2\times S_3^2,\\
C_W(\omega_1\,+\,\omega_2)\,\cong\,C_W(\omega_1\,+\,\omega_3)\,\cong\,
C_W(\omega_2\,+\,\omega_6)\,\cong\,C_W(\omega_5\,+\,\omega_6)\,\cong\,S_5,\\
C_W(\omega_1\,+\,\omega_4)\,\cong\,C_W(\omega_3\,+\,\omega_4)\,\cong\,
C_W(\omega_3\,+\,\omega_5)\,\cong\,C_W(\omega_4\,+\,\omega_5)\\
\cong\,C_W(\omega_4\,+\,\omega_6)\,\cong\,S_2^2\times S_3,\\
C_W(\omega_1\,+\,\omega_5)\,\cong\,C_W(\omega_2\,+\,\omega_3)\,\cong\,
C_W(\omega_2\,+\,\omega_5)\,\cong\,C_W(\omega_3\,+\,\omega_6)\,\cong\,
S_2\times S_4,\\
C_W(\omega_1\,+\,\omega_6)\,\cong\,W(D_4),\quad
C_W(\omega_2\,+\,\omega_4)\,\cong\,S_3^2,\quad
C_W(\omega_2\,+\,\omega_3\,+\,\omega_5)\,\cong\,S_2^3.
\end{gather*}
(2) Let $\,G\,$ be of type $\,E_7.\,$ Then
$\,|W|\,=\,2^{10}\cdot 3^4\cdot 5\cdot 7\,$,
\begin{gather*}
C_W(\omega_1)\,\cong\,W(D_6),\quad C_W(\omega_2)\,\cong\,S_7,\quad
C_W(\omega_3)\,\cong\,S_2\times S_6,\\
C_W(\omega_4)\,\cong\,S_2\times S_3\times S_4,\quad
C_W(\omega_5)\,\cong\,S_3\times S_5,\quad
C_W(\omega_6)\,\cong\,S_2\times W(D_5),\\
C_W(\omega_7)\,\cong\,W(E_6),\quad
C_W(\omega_1\,+\,\omega_2)\,\cong\,C_W(\omega_1\,+\,\omega_3)\,\cong\,
C_W(\omega_2\,+\,\omega_7)\,\cong\,S_6,\\
C_W(\omega_1\,+\,\omega_4)\,\cong\,S_2^2\times S_4,\quad
C_W(\omega_1\,+\,\omega_5)\,\cong\,C_W(\omega_2\,+\,\omega_4)\,\cong\,
S_3\times S_4,\\
C_W(\omega_1\,+\,\omega_6)\,\cong\,S_2\times W(D_4),\quad
C_W(\omega_1\,+\,\omega_7)\,\cong\,
C_W(\omega_6\,+\,\omega_7)\,\cong\,W(D_5),\\
C_W(\omega_3\,+\,\omega_7)\,\cong\,C_W(\omega_5\,+\,\omega_7)\,\cong\,
S_2\times S_5,\quad
C_W(\omega_4\,+\,\omega_6)\,\cong\,S_2^3\times S_3,\\
C_W(\omega_4\,+\,\omega_7)\,\cong\,S_2\times S_3^2,\quad
C_W(\omega_2\,+\,\omega_3\,+\,\omega_5)\,\cong\,S_2^2\times S_3.
\end{gather*}
(3) Let $\,G\,$ be of type $\,E_8.\,$ Then
$\,|W|\,=\,2^{14}\cdot 3^5\cdot 5^2\cdot 7\,$,
\begin{gather*}
C_W(\omega_1)\,\cong\,W(D_7),\quad C_W(\omega_2)\,\cong\,S_8,\quad
C_W(\omega_3)\,\cong\,S_2\times S_7,\\
C_W(\omega_4)\,\cong\,S_2\times S_3\times S_5,\quad
C_W(\omega_5)\,\cong\,S_4\times S_5,\\
C_W(\omega_6)\,\cong\,S_3\times W(D_5),\quad
C_W(\omega_7)\,\cong\,S_2\times W(E_6),\\
C_W(\omega_8)\,\cong\,W(E_7),\quad
C_W(\omega_1\,+\,\omega_4)\,\cong\,C_W(\omega_5\,+\,\omega_7)\,\cong\,
\,S_2^2\times S_5,\\
C_W(\omega_1\,+\,\omega_8)\,\cong\,W(D_6),\quad
C_W(\omega_2\,+\,\omega_8)\,\cong\,S_7,\quad
C_W(\omega_3\,+\,\omega_8)\,\cong\,S_2\times S_6,
\end{gather*}
\vspace*{-2ex}
\begin{gather*}
C_W(\omega_4\,+\,\omega_6)\,\cong\,S_2^2\times S_3^2,\quad
C_W(\omega_4\,+\,\omega_8)\,\cong\,S_2\times S_3\times S_4,\\
C_W(\omega_5\,+\,\omega_8)\,\cong\,S_3\times S_5,\quad
C_W(\omega_6\,+\,\omega_8)\,\cong\,S_2\times W(D_5),\\
C_W(\omega_7\,+\,\omega_8)\,\cong\,W(E_6),\quad
C_W(\omega_2\,+\,\omega_3\,+\,\omega_5)\,\cong\,S_2^2\times S_4.
\end{gather*}
(4) Let $\,G\,$ be of type $\,F_4.\,$ Then $\,|W|\,=\,2^{7}\cdot 3^2\,$,
\begin{gather*}
C_W(\omega_1)\,\cong\,C_W(\omega_4)\,\cong\,W(C_3),\quad
C_W(\omega_2)\,\cong\,C_W(\omega_3)\,\cong\,S_2\times S_3,\\
C_W(\omega_1\,+\,\omega_2)\,\cong\,C_W(\omega_3\,+\,\omega_4)\,\cong\,
\,S_3,\quad
C_W(\omega_1\,+\,\omega_4)\,\cong\,W(C_2),\\
C_W(\omega_1\,+\,\omega_3)\,\cong\,C_W(\omega_2\,+\,\omega_3)\,\cong\,
C_W(\omega_2\,+\,\omega_4)\,\cong\,\,S_2^2, \\
C_W(\omega_1\,+\,\omega_2\,+\,\omega_4)\,\cong\,S_2.
\end{gather*}
(5) Let $\,G\,$ be of type $\,G_2.\,$ Then
$\,|W|\,=\,12,\quad C_W(\omega_1)\,\cong\,C_W(\omega_2)\,\cong\,S_2\,$.
\end{lemma}
\begin{proof}
The information on $\,W(G)\,$ is extracted from Tables 5-8 of~\cite{bourb}.
Centralizers are found by using the remark preceeding
Lemma~\ref{remark}.
\end{proof}
\begin{lemma} \label{lamu}
Let $\,\lambda\in \Lambda_p\,$ and $\,\mu\,\in\,P_{++}\,$ be such that
$\,\mu < \lambda\,$ (that is,
$\,\lambda - \mu\,=\,$ sum of positive roots). Then
$\,{\cal X}_{\mathbb C}(\mu)\,\subseteq\,{\cal X}(\lambda)$, where
$\,{\cal X}_{\mathbb C}(\mu)\,$ denotes the set of weights of a
complex $\,{\mathfrak g}_{\mathbb C}$-module of highest weight $\,\mu\,$.
\end{lemma}\noindent
\begin{proof} As $\,{\cal X}_{\mathbb C}(\mu)\,$ and $\,{\cal
X}(\lambda)\,$ are $\,W\,$-invariant, it is enough to prove that
\mbox{$\,{\cal X}_{++,\mathbb C}(\mu)\,\subseteq\,{\cal X}_{++}(\lambda)$}.
Let $\,\gamma\,\in\,{\cal X}_{++,\mathbb C}(\mu)\,=\,(\mu - Q_+) \cap
P_{++}$. Then $\,\mu\,-\,\gamma \,\in\,Q_+\,$ and $\,\gamma\in P_{++}$.
By hypothesis, $\,\lambda\,-\,\mu\,\in\,Q_+\,$ so
$\,\lambda - \gamma + \gamma - \mu\,\in\,Q_+\,$ implying that
$\,\lambda - \gamma\,\in\,Q_+$. Hence,
$\,\gamma\,\in\,(\lambda - Q_+) \cap P_{++}\,=\,\,{\cal
X}_{++}(\lambda)$, proving the lemma.
\end{proof}
\bigskip
\noindent
The following lemma is a very useful fact on multiplicities.
\begin{lemma}\label{mulmin}
Let $\,V\,$ be an $\,A_{\ell}(K)$-module of highest weight
$\,\lambda\,=\,\omega_i\,+\,\omega_j\,$, where $\,1\leq i<j \leq
\ell$. Then the weight $\,\mu\,=\,\omega_{i-1}\,+\,\omega_{j+1}\,\in
\,{\cal X}_{++}(V)\,$ has multiplicity $\,j-i+1\,$
if $\,p\,$ does not divide $\,j-i+2,\,$ and $\,j-i\,$ otherwise.
\end{lemma}\noindent
\begin{proof}
Let $\,A\,$ be the Lie algebra (or group) generated by
$\,\{\,e_{\alpha_i=\alpha_1'},\ldots,\,e_{\alpha_j=\alpha_{j-i+1}'}\,\}.\,$
Note that $\,A\,$ is isomorphic to a Lie algebra (or group) of type
$\,A_{j-i+1}\,$.
Consider the $\,A$-module $\,U\,=\,V|_A$. $\,U\,$ is a nontrivial
homomorphic image of $\,V(\tilde{\alpha}')\,\cong\,\mathfrak{sl}(j-i+2),\,$
the adjoint module with respect to $\,A$.
The minuscule weight of $\,U\,$ is
$\,\mu\,=\,\omega'_{i-1}\,+\,\omega'_{j+1}\,=\,0'\,$. It
has multiplicity $\,j-i+1\,$ if $\,p\,$ does not divide $\,j-i+2,\,$
and multiplicity $\,j-i\,$ otherwise (for
the Lie algebra of type $\,A_n\,$ is simple if $\,p\,$ does not divide
$\,n+1\,$ and has one dimensional centre otherwise (in this case the
quotient algebra is simple)).
Now, by Smith's Theorem~\cite{smith}, the result follows.
\end{proof}
\subsubsection{Realizing $\,E(2\omega_1)\,$}\label{roapm}
In this subsection we describe the realization of the irreducible
$\,B_{\ell}(K)$- or $\,D_{\ell}(K)$-module of highest weight $\,2\omega_1\,$,
denoted by $\,E(2\omega_1)\,$. We need the following definition
due to Donkin and Jantzen.
\begin{definition}\label{weylfilt}{\rm\cite[11.5]{donk1}}
A descending chain $\,V\,=\,V_1\supset V_2\supset V_3\supset \cdots\,$
of submodules of a $\,G$-module $\,V\,$ is called a {\bf Weyl filtration}
if each $\,V_i/V_{i+1}\,$ is isomorphic to some $\,V_K(\lambda_i)\,$ with
$\,\lambda_i\,\in\,P_{++}\,$.
\end{definition}
\begin{proposition}\label{filttensor}{\rm\cite[11.5.2]{donk1},\cite{mathi}}
Let $\,\lambda,\,\mu\,\in\,P_{++}\,$. Then $\,V_K(\lambda)\otimes V_K(\mu)\,$
has a Weyl filtration.
\end{proposition}
Let $\,V=E(\omega_1)\,$ be the irreducible module of highest
weight $\,\omega_1\,$ and \mbox{$\,\dim\,V = n$}. Let $\,f\,$ be a
non-degenerate quadratic form on $\,V\,$. There exist a basis of
$\,V\,$ with respect to which $\,f\,$ has the form
$\,f\,=\,x_1^2\,+\,x_2^2\,+\,\cdots\,+\,x_n^2\,$, where the $\,x_i$'s
are indeterminates.
Let $\,\mathfrak g\,=\,\mathfrak{so}(f)\,=\,\{ x\in\,{\rm End}(V)\,/\,
f(x\cdot v,\,w)\,+\,f(v,\,x\cdot w)\,=\,0,\;\mbox{for}\;v,\,w\in V\}$.
For $\,n\,=\,2\ell+1\,$, $\,\mathfrak g\,$ is a Lie algebra of type
$\,B_{\ell}\,$ and for $\,n\,=\,2\ell\,$, $\,\mathfrak g\,$ is a Lie
algebra of type $\,D_{\ell}\,$.
Identify $\,\mathfrak{so}(f)\,$ with the space $\,{\rm Skew}_n\,$
of all $\,n\times n\,$ skew-symmetric matrices over $\,K$.
Let $\,G\,=\,\SO (f)\,$. Denote by $\,{\rm Symm}_n\,$ the space of
all $\,n\times n\,$ symmetric matrices over $\,K$. Then $\,{\rm Symm}_n\,$
and $\,S^2(V)\,$ are isomorphic as $\,G$-modules. Hence, we can
identify $\,S^2(V)\,$ with $\,{\rm Symm}_n\,$.
The Lie algebra $\,\mathfrak{so}(f)\,$ acts on the space $\,{\rm Symm}_n\,$ as
follows. Given $\,x\,\in\,\mathfrak{so}(f)\,$ and $\,s\,\in\,{\rm Symm}_n\,$
one has $\,x(s)\,=\,xs\,-\,sx\,$. It is easy to see that $\,S^2(V)\,$ is an
$\,\mathfrak{so}(f)$-module containing a highest
weight vector of weight $\,2\omega_1\,$. Hence, so does $\,{\rm Symm}_n\,$.
Let $\,U\,=\,{\rm Symm}_n\cap\mathfrak{sl}(n)\,$. Then $\,\mathfrak{so}(f)\,$
acts on $\,U\,$. Since $\,\dim\,({\rm Symm}_n/U)= 1$,
$\,U\,$ also has a highest weight vector of weight $\,2\omega_1\,$.
By~Proposition~\ref{filttensor}, $\,{\rm End}(V)\,\cong\,V\otimes V\,
\cong\,{\rm Symm}_n\oplus {\rm Skew}_n\,$ has Weyl filtration. It follows
that $\,S^2(V)\,$ has Weyl filtration as well~\cite[Chapter 4]{andjan}.
Let $\,V(2\omega_1)\,$ be the Weyl module with highest weight $\,2\omega_1\,$.
Then, by Weyl's dimension formula \cite[24.3]{hum1},
$\,\dim\,V(2\omega_1)=\displaystyle\frac{(n+2)(n-1)}{2}=\mbox{$\dim\,
{\rm Symm}_n -1$}$. As $\,\dim\,S^2(V)\,=\,\displaystyle
\frac{n\,(n+1)}{2}\,$, the members of the Weyl filtration of
$\,S^2(V)\,$ are $\,V(2\omega_1)\,$ and $\,V(0),\,$ the
trivial module (by dimension reasons). \vspace{2ex}
{\bf Claim 1}: $\,V(2\omega_1)\,$ is a submodule of $\,S^2(V).\,$
Indeed, suppose $\,V(2\omega_1)\,\cong\,\displaystyle\frac{S^2(V)}{V(0)}\,$.
Let $\,\bar v\,$ be the highest weight vector of weight $\,2\omega_1\,$
in $\,\displaystyle\frac{S^2(V)}{V(0)}$. Let $\,v\,$ be
the preimage of $\,\bar v\,$ in $\,S^2(V)_{2\omega_1}\,$. All
$\,T$-weights of $\,S^2(V)\,$ are $\,\leq\,2\omega_1\,$ with respect
to our partial ordering. Hence, $\,v\,$ is a highest weight vector
of $\,S^2(V)$. Let $\,\tilde V\,$ denote the $\,G$-submodule of
$\,S^2(V)\,$ generated by $\,v\,$. By the universal property of Weyl
modules (see Theorem~\ref{univprop}(3)), there exists
a $\,G$-epimorphism $\,\varphi\,:\,V(2\omega_1)\longrightarrow \tilde
V\,$. On the other hand, as $\,V(2\omega_1)\,\cong\,\displaystyle
\frac{S^2(V)}{V(0)}\,$, the image of $\,\tilde V\,$ in
$\,\displaystyle\frac{S^2(V)}{V(0)}\,$ is the whole $\,\displaystyle
\frac{S^2(V)}{V(0)}$. Thus, there is a $\,G$-epimorphism $\,\psi\,:\,
\tilde V\longrightarrow V(2\omega_1)\,$. It follows that
$\,\tilde V\,\cong\,V(2\omega_1)\,$ by Theorem~\ref{univprop}(3),
whence the claim. \vspace{2ex}
{\bf Claim 2}: $\,V(2\omega_1)\,\cong\,{\rm Symm}_n\cap\mathfrak{sl}(n)\,$.
First note that $\,G\,=\,\SO (f)\,$ acts by conjugation on the space
$\,{\rm Mat}_n\,$ of all $\,n\times n\,$ matrices over $\,K$, and
$\,{\rm Symm}_n\,$ is invariant under this action. Hence \mbox{$\,U\,=\,
{\rm Symm}_n\cap\mathfrak{sl}(n)\,$} is a $\,G$-submodule of $\,{\rm
Symm}_n\,$.
Now $\,U\,$ contains all vectors of nonzero weight (otherwise
$\,{\cal X}({\rm Symm}_n/U)\,$ would contain a nonzero weight,
contradicting $\,\dim\,({\rm Symm}_n/U)\,=1\,$). Moreover, each weight
of $\,U\,$ is $\,\leq\,2\omega_1\,$, and $\,2\omega_1\,$ has
multiplicity $\,1\,$ (as this is true for $\,S^2(V)\,$). As
$\,{\rm Symm}_n\,\cong\,S^2(V)\,$ as $\,G$-modules, Claim 1 now shows
that $\,U\,\cong\,V(2\omega_1)\,$, proving Claim 2.
{\bf Claim 3}: $\,V(2\omega_1)\,$ is an irreducible
$\,\mathfrak{so}(f)$-module if and only if $\,p\nmid n\,$.
If $\,p\mid n\,$, then $\,E(2\omega_1)\,\cong\,\displaystyle
\frac{{\rm Symm}_n\cap\mathfrak{sl}(n)}{scalars}\,$.
Let $\,\Phi(2\omega_1)\,$ be the maximal submodule of $\,V(2\omega_1)$.
Recall that $\,\displaystyle\frac{V(2\omega_1)}{\Phi(2\omega_1)}\,\cong
\,E(2\omega_1)\,$. By Theorem~\ref{theop1}, $\,{\cal X}(E(2\omega_1))\,=\,
{\cal X}(V(2\omega_1))\,$. One can show that each nonzero
$\,\mu\,\in\,{\cal X}(V(2\omega_1))\,$ has multiplicity $\,1$. This implies
$\,\Phi(2\omega_1)\,=\,a\,E(0)\,$, for some $\,a\geq 0$.
The group $\,G\,$ acts trivially on $\,\Phi(2\omega_1)\,$.
As $\,\Phi(2\omega_1)\subseteq {\rm Symm}_n\cap
\mathfrak{sl}(n)\,$ and $\,G\,$ acts on$\,{\rm Symm}_n\,$ by
conjugation, $\,\Phi(2\omega_1)\,$ consists of scalar matrices,
by Schur's Lemma~\cite[1.5]{isaacs}. Hence,
$\,\dim\,\Phi(2\omega_1)\,\leq\,1\,$. As $\,\Phi(2\omega_1)\subseteq
\mathfrak{sl}(n)\,$, $\,\Phi(2\omega_1)\,\neq\,0\,$ if and only if $\,p\mid n$.
Hence, if $\,p\nmid n$, then $\,V(2\omega_1)\,\cong\,E(2\omega_1)\,$
is irreducible. If $\,p\mid n$, then $\,E(2\omega_1)\,\cong\,\displaystyle
\frac{{\rm Symm}_n\cap\mathfrak{sl}(n)}{scalars}\,$, as claimed.
\subsection{The Procedure}\label{proced}
Let $\,V\,$ be a rational $\,G$-module.
Our objective throughout this work is to classify the exceptional
$\,\mathfrak g$-modules.
The procedure used in the classification relies heavily on the
necessary conditions given by Theorem~\ref{???}. We deal with the set
$\,{\cal X}_{++}(V)\,$ of weights of the module $\,V$. Hence we need
to guarantee that weights really occur in $\,{\cal X}_{++}(V)$.
{\bf From now on assume that $\,p\,$ is non-special for $\,G\,$ and
also that $\,p\neq 2\,$ if $\,V\,$ has highest weight $\,\omega_1\,$
for groups of type $\,G_2\,$}.
Under these assumptions, Theorem~\ref{premprin} says that {\it the system
of weights of an infinitesimally irreducible representation
$\,\pi\,:\,G\longrightarrow \GL(V),\,$ with highest weight
$\,\lambda\in\Lambda_p\,$ coincides
with the system of weights of an irreducible complex representation
$\,\pi_{\mathbb C}\,$ of a Lie algebra $\,{\mathfrak g}_{\mathbb C}\,$
with the same highest weight.} In particular, the set of dominant
weights of the representation $\pi$ is
$\,{\cal X}_{++}(\lambda)=\,(\lambda - Q_+)\,\cap\,P_{++}$.
Thus, in the sequel {\bf module} means {\bf an infinitesimally irreducible
finite dimensional rational
$\,G$-module $\,V\,$ of highest weight $\,\lambda\in\Lambda_p\,$}.
In particular, $\,V\,$ is an irreducible (restricted) $\,\mathfrak g$-module
(cf. Theorem~\ref{curt}). \vspace{2ex}
We now describe our procedure.
Recall that by Theorem~\ref{???}, an exceptional
$\,\mathfrak g$-module $\,V\,$ satisfies the inequalities
\begin{gather*}
s(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\,good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,|W\mu|\,\leq\,\mbox{{\bf limit}}\qquad\qquad\mbox{and}
\vspace{1.3ex}\\
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\;good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,\frac{|W\mu|}{|R_{long}|}\,
|R_{long}^+-R^+_{\mu,p}|\,\leq\,|R|\,.
\end{gather*}
Thus, we start by eliminating all possible modules of highest weight
$\,\lambda\,$ for which
\begin{equation}\label{?ref}
s(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\,good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,|W\mu|\;>\;\mbox{{\bf limit}}\,,
\end{equation}
since by Theorem~\ref{???}, $\,G$-modules
$\,V\,$ satisfying~\eqref{?ref} are {\bf not} exceptional.
In general, we proceed as follows:
Given a weight $\,\mu\in{\cal X}_{++}(V)\,$ we produce
weights $\,\mu =\mu_0,\,\mu_1,\,\mu_2,\,\ldots\,\in\,{\cal
X}_{++,\mathbb C}(\mu)\,\subseteq\,{\cal X}_{++}(V)\,$ and calculate
the sum of the orbit sizes
\[
s_1(\mu)\,=\,\sum_{\stackrel{\scriptstyle \mu_i\in{\cal X}_{++,\mathbb
C}(\mu)}{\mu_i\,good}}\,|W\mu_i|\,.
\]
If this sum is already bigger than the respective {\bf limit}, then we apply:
\begin{proposition}\label{criteria}
Let $\,\lambda\,$ be the highest weight of $\,V.\,$
If $\,\mu\,\in\,{\cal X}_{++}(V)\,$ and
$\,s_1(\mu)\,=\,\displaystyle\sum_{\stackrel{\scriptstyle \mu_i\,good}{\mu_i
\in{\cal X}_{++,\mathbb C}(\mu)}}\,|W\mu_i|\,>\,\mbox{{\bf limit}},\,$ then
$\,s(V)\,>\,\mbox{{\bf limit}}\,$. Hence $\,V\,$ is not an exceptional
$\,\mathfrak g$-module. In particular, if $\,\mu\,$ is a good weight
in $\,{\cal X}_{++}(V)\,$ and $\,|W\mu|\,>\,\mbox{{\bf limit}},\,$
then $\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{proposition}\noindent
\begin{proof}
Just note that $\,{\cal X}_{++,\mathbb C}(\mu)\,\subseteq\,{\cal X}_{++}(V)\,$
and use~(\ref{?ref}).
\end{proof}
In the process, we usually obtain first certain reduction lemmas,
which in most cases say that if a module $\,V\,$ is such
that $\,{\cal X}_{++}(V)\,$ contains good weights with many (at least
$\,3\,$ or $\,4\,$) nonzero coefficients, then $\,V\,$ is not an exceptional
module. See, for example, Lemmas~\ref{4coef},~\ref{3orless},~\ref{3coefbn},
\ref{3coefcn},~\ref{3cdn},~\ref{e61},~\ref{e71},~\ref{e81}.
In some cases we calculate the quantities
$\,|R_{long}^+-R^+_{\mu_i,p}|\,$ for each of the weights
$\,\mu =\mu_0,\,\mu_1,\,\mu_2,\,\ldots\,\in\,{\cal
X}_{++,\mathbb C}(\mu)\,\subseteq\,{\cal X}_{++}(V)\,$ and prove that
\begin{equation}\label{rpv2}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\;good}{\mu\in{\cal X}_{++}(V)}}
\,m_{\mu}\,\frac{|W\mu|}{|R_{long}|}\,
|R_{long}^+-R^+_{\mu,p}|\,>\,|R|\,.
\end{equation}
If \eqref{rpv2} holds for $\,V\,$ then, again by Theorem~\ref{???},
$\,V\,$ is not exceptional.
Sometimes we assume that $\,V\,$ has highest weight $\,\mu\,$. Then we use
information on the multiplicities of weights in $\,{\cal X}_{++}(V)\,$
to get the inequality~(\ref{?ref}) or \eqref{rpv2}.
We frequently use the works of Gilkey-Seitz~\cite{gise} and
Burgoyne-Williamson~\cite{buwil}, which give
tables of multiplicities of weights for representations of exceptional
and low-rank classical Lie algebras, respectively.
In some cases, we calculate multiplicities by
using some Linear Algebra and the fact that
$\,V\,$ has a basis consisting of the elements $\,f_{i_1}^{m_{i_1}}\,
f_{i_2}^{m_{i_2}}\,\cdots\,f_{i_k}^{m_{i_k}}\cdot v_0\,$, where
$\,v_0\,$ is a highest weight vector of $\,V\,$ (cf. Lemma~\ref{bor42}).
Using this procedure we end up with a small list of weights, whose
corresponding modules may or may not be exceptional.
Then we have to decide precisely which ones are exceptional.
To prove that a module $\,V\,$ is not exceptional it is enough to
exhibit a nonzero vector $\,v\in V\,$ for which $\,\mathfrak
g_v\,\subset\,\mathfrak z(\mathfrak g)\,$ or $\,\mathfrak g_v\,=\,0$,
but this is not an easy task as it depends on how the
representation is realized.
The following argument relates to the dimension of $\,V$.
Let $\,\varepsilon\,=\,\dim\,\mathfrak z(\mathfrak g)\,$.
\begin{proposition}\label{dimcrit}
Let $\,V\,$ be an irreducible $\,\mathfrak g$-module.
If $\,\dim\,V\,<\,\dim\,{\mathfrak g}\,-\,\varepsilon,\,$
then $\,V\,$ is an exceptional module.
\end{proposition}\noindent
\begin{proof} In proving the proposition, we may assume that $\,\dim\,V\,>\,1$.
By the irreducibility, $\,V\,$ does not
have $\,1$-dimensional invariant subspaces, so that $\,\dim\,
{\mathfrak g}_v\,<\,\dim\,{\mathfrak g}.\,$
Suppose $\,V\,$ is not exceptional. Then there exists $\,0\neq v\,\in\,V\,$
such that $\,{\mathfrak g}_v\,\subseteq\,\mathfrak z(\mathfrak g)$.
Consider the linear map $\,\psi_v\,:\,\mathfrak g\longrightarrow V\,$
given by $\,x\longmapsto x\cdot v\,$. Thus,
$\,\ker\,\psi_v\subseteq \mathfrak z(\mathfrak g)\,$ and, by Linear Algebra,
$\,\dim\,\psi_v(\mathfrak g)\,\geq \dim\,\mathfrak g\,-\,\dim\,
\mathfrak z(\mathfrak g)$. As \mbox{$\,\psi_v(\mathfrak g)\subseteq V\,$},
by contradiction, the result follows.
\end{proof}
\vspace{2ex}\noindent
\begin{remark} The dimension criteria for a module to be exceptional
(given by Proposition~\ref{dimcrit}) is independent of $\,p$.
\end{remark}
\medskip
The following proposition gives the list of all rational irreducible
representations of Chevalley groups whose weight multiplicities are
equal to $\,1$.
\begin{proposition}{\rm\cite[p. 13]{supruzal}}\label{supruzal}
Let $\,G\neq A_1(K)\,$ be a simply connected simple algebraic group over an
algebraically closed field $\,K\,$ of characteristic $\,p>0$. Define
the set of weights $\,\Omega=\Omega(G)\,$ as follows:
$\,\Omega(A_{\ell}(K))\,=\,\{\,\omega_i,\,a\,\omega_1,\,b\,\omega_{\ell},\,
c\,\omega_j\,+\,(p-1-c)\,\omega_{j+1};\;1\leq i\leq \ell,
\;1\leq j < \ell,\;\,0\leq a,\,b,\,c\,<\,p\,\}\,$;
$\,\Omega(B_{\ell}(K))\,=\,\{\,\omega_1,\,\omega_{\ell}\,\}\,$ for
$\,\ell\geq 3,\;p>2\,$;
$\,\Omega(C_{\ell}(K))\,=\,\{\,\omega_\ell,\;\mbox{for}\;\ell=2,\,3;\;\;
\omega_1,\,\omega_{\ell-1}+\frac{(p-3)}{2}\omega_{\ell},\;
\frac{(p-1)}{2}\omega_{\ell}\,\}\,$ for $\,p>2\,$ and
$\,\Omega(C_{\ell}(K))\,=\,\{\,\omega_1,\,\omega_\ell\,\}\,$ for
$\,p=2\,$;
$\,\Omega(D_{\ell}(K))\,=\,\{\,\omega_1,\,\omega_{\ell-1},\,\omega_\ell\,\}\,$;
\qquad
$\,\Omega(E_6(K))\,=\,\{\,\omega_1,\,\omega_6\,\}\,$;
$\,\Omega(E_7(K))\,=\,\{\,\omega_7\,\}\,$;
\qquad
$\,\Omega(F_4(K))\,=\,\{\,\omega_4\,\}\,$ for $\,p=3\,$ and
$\,\emptyset\,$ for $\,p\neq 3\,$;
$\,\Omega(G_2(K))\,=\,\{\,\omega_1\,\}\,$ for $\,p\neq 3\,$ and
$\,\Omega(G_2(K))\,=\,\{\,\omega_1,\omega_2\,\}\,$ for $\,p=3\,$.
Let $\,\varphi\,$ be an irreducible rational representation of $\,G\,$
with highest weight
$\,\omega\,=\,\displaystyle\sum_{i=0}^k\,p^i\lambda_i$,
where $\,\lambda_i\,$ are the highest weights of the representations
from $\,M(G)$. \linebreak The multiplicities of all the weights of the given
representation $\,\varphi\,$ are equal to $\,1\,$ if and only if
$\,\lambda_i=0\,$ or $\,\lambda_i\in\Omega(G)\,$ for $\,0\leq i\leq
k$, $\,\lambda_{i+1}\neq\omega_1\,$ for $\,p=2$,
$\,G=C_{\ell}(K),\,$$\,\lambda_i=\omega_{\ell}\,$ and for
$\,G=G_2(K),\,p=2,\,$ $\,\lambda_i=\omega_1\,$ or
$\,p=3,\,\lambda_i=\omega_2$.
\end{proposition}
\begin{proposition}\label{twistexc}
Let $\,\hat{\sigma}\,$ be a graph automorphism of $\,\mathfrak g\,$ induced
by a graph automorphism $\,\sigma\,$ of $\,G$. If $\,V\,$ is an
exceptional $\,\mathfrak g$-module, then so is $\,V^{\hat{\sigma}}\,$.
\end{proposition}\noindent
\begin{proof}
The action of the graph automorphism on $\,G\,$ and $\,\mathfrak g\,$
is described in Section~\ref{graphaut}. Denote by $\,\rho_{\sigma}\,$
(respectively, $\,\rho_{\hat{\sigma}}\,$) the twisted representation
of $\,G\,$ (respectively, $\,\mathfrak g\,$). We have
$\,\rho_{\hat{\sigma}}(x)\cdot v\,=\,\rho(x^{\hat{\sigma}})\cdot v\,$ for all
$\,x\in\mathfrak g,\;v\in V\,$.
As $\,\sigma\,$ preserves positive roots, it preserves the
Borel subgroup of $\,G$. Hence, $\,V\,$ and $\,V^{\hat{\sigma}}\,$
have the same highest weight vector, but $\,V^{\hat{\sigma}}\,$ has
highest weight $\,\lambda^{\sigma}\,$, where $\,\lambda\,$ is the
highest weight of $\,V$.
Now, let $\,^{\sigma}\mathfrak g_v\,$ denote the isotropy subalgebra
of $\,v\,$ with respect to the representation $\,\rho_{\hat{\sigma}}\,$.
If $\,\rho(x)\cdot v\,=\,0,\,$ then
$\,\rho_{\hat{\sigma}}(x^{{\hat{\sigma}}^{-1}})\cdot v\,=\,
\rho((x^{{\hat{\sigma}}^{-1}})^{\hat{\sigma}})\cdot v\,=\,0$.
Hence, $\,\mathfrak g_v\,\subseteq\,^{\sigma}\mathfrak g_v\,$.
By symmetry, equality holds and the result follows.
\end{proof}
\vspace{2ex}\noindent
\begin{remark}
Proposition~\ref{twistexc} implies that if $\,V\,$ is not an
exceptional module, then $\,V^{\hat{\sigma}}\,$ is also not
exceptional.
\end{remark}
We call $\,V\,$ and $\,V^{\hat{\sigma}}\,$ {\bf graph-twisted}
modules. If the graph automorphism has order $\,2$, then we also say
that $\,V\,$ and $\,V^{\hat{\sigma}}\,$ are {\bf graph-dual} modules.
\subsection{The Results}\label{resul}
\subsubsection{Lie Algebras of Exceptional Type}
\begin{theorem}\label{frlaet}
Let $\,G\,$ be a simply connected simple algebraic group of exceptional
type, and $\,\mathfrak g\,=\,{\cal L}(G).\,$
Let $\,V\,$ be an infinitesimally irreducible $\,G$-module.
If the highest weight of $\,V\,$ is listed in Table~\ref{table3},
then $\,V\,$ is an exceptional $\,\mathfrak g$-module. If $\,p\,$
is non-special for $\,G\,$, then the modules listed
in Table~\ref{table3} are the only exceptional $\,\mathfrak g$-modules.
\end{theorem}
The proof of this theorem will be given in Section~\ref{proofexcp}.
\subsubsection{Lie Algebras of Classical Type}\label{lact}
In this case, after using the procedure described in
Section~\ref{proced}, for each
algebra type we end up with a short list of highest weights.
From these lists we have to decide, using further criteria, whether the
corresponding modules are exceptional or not.
Most of the modules that are proved to be exceptional satisfy
Proposition~\ref{dimcrit}.
We have reduction lemmas for each group/algebra type
which are proved in this section.
The proofs of Theorems~\ref{anlist},~\ref{listbn},~\ref{listcn},
and~\ref{dnlist} are rather technical and are given in
Appendix~\ref{appendixa} of this work.
\subsubsection*{\ref{lact}.1. Algebras of Type $\,A\,$}\label{alresults}
\addcontentsline{toc}{subsubsection}{\protect\numberline{\ref{lact}.1}
Algebras of Type $\,A\,$}
By Theorem~\ref{???}, any $\,A_{\ell}(K)$-module $\,V\,$ such that
\begin{equation}\label{allim}
s(V)\,=\,\displaystyle\sum_{\stackrel{\scriptstyle\mu\,good}
{\scriptstyle \mu\in{\cal X}_{++}(V)}}\,m_{\mu}\,|W\mu|\,
>\,\ell\,(\,\ell\,+\,1\,)^2
\end{equation}
or
\begin{equation}\label{rral}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\;good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,\ell\,(\ell+1)
\end{equation}
is not an exceptional $\,\mathfrak g$-module.
For groups of type $\,A_{\ell}$, $\,|W|\,=\,(\ell+1)!\,$,
$\,|R_{long}|\,=\,|R|\,=\,\ell\,(\ell+1)\,$ and
$\,|R_{long}^+|\,=\,\displaystyle\frac{\ell\,(\ell+1)}{2}.\,$
Let $\,V\,$ be an $\,A_{\ell}(K)$-module. Here we prove some
reduction lemmas.
\begin{lemma}\label{4coef} Let $\,\ell\geq 4.\,$
If $\,{\cal X}_{++}(V)\,$ contains a weight
$\,\mu\,=\,\displaystyle\sum_{i=1}^{\ell}\,b_i\,\omega_i\,$ with 4 or
more nonzero coefficients, then $\,V\,$ is not an exceptional
$\,A_{\ell}(K)$-module.
\end{lemma}\noindent
\begin{proof}
Let $\,\mu=b_1\,\omega_1\,+\,\cdots\,+\,b_{\ell}\,\omega_{\ell}\,\in\,
{\cal X}_{++}(V)\,$ have at least 4 nonzero coefficients.
We want to prove that $\,s(V)\,>\,\ell(\ell + 1)^2$.
\underline{Step 1}: {\bf Claim}:
If $\,\nu=c_1\,\omega_1\,+\,\cdots\,+\,c_{\ell}\,
\omega_{\ell}\,$ is a weight with exactly 4 nonzero coefficients, say
$\,\nu=c_1\,\omega_{i_1}\,+\,c_{2}\,\omega_{i_2}\,
+\,c_{3}\,\omega_{i_3}\,+\,c_{4}\,\omega_{i_4}$, with
$\,i_1<i_2<i_3<i_4\,$ and $\,c_{j}\neq 0$, then
$\,|W\nu|\,>\,\ell(\ell + 1)^2$.
First we prove that
if $\,\bar{\mu}\,=\,d_{1}\,\omega_{1}\,+\,d_{2}\,\omega_{2}\,
+\,d_{3}\,\omega_{3}\,+\,d_{4}\,\omega_{4}\,$ (or graph-dually
$\,\bar{\mu}\,=\,d_{\ell -3}\,\omega_{\ell -3}\,+\,d_{\ell-2}\,
\omega_{\ell -2}\,+\,d_{\ell -1}\,\omega_{\ell -1}\,+\,d_{\ell}\,
\omega_{\ell}\,$), with all $\,d_i\neq 0$, then
$\,|W\nu|\,>\,|W\bar{\mu}|$, that is,
$\,W\bar{\mu}\,$ is the smallest possible orbit for weights
having exactly 4 nonzero coefficients.
Indeed, by Lemma~\ref{orban},
\begin{eqnarray} \label{an1}
|W\bar{\mu}| & = & \frac{(\ell +1)!}{(\ell -3)!}\,=\,(\ell +1)\,\ell
\,(\ell -1)\,(\ell -2)\qquad \mbox{and}\nonumber
\end{eqnarray}
\begin{eqnarray} \label{an2}
|W\nu| & = & \frac{(\ell + 1)!}
{i_1!\,(i_2-i_1)!\,(i_3-i_2)!\,(i_4-i_3)!
\,(\ell-i_4+1)!}.\nonumber
\end{eqnarray}
Thus, as we prove in Appendix~\ref{appineq1}, for all $\,\ell\,\geq\,4$.
\begin{eqnarray}\label{ineq1}
\frac{|W\nu|}{|W\bar{\mu}|}\,=\,\frac{(\ell - 3)!}
{i_1!\,(i_2-i_1)!\,(i_3-i_2)!\,(i_4-i_3)!\,(\ell-i_4+1)!}
\,\geq\,1\,.
\end{eqnarray}
Now for $\,\ell\geq 4$,
$\,|W\bar{\mu}|\,-\,\ell(\,\ell\,+\,1)^2\,=\,
\ell\,(\ell^3 -2\ell^2 -\ell +2)\,-\,\ell(\,\ell\,+\,1)^2\,=\,
\ell\,[\ell(\ell^2 -3\ell -3)\,+\,1]\,>\,0\,$
(as $\,\ell^2 -3\ell -3\,$ is a
positive increasing function on $\,\ell$, provided $\,\ell\,\geq\,4\,$).
Therefore, $\,|W\nu|\,\geq\,|W\bar{\mu}|\,>\,\ell(\ell + 1)^2$,
proving the claim.
\underline{Step 2}: As $\,\mu\,$ has at least
4 nonzero coefficients, by Lemma~\ref{remark},
\mbox{$\,|W\mu|\,>\,|W\nu|\,$} for some weight $\,\nu\,$ with exactly
4 nonzero coefficients. Hence, by Step 1, $\,|W\mu|\,>\,\ell(\ell + 1)^2$.
Therefore, if $\,\mu\,\in\,{\cal X}_{++}(V)\,$ is a good weight with
at least 4 nonzero coefficients,
then $\,s(V)\,\geq\,|W\mu|\,>\,\ell(\ell + 1)^2\,$
and, by~(\ref{allim}), $\,V\,$ is not an exceptional module.
If $\,\mu\,\in\,{\cal X}_{++}(V)\,$ is a bad weight, then write
$\,\mu\,=\,b_{1}\,\omega_{i_1}\,+\,b_{2}\,\omega_{i_2}\,
+\,b_{3}\,\omega_{i_3}\,+\,b_{4}\,\omega_{i_4}\,+\,\cdots$, (where
$\,b_1,\,b_2,\,b_3,\,b_4\,\geq 2\,$ are the first 4 nonzero coefficients of
$\,\mu\,$ and $\,i_1<i_2<i_3<i_4$). Thus, the good weight
$\,\mu_1 \,=\,\mu-(\alpha_{i_2}+\cdots+\alpha_{i_3})\,=\,
b_{1}\,\omega_{i_1}\,+\,\omega_{i_2-1}\,+\,(b_{2}-1)\,\omega_{i_2}\,
+\,(b_{3}-1)\,\omega_{i_3}\,+\,\omega_{i_3+1}\,+\,b_{4}\,\omega_{i_4}\,+
\,\cdots\,\in\,{\cal X}_{++}(V)$. As $\,\mu_1\,$ has at least 4
nonzero coefficients, by Step 2, $\,s(V)\,\geq\,|W\mu_1|\,>\,\ell(\ell +
1)^2\,$. Hence, again by~(\ref{allim}), $\,V\,$ is not an exceptional
module. This proves the lemma.
\end{proof}
Now we can prove \\
{\bf Lemma~\ref{nonexcepnonfree}.} {\it
Let $\,\mathfrak g\,=\,\mathfrak{sl}(np,K),\,$ where
$\,p>2\,$ and $\,np\geq 4.\,$ Then the Steinberg module
$\,V\,=\,E((p-1)\rho)\,$ is not an exceptional module, but it is non-free.}
\\ \noindent
\begin{proof}By Lemma~\ref{4coef}, $\,V\,$ is not exceptional. To prove that
$\,V\,$ is non-free, let $\,z\,=\,h_{\alpha_1}\,+\,2\,h_{\alpha_2}\,+
\,3\,h_{\alpha_3}\,+\,\cdots\,+\,(np-1)\,h_{\alpha_{np-1}}\,$.
First we claim that $\,z\,$ is a nonzero central element of $\,\mathfrak g.\,$
Indeed, note that
$\,\alpha_j(z)\,=\,(j-1)\cdot(-1)\,+\,j\cdot 2\,+\,(j+1)\cdot(-1)\,
=\,0\;$ for all $\,j\,=\,2,\ldots,np-2\,$.
Also $\,\alpha_1(z)\,=\,\alpha_{np-1}(z)\,=\,0.\,$ Hence,
$\,[z,\,e_{\alpha_j}]\,=\,\alpha_j(z)\,e_{\alpha_j}\,=\,0$,
for all $\,j\,$. Therefore, $\,[z,\,e_{\alpha}]\,=\,0\,$ for
all $\,\alpha\in R$. But $\,\{ e_{\alpha}\,/\,\alpha\in R\}\,$ generate
$\,\mathfrak g\,$, hence $\,[z,\,y]\,=\,0\,$ for all $\,y\in\mathfrak
g\,$, proving the claim.
Now we prove that $\,z\in \mathfrak g_v\,$ for all $\,v\in V$.
We have
\begin{align*}
(p-1)\rho(z)\, &=\,(p-1)\,\sum_{i=1}^{np-1}\,\omega_i(z)\,=\,
(p-1)\,\sum_{i=1}^{np-1}\,i\,\frac{2(\omega_i,\,{\alpha_i})}
{(\alpha_i,\,\alpha_i)}\\
&=\,(p-1)\,\sum_{i=1}^{np-1}\,i\,=\,\frac{(p-1)\,(np-1)\,np}{2}\,\equiv\,0
\pmod p.
\end{align*}
The other weights of $\,V\,=\,E((p-1)\rho)\,$ are of the form
$\,(p-1)\rho\,-\,\displaystyle\sum_{\beta>0}\,n_{\beta}\beta\,$, where
$\,n_{\beta}\in\mathbb Z^+\,$. Hence
$\,\mu(z)\,=\,0,\,$ for each $\,\mu\in{\cal X}_{++}(V)$.
The vectors $\,v\in V\,$ are of the form
$\,v\,=\,\sum\,v_{\mu},\,$ with $\,v_{\mu}\in V_{\mu}\,$. Thus
$\,z\cdot v\,=\,\sum\,z\cdot v_{\mu}\,=\,\sum\mu(z)\,v_{\mu}\,=\,0,\,$
whence $\,z\in \mathfrak g_v\,$ for all $\,v\in V$.
Thus, $\,V\,$ is a non-free module.
\end{proof}
{\bf From now on, by Lemma~\ref{4coef}, it suffices to consider
$\,A_{\ell}(K)$-modules $\,V\,$ such that $\,{\cal X}_{++}(V)\,$
contains only weights with at most $\,3\,$ nonzero coefficients.}
The following lemmas reduce this assumption even further.
\begin{lemma}\label{3orless}
Let $\,\ell\,\geq\,4$. If $\,V\,$ is an $\,A_{\ell}(K)$-module such
that $\,{\cal X}_{++}(V)\,$ satisfies
any of the following conditions, then $\,V\,$ is not an exceptional
$\,\mathfrak g$-module.
(a) $\,{\cal X}_{++}(V)\,$ contains at least $2$ good weights
with $3$ nonzero coefficients.
(b) $\,{\cal X}_{++}(V)\,$ contains a bad weight
with $3$ nonzero coefficients.
(c) $\,{\cal X}_{++}(V)\,$ contains just one good weight
with $3$ nonzero coefficients, (at least) $2$ good weights
with $2$ nonzero coefficients and a nonzero minimal
weight or any other good weight.
\end{lemma}\noindent
\begin{proof}
Let
$\,\mu\,=\,a\,\omega_{i_1}\,+\,b\,\omega_{i_2}\,+\,c\,\omega_{i_3}\,$
be a weight with $\,i_1<i_2<i_3\,$ and $\,a,\,b,\,c\,\geq\,1$.
For $\,\ell\geq 4,\,$ the orbit size of $\,\mu\,$ is
\begin{equation} \label{orb3c}
|W\mu| =\frac{(\ell + 1)!}{i_1!\,(i_2-i_1)!\,
(i_3-i_2)!\,(\ell-i_3+1)!}\,\geq \, \frac{(\ell\,+\,1)!}{(\ell\,-\,2)!}\,
=\,\ell\,(\ell^2\,-\,1).\;
\end{equation}
This inequality is proved in Appendix~\ref{apporb3c}.
(a) If $\,{\cal X}_{++}(V)\,$ has at least $2$ good weights with exactly
$\,3\,$ nonzero coefficients, then
$\,s(V)\,-\,\ell(\ell+1)^2\,\geq\,2\,\ell\,(\ell^2\,-\,1)\,-\,
(\ell^3\,+\,2\ell^2\,+\,\ell)\,=\,\ell\,(\ell^2\,-\,2\ell\,-\,3)\,>\,0$,
for all $\,\ell\,\geq\,4$. Hence, by~\eqref{allim}, $\,V\,$ is not exceptional.
(b) If $\,\mu\,=\,a\,\omega_{i_1}\,+\,b\,\omega_{i_2}\,+\,c\,\omega_{i_3}\,
\in\,{\cal X}_{++}(V)\,$ is bad, then $\,a,\,b,\,c\,\geq 2.\,$
Thus,
\mbox{$\mu_1=\mu-(\alpha_{i_1}+\cdots
+\alpha_{i_2})\,=\,\omega_{i_1-1}\,+\,(a-1)
\omega_{i_1}\,+\,(b-1)\omega_{i_2}\,+\,\omega_{i_2+1}\,
+\,c\,\omega_{i_3}\,$,}\linebreak
\mbox{$\mu_2=\mu-(\alpha_{i_2}+\cdots +\alpha_{i_3})\,=\,a\omega_{i_1}\,+
\,\omega_{i_2-1}\,+\,(b-1)\omega_{i_2}\,+\,(c-1)\omega_{i_3}\,
+\,\omega_{i_3+1}\,$}\linebreak
are good weights in $\,{\cal X}_{++}(V)$,
both with at least $3$ nonzero coefficients, hence satisfying (a).
(c)
If $\,\nu=a\omega_{i_1}\,+\,b\omega_{i_2}\,$ is a weight
with $2$ nonzero coefficients, then $\,|W\nu|\,\geq\,
\ell\,(\ell\,+\,1)$, for all $\,\ell\geq 4$. Indeed,
by Appendix~\ref{apporb2c},
\begin{eqnarray} \label{orb2c}
|W\nu| & = & \frac{(\ell + 1)!}{i_1!\,(i_2-i_1)!\,
(\ell-i_2+1)!}\,\geq\,\frac{(\ell\,+\,1)!}{(\ell\,-\,1)!}
\,=\,\ell\,(\ell\,+\,1).
\end{eqnarray}
Thus, if $\,{\cal X}_{++}(V)\,$ contains just one good weight with $3$
nonzero coefficients, at least $2$ good weights with
$2$ nonzero coefficients and a minimal weight $\,\omega_i\neq 0\,$
or any other good weight (whose orbit size is $\,\geq 1$) then,
by~\eqref{orb3c} and~\eqref{orb2c}, $\,s(V)\,\geq
\,\ell\,(\ell^2\,-\,1)\,+\,2\,\ell\,(\ell\,+\,1)\,+\,1
\, =\,\ell\,(\,\ell\,+\,1\,)^2\,+\,1\,>\,\ell\,(\,\ell\,+\,1\,)^2\,$
and, by~\eqref{allim}, $\,V\,$ is not an exceptional module. This
proves the lemma.
\end{proof}
\begin{lemma}\label{nlf3c}
Let $\,\ell\,\geq\,4.\,$ If $\,V\,$ is an $\,A_{\ell}(K)$-module such
that $\,{\cal X}_{++}(V)\,$ contains a good
weight with $3$ nonzero coefficients, then $\,V\,$ is not an exceptional
$\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
Let $\,\mu\,=\,a\,\omega_{i_1}\,+\,b\,\omega_{i_2}\,+\,c\,\omega_{i_3},\,$
with $\,1\,<\,i_1\,<\,i_2\,<\,i_3\,$ and $\,a,\,b,\,c\,$ nonzero, be a
good weight in $\,{\cal X}_{++}(V)$.
\underline{Case 1}: Suppose $\,1<i_1\,<\,i_2\,<\,i_3\,<\,\ell\,$.\\
For $\,a\geq 1,\,b\geq 1,\,c\geq 1,\,$
$\,\mu\,=\,a\omega_{i_1}\,+\,b\omega_{i_2}\,+\,c\omega_{i_3}\;$ and
$\,\mu_1\,=\mbox{$\mu-(\alpha_{i_1} +\cdots +\alpha_{i_3})$}
=\,\omega_{i_1-1}\,+\,(a-1)\,\omega_{i_1}\,+\,b\,\omega_{i_2}\,+
\,(c-1)\omega_{i_3}\,+\,\omega_{i_3+1}\;$
are good weights in $\,{\cal X}_{++}(V)$.
Hence, Lemma~\ref{3orless}(a) applies.
\underline{Case 2}: Suppose $\,1<i_1 < i_2 < i_3=\ell\;$ (or graph-dually
$\,1=i_1 < i_2 < i_3<\ell\,$).
(i) For $\,a\geq 2,\,b\geq 1,\,c \geq 1\,$ (or
$\,a\geq 1,\,b\geq 1,\,c\geq 2\,$),
$\,\mu\,=\,a\omega_{i_1}\,+\,b\omega_{i_2}\,+\,c\omega_{\ell}\;$ and
$\,\mu_1\,=\,\mu-(\alpha_{i_1} +\cdots +\alpha_{\ell})\,
=\,\omega_{i_1-1}\,+\,(a-1)\,\omega_{i_1}\,+\,b\,\omega_{i_2}\,+
\,(c-1)\omega_{\ell}\;$ are good weights in $\,{\cal X}_{++}(V)$.
Hence, Lemma~\ref{4coef} or~\ref{3orless}(a) applies.
(ii) For $\,a=c=1,\,b\geq 2$,
$\,\mu\,=\,\omega_{i_1}\,+\,b\omega_{i_2}\,+\,\omega_{\ell}\;$ and
$\,\mu_1=\mbox{$\mu-(\alpha_{i_1} +\cdots +\alpha_{i_2})$}
=\,\omega_{i_1-1}\,+\,(b-1)\,\omega_{i_2}\,+\,\omega_{i_2+1}\,+\,
\omega_{\ell}\;$ are both good weights in $\,{\cal X}_{++}(V)$.
Thus, Lemma~\ref{4coef} or~\ref{3orless}(a) applies.
(iii) For $\,a=b=c=1,\,$
$\,\mu\,=\,\omega_{i_1}\,+\,\omega_{i_2}\,+\,\omega_{\ell}\,$ and
$\,\mu_1\,=\,\mu\,-\,(\alpha_{i_1} +\cdots +\alpha_{i_2})\,
=\,\omega_{i_1-1}\,+\,\omega_{i_2+1}\,+\,\omega_{\ell}\,\in\,
{\cal X}_{++}(V)$.
For $\,i_2+1\,<\,\ell,\,$ $\,\mu_1\,$ has 3 nonzero coefficients and
Lemma~\ref{3orless}(a) applies. For $\,i_2+1\,=\,\ell,\,$
$\,\mu =\omega_{i_1}\,+\,\omega_{\ell-1}\,+\,\omega_{\ell},\;$
$\,\mu_1=\omega_{i_1-1}\,+\,2\omega_{\ell},\,$
$\,\mu_2=\mu_1\,-\,\alpha_{\ell}\,
=\,\omega_{i_1-1}\,+\,\omega_{\ell-1},\;$
$\,\mu_3\,=\,\mu_2\,-\,(\alpha_{i_1-1} +\cdots +\alpha_{\ell-1})\,
=\,\omega_{i_1-2}\,+\,\omega_{\ell}\,\in\,{\cal X}_{++}(V)\,$ and
Lemma~\ref{3orless}(c) applies.
In any of that above cases, $\,V\,$ is not an exceptional module.
\underline{Case 3}: Suppose
$\,i_1=1,\;i_3\,=\,\ell\;$ and $\,2<\,i_2\,<\ell-1\,$.
(i) For $\,a\,\geq\,1,\,b\,\geq\,2,\,c\,\geq\,1,\;$
$\,\mu\,=\,a\,\omega_{1}\,+\,b\,\omega_{i_2}\,+\,c\,\omega_{\ell}\;$ and
$\,\mu_1=\mu\,-\,\alpha_{i_2}\,=\,a\,\omega_{1}\,+\,\omega_{i_2-1}\,
+\,(b-2)\,\omega_{i_2}\,+\,\omega_{i_2+1}\,+\,c\,\omega_{\ell}
\,\in\,{\cal X}_{++}(V)\,$ and Lemma~\ref{4coef} applies.
(ii) For $\,a\,\geq\,2,\,b\,=\,1,\,c\,\geq\,1\,$ (or
$\,a\,\geq\,1,\,b\,=\,1,\,c\,\geq\,2\,$),
$\,\mu =a\,\omega_{1}\,+\,\omega_{i_2}\,+\,c\omega_{\ell},\;$
$\,\mu_1\,=\,\mu\,-\,(\alpha_{1}+\cdots+\alpha_{i_2})\,
=\,(a-1)\,\omega_{1}\,+\,\omega_{i_2+1}\,+\,c\,\omega_{\ell},\;$
$\,\mu_2\,=\,\mu\,-\,(\alpha_{i_2}+\cdots+\alpha_{\ell})\,
=\,a\,\omega_{1}\,+\,\omega_{i_2-1}\,+\,(c-1)\,\omega_{\ell}\;$ and
$\,\mu_3\,=\,\mu\,-\,(\alpha_{1}+\cdots+\alpha_{\ell})\,
=\,(a-1)\,\omega_{1}\,+\,\omega_{i_2}\,+\,(c-1)\,\omega_{\ell}
\,\in\,{\cal X}_{++}(V).\,$ Thus,
$\,{\cal X}_{++}(V)\,$ has at least 2 good weights with 3 nonzero
coefficients and Lemma~\ref{3orless}(a) applies.
(iii) For $\,a=b=c=1,\;$
$\,\mu\,=\,\omega_{1}\,+\,\omega_{i_2}\,+\,\omega_{\ell},\;$
$\,\mu_1\, =\,\mbox{$ \mu-(\alpha_{1}+\cdots+\alpha_{i_2})$}\,
=\,\omega_{i_2+1}\,+\,\omega_{\ell},\;$
$\,\mu_2\, =\,\mu-(\alpha_{i_2}+\cdots+\alpha_{\ell})\,
=\,\omega_1\,+\,\omega_{i_2-1}\;$ and
$\mu_3\,=\,\mbox{$\mu_2-(\alpha_1+\cdots+$}\alpha_{i_2-1})\,
=\,\omega_{i_2}\;$
are all good weights in $\,{\cal X}_{++}(V).\,$ Hence, by
Lemma~\ref{3orless}(c), $\,V\,$ is not an exceptional module.
{\it Hence, if $\,{\cal X}_{++}(V)\,$ contains a good weight
$\,\mu\,=\,a\,\omega_{i_1}\,+\,b\,\omega_{i_2}\,+\,c\,\omega_{i_3}\,$
(with $\,3\,$ nonzero coefficients), then we can assume
$\,i_1=1,\;i_2=2,\;i_3\,=\,\ell\,$ (or graph-dually
$\,i_1=1,\;i_2=\ell-1,\;i_3=\ell\,$).} This case is treated as follows.
\underline{Case 4}: $\,i_1=1,\;i_2=2,\;i_3\,=\,\ell\;$ (or graph-dually
$\,i_1=1,\;i_2=\ell-1,\;i_3=\ell\,$).
(i) For $\,a\geq 1,\,b\geq 2,\,c\geq 1\,$ (or
$\,a\geq 2,\,b \geq 1,\,c\geq 1\,$),
$\,\mu\,=\,a\omega_{1}\,+\,b\omega_{2}\,+\,c\omega_{\ell},\;$
$\mu_1\,=\,\mu\,-\,(\alpha_1 +\alpha_2)\,=\,(a-1)\,\omega_{1}\,+\,
\mbox{$(b-1)\omega_2$}\,+\,\omega_{3}\,+\,c\omega_{\ell}\,
\in\,{\cal X}_{++}(V)$.
For $\,a\geq 1,\,b\geq 1,\,c \geq 2,\,$ also
$\,\mu_2\,=\,\mu-\alpha_{\ell}\,=\,a\,\omega_{1}\,+\,b\omega_2\,
+\,\omega_{\ell-1}\,+\,(c-2)\omega_{\ell}\,\in\,{\cal
X}_{++}(V)$. In both cases, as $\,\ell\geq
4$, Lemma~\ref{4coef} or~\ref{3orless}(a) applies.
{\it Therefore, if $\,{\cal X}_{++}(V)\,$ contains a good weight
$\,\mu\,$ with $\,3\,$ nonzero coefficients, then we can assume that
$\,\mu\,=\,\omega_{1}\,+\,\omega_{2}\,+\,\omega_{\ell}\,$.}
(ii) Let $\,\mu\,=\,\mu_0\,=\,\omega_{1}\,+\,\omega_{2}\,+\,\omega_{\ell}\in
{\cal X}_{++}(V)$. Then $\,\mu_1\,=\,\omega_3\,+\,\omega_{\ell}\,$,
$\,\mu_2\,=\,\mu_1-(\alpha_3\,+\,\cdots\,+\,\alpha_{\ell})\,=\,\omega_2\in
{\cal X}_{++}(V)$. For $\,\ell\,\geq\,6\,$ and any prime $\,p$,
\begin{align*}
s(V) &\geq\,(\ell + 1)\,\ell\,(\ell -1)\,+\,\displaystyle\frac{(\ell +
1)\,\ell\,(\ell-1)(\ell-2)}{3!}\,+\,\frac{(\ell\,+\,1)\,\ell}{2!}\\
& =\,\displaystyle\frac{(\ell + 1)\,\ell\,(\ell^2+\ell + 3)}{6}\,>\,
\ell\,(\ell+1)^2\,.
\end{align*}
Hence, by~\eqref{allim}, $\,V\,$ is not an exceptional module.
For $\,\ell=5\,$ and $\,p\geq 2\,$, $\,|R_{long}^+-R_{\mu,p}^+|\geq
8\,$. Thus $\,r_p(V)\,\geq\,\displaystyle\frac{120\cdot
8}{30}\,=\,32\,>\,30\,$.
For $\,\ell=4$, $\,\mu\,=\,\omega_1\,+\,\omega_2\,+\,\omega_4\,$,
$\,\mu_1\,=\,\omega_3\,+\,\omega_{4}\,$. For $\,p\geq 2\,$,
$\,|R_{long}^+-R_{\mu,p}^+|\geq 6,\;|R_{long}^+-R_{\mu,p}^+|\geq 4$.
Thus $\,r_p(V)\,\geq\,\displaystyle\frac{60\cdot
6}{20}\,+\,\frac{20\cdot 4}{20}\,=\,22\,>\,20\,$.
Hence, for $\,\ell=4,\,5,\,$ by~\eqref{rral}, $\,V\,$ is not an
exceptional module. This proves the lemma.
\end{proof}
The main theorem for groups of type $\,A_{\ell}\,$ is as follows. Its proof
is given in Appendix~\ref{appal}
\begin{theorem}\label{anlist}
If $\,V\,$ is an infinitesimally irreducible
$\,A_{\ell}(K)$-module with highest weight listed
in Table~\ref{tablealall} (p. \pageref{tablealall}),
then $\,V\,$ is an exceptional $\,\mathfrak g$-module.
If $\,V\,$ has highest weight different from the ones listed in Tables
~\ref{tablealall} or~\ref{leftan} (p. \pageref{leftan}), then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{theorem}
\subsubsection*{\ref{lact}.2. Algebras of Type $\,B\,$}\label{blresults}
\addcontentsline{toc}{subsubsection}{\protect\numberline{\ref{lact}.2}
Algebras of Type $\,B\,$}
By Theorem~\ref{???}, any $\,B_{\ell}(K)$-module $\,V\,$ such that
\begin{equation}\label{bllim}
s(V)\,=\,\displaystyle\sum_{\stackrel{\scriptstyle\mu\,good}
{\scriptstyle \mu\in{\cal X}_{++}(V)}}\,m_{\mu}\,|W\mu|\,
>\,\left\{
\begin{array}{ll}
2\,\ell^3 & \mbox{if}\;\ell\geq 5\\
8\,\ell^2 & \mbox{if}\;\ell\leq 4.
\end{array}\right.
\end{equation}
or
\begin{equation}\label{rrbl}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\;good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,2\,\ell^2
\end{equation}
is not an exceptional $\,\mathfrak g$-module.
Note that for $\,\ell =4$, $\,8\ell^2=2\ell^3\,$.
For groups of type $\,B_{\ell}$, $\,|W|\,=\,2^{\ell}\,\ell !$,
$\,|R_{long}|\,=\,|D_{\ell}|\,=\,2\,\ell\,(\ell-1)\,$ and
$\,|R_{long}^+|\,=\,\ell\,(\ell-1).\,$ Recall that $\,p\geq 3$.
The orbit sizes of weights are given by Lemma~\ref{orbbncndn}.
Let $\,V\,$ be a $\,B_{\ell}(K)$-module. We prove a reduction lemma.
\begin{lemma}\label{3coefbn}
Let $\,\ell\,\geq\,4$. If $\,{\cal X}_{++}(V)\,$ contains a
weight with 3 or more nonzero coefficients, then $\,V\,$
is not an exceptional $\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
I) Let $\,\mu=b_1\,\omega_1\,+\,\cdots\,+\,b_{\ell}\,\omega_{\ell}\,\in
\,{\cal X}_{++}(V)\,$ be a good weight with at least 3 nonzero coefficients.
We claim that $\,s(V)\,>\,2\ell^3\,$.
\underline{Step 1}:
Write $\,\mu\,=\,b_1\,\omega_{i_1}\,+\,b_2\,\omega_{i_2}\,+\,
b_3\,\omega_{i_3}\,+\cdots\,$, where $\,1\leq\,i_i<i_2<i_3\,$ are the
first 3 nonzero coefficients of $\,\mu$. By
Lemma~\ref{remark}, $\,|W\mu|\,\geq\,|W\bar{\mu}|,\,$ where
$\,\bar{\mu}\,=\,d_1\,\omega_{i_1}\,+
\,d_2\,\omega_{i_2}\,+\,d_3\,\omega_{i_3}\,$ with $\,d_j\neq 0\,$ for
all $\,1\leq j\leq 3$. Therefore, to prove the claim, it
suffices to prove that $\,|W\bar{\mu}|\,>\,2\,\ell^3$, for any weight
$\,\bar{\mu}\,$ with exactly 3 nonzero coefficients.
\underline{Step 2}: Let $\,\bar{\mu}\,=\,d_1\,\omega_{i_1}\,+
\,d_2\,\omega_{i_2}\,+\,d_3\,\omega_{i_3}\,$ with $\,d_j\neq 0\,$ for
$\,1\leq j\leq 3\,$. We prove that $\,|W\bar{\mu}|\,>\,2\,\ell^3\,$ by
considering the different possibilities for $\,(i_1,\,i_2,\,i_3)$.
(a) Let $\,(i_1,\,i_2,\,i_3)\,=\,(1,\,2,\,3)\,$. In this case denote
$\,\bar{\mu}\,$ by $\,\nu=c_1\,\omega_1\,+\,c_2\omega_2\,+\,c_{3}\,
\omega_{3}\,$, with $\,c_1,\,c_2,\,c_3\,$ all nonzero. Then
$\,|W\nu|\,=\,2^3\,\ell\,(\ell -1)\,(\ell -2)\,$. Thus, for
$\,\ell\geq 4$, $\,|W\nu|\,-\,2\,\ell^3 \,=\,6\ell^2\,(\ell-4)\,
+\,16\ell\,>\,0\,$ whence $\,|W\nu|\,>\,2\,\ell^3 \,$.
{\bf Note}: For $\,\ell =4,\,$ any choice of $\,(i_1,\,i_2,\,i_3)\,$ implies
$\,|W\bar{\mu}|\,=\,2^6\cdot
3\,>\,2^7=8\cdot 4^2$. Thus, from now on let $\,\ell\geq 5\,$ and
$\,{\mathbf\nu=c_1\,\omega_1\,+\,c_2\omega_2\,+\,c_{3}\,
\omega_{3}}\,$, with $\,c_i\neq 0$.
(b)i) Let $\,1\leq i_1\,<\,i_2\,<\,i_3\,<\,\ell -1\,$ and
$\,\bar{\mu}\,=\,d_1\,\omega_{i_1}\,+\,d_{2}\,\omega_{i_2}\,
+\,d_{3}\,\omega_{i_3}$, with $\,d_{j}\neq 0$. Then
$\,|W\bar{\mu}|\, =\, \displaystyle 2^{i_3}\,\binom{i_2}{i_1}\,
\binom{i_3}{i_2}\,\binom{\ell}{i_3}\,$.
Note that for $\,1\leq i_1< i_2 < i_3,\;3\leq i_3\leq\ell-3\,$ and
$\,\ell\geq 6,\,$ we have $\,\displaystyle\binom{\ell}{i_3}\geq
\binom{\ell}{3}\,$ and $\,\displaystyle\binom{i_2}{i_1}\,
\binom{i_3}{i_2}\geq 6\,$. Thus,
\[
\frac{|W\bar{\mu}|}{|W{\nu}|} \,= \,
\frac{2^{i_3-3}}{\ell (\ell - 1)(\ell - 2)}\,\binom{i_2}{i_1}\,
\binom{i_3}{i_2}\,\binom{\ell}{i_3}
\,\geq\,\frac{2^{i_3-3}\cdot 6}{\ell (\ell - 1)(\ell -2)}\,\binom{\ell}{3}
\, =\, 2^{i_3-3}\geq 1\,.
\]
Hence,
for $\,\ell\geq 6,\,$ $\,|W\bar{\mu}|\,\geq\,|W{\nu}|\,>\,2\ell^3\,$.
ii) If $\,1\leq i_1< i_2 < i_3=\ell -2\,$ then, for $\,2\leq i_2\leq
\ell-3\,$ and $\,\ell\geq 5,\,$
\begin{eqnarray}
\frac{|W\bar{\mu}|}{|W{\nu}|} & = &\frac{2^{\ell
-6}}{(\ell-2)}\,\binom{i_2}{i_1}\,\binom{\ell -2}{i_2}\,
\geq\, \frac{2^{\ell
-6}}{(\ell-2)}\,\binom{i_2}{i_1}\,\binom{\ell-2}{2}
\geq 2^{\ell-6}\,(\ell-3) \,\geq\,1\,.\nonumber
\end{eqnarray}
{\bf Therefore, if $\,\bar{\mu}\,=\,d_1\,\omega_{i_1}\,+\,d_{2}\,\omega_{i_2}\,
+\,d_{3}\,\omega_{i_3}$, with $\,d_{j}\neq 0\,$ and
$\,1\leq i_1\,<\,i_2\,<\,i_3\,\leq\,\ell -2\,$, then
$\,|W\bar{\mu}|\,\geq\,|W{\nu}|\,>\,2\ell^3\,$. Hence, we can assume
$\,\ell -1\leq \,i_3\,$.} We deal with this case in the sequel.
(c) For $\,1\leq i_1< i_2 <i_3\,=\,\ell -1,\,$
$\,\bar{\mu}\,=\,b_1\,\omega_{i_1}\,+\,b_{2}\,\omega_{i_2}\,
+\,b_{3}\,\omega_{\ell -1}\,$ and \\
$\,|W\bar{\mu}|\, = \,\displaystyle 2^{\ell-1}\,\ell\,\binom{i_2}{i_1}\,
\binom{\ell -1}{i_2}\,$.
Thus, for $\,2\leq i_2\leq \ell-2\,$ and $\,\ell\geq 4,\,$
\begin{eqnarray}
\frac{|W\bar{\mu}|}{|W{\nu}|} & = & \frac{2^{\ell-4}\;\ell}{\ell\,(\ell -
1)\,(\ell -2)}\,\binom{i_2}{i_1}\,\binom{\ell -1}{i_2} \geq
2^{\ell-4}\,\geq\,1\,.\nonumber
\end{eqnarray}
d)i) Let $\,1\leq i_1< i_2 <i_3\,=\,\ell$. Then
$\,\mu\,=\,\bar{\mu}\,=\,b_1\,\omega_{i_1}\,+\,b_{2}\,\omega_{i_2}\,
+\,b_{3}\,\omega_{\ell}\;$ and
$\,|W\bar{\mu}|\, = \,\displaystyle 2^{\ell}\,\ell\,\binom{i_2}{i_1}\,
\binom{\ell -1}{i_2}\,$.
Thus, for $\,3\leq i_2\leq\ell-2\,$ and $\,\ell\geq 5,\,$
\begin{eqnarray}
\frac{|W\bar{\mu}|}{|W{\nu}|} & = & \frac{2^{\ell-3}}{\ell(\ell - 1)
(\ell -2)}\,\binom{i_2}{i_1}\,\binom{\ell}{i_2} \geq 2^{\ell-4}\, \geq\,1\,.
\nonumber
\end{eqnarray}
ii) For $\,i_2\,=\,2\,$ (hence $\,i_1=1\,$) and $\,i_3=\ell$,
$\,\mu\,=\,\bar{\mu}\,=\,b_1\,\omega_{1}\,+\,b_{2}\,\omega_{2}\,
+\,b_{3}\,\omega_{\ell}\,$. Then, for $\,\ell\geq 4,\,$
$\,|W\mu|\,=\,2^{\ell}\,\ell(\ell-1)\,>\,2\ell^3\,$.
iii) For $\,1\leq i_1 < i_2\,=\,\ell -1,\,i_3=\ell$,
$\,\mu\,=\,\bar{\mu}\,=\,b_1\,\omega_{i_1}\,+\,b_{2}\,\omega_{\ell -1}\,
+\,b_{3}\,\omega_{\ell}\;$ and
$\,|W\bar{\mu}|\,=\,2^{\ell}\,\ell\,\displaystyle\binom{\ell-1}{i_1}\,$.
Hence, for $\,1\leq i_1 \leq \ell-2\,$ and $\,\ell\geq 4,\,$
$\,|W\bar{\mu}|\,\geq\,2^{\ell}\,\ell\,(\ell-1)\,>\,2\ell^3\,$.
{\bf Therefore, if $\,{\cal X}_{++}(V)\,$ contains a good weight $\,\mu\,$
with at least 3 nonzero coefficients, then
$\,s(V)\,\geq\,|W\mu|\,>\,2\ell^3\,$. Hence, by~\eqref{bllim}, $\,V\,$
is not an exceptional module.}
II) Suppose that $\,\mu\,$ is a bad weight in $\,{\cal X}_{++}(V)$.
This implies $\,b_j\geq 3\,$ for all nonzero coefficients.
We claim that it is possible to produce, from $\,\mu,\,$ a
good weight $\,\mu_1\in{\cal X}_{++}(V)\,$ having 3 or more nonzero
coefficients. Indeed:
(a) if $\,1\leq\,i_i<i_2<i_3\,<\,\ell -1$, then
$\,\mu\,=\,b_1\,\omega_{i_1}\,+\,b_2\,\omega_{i_2}\,+\,
b_3\,\omega_{i_3}\,+\cdots\;$ and
$\,\mu_1 = \mu-(\alpha_{i_2}+\cdots +\alpha_{i_3})
= b_1\,\omega_{i_1}\,+\,\omega_{i_2-1}\,+\,(b_2-1)\,\omega_{i_2}\,+\,
(b_3-1)\,\omega_{i_3}\,+\,\omega_{i_3+1}\,+\,\cdots\;\in\,{\cal
X}_{++}(V)$.
(b) if $\,1\leq\,i_i<i_2<i_3\,=\,\ell -1$, then
$\,\mu\,=\,b_1\,\omega_{i_1}\,+
\,b_2\,\omega_{i_2}\,+\,b_3\,\omega_{\ell -1}\,+\,\cdots\;$ and
$\,\mu_1 = \mu-(\alpha_{i_1}+\cdots +\alpha_{i_2})
= \omega_{i_1-1}\,+\,(b_1-1)\,\omega_{i_1}\,+\,(b_2-1)\omega_{i_2}\,+\,
\omega_{i_2+1}\,+\,b_3\,\omega_{\ell -1}\,+\,\cdots\;\,\in\,{\cal
X}_{++}(V)$.
(c) if $\,1\leq\,i_i<i_2<i_3\,=\,\ell$, then $\,\mu\,=\,b_1\,\omega_{i_1}\,+
\,b_2\,\omega_{i_2}\,+\,b_3\,\omega_{\ell}\;$ and\linebreak
$\,\mu_1\,=\,\mu-(\alpha_{i_1}+\cdots +\alpha_{\ell})\,=\,
\omega_{i_1-1}\,+\,(b_1-1)\,\omega_{i_1}\,+\,b_2\omega_{i_2}\,+\,
b_3\,\omega_{\ell}\,\in\,{\cal X}_{++}(V)$.
In all these cases, $\,\mu_1\,$ is a good weight with at least 3 nonzero
coefficients, since $\,b_1,\,b_2,\,b_3\,\geq 3$. Hence, by Part I of
this proof, $\,s(V)\,\geq\,|W\mu_1|\,>\,2\ell^3\,$ and,
by~\eqref{bllim}, $\,V\,$ is not an exceptional module.
This proves the lemma.
\end{proof}
The main theorem for groups of type $\,B$ is as follows. Its proof is given
in Appendix~\ref{appbl}.
\begin{theorem}\label{listbn}
Suppose $\,p\,>\,2.\,$
If $\,V\,$ is an infinitesimally irreducible
$\,B_{\ell}(K)$-module with highest weight listed in
Table~\ref{tableblall} (p. \pageref{tableblall}),
then $\,V\,$ is an exceptional $\,\mathfrak g$-module.
If $\,V\,$ has highest weight different from the ones listed in Tables
~\ref{tableblall} or~\ref{leftbn} (p. \pageref{leftbn}), then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{theorem}
\subsubsection*{\ref{lact}.3. Algebras of Type $\,C\,$}\label{clresults}
\addcontentsline{toc}{subsubsection}{\protect\numberline{\ref{lact}.3}
Algebras of Type $\,C\,$}
By Theorem~\ref{???}, any $\,C_{\ell}(K)$-module $\,V\,$ such that
\begin{equation}\label{cllim}
s(V)\,=\,\displaystyle\sum_{\stackrel{\scriptstyle\mu\,good}
{\scriptstyle \mu\in{\cal X}_{++}(V)}}\,m_{\mu}\,|W\mu|\,
>\,4\,\ell^3
\end{equation}
or
\begin{equation}\label{rrcl}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\;good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,2\,\ell^2
\end{equation}
is not an exceptional $\,\mathfrak g$-module. For groups of
type $\,C_{\ell}$, $\,|W|\,=\,2^{\ell}\,\ell !$,
$\,|R_{long}|\,=\,|A_1^{\ell}|\,=\,2\,\ell\,$ and
$\,|R_{long}^+|\,=\,\ell.\,$ Recall that $\,p\geq 3.\,$
The orbit sizes of weights are given by Lemma~\ref{orbbncndn}.
Let $\,V\,$ be a $\,C_{\ell}(K)$-module. We prove a reduction lemma.
\begin{lemma}\label{3coefcn}
Let $\,\ell\,\geq\,6$. If $\,{\cal X}_{++}(V)\,$ contains a
weight with 3 or more nonzero coefficients, then $\,V\,$
is not an exceptional $\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
I) Let $\,\mu=b_1\,\omega_1\,+\,\cdots\,+\,b_{\ell}\,\omega_{\ell}\,\in\,
{\cal X}_{++}(V)\,$ be a good weight with at least 3 nonzero coefficients.
We claim that $\,s(V)\,>\,4\ell^3\,$, for $\,\ell\geq 6$.
\underline{Step 1}:
Write $\,\mu\,=\,b_1\,\omega_{i_1}\,+\,b_2\,\omega_{i_2}\,+\,
b_3\,\omega_{i_3}\,+\cdots\,$, where $\,1\leq\,i_i<i_2<i_3\,$ are the
first 3 nonzero coefficients of $\,\mu$. By
Lemma~\ref{remark}, $\,|W\mu|\,\geq\,|W\bar{\mu}|,\,$ where
$\,\bar{\mu}\,=\,d_1\,\omega_{i_1}\,+
\,d_2\,\omega_{i_2}\,+\,d_3\,\omega_{i_3}\,$ with $\,d_j\neq 0\,$ for
all $\,1\leq j\leq 3$. Therefore, to prove the claim, it
suffices to prove that $\,|W\bar{\mu}|\,>\,4\,\ell^3$, for any weight
$\,\bar{\mu}\,$ with exactly 3 nonzero coefficients.
\underline{Step 2}: Let $\,\bar{\mu}\,=\,d_1\,\omega_{i_1}\,+
\,d_2\,\omega_{i_2}\,+\,d_3\,\omega_{i_3}\,$ with $\,d_j\neq 0\,$ for
$\,1\leq j\leq 3\,$. We prove that $\,|W\bar{\mu}|\,>\,4\,\ell^3\,$ by
considering the different possibilities for $\,(i_1,\,i_2,\,i_3)$.
(a) Let $\,(i_1,\,i_2,\,i_3)\,=\,(1,\,2,\,3)\,$ and denote
$\,\bar{\mu}\,$ by $\,\nu=c_1\,\omega_1\,+\,c_2\omega_2\,+\,c_{3}\,
\omega_{3}\,$, with $\,c_1,\,c_2,\,c_3\,$ all nonzero. Then
$\,|W\nu|\,=\,\displaystyle\frac{2^{\ell}\,\ell !}
{2^{(\ell -3)}\,(\ell -3)!}\,=\,2^3\,\ell\,(\ell -1)\,(\ell
-2)\,$. Thus, for $\,\ell\geq 6$, $\,|W\nu|\,-\,4\,\ell^3 \,=\,
\,4\ell^2\,(\ell-6)\,+\,16\ell\,>\,0\,$ whence $\,|W\nu|\,>\,4\,\ell^3\,$.
(b) Let $\,1\leq i_1\,<\,i_2\,<\,i_3\,<\,\ell -1\,$ and
$\,\bar{\mu}\,=\,d_1\,\omega_{i_1}\,+\,d_{2}\,\omega_{i_2}\,
+\,d_{3}\,\omega_{i_3}$, with $\,d_{j}\neq 0$. Then
$\,|W\bar{\mu}|\, =\, \displaystyle 2^{i_3}\,\binom{i_2}{i_1}\,
\binom{i_3}{i_2}\,\binom{\ell}{i_3}\,$. Thus, for $\,3\leq i_3\leq
\ell-2\,$ and $\,\ell\geq 6,\,$ $\,|W\bar{\mu}|\,-\,4\,\ell^3
\,\geq\,2^3\,\ell\,(\ell -1)\,(\ell-2)\,-\,4\,\ell^3\,\geq
\,4\ell^2\,(\ell-6)\,+\,16\ell\,>\,0\,$.
{\bf Therefore, if $\,\bar{\mu}\,=\,d_1\,\omega_{i_1}\,+\,d_{2}\,\omega_{i_2}\,
+\,d_{3}\,\omega_{i_3}$, with $\,d_{j}\neq 0\,$ and
$\,1\leq i_1\,<\,i_2\,<\,i_3\,\leq\,\ell -2\,$, then
$\,|W\bar{\mu}|\,>\,2\ell^3\,$. Hence, we can assume
$\,\ell -1\leq \,i_3\,$.} We deal with this case in the sequel.
(c) For $\,1\leq i_1\,<\,i_2\,<i_3\,=\,\ell -1,\,$
$\,\bar{\mu}\,=\,b_1\,\omega_{i_1}\,+\,b_{2}\,\omega_{i_2}\,
+\,b_{3}\,\omega_{\ell -1}\;$ and \\
$\,|W\bar{\mu}|\, = \,\displaystyle 2^{\ell-1}\,\ell\,\binom{i_2}{i_1}
\binom{\ell -1}{i_2}\,$. Thus, for $\,2\leq i_2\leq \ell-2\,$ and
$\,\ell\geq 5$, $\,|W\bar{\mu}|\,\geq\,2^{\ell-1}\,\ell\,
(\ell-1)\,(\ell-2)\,>\,4\ell^3\,$.
d) Let $\,1\leq i_1\,<\,i_2\,<i_3\,=\,\ell\,$ and
$\,\mu\,=\,\bar{\mu}\,=\,b_1\,\omega_{i_1}\,+\,b_{2}\,\omega_{i_2}\,
+\,b_{3}\,\omega_{\ell}$. Then,
$\,|W\bar{\mu}|\, = \,\displaystyle 2^{\ell}\,\binom{i_2}{i_1}\,
\binom{\ell}{i_2}\,$. Hence, for $\,2\leq i_2\leq\ell -1\,$ and
$\,\ell\geq 5,\,$ we have $\,|W\bar{\mu}|\,\geq\,2^{\ell}\,\ell\,(\ell-1)\,
>\,4\ell^3\,$.
{\bf Therefore, if $\,{\cal X}_{++}(V)\,$ contains a good weight $\,\mu\,$
with at least 3 nonzero coefficients, then
$\,s(V)\,\geq\,|W\mu|\,>\,2\ell^3\,$. Hence, by~\eqref{bllim}, $\,V\,$
is not an exceptional module.}
II) Suppose that $\,\mu\,$ is a bad weight in $\,{\cal X}_{++}(V)\,$
with at least 3 nonzero coefficients. As $\,p\geq 3$,
we have $\,b_j\geq 3\,$ for all nonzero coefficients.
We claim that it is always possible to produce, from $\,\mu,\,$ a
good weight $\,\mu_1\in{\cal X}_{++}(V)\,$ having 3 or more nonzero
coefficients. Indeed:
(a) if $\,1\leq\,i_i<i_2<i_3\,\leq\,\ell -1$, then
$\,\mu\,=\,b_1\,\omega_{i_1}\,+\,b_2\,\omega_{i_2}\,+\,
b_3\,\omega_{i_3}\,+\cdots\;$ and
$\,\mu_1 = \mu-(\alpha_{i_2}+\cdots +\alpha_{i_3})
= b_1\,\omega_{i_1}\,+\,\omega_{i_2-1}\,+\,(b_2-1)\,\omega_{i_2}\,+\,
(b_3-1)\,\omega_{i_3}\,+\,\omega_{i_3+1}\,+\,\cdots\,\in\,{\cal
X}_{++}(V)$.
(b) if $\,1\leq\,i_i<i_2<i_3\,=\,\ell$, then $\,\mu\,=\,b_1\,\omega_{i_1}\,+
\,b_2\,\omega_{i_2}\,+\,b_3\,\omega_{\ell}\;$ and
$\,\mu_1\,=\,\mu-(\alpha_{i_1}+\cdots +\alpha_{\ell})\,=\,
\omega_{i_1-1}\,+\,(b_1-1)\,\omega_{i_1}\,+\,b_2\omega_{i_2}\,+\,
\omega_{\ell-1}\,+\,(b_3-1)\,\omega_{\ell}\,\in\,{\cal X}_{++}(V)$.
In all these cases, $\,\mu_1\,$ is a good weight with at least 3 nonzero
coefficients, since $\,b_1,\,b_2,\,b_3\,\geq 3$. Hence, by Part I of
this proof, $\,s(V)\,\geq\,|W\mu_1|\,>\,4\ell^3\,$ and,
by~\eqref{bllim}, $\,V\,$ is not an exceptional module.
This proves the lemma.
\end{proof}
The main theorem for groups of type $\,C\,$ is as follows. Its proof is
given in Appendix~\ref{appcl}.
\begin{theorem}\label{listcn}
Suppose $\,p\,>\,2.\,$
If $\,V\,$ is an infinitesimally irreducible
$\,C_{\ell}(K)$-module with highest weight listed in
Table~\ref{tableclall} (p. \pageref{tableclall}),
then $\,V\,$ is an exceptional $\,\mathfrak g$-module.
If $\,V\,$ has highest weight different from the ones listed in Tables
\ref{tableclall} or~\ref{leftcn} (p. \pageref{leftcn}), then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{theorem}
\newpage
\subsubsection*{\ref{lact}.4. Algebras of Type $\,D\,$}\label{resuldl}
\addcontentsline{toc}{subsubsection}{\protect\numberline{\ref{lact}.4}
Algebras of Type $\,D\,$}
By Theorem~\ref{???}, any $\,D_{\ell}(K)$-module $\,V\,$ such that
\begin{equation}\label{dllim}
s(V)\,=\,\displaystyle\sum_{\stackrel{\scriptstyle\mu\,good}
{\scriptstyle \mu\in{\cal X}_{++}(V)}}\,m_{\mu}\,|W\mu|\,
>\,\left\{
\begin{array}{ll}
2\,\ell^2\,(\ell-1) & \mbox{if}\;\ell\geq 5\\
\frac{8\,\ell^2\,(\ell-1)}{(\ell-2)} & \mbox{if}\;\ell\leq 4.
\end{array}\right.
\end{equation}
or
\begin{equation}\label{rrdl}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\;good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,2\,\ell\,(\ell-1)
\end{equation}
is not an exceptional $\,\mathfrak g$-module.
For groups of type $\,D_{\ell}\,$, $\,|W|\,=\,2^{\ell-1}\,\ell !$,
$\,|R|\,=\,|R_{long}|\,=\,2\ell(\ell-1)$. The orbit sizes of weights
are given by Lemma~\ref{orbbncndn}.
Let $\,V\,$ be a $\,D_{\ell}(K)$-module. We prove a reduction lemma.
\begin{lemma}\label{3cdn}
Let $\,\ell\,\geq\,5$. If $\,{\cal X}_{++}(V)\,$ contains a
weight with 3 or more nonzero coefficients, then $\,V\,$
is not an exceptional $\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
I) Let $\,\mu=b_1\,\omega_1\,+\,\cdots\,+\,b_{\ell}\,\omega_{\ell}\,\in\,
{\cal X}_{++}(V)\,$ be a good weight with at least 3 nonzero coefficients.
We claim that $\,s(V)\,>\,2\ell^2\,(\ell-1)\,$.
\underline{Step 1}:
Write $\,\mu\,=\,b_1\,\omega_{i_1}\,+\,b_2\,\omega_{i_2}\,+\,
b_3\,\omega_{i_3}\,+\cdots\,$, where $\,1\leq\,i_i<i_2<i_3\,$ are the
first 3 nonzero coefficients of $\,\mu$. By
Lemma~\ref{remark}, $\,|W\mu|\,\geq\,|W\bar{\mu}|,\,$ where
$\,\bar{\mu}\,=\,a\omega_{i_1}\,+\,b\omega_{i_2}\,+\,c\omega_{i_3}\,$
with $\,a,\,b,\,c\,\neq 0$. Therefore, to prove the claim, it
suffices to prove that $\,|W\bar{\mu}|\,>\,2\,\ell^2\,(\ell-1)$, for any weight
$\,\bar{\mu}\,$ with exactly 3 nonzero coefficients.
\underline{Step 2}: Let $\,\bar{\mu}\,=\,a\omega_{i_1}\,+
\,b\omega_{i_2}\,+\,c\omega_{i_3}\,$ with $\,a,\,b,\,c\,\neq 0$.
We prove that $\,|W\bar{\mu}|\,>\,2\,\ell^2\,(\ell-1)\,$ by
considering the different possibilities for $\,(i_1,\,i_2,\,i_3)$.
(a) Let $\,1\leq i_1\,<\,i_2\,<\,i_3\,\leq\,\ell-3\,$ (hence $\,\ell
\geq 6\,$) and $\,\bar{\mu}\,=\,a\omega_{i_1}\,+\,b\omega_{i_2}\,
+\,c\omega_{i_3}\,$. Then for $\,3\leq i_3\leq \ell-3\,$ and $\,\ell\geq 6$,
$\,\displaystyle\binom{\ell}{i_3}\geq\binom{\ell}{3}\,$ and
$\,\displaystyle\binom{i_2}{i_1}\,\binom{i_3}{i_2}\geq 6\,$. Hence
\[
|W\bar{\mu}|\,=\, \displaystyle 2^{i_3}\,\binom{i_2}{i_1}\,\binom{i_3}{i_2}\,
\binom{\ell}{i_3}\,\geq\,2^{3}\,\ell\,(\ell-1)\,(\ell-2)\,>\,2\,
\ell^2\,(\ell-1)\,.
\]
(b) For $\,1\leq i_1\,<\,i_2\,<i_3=\ell-2,\,$ $\,\bar{\mu}\,=\,a\omega_{i_1}\,+
\,b\omega_{i_2}\,+\,c\omega_{\ell-2}\,$. Thus, for $\,2\leq i_2\leq \ell-3\,$
and $\,\ell\geq 5$,
\[
\begin{array}{lcl}
|W\bar{\mu}|\, & = & \displaystyle
\frac{2^{\ell-1}\,\ell !}{i_1!\,(i_2-i_1)!\,(\ell-2-i_2)!\,
2!\,2!}\, =\,2^{\ell -3}\,\ell\,(\ell-1)\,\binom{i_2}{i_1}\,
\binom{\ell-2}{i_2}\vspace{.5ex}\\
& \geq & 2^{\ell -3}\,\ell\,(\ell-1)\,(\ell-2)\, >\,
2\,\ell^2\,(\ell-1) \,.
\end{array}
\]
(c) For $\,1\leq i_1\,<\,i_2\,\leq \ell-2,\,i_3=\ell-1\,$ (or by a
graph-twist $\,1\leq i_1\,<\,i_2\,\leq \ell-2,\,i_3\,=\,\ell\,$),
$\,\bar{\mu}\,=\,a\omega_{i_1}\,+\,b\omega_{i_2}\,+\,c\omega_{\ell-1}\;$
(or $\,\bar{\mu}\,=\,a\omega_{i_1}\,+\,b\omega_{i_2}\,+\,c\omega_{\ell}\,$).
Thus, for $\,2\leq i_2\leq \ell-2\,$ and $\,\ell\geq 5,\,$
\[
\begin{array}{lcl}
|W\bar{\mu}|\, & = & \displaystyle 2^{\ell -1}\,\binom{i_2}{i_1}\,
\binom{\ell}{i_2}\,\geq\,2^{\ell -1}\cdot 2\binom{\ell}{2}\,\geq\,
2^{\ell-1}\,\ell\,(\ell-1)\,>\,2\,\ell^2\,(\ell-1) \,.
\end{array}
\]
(d) For $\,1\leq i_1\leq \ell-2,\,i_2\,=\,\ell-1\,$ and $\,i_3\,=\,\ell,\;$
$\,\bar{\mu}\,=\,a\omega_{i_1}\,+\,b\omega_{\ell-1}\,+\,c\omega_{\ell}\,$.
Thus, for $\,\ell\geq 5$,
$\,|W\bar{\mu}|\, = \displaystyle 2^{\ell-1}\,\ell\,\binom{\ell-1}{i_1}\geq
2^{\ell-1}\,\ell\,(\ell-1)\,>\,2\,\ell^2\,(\ell-1) \,$.
{\bf Therefore, if $\,{\cal X}_{++}(V)\,$ contains a good weight $\,\mu\,$
with at least 3 nonzero coefficients, then $\,s(V)\,\geq\,|W\mu|\,>\,
2\ell^2\,(\ell-1)\,$. Hence, \mbox{{\bf by}~\eqref{dllim},} $\,V\,$
is not an exceptional module.}
II) Let $\,\mu\,$ be a bad weight in $\,{\cal X}_{++}(V)\,$ with at
least 3 nonzero coefficients. We have $\,b_j\geq 2\,$ for all nonzero
coefficients. We claim that it is possible to produce, from $\,\mu,\,$ a
good weight $\,\mu_1\in{\cal X}_{++}(V)\,$ having 3 or more nonzero
coefficients. Indeed,
(a) if $\,1\leq\,i_1<i_2<i_3\,<\,\ell -2$, then
$\,\mu\,=\,b_1\,\omega_{i_1}\,+\,b_2\,\omega_{i_2}\,+\,
b_3\,\omega_{i_3}\,+\cdots\;$ and
$\,\mu_1\,=\,\mu-(\alpha_{i_2}+\cdots +\alpha_{i_3})\,=\,
b_1\,\omega_{i_1}\,+\,\omega_{i_2-1}\,+\,(b_2-1)\,\omega_{i_2}\,+\,
(b_3-1)\,\omega_{i_3}\,+\,\omega_{i_3+1}\,+\,\cdots\,\in\,{\cal
X}_{++}(V).\,$
(b) if $\,1\leq\,i_1<i_2<i_3\,=\,\ell -2$. Then
$\,\mu\,=\,b_1\,\omega_{i_1}\,+
\,b_2\,\omega_{i_2}\,+\,b_3\,\omega_{\ell -2}\,+\,\cdots\;$ and
$\,\mu_1\,=\,\mu-(\alpha_{i_2}+\cdots +\alpha_{\ell -2})\,=\,
b_1\,\omega_{i_1}\,+\,\omega_{i_2-1}\,+\,
(b_2-1)\omega_{i_2}\,+\,(b_3-1)\,\omega_{\ell -2}\,+\,
\omega_{\ell-1}\,+\,\omega_{\ell}\,\in\,{\cal X}_{++}(V).\,$
(c) if $\,1\leq\,i_1<i_2<i_3\,=\,\ell-1\,$ (or by a graph-twist
$\,i_3\,=\,\ell\,$), then $\,\mu\,=\,b_1\,\omega_{i_1}\,+
\,b_2\,\omega_{i_2}\,+\,b_3\,\omega_{\ell-1}\,+\,\cdots\;\,$
(or $\,\mu\,=\,b_1\,\omega_{i_1}\,+
\,b_2\,\omega_{i_2}\,+\,b_3\,\omega_{\ell}\,$).
i) For $\,1\leq\,i_1<i_2<\ell-2,\,$ $\,{\cal X}_{++}(V)\,$ contains
$\,\mu_1\,=\,\mu-(\alpha_{i_1}+\cdots +\alpha_{i_2})\,=\,
\omega_{i_1-1}\,+\,(b_1-1)\,\omega_{i_1}\,+\,(b_2-1)\omega_{i_2}\,+\,
\omega_{i_2+1}\,+\,b_3\,\omega_{\ell-1}\,+\,\cdots$
(or $\,\mu_1\,=\,\mu-(\alpha_{i_1}+\cdots +\alpha_{i_2})\,=\,
\omega_{i_1-1}\,+\,(b_1-1)\,\omega_{i_1}\,+\,b_2\omega_{i_2}\,+\,
\omega_{i_2+1}\,+\,b_3\,\omega_{\ell}$.)
ii) For $\,i_2\,=\,\ell-2,\;$
$\,\mu\,=\,b_1\,\omega_{i_1}\,+
\,b_2\,\omega_{\ell-2}\,+\,b_3\,\omega_{\ell-1}\,+\,\cdots\;\,$
(or $\,\mu\,=\,b_1\,\omega_{i_1}\,+
\,b_2\,\omega_{\ell-2}\,+\,b_3\,\omega_{\ell}\,$) and
$\,\mu_1\,=\,\mu-(\alpha_{i_1}+\cdots +\alpha_{\ell-2})\,=\,
\omega_{i_1-1}\,+\,(b_1-1)\,\omega_{i_1}\,+\,(b_2-1)\omega_{\ell-2}\,+
\,(b_3+1)\,\omega_{\ell-1}\,+\,\omega_{\ell}\;$
(or
$\,\mu_1\,=\,\mu-(\alpha_{i_1}+\cdots +\alpha_{\ell-2})\,=\,
\omega_{i_1-1}\,+\,(b_1-1)\,\omega_{i_1}\,+\,(b_2-1)\omega_{\ell-2}\,+\,
\omega_{\ell-1}\,+\,(b_3+1)\,\omega_{\ell}$) $\,\in\,{\cal X}_{++}(V)$.
In all these cases, $\,\mu_1\,$ is a good weight with at least 3 nonzero
coefficients, since $\,b_1,\,b_2,\,b_3\,\geq 2$. Hence, by Part I of
this proof, $\,s(V)\,\geq\,|W\mu_1|\,>\,2\ell^3\,$ and,
by~\eqref{dllim}, $\,V\,$ is not an exceptional module.
This proves the lemma.
\end{proof}
The main theorem for groups of type $\,D\,$ is as follows.
Its proof is given in Appendix~\ref{appdl}.
\begin{theorem}\label{dnlist}
If $\,V\,$ is an infinitesimally irreducible
$\,D_{\ell}(K)$-module with highest weight listed
in Table~\ref{tabledlall} (p. \pageref{tabledlall}),
then $\,V\,$ is an exceptional $\,\mathfrak g$-module.
If $\,V\,$ has highest weight different from the ones listed in Tables
~\ref{tabledlall} or~\ref{leftdn} (p. \pageref{leftdn}), then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{theorem}
\newpage
\part*{Chapter 5}
\part*{The Proofs}
\addcontentsline{toc}{section}{\protect\numberline{5}{The Proofs}}
\setcounter{section}{5}
\setcounter{subsection}{0}
In this Chapter, Theorem~\ref{frlaet} (p. \pageref{frlaet})
is proved. The proof is divided into
lemmas for each group or algebra type.
Recall that $\,\dim\,\mathfrak g\,=\,|R|+\ell\,$ and
$\,\varepsilon=\dim\,\mathfrak z(\mathfrak g)\,$.
We use Lemma~\ref{orbexcp} for sizes of centralizers and orbits of weights.
The notation for roots is as follows. For $\,\gamma=\displaystyle
\sum_{j=1}^{\ell}\,b_j\,\alpha_j$, we write $\,\gamma =
(b_1\,b_2\,\cdots\,b_{\ell})\,$.
\subsection{Groups or Lie Algebras of Exceptional Type}
\label{proofexcp}
\begin{remark}\label{remA}
Let $\,R\,$ be of type $\,E_6,\,E_7\,$ or $\,E_8$. Let
$\,\lambda = \displaystyle\sum_{i=1}^{\ell}\,a_i\,\omega_i\,$ be a
dominant weight such that $\,a_k\neq 0$, for some $\,1\leq k\leq\ell$. Then
$\,C_W(\lambda)\,\subseteq\,C_W(\omega_k).\,$ Hence, by Lemma~\ref{remark},
$\,|W\lambda|\,\geq\,|W\omega_k|$.
\end{remark}
\subsubsection{Type $E_6$}
In this case the orbit sizes of the fundamental weights are as follows.
\vspace{2ex}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$\omega_i$ & $\omega_1$ & $\omega_2$ & $\omega_3$ & $\omega_4$
& $\omega_5$ & $\omega_6$ \\ \hline
$|W\omega_i|$ & $ 27 $ & $72$ & $ 216$ & $ 720$ & $216$ & $27$ \\ \hline
\end{tabular}
\end{center}
\vspace{2ex}
By Theorem~\ref{???}, any $\,E_6(K)$-module $\,V\,$ such that
\begin{equation} \label{324}
s(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\in{\cal X}_{++}(V)}
{\scriptstyle\mu\;good}}\,m_{\mu}\,|W\mu|\,>\,324
\end{equation}
or
\begin{equation}\label{rr72}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\;good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,72
\end{equation}
is not an exceptional $\,\mathfrak g$-module.
For groups of type $\,E_6\,$, $\,|W|\,=\,2^7\cdot 3^4\cdot 5\,$ and
$\,|R_{long}|\,=\,|R|\,=\,72\,=\,2^3 \cdot 3^2\,$.
The following are reduction lemmas.
\begin{lemma}\label{e61}
Let $\,V\,$ be an $\,E_6(K)$-module. If $\,{\cal X}_{++}(V)\,$ contains:
(a) a good weight with at least $3$ nonzero coefficients or
(b) a (good or bad) weight with nonzero coefficient for $\,\omega_4\,$
(this will be called the {\bf $\,\omega_4$-argument} hereafter),\\
then $\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
(a) Let $\,\mu= \displaystyle\sum_{i=1}^6\,a_i\,\omega_i\,$ (with at
least $3$ nonzero coefficients) be a good weight in $\,{\cal X}_{++}(V)$.
Then $\,s(V)\,\geq\,|W\mu|\,\geq\,2^4\cdot 3^3\cdot 5\,>\,324$.
Hence, by (\ref{324}), $\,V\,$ is not an exceptional $\,\mathfrak g$-module.
(b) Let $\,\mu= \displaystyle\sum_{i=1}^{6}\,a_i\,\omega_i\,$ be
such that $\,a_4\neq 0$. By Remark~\ref{remA}, $\,|W\mu|\,\geq\,720\,$.
If $\,\mu\,$ is a good weight in $\,{\cal X}_{++}(V)$, then
$\,s(V)\,\geq\,720\,>\,324$. Hence, by (\ref{324}), $\,V\,$ is not exceptional.
If $\,\mu\,$ is a bad weight in $\,{\cal X}_{++}(V)\,$ (with
$\,a_4\neq 0\,$, hence $\,a_4\geq 2\,$), then $\,\mu_1\,=\,\mu-\alpha_4\,
=\,a_1\omega_1\,+\,(a_2+1)\omega_2\,+\,(a_3+1)\omega_3\,+\,
(a_4-2)\omega_4\,+\,(a_5+1)\omega_5\,+\,a_6\omega_6\,\in\,{\cal X}_{++}(V)\,$
is a good weight with (at least) $3$ nonzero coefficients.
Hence, by (a), $\,V\,$ is not exceptional. This proves the lemma.
\end{proof}
\begin{lemma}\label{e6r2}
Let $\,V\,$ be an $\,E_6(K)$-module. If $\,{\cal X}_{++}(V)\,$
contains a good weight with $\,2\,$ nonzero coefficients, then
$\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
Let $\,\mu\,=\,a\omega_j\,+\,b\omega_k\,$ (with $\,1\leq j<k\leq 6\,$
and $\,a\geq 1,\,b\geq 1\,$) be a good weight in $\,{\cal X}_{++}(V)$.
(a) If $\,j=4\,$ or $\,k=4$, then the $\,\omega_4$-argument
(Lemma~\ref{e61}(b)) applies.
(b) Let $\,\mu\in\{\,a\omega_1\,+\,b\omega_2\,,\;
a\omega_1\,+\,b\omega_3\,,\;a\omega_2\,+\,b\omega_6\,,\;
a_5\omega_5 + a_6\omega_6\,,\;\}\,$. Then $\,s(V)\,\geq\,|W\mu|\geq
\displaystyle\frac{2^7\cdot 3^4\cdot 5}{5!}=432\,>\,324\,$. Hence,
by~(\ref{324}), $\,V\,$ is not exceptional.
(c) Let $\,\mu\in\{\,a\omega_1\,+\,b\omega_5\,,
\;a\omega_2\,+\,b\omega_3\,,\;a\omega_2\,+\,b\omega_5\,,
\;a\omega_3\,+\,b\omega_6\,\}$.
Then $\,s(v)\,\geq\,|W\mu|\,=\,2^3\cdot 3^3\cdot 5=1080>324\,$. Hence,
by (\ref{324}), $\,V\,$ is not exceptional.
(d) Let $\,\mu\,=\,a\omega_1\,+\,b\omega_6\,\in\,{\cal X}_{++}(V)\,$.
For $\,a\geq 2,\,b\geq 1\,$ (resp., $\,a\geq 1,\,b\geq 2\,$),
$\,\mu_1=\mu -\alpha_1=\,(a-2)\omega_1\,+\,\omega_3\,+\,b\omega_6\,$
(resp., $\,\mu_1=\mu -\alpha_6\,=\,a\omega_1\,+\,\omega_5\,+\,
(b-2)\omega_6\,$) $\,\in\,{\cal X}_{++}(V)\,$.
For $\,a\geq 3\,$ (resp., $\,b\geq 3\,$), Lemma~\ref{e61}(a) applies.
For $\,a=2\,$ (resp., $\,b=2\,$), $\,\mu_1\,$ satisfies case (c)
(resp., (b)) of this proof. In any case $\,V\,$ is not exceptional.
Let $\,\mu= \omega_1+\omega_6\,\in{\cal X}_{++}(V)$. Then
$\,\mu_1=\mu-(1\,0\,1\,1\,1\,1)\,=\,\omega_2\,\in\,{\cal X}_{++}(V)$.
For $\,p\geq 2$, $\,|R_{long}^+-R^+_{\mu,p}|\,\geq\,16\,$ and
$\,|R_{long}^+-R^+_{\mu_1,p}|\,\geq\,20$. Thus
$\,r_p(V)\,\geq\,\displaystyle\frac{2\cdot 3^3\cdot 5\cdot 16}{2^3\cdot 3^2}
\,+\,\frac{2^3\cdot 3^2\cdot 20}{2^3\cdot 3^2}\,=\,80\,>\,72\,$.
Hence, by~\eqref{rr72}, $\,V\,$ is not exceptional.
(e) Let $\,\mu\,=\,a\omega_3\,+\,b\omega_5\,$. Then $\,s(V)\,\geq\,|W\mu|
\,=\,2^3\cdot 3^3\cdot 5\,=\,432\,>\,324\,$. Hence, by (\ref{324}), $\,V\,$
is not exceptional. This proves the lemma.
\end{proof}
\vspace{1ex}
{\bf Claim 1}: ({\bf The $\,\omega_3$-(or $\,\omega_5$)-argument}).
{\it Let $\,p\geq 3$. If $\,V\,$ is an $\,E_6(K)$-module such that
$\,\omega_3\,$ (or graph-dually $\,\omega_5\,$)
$\,\in\,{\cal X}_{++}(V)$, then $\,V\,$ is not an exceptional
$\,\mathfrak g$-module.}
Indeed, $\,|W\omega_3|\,=\,2^3\cdot 3^3\,$ and, for $\,p\geq 3$,
$\,|R_{long}^+-R^+_{\omega_3,p}|\,=\,25$. Thus
$\,r_p(V)\,\geq\,\displaystyle\frac{2^3\cdot 3^3\cdot 25}{2^3\cdot 3^2}\,=
\,75\,>\,72\,$. Hence, by~\eqref{rr72}, $\,V\,$ is not exceptional, proving
the claim. \hfill $\Box$ \vspace{1ex}
\begin{lemma}\label{e6nlf}
Let $\,G\,$ be a simply connected simple algebraic group of type
$\,E_6$. A $\,\mathfrak g$-module $\,V\,$ is exceptional if and only if
its highest weight is $\,\omega_1,\,\omega_2\,$ or $\,\omega_6$.
\end{lemma}\noindent
\begin{proof}
($\,\Longleftarrow\,$)
If $\,V\,$ has highest weight $\lambda= \omega_1\,$ (or graph-dually
$\lambda= \omega_6\,$), then $\,\dim\,V\,=\,27\,<\,78\,-\,\varepsilon\,$.
Hence, by Proposition~\ref{dimcrit}, $\,V\,$ is an exceptional module.
If $\,V\,$ has highest weight $\,\omega_2\,=\,\tilde{\alpha}\,$
then $\,V\,$ is the adjoint module, which is exceptional
by Example~\ref{adjoint}.
($\,\Longrightarrow\,$) Let $\,V\,$ be an $\,E_6(K)$-module.
{\bf Claim 2}: {\it If $\,{\cal X}_{++}(V)\,$ contains a nonzero bad
weight, then $\,V\,$ is not an exceptional $\,\mathfrak g$-module.}
Indeed, let $\,\nu = \sum_{i=0}^6\,a_i\,\omega_i\,$ be a nonzero bad
weight in $\,{\cal X}_{++}(V)$. Hence $\,a_i\equiv 0\pmod p,\,$ for all $\,i\,$
and at least one $\,a_i\neq 0\,$ (in which case $\,a_i\geq 2$.)
(i) $\,a_4\neq 0\,\Longrightarrow\,\nu\,$ satisfies the
$\,\omega_4$-argument (Lemma~\ref{e61}(b)).
(ii) $\,a_3\neq 0\,$ (or graph-dually $\,a_5\neq 0\,$)
$\,\Longrightarrow\,\mu=\nu - \alpha_3 =
(a_1+1)\omega_1 + a_2\omega_2 +(a_3-2)\omega_3 +(a_4+1)\omega_4 +
a_5\omega_5 +a_6\omega_6\in{\cal X}_{++}(V)$. Hence
the $\,\omega_4$-argument applies.
(iii) $\,a_2\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_2 =
a_1\omega_1 + (a_2-2)\omega_2 +a_3\omega_3 +(a_4+1)\omega_4 +
a_5\omega_5 +a_6\omega_6\in{\cal X}_{++}(V)$. Hence
the $\,\omega_4$-argument applies.
(iv) $\,a_1\neq 0\,$(or graph-dually $\,a_6\neq 0\,$) $\,\Longrightarrow\,
\mu=\nu - \alpha_1 = (a_1-2)\omega_1 + a_2\omega_2 +
(a_3+1)\omega_3 +a_4\omega_4 + a_5\omega_5 +a_6\omega_6\in{\cal X}_{++}(V)$.
For $\,p\geq 3,\,$ $\,|R_{long}^+-R^+_{\mu,p}|\,=\,21\,$.
Thus, as $\,|W\mu|\geq 2^4\cdot 3^3\,$, $\,r_p(V)\,\geq\,\displaystyle
\frac{2^4\cdot 3^3\;\;21}{2^3\cdot 3^2}\,=\,126\,>\,72$. Hence
by~\eqref{rr72}, $\,V\,$ is not exceptional.
For $\,p=2\,$ and $\,a_1>2$, $\,\mu\,$ does not occur in $\,{\cal X}_{++}(V)$.
For $\,p= 2\,$ we can assume $\,a_1=2\,$ and $\,a_i=0\,(\,2\leq i\leq 6)$.
Then $\,\nu=2\omega_1\,\in{\cal X}_{++}(V)$ only if $\,V\,$ has
highest weight $\,\lambda= \omega_1\,+\,\omega_5\,$ (for instance).
Hence, by Lemma~\ref{e6r2}, $\,V\,$ is not exceptional. This proves Claim 2.
{\bf Hence, by Lemma~\ref{e6r2} and Claim 2,
if $\,{\cal X}_{++}(V)\,$ contains a weight with
$2$ nonzero coefficients, then $\,V\,$ is not an exceptional module.}
{\bf Therefore, by Claim 2, we can assume
that $\,{\cal X}_{++}(V)\,$ contains only good weights with at most one
nonzero coefficient.}
Let $\,\lambda\, = \,a\,\omega_i\,$ (with $\,a\geq 1\,$)
be a good weight in $\,{\cal X}_{++}(V)$.
(a) Let $\,a\geq 1\,$ and $\,\lambda\,=\,a\,\omega_4\,$.
Then the $\,\omega_4$-argument (Lemma~\ref{e61}(b)) applies.
(b) Let $\,a\geq 1\,$ and $\,\lambda\,=\,a\,\omega_3\,$
(or graph-dually $\,\lambda\,=\,a\,\omega_5\,$).
For $\,a\geq 2$, $\,\mu\,=\,\lambda - \alpha_3\,=\,\omega_1\,+\,(a-2)
\omega_3\,+\,\omega_4\,\in\,{\cal X}_{++}(V)$. Hence,
the $\,\omega_4$-argument applies.
For $\,a=1\,$ and $\,p\geq 3\,$, $\,\lambda\,=\,\omega_3\,$ satisfies Claim 1.
For $\,p=2$, we may assume that $\,V\,$ has highest weight $\,\mu\,=\,
\omega_3\,$. Then $\,\mu_1\,=\,\mu-(112210)\,=\,\omega_6\,\in\,
{\cal X}_{++}(V)$. By~\cite[p. 414]{gise}, $\,m_{\mu_1}=4$. As
$\,|R_{long}^+-R^+_{\mu,2}|\,=\,20\,$ and
$\,|R_{long}^+-R^+_{\mu_1,2}|\,=\,16$, $\,r_2(V)\,\geq\,\displaystyle
\frac{2^3\cdot 3^3\cdot 20}{2^3\cdot 3^2}\,+\,4\cdot
\frac{3^3\cdot 16}{2^3\cdot 3^2}\,=\,84\,>\,72\,$. Hence,
by~\eqref{rr72}, $\,V\,$ is not exceptional.
(c) Let $\,a\geq 1\,$ and $\,\lambda\,=\,a\,\omega_1\,$
(or graph-dually $\,\mu\,=\,a\,\omega_6\,$) be a good weight in
$\,{\cal X}_{++}(V)$.
For $\,a\geq 2$, $\,\mu_1\,=\,\lambda-\alpha_1\,=\,(a-2)\omega_1\,+\,\omega_3\,
\in\,{\cal X}_{++}(V)$. For $\,a\geq 3$, $\,\mu_1\,$ satisfies
Lemma~\ref{e6r2}. For $\,a=2\,$ (hence $\,p\geq 3\,$), $\,\mu_1\,$
satisfies Claim 1. In these cases $\,V\,$ is not exceptional.
For $\,a=1$, we may assume that $\,V\,$ has highest weight $\,\mu=
\omega_1\,$ (or graph-dually $\,\mu\,=\,\omega_6\,$).
Then $\,\dim\,V\,=\,27\,<\,78\,-\,\varepsilon\,$. Hence,
by Proposition~\ref{dimcrit}, $\,V\,$ is an exceptional module.
(d) Let $\,a\geq 1\,$ and $\,\mu\,=\,a\,\omega_2\,$ be a good weight in
$\,{\cal X}_{++}(V)$.
For $\,a\geq 2$, $\,\mu_1\,=\,\mu-\alpha_2\,=\,(a-2)\omega_2\,+\,\omega_4\,
\in\,{\cal X}_{++}(V)\,$ satisfies the $\,\omega_4$-argument
(Lemma~\ref{e61}(b)). For $\,a=1\,$, we may assume that $\,V\,$
has highest weight $\,\omega_2\,=\,\tilde{\alpha}\,$. Then $\,V\,$ is
the adjoint module, which is exceptional by Example~\ref{adjoint}.
This proves the lemma.
\end{proof}
\subsubsection{Type $\,E_7$}
The orbit sizes of the fundamental weights for groups of type
$\,E_7\,$ are as follows.
\vspace{2ex}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$\omega_i$ & $\omega_1$ & $\omega_2$ & $\omega_3$ & $\omega_4$
& $\omega_5$ & $\omega_6$ & $\omega_7$ \\ \hline
$|W\omega_i|$ & $ 126 $ & $576$ & $ 2016$ & $10080$ & $4032$ & $756$
& $56$ \\ \hline
\end{tabular}
\end{center}
\vspace{2ex}
By Theorem~\ref{???}, any $\,E_7(K)$-module $\,V\,$ such that
\begin{equation} \label{588}
s(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\in{\cal X}_{++}(V)}
{\scriptstyle \mu\; good}}\,m_{\mu}\,|W\mu|\,>\,588
\end{equation}
or
\begin{equation}\label{rr126}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\; good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,126
\end{equation}
is not an exceptional $\,\mathfrak g$-module. For groups of type $\,E_7\,$,
$\,|W|\,=\,2^{10}\cdot 3^4\cdot 5\cdot 7\,$ and
$\,|R_{long}|\,=\,|R|\,=\,126\,=\,2\cdot 3^2\cdot 7\,$.
\begin{lemma}\label{e71}
Let $\,V\,$ be an $\,E_7(K)$-module. If $\,{\cal X}_{++}(V)\,$ contains:
(a) a good weight with nonzero coefficient
for $\,\omega_i\,$ for some $\,i\in\{ 3,\,4,\,5,\,6\}\,$ or
(b) a good weight with at least $2$ nonzero coefficients or
(c) a nonzero bad weight,\\
then $\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
(a) Let $\,\mu\,=\,\displaystyle\sum_{i=1}^7\,a_i\,\omega_i\,$ be a good
weight in $\,{\cal X}_{++}(V)$. If $\,a_k\neq 0\,$ then, by
Remark~\ref{remA}, $\,|W\mu|\geq |W\omega_k|\,$. Thus, if $\,a_k\neq
0\,$ for $\,k\in\{ 3,\,4,\,5,\,6\,\}$, then $\,s(V)\,\geq\,756\,>\,588\,$.
Hence, by (\ref{588}), $\,V\,$ is not an exceptional module.
(b) Let $\,\mu\,= \,a\,\omega_j\,+\,b\,\omega_k\,$ (with
$\,1\leq j<k\leq 7\,$ and $\,a\geq 1,\,b\geq 1\,$) be a good weight in
$\,{\cal X}_{++}(V)$. If $\,j\,$ or $\,k\;\in\,\{ 3,4,5,6\,\}$,
then case (a) of this lemma applies.
If $\,\mu\,=\,a\,\omega_1\,+\,b\,\omega_2\,$ or
$\,\mu\,=\,a\,\omega_2\,+\,b\,\omega_7 \,$, then
$\,|W\mu|\,=\,2^6\cdot 3^2\cdot 7\,=\,4032\,$.
If $\,\mu\,=\,a\,\omega_1\,+\,b\,\omega_7\,$, then
$\,|W\mu|\,=\,2^3\cdot 3^3\cdot 7\,=\,1512.\,$ In these cases,
$\,s(V)\,>\,588\,$. Hence, by (\ref{588}), $\,V\,$ is not exceptional.
(c) Let $\,\nu\,=\,\displaystyle\sum_{i=1}^7\,a_i\,\omega_i\,$ be
a nonzero bad weight in $\,{\cal X}_{++}(V)\,$ (that is,
$\,a_i\equiv 0\pmod p,\,$ for all $\,i\,$, but at least one $\,a_i\neq 0$.
(Note that $\,a_i\neq 0\,\Longrightarrow\,a_i\geq 2$.)
Write $\,\Omega_{j,k,\ldots}\,=\,\displaystyle
\sum_{r\neq j,k,\ldots}\,a_r\,\omega_r\,$.
i) $\,a_1\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_1 =
(a_1-2)\omega_1 + (a_3+1)\omega_3 + \Omega_{1,3}\,\in{\cal X}_{++}(V)$.
ii) $\,a_2\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_2 =
(a_2-2)\omega_2 +(a_4+1)\omega_4 + \Omega_{2,4}\,
\in{\cal X}_{++}(V).\,$
iii) $\,a_3\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_3 =
(a_1+1)\omega_1 + (a_3-2)\omega_3 +(a_4+1)\omega_4 + \Omega_{1,3,4}\,
\in{\cal X}_{++}(V).\,$
iv) $\,a_4\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_4 =
(a_2+1)\omega_2 +(a_3+1)\omega_3 +(a_4-2)\omega_4 +\mbox{$(a_5 +1)\omega_5$}
+\Omega_{2,3,4,5}\,\in{\cal X}_{++}(V).\,$
v) $\,a_5\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_5 =
(a_4+1)\omega_4 +(a_5-2)\omega_5 +(a_6+1)\omega_6 + \Omega_{4,5,6}\,
\in{\cal X}_{++}(V).\,$
vi) $\,a_6\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_6 =
(a_5+1)\omega_5 +(a_6-2)\omega_6 +(a_7+1)\omega_7 + \Omega_{5,6}\,
\in{\cal X}_{++}(V).\,$
vii) $\,a_7\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_7 =
(a_6+1)\omega_6 +(a_7-2)\omega_7 + \Omega_{6,7}\,\in{\cal X}_{++}(V).\,$
In all these cases, $\,\mu\,$ satisfies part (a) of this lemma. Hence
$\,V\,$ is not an exceptional module. This proves the lemma.
\end{proof}
\begin{lemma}\label{e72}
Let $\,G\,$ be a simply connected simple algebraic group of
type $\,E_7$. A $\,\mathfrak g$-module $\,V\,$ is exceptional
if and only if its highest weight is $\,\omega_1\,$ or $\,\omega_7$.
\end{lemma}\noindent
\begin{proof}
($\,\Longleftarrow\,$)
If $\,V\,$ is an $\,E_7(K)$-module of highest weight $\lambda= \omega_1\,$,
then $\,V\,$ is the adjoint module, which is exceptional
by Example~\ref{adjoint}.
If $\,V\,$ is an $\,E_7(K)$-module of highest weight $\,\lambda\,=\,
\omega_7\,$, then $\,\dim\,V\,=\,56\,<\,133\,-\,\varepsilon\,$.
Hence, by Proposition~\ref{dimcrit}, $\,V\,$ is an exceptional module.
($\,\Longrightarrow\,$) Let $\,V\,$ be an $\,E_7(K)$-module.
By Lemma~\ref{e71}, {\bf it suffices to consider modules $\,V\,$ such that
$\,{\cal X}_{++}(V)\,$ contains good weights with at most one nonzero
coefficient.}
Let $\,\lambda\,=\,a\,\omega_i\,$ (with $\,a\geq 1\,$ and $\,1\leq i\leq 7\,$)
be a good weight in $\,{\cal X}_{++}(V)$.
For $\,i\in\{ 3,\,4,\,5,\,6\}\,$, Lemma~\ref{e71}(a) applies.
Hence we can assume $\,i\in\{ 1,\,2,\,7\}\,$. These cases are treated in the
sequel.
(i) Let $\,a\geq 1\,$ and $\,\lambda\,=\,a_1\,\omega_1\,$.
For $\,a\geq 2$, $\,\mu=\lambda - \alpha_1=(a-2)\omega_1 +\omega_3
\in{\cal X}_{++}(V)\,$ satisfies Lemma~\ref{e71}(a).
For $\,a=1$, we may assume that $\,V\,$ has highest weight
$\,\lambda =\omega_1\,$. Then $\,V\,$ is the adjoint module,
which is exceptional by Example~\ref{adjoint}.
(ii) Let $\,a\geq 1\,$ and $\,\lambda\,=\,a\,\omega_2\,$.
For $\,a\geq 2$, $\,\mu=\lambda - \alpha_2=(a-2)\omega_2 +\omega_4
\in{\cal X}_{++}(V)\,$ satisfies Lemma~\ref{e71}(a).
For $\,a=1$, $\,\lambda =\omega_2\,$ and $\,\mu\,=\,\lambda -
(1223210)\,=\,\omega_7\,\in\,{\cal X}_{++}(V)$. Thus
$\,s(V)\,\geq\,576 + 56\,=\,632\,>\,588\,$. Hence, by~\eqref{588},
$\,V\,$ is not exceptional.
(iii) Let $\,a\geq 1\,$ and $\,\lambda\,=\,a\,\omega_7\,$.
For $\,a\geq 2$, $\,\mu=\lambda - \alpha_7=\omega_6 +(a-2)\omega_7
\in{\cal X}_{++}(V)\,$ satisfies Lemma~\ref{e71}(a).
For $\,a=1$, we may assume that $\,V\,$ has highest weight $\,\lambda
=\omega_7$. Then $\,\dim\,V\,=\,56\,<\,133\,-\varepsilon\,$. Hence,
by Proposition~\ref{dimcrit}, $\,V\,$ is an exceptional module.
This proves the lemma.
\end{proof}
\subsubsection{Type $\,E_8$}
In this case the orbit sizes of the fundamental weights are as follows.
\vspace{2ex}
\begin{center}
\begin{tabular}{|c|c|}
\hline
$\omega_i$ & $|W\omega_i|$ \\ \hline
$\omega_1$ & $ 1080 $ \\ \hline
$\omega_2$ & $17280$ \\ \hline
$\omega_3$ & $2^9\cdot 3^3\cdot 5 $ \\ \hline
$\omega_4$ & $ 2^9\cdot 3^3\cdot 5\cdot 7 $ \\ \hline
$\omega_5$ & $2^8\cdot 3^3\cdot 5 \cdot 7 $ \\ \hline
$\omega_6$ & $ 2^6\cdot 3^3\cdot 5 \cdot 7 $ \\ \hline
$\omega_7$ & $ 2^6\cdot 3\cdot 5 \cdot 7 $\\ \hline
$\omega_8$ & $120$
\\ \hline
\end{tabular}
\end{center}
\vspace{2ex}
By Theorem~\ref{???}, any $\,E_8(K)$-module $\,V\,$ such that
\begin{equation} \label{1011}
s(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\in{\cal X}_{++}(V)}
{\scriptstyle \mu\;good}}\,m_{\mu}\,|W\mu|\,>\,1011
\end{equation}
or
\begin{equation}\label{rr240}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\;good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,240
\end{equation}
is not an exceptional $\,\mathfrak g$-module.
For groups of type $\,E_8\,$, $\,|W|\,=\,2^{14}\cdot 3^5\cdot 5^2\cdot
7\,$ and $\,|R_{long}|\,=\,|R|\,=\,240\,=\,2^4\cdot 3\cdot 5$.
\begin{lemma}\label{e81}
Let $\,V\,$ be an $\,E_8(K)$-module. If $\,{\cal X}_{++}(V)\,$ contains:
(a) a good weight with nonzero coefficient for $\,\omega_i\,$, for some
$\,i\in\{ 1,\ldots,7\}\;$ or
(b) a nonzero bad weight,\\
then $\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
(a) Let $\,\mu\,=\,\displaystyle\sum_{i=1}^8\,a_i\,\omega_i\,$ be a good
weight in $\,{\cal X}_{++}(V)$. If $\,a_k\neq 0\,$ then, by
Remark~\ref{remA}, $\,|W\mu|\geq |W\omega_k|\,$. Thus, if $\,a_k\neq
0\,$ for $\,k\in\{ 1,\,2,\,3,\,4,\,5,\,6,\,7\,\}$ then, by table above,
$\,s(V)\,\geq\,1080\,>\,1011\,$. Hence, by (\ref{1011}), $\,V\,$ is not an
exceptional module.
(b) Let $\,\nu\,=\,\displaystyle\sum_{i=1}^8\,a_i\,\omega_i\,$ be
a nonzero bad weight in $\,{\cal X}_{++}(V)\,$ (that is,
$\,a_i\equiv 0\pmod p,\,$ for all $\,i\,$, but at least one $\,a_i\neq 0$).
(Note that $\,a_i\neq 0\,\Longrightarrow\,a_i\geq 2$.)
Write $\,\Omega_{j,k,\ldots}\,=\,\displaystyle
\sum_{r\neq j,k,\ldots}\,a_r\,\omega_r\,$.
i) $\,a_1\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_1 =
(a_1-2)\omega_1 + (a_3+1)\omega_3 + \Omega_{1,3}\in{\cal X}_{++}(V)$.
ii) $\,a_2\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_2 =
(a_2-2)\omega_2 +(a_4+1)\omega_4 + \Omega_{2,4}
\in{\cal X}_{++}(V).\,$
iii) $\,a_3\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_3 =
(a_1+1)\omega_1 + (a_3-2)\omega_3 +(a_4+1)\omega_4 + \Omega_{1,3,4}
\in{\cal X}_{++}(V).\,$
iv) $\,a_4\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_4 =
(a_2+1)\omega_2 +(a_3+1)\omega_3 +(a_4-2)\omega_4 + (a_5 +1)\omega_5 +
\Omega_{2,3,4,5}\in{\cal X}_{++}(V).\,$
v) $\,a_5\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_5 =
(a_4+1)\omega_4 +(a_5-2)\omega_5 +(a_6+1)\omega_6 + \Omega_{4,5,6}
\in{\cal X}_{++}(V).\,$
vi) $\,a_6\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_6 =
(a_5+1)\omega_5 +(a_6-2)\omega_6 +(a_7+1)\omega_7 + \Omega_{5,6}
\in{\cal X}_{++}(V).\,$
vii) $\,a_7\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_7 =
(a_6+1)\omega_6 +(a_7-2)\omega_7 + (a_8+1)\omega_8 +
\Omega_{6,7}\in{\cal X}_{++}(V).\,$
viii) $\,a_8\neq 0\,\Longrightarrow\,\mu=\nu - \alpha_8 =
(a_7+1)\omega_7 +(a_8-2)\omega_8 +\Omega_{7,8}\in{\cal X}_{++}(V).\,$\\
In all these cases, $\, \mu\,$ satisfies part (a) of this lemma. Hence
$\,V\,$ is not an exceptional module. This proves the lemma.
\end{proof}
\begin{lemma}\label{nlfe8}
The only exceptional $\,E_8(K)$-module is the adjoint module.
\end{lemma}\noindent
\begin{proof}
Let $\,V\,$ be an $\,E_8(K)$-module. By Lemma~\ref{e81}, {\bf we can assume
that $\,{\cal X}_{++}(V)\,$ contains only good weights of the form $\,\lambda
= a\,\omega_8\,$ (with $\,a\geq 1\,$).}
For $\,a\geq 2$, $\,\mu=\lambda - \alpha_8=\omega_7 + (a-2)\omega_8\,
\in{\cal X}_{++}(V)$. Hence Lemma~\ref{e81}(a) applies.
For $\,a=1\,$, we can assume that $\,V\,$ has highest weight
$\,\lambda = \omega_8\,$. Then $\,V\,$ is the adjoint module,
which is exceptional by Example~\ref{adjoint}.
This proves the lemma.
\end{proof}
\subsubsection{Type $\,F_4\,$}
For groups of type $\,F_4\,$ the orbit sizes for the fundamental
weights are as follows.
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$\omega_i$ & $\omega_1$ & $\omega_2$ & $\omega_3$ & $\omega_4$
\\ \hline
$|W\omega_i|$ & $ 24 $ & $96$ & $ 96$ & $ 24$ \\ \hline
\end{tabular}
\end{center}
\vspace{2ex}
By Theorem~\ref{???}, any $\,F_4(K)$-module $\,V\,$ such that
\begin{equation} \label{192}
s(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\in{\cal X}_{++}(V)}
{\scriptstyle\mu\;is good}}\,m_{\mu}\,|W\mu|\,>\,192
\end{equation}
or
\begin{equation}\label{rr48}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\; good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,48
\end{equation}
is not an exceptional $\,\mathfrak g$-module. For groups of type $\,F_4\,$,
$\,p\geq 3$, $\,|W|\,=\,2^7\cdot 3^2\,$,
$\,|R|\,=\,48\,,\;\,|R_{long}|\,=\,|R(D_4)|\,=\,2^3\cdot 3$.
$\,R_{long}^+\,=\,\{ \varepsilon_i\pm\varepsilon_j\,/\,1\leq i<j\leq 4\}\,
=\,\{ (2 3 4 2),\,(1 3 4 2),\,(1 2 4 2),\,(1 2 2 0),\,(1 1 2 0),\,(0 1 2 0),
\,(0 1 2 2),\,(1 1 2 2),\,(1 2 2 2),\,(1 0 0 0)$, \linebreak
$\,(1 1 0 0),\,(0 1 0 0)\}$.
Furthermore,
\begin{eqnarray}
\omega_1 & = & 2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4 = (2342)
\nonumber \\
\omega_2 & = & 3\alpha_1 + 6\alpha_2 + 8\alpha_3 + 4\alpha_4 = (3684)
\nonumber \\
\omega_3 & = & 2\alpha_1 + 4\alpha_2 + 6\alpha_3 + 3\alpha_4 = (2463)
\nonumber \\
\omega_4 & = & 1\alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4 = (1232)
\nonumber
\end{eqnarray}
\begin{lemma}\label{fared}
Let $\,V\,$ be an $\,F_4(K)$-module. If $\,{\cal X}_{++}(V)\,$ contains
(a) a good weight with $3$ nonzero coefficients or
(b) a good weight with $2$ nonzero coefficients or
(c) any nonzero bad weight, \\
then $\,V\,$ is not an exceptional $\,\mathfrak g$-module.
\end{lemma}\noindent
\begin{proof}
(a) Let $\,\mu\in{\cal X}_{++}(V)\,$ be a good weight (with $3$ nonzero
coefficients). Then $\,s(V)\,\geq\,|W\mu|\,=\,576\,>\,192\,$.
Hence, by \eqref{192}, $\,V\,$ is not exceptional.
(b) Let $\,\mu\,=\,a\,\omega_i\,+\,b\,\omega_j\,$ (with $\,a\geq 1,\,b\geq
1\,$ and $\,1\leq i<j\leq 4\,$) be a good weight in $\,{\cal X}_{++}(V)$.
i) Let $\,\mu\in\{\,a\omega_1\,+\,b\omega_3\,,\;a\omega_2\,+\,b\omega_3\,,\;
a\omega_2\,+\,b\omega_4\,\}$. Then $\,s(V)\,\geq\,|W\mu|\geq\displaystyle
\frac{2^7\cdot 3^2}{2!\cdot 2!}=2^5\cdot 3^2 =288>192\,$.
ii) Let $\,\mu\,=\,a\omega_1\,+\,b\omega_2\;$ or
$\;\mu\,=\,c\omega_3\,+\,d\omega_4\,$. Then $\,|W\mu|\,=\,2^6\cdot 3=192$.
If $\,\mu\,=\,a\omega_1\,+\,b\omega_2\,=\,a(2342)\,+\,b(3684)\,$,
then $\,\mu_1=\mu - (1221)\,=\,a(2342) +(b-1)(3684)+(2463)\,=\,
a\omega_1\,+\,(b-1)\omega_2\,+\,\omega_3 \in {\cal X}_{++}(V)$.
If $\,\mu\,=\,c\omega_3\,+\,d\omega_4\,=\,c(2463)\,+\,d(1232)\,$, then
$\,\mu_1=\mu - (0011)\,=\,(c-1)(2463) +(d-1)(1232)+(3684)\,=\,\omega_2\,+
\,(c-1)\omega_3\,+\,(d-1)\omega_4\,\in{\cal X}_{++}(V)$.
In both cases, $\,s(V)\,\geq\,|W\mu|\,+\,|W\mu_1|\,\geq\,288\,>\,192\,$.
Hence, by~\eqref{192}, $\,V\,$ is not exceptional.
iii) Let $\,\mu\,=\,a\omega_1\,+\,b\omega_4\,$ (with $\,a\geq 1,\,b\geq 1\,$)
be a good weight in $\,{\cal X}_{++}(V)$. As $\,\omega_1=
\tilde{\alpha}=(2342)\,$ and $\,\omega_4=(1232)\,$ are also roots,
$\,\mu_1=\omega_1 +\omega_4\,\in\,{\cal X}_{++}(V)$. Also
$\,\mu_2=\mu_1-(1111)\,=\,\omega_3\,\in\,{\cal X}_{++}(V)$. Thus
$\,s(V)\,\geq\,|W\mu|\,+\,|W\mu_1|\,\geq 144 +96 =240>\,192$.
Hence, by~\eqref{192}, $\,V\,$ is not exceptional. This proves (b).
(c) Let $\,\mu = \displaystyle\sum_{i=1}^4\,a_i\,\omega_i\,$ be a nonzero
bad weight in $\,{\cal X}_{++}(V)$. Hence, as $\,p\geq 3$,
the nonzero coefficients of $\,\mu\,$ are $\,\geq 3\,$.
i) $\,a_1\neq 0\,\Longrightarrow\,\mu_1\,=\,\mu -(1110)\,=\,(a_1-1)\omega_1
+(a_4+1)\omega_4 +\Omega_{1,4}\,\in{\cal X}_{++}(V)$.
ii) $\,a_2\neq 0\,\Longrightarrow\,\mu_1=\mu -(1221)=(a_2-1)\omega_2 +
(a_3+1)\omega_3+\Omega_{2,3}\in{\cal X}_{++}(V)$.
iii) $\,a_3\neq 0\,\Longrightarrow\,\mu_1=\mu -(0121)=(a_1+1)\omega_1 +
(a_3-1)\omega_3+\Omega_{1,3}\in{\cal X}_{++}(V)$.
In all these cases, $\,\mu_1\,$ is a good weight in $\,{\cal X}_{++}(V)\,$
with $2$ nonzero coefficients. Hence, by part (b) of this lemma,
$\,V\,$ is not exceptional.
iv) We can assume that $\,a_4\,$ is the only nonzero coefficient of
$\,\mu\,$. As $\,\mu\,$ is bad, $\,a_4\geq 3\,$. Thus
$\,\mu_1\,=\mu-\alpha_4=\omega_3 +(a_4-2)\omega_4\in {\cal
X}_{++}(V)\,$ is a good weight with $2$ nonzero coefficients. Hence, by
part (b) of this proof, $\,V\,$ is not exceptional.
This proves the lemma.
\end{proof}
\begin{lemma}\label{f4nlf}
Let $\,V\,$ be an $\,F_4(K)$-module. Then $\,V\,$ is an exceptional
$\,\mathfrak g$-module if and only if $\,V\,$ has highest weight
$\,\omega_1\,$ (that is, the adjoint module) or $\,\omega_4\,$.
\end{lemma}\noindent
\begin{proof}
($\,\Longleftarrow\,$)
If $\,V\,$ has highest weight $\,\omega_1\,$, then $\,V\,$ is the
adjoint module, which is exceptional by Example~\ref{adjoint}.
If $\,V\,$ has highest weight $\,\omega_4\,$ then
$\,\dim\,V\,=\,26\,$ (for $\,p\neq 3\,$) or $\,\dim\,V\,=\,25\,$ (for
$\,p= 3\,$). In any case $\,\dim\,V\,<\,52\,-\,\varepsilon\,$.
Hence, by Proposition~\ref{dimcrit}, $\,V\,$ is an exceptional module.
($\,\Longrightarrow\,$)
Recall that $\,p\geq 3\,$ for groups of type $\,F_4\,$.
By Lemma~\ref{fared}, {\bf we can assume that $\,{\cal X}_{++}(V)\,$
contains only good weights with at most one nonzero coefficient.}
Let $\,\mu\,=\,a\,\omega_i\,$ (with $\,a\geq 1\,$ and $\,1\leq i\leq 4\,$)
be a good weight in $\,{\cal X}_{++}(V)$.
(a) Let $\,a\geq 1\,$ and $\,\mu\,=\,a\,\omega_2\,$. Then
$\,\mu_1\,=\mu -(1221)=\,(a-1)\omega_2\,+\,\omega_3\,\in\,
{\cal X}_{++}(V)$. For $\,a_2\geq 2,\,$ $\,\mu_1\,$ satisfies
Lemma~\ref{fared}(b). For $\,a=1,\,$ $\,\mu\,=\,\omega_2\,$,
$\,\mu_1\,=\,\omega_3\,$ and $\,\mu_2\,=\mu -(1342)=\,\omega_1
\,\in{\cal X}_{++}(V)$. For $\,p\geq 3,\,$ these are all good weights.
Thus $\,s(V)\,\geq\,|W\mu|\,+\,|W\mu_1|\,+\,|W\mu_2|\,=\,
216\,>\,192\,$. Hence, \mbox{by \eqref{192},} $\,V\,$ is not exceptional.
(b) Let $\,a\geq 1\,$ and $\,\mu\,=\,a\omega_3\,$. Then
$\,\mu_1=\mu -(1231)=\,(a-1)\omega_3\,+\,\omega_4\,\in\,{\cal X}_{++}(V)$.
For $\,a\geq 2,\,$ Lemma~\ref{fared}(b) applies.
For $\,a=1,\,$ $\,\mu=\omega_3\,$,
$\,\mu_1=\omega_4\,$ and $\,\mu_2=\,\mu-(0 1 2 1)\,=\,\omega_1\,\in
\,{\cal X}_{++}(V)$. For $\,p\geq 3$, these are all good weights and
$\,|R_{long}^+-R^+_{\omega_3,3}|\,=\,|R_{long}^+-R^+_{\omega_1,3}|=9,\,$ and
$\,|R_{long}^+-R^+_{\omega_4,3}|=6\,$. Thus $\,r_p(V)\,\displaystyle
=\,\frac{96\cdot 9}{24}\,+\,\frac{24\cdot 9}{24}\,+\,\frac{24\cdot 6}{24}
\,=\,51\,>\,48\,$. Hence, \mbox{by~(\ref{rr48}),} $\,V\,$ is not an exceptional
module.
(c) Let $\,a\geq 1\,$ and $\,\mu\,=\,a\,\omega_1\,$. Then
$\,\mu_1=\mu -(1110)=\,(a-1)\omega_1\,+\,\omega_4\,\in{\cal X}_{++}(V)$.
For $\,a\geq 2$, Lemma~\ref{fared}(b) applies. For $\,a=1$, we may
assume that $\,V\,$ has highest weight $\,\omega_1\,$, then $\,V\,$ is the
adjoint module, which is exceptional by Example~\ref{adjoint}.
(d) Let $\,a\geq 1\,$ and $\,\mu\,=\,a\,\omega_4\,$.
For $\,a\geq 2$, $\,\mu_1=\mu -\alpha_4=\,\omega_3\,+\,(a-2)\omega_4\,
\in{\cal X}_{++}(V)$. For $\,a\geq 3$, Lemma~\ref{fared}(b) applies.
For $\,a=2\,$, $\,\mu_1\,=\,\omega_3\,$ satisfies case (b) above.
Hence we can assume that $\,V\,$ has highest weight $\,\omega_4$,
then $\,\dim\,V\,=\,25\,$ (for $\,p=3\,$) or $\,26\,$ otherwise.
In both cases $\,\dim\,V\,<\,52\,-\,\varepsilon\,$. Hence,
by Proposition~\ref{dimcrit}, $\,V\,$ is an exceptional module.
This proves the lemma.
\end{proof}
\subsubsection{Type $\,G_2\,$}
By Theorem~\ref{???}, any $\,G_2(K)$-module $\,V\,$ such that
\begin{equation} \label{36}
s(V)\,=\,\sum_{\stackrel{\scriptstyle\mu\in{\cal X}_{++}(V)}
{\scriptstyle\mu\;good}}\,m_{\mu}\,|W\mu|\,>\,36
\end{equation}
or
\begin{equation}\label{rr12}
r_p(V)\,=\,\sum_{\stackrel{\scriptstyle{\mu\in{\cal X}_{++}(V)}}
{\scriptstyle \mu\;good}}\,m_{\mu}\,
\frac{|W\mu|}{|R_{long}|}\,|R_{long}^+-R^+_{\mu,p}|\,>\,12
\end{equation}
is not an exceptional $\,\mathfrak g$-module.
For groups of type $\,G_2\,$, $\,|W|\,=\,12\,$,
$\,|R|\,=\,12\,=\,2^2\cdot 3\,$, $\,|R_{long}|\,=\,6\,$ and
$\,R_{long}^+\,=\,\{ \alpha_2,\,3\alpha_1\,+\,\alpha_2,\,3\alpha_1\,
+\,2\alpha_2 \}$. Furthermore, $\,|W\omega_1|\,=\,|W\omega_2|\,=\,6\,$ and
$\,|W(\omega_1 +\omega_2)|\,=\,12\,$, where
$\,\omega_1 \, =\, 2\alpha_1 + \alpha_2\;$ and
$\;\omega_2 \, =\, 3\alpha_1 + 2\alpha_2\,$.
\vspace{2ex}\\
\begin{remark}\label{g2rem}
If $\,\mu= a\,\omega_1\,+\,b\,\omega_2\,$ and
$\,\Lambda\,=\,A\,\omega_1\,+\,B\,\omega_2\,$ are
dominant weights such that $\,A\geq a\,$ and $\,B\geq b$, then
$\,{\cal X}_{++,\mathbb C}(\mu)\,\subseteq\,{\cal X}_{++,\mathbb
C}(\Lambda)$. For $\,{\cal X}_{++,\mathbb C}(\Lambda)\,=\,(\Lambda\,
-\,Q_+)\cap P_{++}\,$ and
$\,\Lambda\, -\,Q_+\,=\,(\Lambda\,-\,\mu)\,+(\mu -\,Q_+)$, where
$\,\Lambda\,-\,\mu\,\in\,Q_+$. Hence $\,(\mu\,-\,Q_+)\,\subseteq\,(\Lambda\,
-\,Q_+),\,$ which implies the claim.
\end{remark}
\begin{lemma}\label{g2nlf}
Let $\,V\,$ be a $\,G_2(K)$-module. Then $\,V\,$ is an exceptional
$\,\mathfrak g$-module if and only if $\,V\,$ has highest weight
$\,\omega_1\,$ or $\,\omega_2.\,$
\end{lemma}\noindent
\begin{proof}
Recall that $\,p\neq 3\,$ for groups of type $\,G_2\,$ and also $\,p\neq
2\,$ if $\,V\,$ has highest weight $\,\omega_1\,$.
($\,\Longleftarrow\,$)
If $\,V\,$ has highest weight $\,\omega_1\,$, then $\,\dim\,V\,=\,7\,$
(for $\,p\neq 2\,$) or $\,\dim\,V\,=\,6\,$ (for $\,p= 2\,$). In both
cases $\,\dim\,V\,<\,14\,-\,\varepsilon\,$. Hence, by
Proposition~\ref{dimcrit}, $\,V\,$ is an exceptional module.
If $\,V\,$ has highest weight $\,\omega_2\,$ then $\,V\,$ is the
adjoint module, which is exceptional by Example~\ref{adjoint}.
($\,\Longrightarrow\,$) Let $\,V\,$ be a $\,G_2(K)$-module.
First observe that bad weights with $2$ nonzero coefficients do not
occur in $\,{\cal X}_{++}(V)$, otherwise it
would contradict Theorem~\ref{curt}(i).
{\bf Therefore we can assume that weights with $2$ nonzero coefficients
occurring in $\,{\cal X}_{++}(V)\,$ are good weights.}
I) Let $\,\mu\,=\,a\,\omega_1\,+\,b\,\omega_2\,$ (with $\,a\geq
1,\,b\geq 1\,$) be a good weight in $\,{\cal X}_{++}(V)$.
For $\,a\geq 2\,$ {\bf or} $\,b\geq 2$, by Remark~\ref{g2rem},
$\,\mu_1=\omega_1\,+\,\omega_2\in{\cal X}_{++}(V)$. Hence also
$\,\mu_2=2\omega_1,\;\mu_3=\omega_2,\;\mu_4=\omega_1,\;\mu_5=0\,\in\,
{\cal X}_{++}(V)$. Thus, for $\,p\geq 5$, $\,s(V)\,\geq\,12+12+6+6+6\,>
\,36\,$. Hence, by (\ref{36}), $\,V\,$ is not exceptional.
For $\,p=2,\,$ a weight of the form $\,a\,\omega_1\,+\,b\,\omega_2\,$,
with $\,a\geq 2\,$ or $\,b\,\geq 2,\,$ does not occur in
$\,{\cal X}_{++}(V)\,$ (otherwise it would contradict
Theorem~\ref{curt}(i)).
{\bf Hence if $\,{\cal X}_{++}(V)\,$ contains a good weight $\,\mu\,$ with $2$
nonzero coefficients, then we can assume that
$\,\mu\,=\,\omega_1\,+\,\omega_2\,$.} We deal with this case in III(a)
below.
II) Let $\,\mu\,=\,a\,\omega_i\,$ (with $\,a\geq 1\,$ and $\,1\leq
i\leq 2\,$) be a weight in $\,{\cal X}_{++}(V)$.
(a) Let $\,a\geq 2\,$ and $\,\mu\,=\,a\omega_2\,$. Then
$\,\mu_1\,=\,\mu-(\alpha_1+\alpha_2)\,=\,\omega_1\,+\,(a-1)\omega_2\,\in\,
{\cal X}_{++}(V)$. For $\,a\geq 3$, $\,\mu_1\,$ satisfies case I) of
this proof. Hence $\,V\,$ is not exceptional.
For $\,a= 2\,$ and $\,p\geq 5,\,$ $\,\mu\,=\,2\omega_2,\,$
$\,\mu_1=\omega_1 +\omega_2\,$, $\,\mu_2\,=\,\mu\,-\,\alpha_2\,=
\,3\omega_1\,$, $\,\mu_3\,=\,\mu_1-(\alpha_1+\alpha_2)\,=\,2\omega_1\,$,
$\,\mu_4\,=\,\mu_3-\alpha_1\,=\,\omega_2\,$ and
$\,\mu_5\,=\,\mu_4-(\alpha_1+\alpha_2)\,=\,\omega_1\,\in\,{\cal X}_{++}(V)$.
Thus $\,s(V)\,\geq\,6\,+\,6\,+\,12\,+\,6\,+\,6\,+\,6\,=\,42\,>\,36\,$.
Hence, by~(\ref{36}), $\,V\,$ is not exceptional.
For $\,p=2,\,$ $\,2\omega_2\,$ does not occur in $\,{\cal X}_{++}(V)$.
{\bf Hence, if $\,{\cal X}_{++}(V)\,$ contains
$\,\mu\,=\,a\omega_2\,$, then we can assume that $\,a=1$.} This case
is treated in III(b) below.
(b) Let $\,a\geq 2\,$ and $\,\mu\,=\,a\omega_1\,$. Then
$\,\mu_1\,=\,\mu-\alpha_1\,=\,(a-2)\omega_1\,+\,\omega_2\,\in\,
{\cal X}_{++}(V)$. For $\,a\geq 4$, $\,\mu_1\,$ satisfies case I) of
this proof. Hence $\,V\,$ is not exceptional.
For $\,a= 3\,$, $\,\mu\,=\,3\omega_1\,$. For $\,p=2\,$, $\,\mu\,$ does
not occur in $\,{\cal X}_{++}(V)$. For $\,p>3$, we may assume that $\,V\,$
has highest weight $\,\mu\,=\,3\omega_1\,$. Then
$\,\mu_1\,=\,\omega_1 +\omega_2\,$, $\,\mu_2\,=\,2\omega_1\,$,
$\,\mu_3\,=\,\omega_2\,$, $\,\mu_4\,=\,\omega_1\,\in{\cal X}_{++}(V)$.
By~\cite[p. 413]{gise}, $\,m_{\mu_1}=1,\,m_{\mu_2}=2,\,m_{\mu_3}=3,\,
m_{\mu_4}=4\,$. Thus $\,s(V)\,\geq\,6 + 12 + 2\cdot 6 + 3\cdot 6 +
4\cdot 6\,=\,72\,>\,36\,$. Hence, by~(\ref{36}), $\,V\,$ is not exceptional.
For $\,a= 2$, we can assume that $\,V\,$ has highest weight
$\,\mu=2\omega_1\,$.
Let $\,U=E(\omega_1)\,$ be the irreducible module of highest
weight $\,\omega_1$. For $\,p\neq 2$, $\,\dim\,U\,=\,7\,$.
By Subsection~\ref{roapm} (p. \pageref{roapm}), \mbox{$\,\mathfrak{so}(f)=
\{x\in {\rm End}\,U\,/\,f(xv,w)\,+\,f(v,xw)=0 \}\,$}
is a Lie algebra of type $\,B_3$. By~\cite[p. 103]{hum1}, it is
possible to describe a Lie algebra of type $\,G_2\,$ as a subalgebra of
$\,\mathfrak{so}(f).\,$ Now, as proved in case III(b)iv)
(p. \pageref{IIIbiv}), $\,E(2\omega_1)\,$ is not an
exceptional $\,\mathfrak{so}(f)$-module. Hence $\,E(2\omega_1)\,$ is also not
exceptional as a $\,G_2(K)$-module.
{\bf Hence, if $\,{\cal X}_{++}(V)\,$ contains
$\,\mu\,=\,a\omega_1\,$, then we can assume that $\,a=1$.} This case
is treated in III(c) below.
III) (a) Let $\,V\,$ be a $\,G_2(K)$-module of highest weight
$\,\lambda\,=\,\omega_1\,+\,
\omega_2\,$. Then $\,\mu_1=2\omega_1\,$, $\,\mu_2=\omega_2\,$,
$\,\mu_3=\omega_1\,\in\,{\cal X}_{++}(V)$. By~\cite[p. 413]{gise},
for $\,p\neq 3,\,7,\,$ $\,m_{\mu_2}\,=\,2\,$ and
$\,m_{\mu_3}\,=\,4\,$. Thus $\,s(V)\,\geq\,12 + 2\cdot 6 + 4\cdot
6\,=\,48\,>\,36\,$. Hence, by~\eqref{36}, $\,V\,$ is not exceptional
For $\,p=7$, $\,m_{\mu_1}\,=\,m_{\mu_2}\,=\,1\,$ and
$\,m_{\mu_3}\,=\,2\,$. As $\,|R_{long}^+-R^+_{2\omega_1,7}|\,=\,
|R_{long}^+-R^+_{\omega_1,7}|\,=\,2,\,$
$\,|R_{long}^+-R^+_{\omega_2,7}|\,=\,|R_{long}^+-R^+_{\lambda,
7}|\,=\,3,\,$ one has $\,r_7(V)\,\displaystyle
=\,\frac{12}{6}\cdot 3\,+\,\frac{6}{6}\cdot 2\,+\,\frac{6}{6}\cdot 3\,+\,
2\cdot\frac{6}{6}\cdot 2\,=\,15\,>\,12\,$. Hence, by~(\ref{rr12}),
$\,V\,$ is not an exceptional module.
(b) Now we may assume that $\,V\,$ has highest weight $\,\omega_2=
\tilde{\alpha}\,$. Then $\,V\,$ is the adjoint module, which is exceptional by
Example~\ref{adjoint}.
(c) Finally, if $\,V\,$ has highest weight $\,\omega_1\,$, then
$\,\dim\,V\,=\,6\,$ (for $\,p=2\,$) or $\,7\,$
(otherwise). In both cases, $\,\dim\,V\,<\,14\,-\,\varepsilon\,$, Hence,
by Proposition~\ref{dimcrit}, $\,V\,$ is an exceptional module.
This proves the lemma.
\end{proof}
\newpage
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 315 |
package azure
import (
"fmt"
"github.com/Azure/packer-azure/packer/builder/azure/common/constants"
"github.com/Azure/packer-azure/packer/builder/azure/smapi/retry"
"github.com/Azure/azure-sdk-for-go/management"
"github.com/Azure/azure-sdk-for-go/management/hostedservice"
"github.com/mitchellh/multistep"
"github.com/mitchellh/packer/packer"
)
type StepUploadCertificate struct {
TmpServiceName string
}
func (s *StepUploadCertificate) Run(state multistep.StateBag) multistep.StepAction {
client := state.Get(constants.RequestManager).(management.Client)
ui := state.Get("ui").(packer.Ui)
errorMsg := "Error Uploading Temporary Certificate: %s"
var err error
ui.Say("Uploading Temporary Certificate...")
certData := []byte(state.Get(constants.Certificate).(string))
if err = retry.ExecuteAsyncOperation(client, func() (management.OperationID, error) {
return hostedservice.NewClient(client).AddCertificate(s.TmpServiceName, certData, hostedservice.CertificateFormatPfx, "")
}); err != nil {
err := fmt.Errorf(errorMsg, err)
state.Put("error", err)
ui.Error(err.Error())
return multistep.ActionHalt
}
state.Put(constants.CertUploaded, 1)
return multistep.ActionContinue
}
func (s *StepUploadCertificate) Cleanup(state multistep.StateBag) {
// do nothing
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,147 |
William McKenzie Morrison produced this cabinet card portrait of actor William Hunter Kendal (1843-1917). Morrison's studio was housed in the Haymarket Theatre building in Chicago, Illinois. Morrison was known for being a photographer who specialized in taking photographs of celebrities. To view other portraits by Morrison, click on the category of "Photographer: Morrison". Kendal's given name was William Hunter Grimston and he was an English actor and manager. He was born in London and had his theatrical debut in Glasgow at age eighteen. Four years later he appeared in London at the Haymarket Theatre. In 1869 he married Madge Robertson (1848-1935) and they performed together for many years. Kendal was a co-partner in managing the St. James Theatre from 1879 through 1888. Between 1889 and 1895, Kendal and his wife toured successfully in the United States and Canada. Their American debut was in "A Scrap of Paper" (1889). The couple retired from acting in 1908.
A young woman is featured in this cabinet card that appears to be a memorial photograph. The image has a musical theme. Note the pictured string instrument and the scrolled sheet music. Perhaps the young woman pictured in the frame was a musician. The photographic studio responsible for this interesting image is the C. S. Roshon studio which was located in Harrisburg, Pennsylvania. The cabinet card gallery has another Roshon photograph in its collection). This second photograph is one of the more controversial images in the gallery's collection because it very well may be a counterfeit cabinet card. The image features a Native American man with a turkey vulture on his head. Click on the category "Photographer: Roshon" to view this photograph.
Two affectionate men pose for their portrait in Newton, Kansas. The men look quite dapper in their suits and with their straw hats. Note that the gentleman wearing the suit and vest has a pocket watch chain visible atop his vest. He is also holding a walking stick.The man standing, and the man sitting on the hammock are showing some shared affection. They could be friends, relatives, or even lovers. It is impossible to guess their relationship. One wonders if homophobia was much of a factor in the cabinet card era in regard to men showing affection to men in public or in photographs. Perhaps a visitor to the cabinet card gallery can competently comment on this issue. The photographer of this image is the Tripp studio in Newton, Kansas. According to print on the reverse of the photograph, the studio was located on the corner of Main Street and Broadway. The photographer, Frank D. Tripp is cited in Anthony's Photographic Bulletin (1896) as the President of the Photographers Association of Kansas. Another source states that Tripp "flourished" as a photographer in Newton during the 1880's. Tripp's obituary appears in The Evening Kansan Republican (1947). He died in Denver, Colorado at age eighty. He was described in the article as a pioneer photographer in Newton. He was an officer in the Newton Masonic Lodge. At some point he moved to Pueblo, Colorado where he was a partner in the Tripp and York photography studio.
Agnes Evans poses for this cabinet card (top) photographed by Newsboy of New York. Agnes Evans was a theatre actress who performed in the Broadway production of the Pit (1904). The actress is wearing a very revealing risque dress. Further research by myself or assistance from visitors to this site will hopefully further illuminate her life and career. Newsboy was a brand of plug tobacco and Newsboy photographs were given away as a premium by tobacconists and drug stores who sold the tobacco. The images were produced by the National Tobacco Works of New York. They were likely produced and issued in the early 1890's. The bottom image features Miss Evans in another Newsboy cabinet card (number 8 in a series). She is wearing a risque costume that includes fingerless gloves.
George Hadley of the Castle Studio produced this portrait of a woman in deep concentration. She is studying what appears to be a small framed photograph. One of her hands rests on a table. The table top holds an inkwell. Printing on the reverse of the cabinet card indicates that the studio had won prizes at exhibitions at Cornwall (1884), London (1884), and Northampton (1884). The studio was located in Lincoln. Lincoln is a cathedral city and county town of Lincolnshire, England.
Johnston Forbes-Robertson (1853-1937) was a celebrated English actor and theater manager. He was considered to be one of the finest actors of his time. He was particularly noted for his portrayal of Hamlet. He did not profess a passion for his acting profession. He was born in London. His father was a journalist and theater critic. He had ten siblings and four of them pursued acting. His original interest was to become an artist, but to support himself financially he entered acting. He worked with Sir Henry Irving for some time as a second lead actor. He then became a lead actor. His starring roles included Dan'l Druce, Blacksmith and The Parvenu (1882). George Bernard Shaw wrote the part of Caesar for him in Caesar and Cleopatra. Forbes Robertson acted in a number of Shakespeare plays and also appeared a number of times with actress Mary Anderson in the 1880's. In 1900 he married the American actress, Gertrude Elliott (1874-1950). In 1930, Forbes Robertson was knighted. This cabinet card portrait was produced by photographer Benjamin Falk who's studio was located in New York City. Forbes Robertson is captured in costume in this image. The reverse of the photo is stamped "J. M. Russell 126 Tremont Street, Boston".
The two young girls in this cabinet card portrait are probably sisters. They are posing with a bale of hay in front of a fake brick house. Both girls are wearing necklaces and the older girl is displaying some religiosity by wearing a cross. The photographer and the location of the photographer's studio are unknown. | {
"redpajama_set_name": "RedPajamaC4"
} | 5,221 |
'use strict';
const path = require('path');
var webpack = require('webpack');
var HtmlPlugin = require('html-webpack-plugin');
var HasteResolverPlugin = require('haste-resolver-webpack-plugin');
const fs = require('fs-extra');
const CUSTOM_NODE_MODULES = fs.readJSONSync(path.resolve(__dirname, '../content/config/custom_node_modules.json'));
var IP = '0.0.0.0';
var PORT = 3000;
var NODE_ENV = process.env.NODE_ENV;
var ROOT_PATH = path.resolve(__dirname, '..');
var PROD = 'production';
var DEV = 'development';
let isProd = NODE_ENV === 'production';
var config = {
paths: {
src: path.join(ROOT_PATH, '.'),
index: path.join(ROOT_PATH, 'index.web'),
},
};
module.exports = {
ip: IP,
port: PORT,
devtool: 'source-map',
resolve: {
alias: {
'react-native': 'ReactWeb',
},
extensions: ['', '.js', '.jsx'],
},
entry: isProd? [
'whatwg-fetch',
'babel-polyfill',
config.paths.index,
]: [
'whatwg-fetch',
'babel-polyfill',
'webpack-dev-server/client?http://' + IP + ':' + PORT,
'webpack/hot/only-dev-server',
config.paths.index,
],
devServer: {
headers: {
'Access-Control-Allow-Origin': 'https://pas-dev.promisefinancial.net:8885',
'customheaderyaw':'etse',
// "Access-Control-Allow-Origin": "http://localhost:4000",
// 'Access-Control-Allow-Credentials': 'true',
},
},
headers: { "X-Custom-Header": "yes" },
output: {
path: path.join(__dirname, './output'),
// path: path.join(__dirname, '../../public/web/output'),
filename: 'bundle.js',
},
plugins: [
new HasteResolverPlugin({
platform: 'web',
nodeModules: ['react-web'],
}),
new webpack.DefinePlugin({
'process.env': {
'NODE_ENV': JSON.stringify(isProd? PROD: DEV),
},
}),
isProd? new webpack.ProvidePlugin({
React: 'react',
}) : new webpack.HotModuleReplacementPlugin({
headers: {
'Access-Control-Allow-Origin': 'https://pas-dev.promisefinancial.net:8885',
'customheaderyaw':'etse',
// "Access-Control-Allow-Origin": "http://localhost:4000",
// 'Access-Control-Allow-Credentials': 'true',
},
}),
// new webpack.NoErrorsPlugin(),
new HtmlPlugin(),
],
module: {
loaders: [{
test: /\.json$/,
loader: 'json',
}, {
test: /\.jsx?$/,
loader: 'react-hot',
include: [config.paths.src],
exclude: [/node_modules/],
}, {
test: /\.js?$/,
loader: 'babel',
query: {
presets: ['es2015', 'react','react-native', 'latest'],
plugins: CUSTOM_NODE_MODULES['babel-plugins'],
},
include: [config.paths.src],
exclude: [/node_modules/],
}, {
test: /\.js?$/,
loader: 'babel',
query: {
presets: ['es2015', 'react','react-native', 'latest'],
plugins: CUSTOM_NODE_MODULES['babel-plugins'],
},
//add your modules here
include: CUSTOM_NODE_MODULES['react-native-components'].map((native_component)=>{
// console.log('list of mods',path.join(ROOT_PATH, `node_modules/${native_component.name}`))
return path.join(ROOT_PATH, `node_modules/${native_component.name}`);
}),
// exclude:[/\.png$/gi]
}, {
test: /\.jsx?$/,
loader: 'babel',
query: {
presets: ['es2015', 'react','react-native', 'latest'],
plugins: CUSTOM_NODE_MODULES['babel-plugins'],
},
//add your modules here
include: CUSTOM_NODE_MODULES['react-native-components'].map((native_component)=>{
// console.log('list of mods',path.join(ROOT_PATH, `node_modules/${native_component.name}`))
return path.join(ROOT_PATH, `node_modules/${native_component.name}`);
}),
// exclude:[/\.png$/gi]
}, {
// Match woff2 in addition to patterns like .woff?v=1.1.1.
test: /\.woff(2)?(\?v=[0-9]\.[0-9]\.[0-9])?$/,
loader: 'url',
query: {
limit: 500000,
mimetype: 'application/font-woff',
name: './fonts/[hash].[ext]',
},
include: [
path.join(ROOT_PATH, 'web/custom_node_modules/react-native-vector-icons'),
],
}, {
// Match woff2 in addition to patterns like .woff?v=1.1.1.
test: /\.ttf(2)?(\?v=[0-9]\.[0-9]\.[0-9])?$/,
loader: 'url',
query: {
limit: 5000000,
// mimetype: 'application/font-woff',
name: './fonts/[hash].[ext]',
},
include: [
path.join(ROOT_PATH, 'web/custom_node_modules/react-native-vector-icons'),
],
}, {
test: /\.(eot|ttf|svg|png|jpg)$/,
loader: 'url-loader?limit=1000000&name=[name]-[hash].[ext]',
include: [
path.join(ROOT_PATH, 'web/custom_node_modules/react-native-vector-icons'),
],
}, {
test: /\.css$/,
loader: 'style-loader!css-loader',
include: [
path.join(ROOT_PATH, 'web/custom_node_modules/react-native-vector-icons'),
],
}, {
test: /\.png$/,
loader: 'url?limit=100000&mimetype=image/png',
// include: config.paths.demo,
}, {
test: /\.jpg$/,
loader: 'file',
// include: config.paths.demo,
},
// {
// test : /\.(ttf|eot|png|jpg|jpeg|gif|svg)(\?[a-z0-9-=&.]+)?$/,
// loader : 'file-loader',
// },
],
},
}; | {
"redpajama_set_name": "RedPajamaGithub"
} | 5,505 |
Lace up a breathable summer style with enough support to scale a mountain in the new Litewave Flow Lace Athletic Shoe from The North Face! Designed to be quick-drying, the Litewave Flow Lace Athletic Shoe is perfect for your summer adventures with its breathable, open mesh uppers, quick-lace closure, OrthoLite® footbed for comfort, and Vibram® XS Trek rubber outsole for traction. Available online at Journeys.com! | {
"redpajama_set_name": "RedPajamaC4"
} | 9,035 |
Q: Undefined method `each' for #Rails 4.1
Ruby 2.0
Credential.rb
class Credential < ActiveRecord::Base
belongs_to :category
has_many :user
validates :name, :login, :password, presence: true
attr_accessor :encryption_key
attr_encrypted :login, key: :encryption_key
attr_encrypted :password, key: :encryption_key
end
User.rb
class User < ActiveRecord::Base
# Include default devise modules. Others available are:
# :confirmable, :lockable, :timeoutable and :omniauthable
devise :database_authenticatable, :registerable,
:recoverable, :rememberable, :trackable, :validatable
has_many :credentials
def you
"You are <b>#{email}</b>"
end
end
CredentialsController.rb
class CredentialsController < ApplicationController
before_filter :authenticate_user!
def create
@credential = current_user.credentials.new
@credential.encryption_key = session[:master_key]
@credential.update(credential_params)
if @credential.save
redirect_to credential_path(@credential), notice: "Password entry created successfully."
else
render "new"
end
end
The line:
@credential.update(credential_params)
throws an exception
undefined method 'each' for #<User:0x4de4f58>
A: You need to edit your associations. You have credentials that has_many :user and users that has_many :credentials. The one with the foreign key should be a belongs_to not has_many. If you're attempting to make a many-to-many relationship, then either use has_many_and_belongs_to or a join table. Further, it should be has_many :users and not has_many :user. That should resolve your error.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,660 |
Sphingomonas aerolata is a Gram-negative and psychrotolerant bacteria from the genus of Sphingomonas.
References
Further reading
External links
Type strain of Sphingomonas aerolata at BacDive - the Bacterial Diversity Metadatabase
aerolata
Bacteria described in 2003 | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,187 |
\section{Introduction}
\label{sec:level1}
In most studies of Cosmic Microwave Background (CMB) anisotropies, the
comparison between observed and theoretical quantities have so far
been done, assuming the linear approximation for cosmological
perturbations. It seems to be successful enough to determine the
cosmological parameters \cite{map,spg,komt}. The present state of our
universe is, however, locally complicated and associated with
nonlinear behavior on various
scales, and so the observed quantities of CMB anisotropies may include
some small effects caused through nonlinear process by various
primordial perturbations. So far the nonlinearity in CMB anisotropies
has been treated as the second-order integrated Sachs-Wolfe effect,
which was analyzed using the three-point correlation and
bispectrum.\cite{mglm,munshi,sg} Recently a general treatment of
second-order temperature fluctuations has been systematically studied
by Bartolo, Matarrese and Riotto\cite{bart1,bart2} due to the
transfer function which was derived from the Boltzmann equation, so as
to be
applied not only to the integrated Sachs-Wolfe effect at the
matter-dominant stage, but also the nonlinear effect at the
recombination epoch and the primordial nonlinear effect.
In recent years, on the other hand, we have studied these nonlinear
effects of inhomogeneities on CMB anisotropies at the matter-dominant
stage, based on the relativistic
second-order theory of cosmological perturbations,\cite{tom1} and on
Mollerach and Matarrese's
second-order formula of CMB anisotropies\cite{cmb}. In previous
papers,\cite{tom2,tom3,tom4,is1,is2} we have studied the
second-order effects of a local special large-scale inhomogeneity
on CMB anisotropies, paying attention to the interaction between it
and primordial random perturbations. In this relativistic theory we
used the first-order and second-order perturbations in pressureless
matter which were derived in compact and analytic forms.
In the present paper we study the
nonlinear effect of only primordial random perturbations on CMB
anisotropies and derive the second-order power spectra, which is given
as the second term in the two-point correlation of CMB
anisotropies. This is a nonlinear correction to the first-order power
spectra and
different from the three-point correlation and bispectrum which have
been used in the above other works to investigate uniquely the
non-Gaussianity in the perturbations generated in various stages.
The second-order random density perturbations themselves are small,
compared with the first-order ones. In the $\Lambda$-dominated model,
however, the second-order power spectra of
CMB anisotropies are not small at all, compared with the
first-order power spectra, because at the early stage the first-order
integrated Sachs-Wolfe effect (ISW) is very small and the second-order
ISW may be dominant over the first-order one.
Especially in the local void model with $\Lambda =
0$\cite{lvm,cele,aln,bis}, the exterior region is described by use of
the Einstein-de Sitter model, in which the first-order ISW vanishes
and the second-order ISW is indispensable.
So their characteristic behaviors may be measured
through the future precise observation and bring useful informations
on the structure and evolution of our universe in the future.
In Sec. II, we show the second-order perturbations in
general flat cosmological models and the
corresponding CMB anisotropies. In Sec. III, we derive the expressions
for the second-order power spectra of CMB anisotropies,
due to primordial random perturbations. Concluding remarks follow in
Section IV.
\section{Second-order metric perturbations and temperature anisotropies}
\label{sec:level2}
First we review the background spacetime and the perturbations which
were derived in the previous paper\cite{tom2}. The background flat
model with dust matter is expressed as
\begin{equation}
\label{eq:m1}
ds^2 = g_{\mu\nu} dx^\mu dx^\nu =
a^2 (\eta) [- d \eta^2 + \delta_{ij} dx^i dx^j],
\end{equation}
where the Greek and Latin letters denote $0,1,2,3$ and $1,2,3$,
respectively, and
$\delta_{ij} (= \delta^i_j = \delta^{ij})$ are the Kronecker
delta. The conformal time $\eta (=x^0)$ is related to the cosmic time
$t$ by $dt = a(\eta) d\eta$. The matter density $\rho$ and the scale
factor $a$ have the relations
\begin{equation}
\label{eq:m2}
\rho a^2 = 3(a'/a)^2 - \Lambda a^2, \quad {\rm and } \quad
\rho a^3 = \rho_0,
\end{equation}
where a prime denotes $\partial/\partial \eta$, $\Lambda$ is the
cosmological constant, $\rho_0$ is an integration constant and the
units $8\pi G = c =1$ are used.
The first-order and second-order metric perturbations
$\mathop{\delta}_1 g_{\mu\nu} (\equiv
h_{\mu\nu})$ and $\mathop{\delta}_2 g_{\mu\nu} (\equiv
\ell_{\mu\nu})$, respectively, were derived explicitly by imposing the
synchronous coordinate condition:
\begin{equation}
\label{eq:m3}
h_{00} = h_{0i} = 0 \quad {\rm and} \quad \ell_{00} = \ell_{0i} = 0.
\end{equation}
Here we show their expressions only in the growing mode:
\begin{eqnarray}
\label{eq:m4}
h^j_i &=& P(\eta) F_{,ij}, \cr
\ell^j_i &=& P(\eta) L^j_i + P^2 (\eta) M^j_i + Q(\eta) N^{|j}_{|i} +
C^j_i,
\end{eqnarray}
where $F$ is an arbitrary potential function of spatial coordinates
$x^1, x^2$ and $x^3, \ h^j_i = \delta^{jl}h_{li}, \ N^{|j}_{|i} =
\delta^{jl} N_{|li} = N_{,ij}, \ F_{,ij}$ means $\partial^2 F/\partial
x^i \partial x^j,$ and $P(\eta)$ and $Q(\eta)$ satisfy
\begin{eqnarray}
\label{eq:m5}
P'' &+& {2a' \over a} P' -1 = 0, \cr
Q'' &+& {2a' \over a} Q' = -\Bigl[P - {5 \over 2} (P')^2\Bigr].
\end{eqnarray}
The three-dimensional covariant derivative $|i$ are defined in the
space with metric $dl^2 = \delta_{ij} dx^i dx^j$ and their suffixes
are raised and lowered by use of $\delta_{ij}$.
The functions $L^j_i$ and $M^j_i$ are defined by
\begin{eqnarray}
\label{eq:m6}
L^j_i &=& {1 \over 2}\Bigl[-3 F_{,i} F_{,j} -2 F \cdot F_{,ij} + {1 \over 2}
\delta_{ij} F_{,l} F_{,l}\Bigr], \cr
M^j_i &=& {1 \over 28}\Big\{19F_{,il} F_{,jl} - 12 F_{,ij} \Delta F -
3\delta_{ij} \Bigl[F_{,kl} F_{,kl} -(\Delta F)^2 \Bigr]\Big\}
\end{eqnarray}
and $N$ is defined by
\begin{equation}
\label{eq:m7}
\Delta N = {1 \over 28} \Bigl[(\Delta F)^2 - F_{,kl}F_{,kl}\Bigr].
\end{equation}
The last term $C^l_i$ satisfies the wave equation
\begin{equation}
\label{eq:m8}
\Box C^j_i = {3 \over 14}(P/a)^2 G^j_i + {1 \over 7}\Bigl[P - {5 \over 2}
(P')^2 \Bigr] \tilde{G}^j_i,
\end{equation}
where $G^j_i$ and $\tilde{G}^j_i$ are second-order traceless and transverse
functions of spatial coordinates, and the operator $\Box$ is defined by
\begin{equation}
\label{eq:m9}
\Box \phi \equiv g^{\mu\nu} \phi_{;\mu\nu} = -a^{-2}
\Bigl(\partial^2/\partial\eta^2 + {2a' \over a}\partial/\partial \eta -
\Delta \Bigr) \phi,
\end{equation}
where $;$ denotes the four-dimensional covariant derivative.
So $C^j_i$ represents the second-order gravitational waves caused by
the first-order density perturbations.
The velocity perturbations $\mathop{\delta}_1 u^\mu$ and
$\mathop{\delta}_2 u^\mu$ vanish, i.e. \ $\mathop{\delta}_1 u^0 =
\mathop{\delta}_1 u^i =0$ and $\mathop{\delta}_2 u^0 =
\mathop{\delta}_2 u^i =0$, and the density perturbations are
\begin{eqnarray}
\label{eq:m10}
\mathop{\delta}_1 \rho/\rho &=& {1 \over \rho a^2} \Bigl({a'\over a}P'
-1\Bigr) \Delta F, \cr
\mathop{\delta}_2 \rho/\rho &=& {1 \over 2\rho a^2}\Bigl\{{1 \over
2}(1 - {a' \over a}P') (3F_{,l}F_{,l} + 8F\Delta F) +{1 \over 2}P
[(\Delta F)^2 + F_{,kl}F_{,kl}] \cr
&+& {1 \over 4}\Bigl[(P')^2 - {2 \over 7}{a'\over a}Q'\Bigr] [(\Delta F)^2 -
F_{,kl}F_{,kl}] - {1 \over 7} {a'\over a}PP' [4F_{,kl}F_{,kl} +
3(\Delta F)^2] \Big\}.
\end{eqnarray}
Next let us consider the CMB temperature $T = T^{(0)} (1 + \delta
T/T)$, in which $T^{(0)}$ is the background temperature and $\delta T/T
(= \mathop{\delta}_1 T/T + \mathop{\delta}_2 T/T)$ is the
perturbations. The
present temperature $T^{(0)}_o$ is related to the emitted background
temperature $T^{(0)}_e$ at the recombination epoch by $T^{(0)}_e = (1+
z_e) T^{(0)}_o$, the temperature perturbation $\tau \equiv (\delta
T/T)_e$ at the recombination epoch is determined by the physical state
before that epoch, and the present temperature perturbations $(\delta
T/T)_o$ is related to $(\delta T/T)_e$ by the gravitational
perturbations along the light ray from the recombination epoch to the
present epoch. The light ray in the unperturbed state is described
using the background wave vector
$k^\mu \ (\equiv dx^\mu/d \lambda)$, where $\lambda$ is the affine
parameter, and its component is $k^{(0)\mu} = (1, -e^i)$, and the ray
is given by $x^{(0)\mu} = [\lambda, (\lambda_0 - \lambda) e^i]$, where
$e^i$ is the directional unit vector. Here and in the following the
suffixes $o$ and $e$ for $\lambda, \eta$ and $r$ denote the present
(observed) epoch and the recombination (emitted) epoch, respectively.
The first-order temperature perturbation is
\begin{equation}
\label{eq:m11}
\mathop{\delta}_1 T/T = \tau + {1 \over 2}\int^{\lambda_e}_{\lambda_o}
d\lambda P'(\eta) F_{,ij} e^i e^j.
\end{equation}
Using the relations $dP/d\lambda = P'$ and $dF/d\lambda = - F_{,i}
e^i$, this equation (\ref{eq:m11}) is expressed as
\begin{equation}
\label{eq:m12}
\mathop{\delta}_1 T/T = \Theta_1 +\Theta_2
\end{equation}
where
\begin{eqnarray}
\label{eq:m13}
\Theta_1 &\equiv& \tau - {1 \over 2}[(P' F_{,i})_e - (P' F_{,i})_o]
e^i, \cr
\Theta_2 &\equiv& {1 \over 2}\int^{\lambda_e}_{\lambda_o}
d\lambda P''(\eta) F_{,i} e^i.
\end{eqnarray}
$\Theta_1$ and $\Theta_2$ represent the intrinsic and Sachs-Wolfe
effects, respectively. The latter can be divided into the ordinary
Sachs-Wolfe effect $\Theta_{sac}$ and the Integrated Sachs-Wolfe effect
$\Theta_{isw}$, where
\begin{eqnarray}
\label{eq:m14}
\Theta_{sac} &\equiv& {1 \over 2} [(P''F)_e - (P''F)_o], \cr
\Theta_{isw} &\equiv& {1 \over 2}\int^{\lambda_e}_{\lambda_o}
d\lambda P'''(\eta) F.
\end{eqnarray}
The second-order temperature perturbation is
\begin{eqnarray}
\label{eq:m15}
\mathop{\delta}_2 T/T &=& I_1 (\lambda_e) \Bigl[{1 \over 2} I_1 (\lambda_e)
- \tau\Bigr] - [{A^{(1)}_e}' + \tau_{,i} e^i] \int^{\lambda_e}_{\lambda_o}
d{\lambda} A^{(1)} \cr
&-& \int^{\lambda_e}_{\lambda_o} d{\lambda} \Big\{{1 \over 2} {A^{(2)}}'
+ A^{(1)}{A^{(1)}}' - {A^{(1)}}''
\int^{\lambda}_{\lambda_o} d\bar{\lambda} A^{(1)}(\bar{\lambda})
\Big\} + {\partial\tau \over \partial d^i} d^{(1)i},
\end{eqnarray}
where $(\eta, x^i) = (\lambda, \lambda_o - \lambda)$ in the
integrands and
\begin{eqnarray}
\label{eq:m16}
I_1 (\lambda_e) &=& - {1 \over 2} \int^{\lambda_e}_{\lambda_o}
d{\lambda} P' F_{,ij} e^i e^j, \cr
A^{(1)} &=& - {1 \over 2} P F_{,ij} e^i e^j , \cr
A^{(2)} &=& - {1 \over 2} [P L^j_i + P^2 M^j_i + Q N_{,ij} +
C^j_i ] e^i e^j.
\end{eqnarray}
These expressions were derived in \S 3 of a previous paper.\cite{tom1}
\section{Power spectra of second-order CMB anisotropies}
\label{sec:level3}
We consider primordial scalar perturbations with $F$ defined by
\begin{equation}
\label{eq:c1}
F = \int d{\bf k} \alpha ({\bf k}) e^{i{\bf kx}},
\end{equation}
where the spatial average for $\alpha ({\bf k})$ is given by
\begin{equation}
\label{eq:c2}
\langle \alpha ({\bf k}) \alpha ({\bf \bar{k}}) \rangle = (2\pi)^{-2}
{\cal P}_F ({\bf k}) \delta ({\bf k} + {\bf \bar{k}}),
\end{equation}
with
\begin{equation}
\label{eq:c3}
{\cal P}_F ({\bf k}) = {\cal P}_{F0} k^{-3} (k/k_0)^{n-1} T_s^2 (k),
\end{equation}
where $T_s (k)$ is the matter transfer function \cite{sugi} and ${\cal
P}_{F0}$ is the normalization constant.
Then the first-order temperature perturbations are
\begin{equation}
\label{eq:c4}
\mathop{\delta}_1 T/T \equiv \Theta_P = -{1 \over 2} \int
d{\bf k} \alpha ({\bf k})
\int^{\lambda_e}_{\lambda_o} d\lambda P' (\eta) (k\mu)^2 e^{i{\bf
kx}},
\end{equation}
where ${\bf x} = r{\bf e}, {\bf kx} = kr\mu, \mu \equiv \cos
\theta_k$ and $\theta_k$ is the angle between the wave vector $k^i$
and a unit vector $e^i$.
This equation can be rewritten as
\begin{equation}
\label{eq:c5}
\Theta_P = \int d{\bf k} \alpha ({\bf k}) \Bigl\{- {1 \over 2}[(P'')_o +
ik (P')_o P_1(\mu)] + {1 \over 2} \sum_l (-i)^l (2l+1) {\Theta}_{P(l)}
P_l (\mu) \Bigr\},
\end{equation}
where $P_l (\mu)$ is the Legendre polynomial and
\begin{equation}
\label{eq:c6}
{\Theta}_{P(l)} = \int^{\lambda_e}_{\lambda_o} d\lambda P'''
(\eta) j_l(kr) - \{k(P')_e (2l+1)^{-1}[(l+1) j_{l+1} (kr_e) - l
j_{l-1}(kr_e)] + j_l (kr_e) (P'')_e \}.
\end{equation}
In these equations, we have $\eta = \lambda$ and $r = \lambda_o -
\lambda$. In the derivation of Eq.(\ref{eq:c6}), we used the
relations\cite{zal,hu}
\begin{equation}
\label{eq:c7}
e^{i{\bf kx}} = e^{ikr\mu} = \sum_l (-i)^l (2l+1) j_l (kr)
P_l (\mu)
\end{equation}
and
\begin{equation}
\label{eq:c8}
(2l +1) \mu P_l(\mu) = (l+1) P_{l+1}(\mu) + l P_{l-1}(\mu).
\end{equation}
The components of the unit vector $e^i$ are expressed as $e^1 = \sin
\theta \cos \phi,
e^2 = \sin \theta \sin \phi, e^3 = \cos \theta$ with respect to $x^1,
x^2, x^3$ axes, respectively.
In order to derive the power spectra, we take
the statistical average $\langle \rangle$ for the primordial
perturbations, and $\langle (\delta T/T)^2
\rangle$ is expressed for the first-order anisotropies as
\begin{equation}
\label{eq:c9}
\langle (\mathop{\delta}_1 T/T)^2 \rangle = \langle (\Theta_P)^2
\rangle = (T_0)^{-2} \sum_l {2l+1 \over 4\pi} C_l.
\end{equation}
The power spectra $C_l$ are
\begin{equation}
\label{eq:c10}
C_l = (T_0)^2 \int dk k^2 {\cal P}_F (k)
|{\cal H}_P^{(l)} (k) |^2,
\end{equation}
where $T_0$ is the present CMB temperature and
\begin{eqnarray}
\label{eq:c10a}
{\cal H}_P^{(0)} (k) &=& -(P'')_o - k(P')_e j_1(kr_e) - (P'')_e j_0(kr_e)
+ \int^{\lambda_e}_{\lambda_o} d\lambda P''' j_0(kr), \cr
{\cal H}_P^{(1)} (k) &=& {1 \over 3}k[(P')_o - (P')_e] j_1^{(1)}(kr_e)
-(P'')_e j_1 (kr_e) + \int^{\lambda_e}_{\lambda_o} d\lambda P''' j_1
(kr).
\end{eqnarray}
For $l \geq 2$, we have
\begin{equation}
\label{eq:c11}
{\cal H}_P^{(l)} (k) = k (P')_e j_l^{(1)} (kr_e) - (P'')_e j_l (kr_e) +
\int^{\lambda_e}_{\lambda_o} d\lambda P''' j_l (kr).
\end{equation}
In the derivation of $C_l$, we used Eq.(\ref{eq:c2}) for $\langle
\alpha ({\bf k}) \alpha ({\bf \bar{k}}) \rangle$ and a mathematical
formula for $P_l (\mu)$:
\begin{equation}
\label{eq:c11a}
\ \int^1_{-1} P_l (\mu) P_{l'} (\mu) d\mu = \ 2/(2l+1), \ 0 \quad
{\rm for} \quad l=l', \ l \ne l',
\end{equation}
respectively.
For the two directions with unit vectors ${\bf e}_{1}$ and ${\bf
e}_{2}$, we have the correlation
\begin{equation}
\label{eq:c12}
(T_0)^{2} \langle \Theta_P({\bf e}_{1}) \Theta_P({\bf e}_{2}) \rangle
= \sum_l {2l+1 \over 4\pi} C_l P_l (\cos \beta).
\end{equation}
where the product ${\bf e}_{1} {\bf e}_{2}$ is equal to $\cos \beta$.
For the second-order temperature anisotropies, we obtain from
Eqs.(\ref{eq:m15}), (\ref{eq:m16}) and (\ref{eq:c1})
\begin{eqnarray}
\label{eq:c13}
\mathop{\delta}_2 T/T &=& \int\int d{\bf k}d\bar{\bf k} \alpha
({\bf k}) \alpha (\bar{\bf k}) \Bigl\{{1\over 8}
\int^{\lambda_e}_{\lambda_o}
\int^{\lambda_e}_{\lambda_o} d\lambda d\bar{\lambda}
P'(\lambda) P'(\bar{\lambda}) (k_e \bar{k}_e)^2 e^{i({\bf k}
{\bf x}+ \bar{\bf k} \bar{\bf x})} \cr
&+& \int^{\lambda_e}_{\lambda_o} d\lambda \Bigl[{1\over 8}
P'(\lambda) \Bigl(3k_e \bar{k}_e +(k_e)^2 +(\bar{k}_e)^2 - {1\over
2}{\bf k} \bar{\bf k}\Bigr) \cr
&+& {1\over 56}
P(\lambda) P'(\lambda)\ \Bigl(19k_e \bar{k}_e {\bf k}\bar{{\bf k}} -
14 (k_e \bar{k}_e)^2 -6(\bar{k}_e k)^2 -6(k_e \bar{k})^2
- 3({\bf k}\bar{\bf k})^2 + 3k^2 (\bar{k})^2
\Bigr) \cr
&+& {1 \over 112} Q'(\lambda) (k_e+\bar{k}_e)^2 \Bigl(k^2\bar{k}^2 -
({\bf k}{\bf \bar{k}})^2\Bigr)/({\bf k}+{\bf \bar{k}})^2 \Bigr]
e^{i({\bf k} +\bar{{\bf k}}){\bf x}} \cr
&+& {1\over 4} \int^{\lambda_e}_{\lambda_o} d \lambda
\Bigl[P''(\lambda) \int^\lambda_{\lambda_o} d\bar{\lambda}
P({\bar{\lambda}}) (k_e \bar{k}_e)^2 \Bigr] e^{i({\bf k}{\bf
x}+ \bar{\bf k} {\bf \bar{x}})} \Bigr\},
\end{eqnarray}
where $k_e$ stands for ${\bf k} {\bf e}, \ P'(\lambda) \equiv
dP(\lambda)
/d\lambda$, and $P'(\bar{\lambda}) \equiv dP(\bar{\lambda})
/d\bar{\lambda}$. Here $\tau$ and ${A_e^{(1)}}'$ in Eq.(\ref{eq:m15})
were neglected, because we pay attentions to the Sachs-Wolfe effect
after the recombination
epoch, and also the term with $C^j_i$ in Eq.(\ref{eq:m16}) was
neglected, because the
contribution of gravitational radiation is very small. It is found
from the above equation that the
average $\langle \mathop{\delta}_2 T/T \rangle$ does not vanish in
contrast to the vanishing first-order one ($\langle \mathop{\delta}_1
T/T \rangle$), and we obtain
\begin{eqnarray}
\label{eq:c14}
\langle \mathop{\delta}_2 T/T \rangle &=& {1\over 8} \int d{\bf k}
(2\pi)^{-2}
{\cal P}_F ({\bf k}) \Bigl\{ \int^{\lambda_e}_{\lambda_o}
\int^{\lambda_e}_{\lambda_o} d\lambda d\bar{\lambda}
P'(\lambda) P'(\bar{\lambda}) (k_e)^4 e^{i{\bf k}({\bf x}
-\bar{{\bf x}})} \cr
&+& \int^{\lambda_e}_{\lambda_o} d\lambda \Bigl[ P'(\lambda)
\Bigl( -(k_e)^2 +
{1\over 2} k^2\Bigr) + P(\lambda) P'(\lambda) \Bigl(k^2
-2(k_e)^2\Bigr)(k_e)^2\Bigr] \cr
&+& 2 \int^{\lambda_e}_{\lambda_o} d\lambda
P''(\lambda) \int^\lambda_{\lambda_o} d\bar{\lambda}
P(\bar{\lambda}) (k_e)^4 e^{i{\bf k}({\bf x}
-\bar{{\bf x}})} \Bigr\},
\end{eqnarray}
where $k_e \equiv {\bf k} {\bf e} = k\mu$ and $d{\bf k} = dk k^2 d\phi
d\mu$. Here we have the
relations ${\bf x} = r{\bf e}, \ \bar{{\bf x}} = \bar{r}{\bf e}$, and
so ${\bf kx} = k_e r = kr \mu, \ - {\bf k\bar{x}} = k \bar{r} (-\mu)$.
Thus $e^{i{\bf k}({\bf x} -\bar{{\bf x}})}$ is expanded as
\begin{eqnarray}
\label{eq:c15}
e^{i{\bf k}({\bf x} -\bar{{\bf x}})} &=& \sum_l (-i)^l (2l+1) j_l (kr)
P_l(\mu) \times \sum_{l'}(-i)^{l'} (2l'+1) j_{l'} (k\bar{r}) P_{l'}
(-\mu)\cr
&=& \sum_{l,l'} (-i)^{l}\ i^{l'}\ (2l+1)(2l'+1)\ j_l (kr) j_{l'}
(k\bar{r}) \ P_l(\mu) P_{l'}(\mu).
\end{eqnarray}
Here for the reduction of Eq.(\ref{eq:c14}), we derive the following
relations using Eq.(\ref{eq:c8})
\begin{equation}
\label{eq:c16}
\sum_l j_l\ \mu P_l = \sum_l j_l^{(1)} P_l, \qquad
\sum_l j_l\ \mu^2 P_l = \sum_l j_l^{(2)} P_l, \qquad {\rm and} \qquad
\sum_l j_l\ \mu^3 P_l = \sum_l j_l^{(3)} P_l,
\end{equation}
where
\begin{eqnarray}
\label{eq:c17}
j_l^{(1)}(kr) &\equiv& {l \over 2l-1} j_{l-1} (kr) + {l+1 \over 2l+3}
j_{l+1} (kr), \cr
j_l^{(2)}(kr) &\equiv& {(l-1)l \over (2l-3)(2l-1)}j_{l-2} (kr) +
{1\over 2l+1} \Bigl[{l^2 \over 2l-1}
+ {(l+1)^2\over (2l+3)}\Bigr] j_l
(kr) + {(l+1)(l+2) \over (2l+3)(2l+5)} j_{l+2} (kr), \cr
j_l^{(3)}(kr) &\equiv& {(l-2)(l-1)l \over (2l-5)(2l-3)(2l-1)} j_{l-3}
(kr) + {l\over 2l-1} \Bigl[{(l-1)^2 \over (2l-3)(2l-1)}\cr
&+& {l^2\over (2l-1)(2l+1)} +{(l+1)^2 \over (2l+1)(2l+3)} \Bigr]
j_{l-1} (kr) + {l+1 \over 2l+3} \Bigl[{l^2 \over (2l-1)(2l+1)}\cr
&+& {(l+1)^2\over (2l+1)(2l+3)} +{(l+2)^2 \over (2l+3)(2l+5)} \Bigr]
j_{l+1} (kr)
+ {(l+1)(l+2)(l+3) \over (2l+3)(2l+5)(2l+7)} j_{l+3} (kr).
\end{eqnarray}
Then we have
\begin{equation}
\label{eq:c18}
\mu^4 e^{i{\bf k}({\bf x} -\bar{{\bf x}})} = \sum_{l,l'} (-i)^l
i^{l'} (2l+1) (2l'+1) j_l^{(2)} (kr) j_{l'}^{(2)} (k\bar{r}) P_l(\mu)
P_{l'} (\mu).
\end{equation}
Using these relations and executing the integrations in
Eq.(\ref{eq:c14}) with respect to $\phi$ and $\mu$, we obtain
\begin{eqnarray}
\label{eq:c19}
\langle \mathop{\delta}_2 T/T \rangle &=& (2\pi)^{-1} \int dk k^2
{\cal P}_F ({\bf k}) \Bigl\{{1\over 4} k^4 \sum_l (2l+1)
\Bigl[\int^{\lambda_e}_{\lambda_o}
d\lambda P'(\lambda) j_l^{(2)} (kr)\Bigr]^2 \cr
&+& {1\over 24} (P_e - P_o) k^2 + {17\over 840} (P_e^2 -P_o^2)
k^4 \cr
&+& {1\over 2} k^4 \sum_l (2l+1) \int^{\lambda_e}_{\lambda_o} d\lambda
P''(\lambda)
j_l^{(2)} (kr) \int^\lambda_{\lambda_o} d\bar{\lambda}
P(\bar{\lambda}) j_l^{(2)}(k\bar{r}) \Bigr\},
\end{eqnarray}
where we used the relation (\ref{eq:c11a}), and $P_e = P(\lambda_e)$ and
$P_o = P(\lambda_o)$.
When we take into account also the third-order anisotropies
$\mathop{\delta}_3 T/T$, we have
\begin{eqnarray}
\label{eq:c20a}
\mathop{\delta} T/T &=& \mathop{\delta}_1 T/T + \mathop{\delta}_2 T/T
+ \mathop{\delta}_3 T/T \cr
&=& \mathop{\delta}_1 T/T + \langle \mathop{\delta}_2 T/T \rangle
+ \Theta_{pp} + \mathop{\delta}_3 T/T,
\end{eqnarray}
where $\Theta_{pp} \equiv \mathop{\delta}_2 T/T - \langle
\mathop{\delta}_2 T/T \rangle$.
Then the total average of $(\delta T/T)^2$ is expressed as
\begin{equation}
\label{eq:c21}
\langle (\mathop{\delta} T/T)^2 \rangle =
\langle \Theta_p^2 \rangle + (\langle \mathop{\delta}_2 T/T
\rangle)^2 + \langle \Theta_{pp}^2 \rangle + 2 \langle
\mathop{\delta}_1 T/T \mathop{\delta}_3 T/T \rangle.
\end{equation}
The last term is of the same order as the second and third terms.
At present we have not obtained any concrete expression of third-order
metric perturbations yet and so the last term is not analyzed,
while the formula for $\mathop{\delta}_3 T/T$ and formal solutions
have recently been derived by
D'Amico et al.\cite{dam} and the perturbative equations to the third
order were studied by Hwang and Noh\cite{hwang} with respect to the
relativistic-Newtonian correspondence.
Here $\langle \mathop{\delta}_2 T/T
\rangle$ is the monopole component without angular dependence and
$\Theta_{pp}$ is the renormalized second-order temperature
fluctuation, which have been discussed by Munshi et al.\cite{munshi}
Now let us make a reduction of $\langle \Theta_{pp}^2 \rangle$. It is
expressed as
\begin{equation}
\label{eq:c22}
\langle \Theta_{pp}^2 \rangle = \langle \Bigl(\mathop{\delta}_2
T/T(\alpha({\bf k})\alpha(\bar{\bf k})) - \langle \mathop{\delta}_2 T/T
\rangle \Bigr) \Bigl(\mathop{\delta}_2 T/T(\alpha(\bar{\bar{\bf
k}})\alpha(\bar{\bar{\bar{\bf k}}})) - \langle \mathop{\delta}_2 T/T
\rangle \Bigr) \rangle,
\end{equation}
where $\mathop{\delta}_2 T/T(\alpha({\bf k})\alpha(\bar{\bf k}))$ is
given
by Eq.(\ref{eq:c13}) and $\mathop{\delta}_2 T/T(\alpha(\bar{\bar{\bf
k}})\alpha(\bar{\bar{\bar{\bf k}}}))$ is obtained from
Eq.(\ref{eq:c13}) by
replacing ${\bf k}, \bar{\bf k}$ by $ \bar{\bar{\bf k}},\
\bar{\bar{\bar{\bf k}}}$. The averaging process in Eq.(\ref{eq:c22})
is performed in the two sets:
$\langle \alpha({\bf k})\alpha(\bar{\bar{\bf k}}) \rangle, \langle
\alpha(\bar{\bf k})\alpha(\bar{\bar{\bar{\bf k}}}) \rangle$ and
$\langle \alpha({\bf
k})\alpha(\bar{\bar{\bar{\bf k}}}) \rangle, \langle \alpha(\bar{\bf k})
\alpha(\bar{\bar{\bf k}}) \rangle$, and the average process for
$\langle \alpha({\bf k})\alpha(\bar{\bf k}) \rangle$ and $\langle
\alpha(\bar{\bar{\bf k}})\alpha(\bar{\bar{\bar{\bf k}}}) \rangle$ is
excluded by subtracting $\langle \mathop{\delta}_2 T/T
\rangle$ from $\mathop{\delta}_2 T/T(\alpha({\bf k})\alpha(\bar{\bf
k}))$ and $\mathop{\delta}_2 T/T(\alpha(\bar{\bar{\bf
k}})\alpha(\bar{\bar{\bar{\bf k}}}))$.
When we sum these average values in the above two sets
and symmetrize the expression with respect to ${\bf k}$ and $\bar{\bf
k}$, we obtain from Eq.(\ref{eq:c22})
\begin{eqnarray}
\label{eq:c23}
\langle \Theta_{pp}^2 \rangle &=& 2 \int\int dk d\bar{k} (k\bar{k})^2
d\phi_k d\bar{\phi}_k d\mu d\bar{\mu} (2\pi)^{-4}{\cal P}_F(k){\cal
P}_F(\bar{k})\cr
&\times& \Bigl\{{1\over 8} \int^{\lambda_e}_{\lambda_o}
\int^{\lambda_e}_{\lambda_o} d\lambda d\bar{\lambda} P'(\lambda)
P'(\bar{\lambda}) (k_e \bar{k}_e)^2 e^{i({\bf kx} +{\bf
\bar{k}\bar{x}})}
+ \int^{\lambda_e}_{\lambda_o} d\lambda \Bigl[{1\over 8} P'(\lambda)
\Bigl(3k_e\bar{k}_e +(k_e)^2+ (\bar{k}_e)^2 -{1\over 2}{\bf
k\bar{k}}\Bigr) \cr
&+& {1\over 56} P(\lambda) P'(\lambda)
\Bigl(19k_e \bar{k}_e {\bf k\bar{k}} - 14(k_e \bar{k}_e)^2
-6(\bar{k}_e k)^2 -6(k_e \bar{k})^2
- 3({\bf k\bar{k}})^2+ 3k^2 (\bar{k})^2\Bigr) \Bigr] e^{i({\bf k}
+{\bf \bar{k}}){\bf x}} \cr
&+& {1\over 4} \int^{\lambda_e}_{\lambda_o} d\lambda P''(\lambda)
\int^\lambda_{\lambda_o}
P(\bar{\lambda}) (k_e \bar{k}_e)^2 e^{i({\bf kx} +{\bf
\bar{k}\bar{x}})} \Bigr\}\cr
&\times& \Bigl\{{1\over 8} \int^{\lambda_e}_{\lambda_o}
\int^{\lambda_e}_{\lambda_o} d\eta d\bar{\eta} P'(\eta)
P'(\bar{\eta}) (k_e \bar{k}_e)^2 e^{-i({\bf k y} +{\bf
\bar{k}\bar{y}})} + \int^{\lambda_e}_{\lambda_o} d\eta \Bigl[{1\over
8} P'(\eta) \Bigl(3k_e\bar{k}_e +(k_e)^2+
(\bar{k}_e)^2 -{1\over 2}{\bf k\bar{k}}\Bigr) \cr
&+& {1\over 56} P(\eta)P'(\eta) \Bigl(19k_e \bar{k}_e {\bf k\bar{k}}
- 14(k_e \bar{k}_e)^2 -6(\bar{k}_e k)^2 -6(k_e \bar{k})^2 - 3({\bf
k\bar{k}})^2
+ 3k^2 (\bar{k})^2\Bigr) \Bigr] e^{-i({\bf k} +{\bf \bar{k}})
{{\bf y}}} \cr
&+& {1\over 4} \int^{\lambda_e}_{\lambda_o} d\eta P''(\eta)
\int^\eta_{\lambda_o} d\bar{\eta}
P(\bar{\eta}) (k_e \bar{k}_e)^2 e^{-i({\bf k y} +{\bf
\bar{k} \bar{y}})} \Bigr\},
\end{eqnarray}
where ${\bf x, \bar{x}}$ and ${\bf y, \bar{y}}$ are functions of
$\lambda, \bar{\lambda}$ and $\eta, \bar{\eta}$, respectively, and we
neglected the terms with $Q$ and $ N^{|j}_{|i}$,
because we have $Q/P^2 < 10^{-2}$ always and they are very small.
Especially $Q$ vanishes in the case $\Lambda = 0$.
Here, when we consider an orthonormal triad vector $e^i_{(1)},
e^i_{(2)}$, and $e^i_{(3)}\ (= e^i)$, the components of ${\bf k}$ and
${\bf \bar{k}}$ with respect to this triad are expressed as ${\bf k} = k
(\sin \theta_k \cos \phi_k, \sin \theta_k \sin \phi_k, \cos
\theta_k)$ and ${\bf \bar{k}} = \bar{k}
(\sin \bar{\theta}_k \cos \bar{\phi}_k, \sin \bar{\theta}_k \sin
\bar{\phi}_k, \cos \bar{\theta}_k)$, and so $\mu = \cos \theta_k$ and
$\bar{\mu} = \cos \bar{\theta}_k$.
Next let us take a notice of terms with ${\bf k\bar{k}}$ and $({\bf
k\bar{k}})^2$ which can be expressed as
\begin{eqnarray}
\label{eq:c24}
{\bf k\bar{k}} &=& k\bar{k} \ [\sin \theta_k \sin \bar{\theta}_k \cos
(\phi_k -\bar{\phi}_k) + \cos \theta_k \cos \bar{\theta}_k], \cr
({\bf k\bar{k}})^2 &=& (k\bar{k})^2 \ \{(\cos \theta_k \cos
\bar{\theta}_k)^2 +{1\over 2}
(\sin \theta_k \sin \bar{\theta}_k)^2 [1 + \cos 2(\phi_k
-\bar{\phi}_k)] \cr
&+& 2 \sin \theta_k \sin \bar{\theta}_k \cos \theta_k \cos \bar{\theta}_k
\cos (\phi_k -\bar{\phi}_k)\}.
\end{eqnarray}
By integrations with respect to $\phi_k$ and
$\bar{\phi}_k$, we obtain $\int\int d\phi_k d\bar{\phi}_k \cos
(\phi_k -\bar{\phi}_k) = \int\int d\phi_k
d\bar{\phi}_k \cos 2(\phi_k -\bar{\phi}_k) = 0$, while $\int\int
\phi_k \bar{\phi}_k [\cos (\phi_k -\bar{\phi}_k)]^2 = \int\int d\phi_k
d\bar{\phi_k} [\cos 2(\phi_k -\bar{\phi}_k)]^2 = (2\pi)^2/2$.
Executing integrations in Eq.(\ref{eq:c23}) with respect to
$\phi_k$ and $\bar{\phi}_k$, and using the relations Eq.(\ref{eq:c7}) and
\begin{equation}
\label{eq:c25}
e^{-i{\bf ky}} = e^{-iks\mu} = \sum_m (-i)^m (2m+1) j_m (ks)
P_m (-\mu) = \sum_m i^m (2m+1) j_m (ks) P_m (\mu),
\end{equation}
we can, therefore, reexpress Eq.(\ref{eq:c23}) as
\begin{equation}
\label{eq:c26}
\langle (\Theta_{pp})^2 \rangle = \sum_l \sum_{l'} \sum_m
\sum_{m'} \Bigl(A^I_{ll'mm'} + A^{II}_{ll'mm'} +
A^{III}_{ll'mm'}\Bigr),
\end{equation}
where the terms $A^I_{ll'mm'}, A^{II}_{ll'mm'}$ and $A^{III}_{ll'mm'}$
come from the terms without $\phi_k$ and $\bar{\phi}_k$, the
coefficients of $\cos^2 (\phi_k -\bar{\phi}_k)$, and the
coefficients of $\cos^2 2(\phi_k -\bar{\phi}_k)$, and their lengthy
expressions are shown in Appendix A.
Moreover let us replace the terms of $\mu j_l P_l, \ \mu^2 j_l P_l$
and $\mu^3 j_l P_l$ by $j_l^{(1)} P_l, \ j_l^{(2)} P_l$ and $j_l^{(3)}
P_l$ using Eqs. (\ref{eq:c16}) and
(\ref{eq:c17}) and execute integrations with respect to $\mu$ and
$\bar{\mu}$ using Eq.(\ref{eq:c11a}). Then we obtain
\begin{equation}
\label{eq:c27}
\langle (\Theta_{pp})^2 \rangle = \sum_l \sum_{l'}
\Bigl(B^I_{ll'} +B^{II}_{ll'}+B^{III}_{ll'}\Bigr),
\end{equation}
where
\begin{eqnarray}
\label{eq:c29}
B^I_{ll'} &=& {1\over 8}(2\pi)^{-2} (2l+1) (2l'+1)
\int\int dk d\bar{k}
(k\bar{k})^2 {\cal P}_F(k){\cal P}_F(\bar{k}) \cr
&\times& \Bigl\{(k\bar{k})^2 \int^{\lambda_e}_{\lambda_o}
\int^{\lambda_e}_{\lambda_o} d\lambda d\bar{\lambda}
P'(\lambda)
P'(\bar{\lambda}) j_l^{(2)}(kr) j_{l'}^{(2)}(\bar{k}\bar{r})
+ 2 (k\bar{k})^2 \int^{\lambda_e}_{\lambda_o} d\lambda P''(\lambda)
\int^{\lambda}_{\lambda_o} d\bar{\lambda} P(\bar{\lambda})j_l^{(2)}(kr)
j_{l'}^{(2)}(\bar{k}\bar{r}) \cr
&+& \int^{\lambda_e}_{\lambda_o} d\lambda \Bigl[P'(\lambda)
\Bigl({5\over 2}k\bar{k} j_l^{(1)}(kr)
j_{l'}^{(1)}(\bar{k}r) + k^2 j_l^{(2)}(kr)
j_{l'} (\bar{k}r) + \bar{k}^2 j_l(kr) j_{l'}^{(2)} (\bar{k}r)\Bigr) \cr
&+& {3\over 14}P(\lambda)P'(\lambda) (k\bar{k})^2 \Bigl(
j_l(kr) j_{l'}(\bar{k}r) -3j_l^{(2)}(kr)
j_{l'}(\bar{k}r) -3j_l(kr) j_{l'}^{(2)}(\bar{k}r) \cr
&+& {1\over 3} j_l^{(2)}(kr) j_{l'}^{(2)}(\bar{k}r) \Bigr) \Bigr]
\Bigr\}^2,
\end{eqnarray}
\begin{eqnarray}
\label{eq:c30}
B^{II}_{ll'} &=& {1\over 32}(2\pi)^{-2} (2l+1) (2l'+1)
\int\int dk d\bar{k} (k\bar{k})^4 {\cal P}_F(k){\cal P}_F(\bar{k})
\cr
&\times& \int^{\lambda_e}_{\lambda_o} d\lambda P'(\lambda)
\Bigl[\Bigl(j_l(kr) -j_l^{(2)}(kr) \Bigr)
-{26\over 7} k\bar{k} P(\lambda) \Bigl(j_l^{(1)}(kr)
-j_l^{(3)}(kr) \Bigr) \Bigl] j_{l'}(kr) \cr
&\times& \int^{\lambda_e}_{\lambda_o} d\eta P'(\eta)
\Bigl[ \Bigl(j_{l'}(\bar{k}s)
-j_{l'}^{(2)}(\bar{k}s) \Bigr) -{26\over 7} k\bar{k} P(\eta)
\Bigl(j_{l'}^{(1)}(\bar{k}s) -j_{l'}^{(3)}(\bar{k}s) \Bigr) \Bigl]
j_l (\bar{k}s),
\end{eqnarray}
\begin{eqnarray}
\label{eq:c31}
B^{III}_{ll'} &=& 2 \Bigl({3\over 56}\Bigr)^2(2\pi)^{-2}
(2l+1) (2l'+1) \int\int dk d\bar{k} (k\bar{k})^6 {\cal
P}_F(k){\cal P}_F(\bar{k}) \cr
&\times& \Big\{\int^{\lambda_e}_{\lambda_o} d\lambda
P(\lambda)P'(\lambda) \Bigl[j_l(kr) j_{l'}(\bar{k}r) -
j_l^{(2)} (kr) j_{l'}(\bar{k}r) -j_l(kr) j_{l'}^{(2)} (\bar{k}r)
+j_l^{(2)}(kr) j_{l'}^{(2)} (\bar{k}r) \Bigr] \Bigr\}^2,
\end{eqnarray}
where $r=\lambda_o -\lambda,\ \bar{r} = \lambda_o -\bar{\lambda},\ s
= \eta_o - \eta, \ \eta_e =
\lambda_e$, \ and \ $\eta_o = \lambda_o$.
Next let us consider two directions with unit directional vectors
${\bf e}_1$ and ${\bf e}_1$. If we define $\mu_1, \mu_2$ and $\beta$
as $ \mu_1 = {\bf k}{\bf e}_1/k,\ \mu_2 = {\bf k}{\bf e}_2/k$ \
and $\cos \beta = {\bf e}_1 {\bf e}_2$, respectively, we have a
mathematical relation
\begin{equation}
\label{eq:c36}
\int \int d \phi_k d\theta_k \sin \theta_k P_l (\mu_1) P_l (\mu_2) =
{4\pi \over 2l+1} P_l (\cos \beta).
\end{equation}
Using Eq.(\ref{eq:c36}) it is found that the correlation between
$\Theta_{pp}$'s in the two directions is expressed as
\begin{equation}
\label{eq:c37}
\langle \Theta_{pp}({\bf e}_1) \Theta_{pp}({\bf e}_2) \rangle
= {1\over 2} \sum_l \sum_{l'} \Bigl(B^I_{ll'} +B^{II}_{ll'}
+B^{III}_{ll'}\Bigr) P_l (\cos \beta) P_{l'} (\cos \beta).
\end{equation}
Here let us expand $P_l P_{l'}$ by $P_n$ as
\begin{equation}
\label{eq:c38}
P_l(\cos \beta) P_{l'}(\cos \beta) = \sum_n b_{ll'n} P_n(\cos \beta).
\end{equation}
The derivation of the coefficient $b_{ll'n}$ is shown in Appendix B.
Then the correlation reduces to
\begin{equation}
\label{eq:c39}
(T_0)^2 \langle \Theta_{pp}({\bf e}_1) \Theta_{pp}({\bf e}_2) \rangle
= \sum_n {2n+1 \over 4\pi} C_n^{(2)} P_n(\cos \beta),
\end{equation}
where
\begin{equation}
\label{eq:c40}
C_n^{(2)} = {4\pi \over 2n+1}(T_0)^2 \sum_{l,l'}
\Bigl(B^I_{ll'} +B^{II}_{ll'} +B^{III}_{ll'} \Bigr) b_{ll'n}.
\end{equation}
The expressions for $\langle \mathop{\delta}_2 T/T \rangle, \ \langle
(\Theta_{pp})^2 \rangle$ and $\langle
\Theta_{pp}({\bf e}_1) \Theta_{pp}({\bf e}_2) \rangle$ in
Eqs.(\ref{eq:c14}), (\ref{eq:c27}) and (\ref{eq:c39}) are our new
result which will
be useful to derive the second-order power spectra.
\section{Concluding remarks}
In this paper we derived the average value $\langle \mathop{\delta}_2
T/T \rangle$ and the second-order power spectra $C_n^{(2)}$ of CMB
anisotropies due to primordial random density perturbations, which
include two random variables $\alpha ({\bf k})$ and $\alpha ({\bf
\bar{k}})$, using the average values of the products of $\alpha
({\bf k})$. The average value $\langle \mathop{\delta}_2
T/T \rangle$ does not vanish and has no angular
dependence. It should be regarded as the monopole component of
temperature fluctuations and the second-order
angular correlation is described by the power spectra $C_n^{(2)}$.
Since we have not derived $\langle \mathop{\delta}_1 T/T
\mathop{\delta}_3 T/T \rangle$ yet, our analysis of second-order power
spectra is incomplete. But we think $\langle \Theta_{pp}^2 \rangle$ may
represent the essential feature of second-order power spectra, that is,
their $l$-dependence. In the next step we will analyze $\langle
\mathop{\delta}_1 T/T \mathop{\delta}_3 T/T \rangle$ with the
third-order metric perturbations for completeness.
In the case $\Lambda = 0$, the first-order temperature fluctuations
due to the integrated Sachs-Wolfe effect vanish, while the second-order
ones do not vanish and play a dominant role in the integrated Sachs-Wolfe
effect, which should not be neglected. In the cosmological local void
model (LVM)\cite{lvm,cele,aln,bis}, the exterior region is
described in terms of the Einstein-de Sitter model, and so the main
part of the integrated Sachs-Wolfe effect in LVM is of second-order.
In the interior region we use open low-density models with
$\Omega_0 < 1$, in which we have nonzero first-order Sachs-Wolfe
effect. It is important for showing the observational reality of LVM
to derive them.
In the case $\Lambda \ne 0$, the first-order temperature fluctuations
due to the integrated Sachs-Wolfe effect decrease rapidly with the
increase of the redshift $z$, but the second-order temperature
fluctuations due to the
integrated Sach-Wolfe effect decrease more slowly, and so the latter
fluctuations may be dominant over the first-order fluctuations at the
early stage. This situation is explained in a separate paper\cite{ti}
using a simple model of density perturbations. Thus a characteristic
behavior of CMB power spectra due to second-order temperature
fluctuations will be measured
through the future precise observation and bring useful informations
on the structure and evolution of our universe in the future.
\begin{acknowledgments}
The author thanks K.T. Inoue and referees for helpful discussions and
comments.
\end{acknowledgments}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,529 |
{"url":"https:\/\/brilliant.org\/problems\/good-one-for-all\/","text":"# Good one for all\n\nThe probability that a student passes at least in one of the three examinations A,B,C is 0.75.\n\nThe probability that he passes in at least two of the exams is 0.5.\n\nAnd the prabability he passes in exactly two of the exams is 0.4.\n\nDenote $$p,q,r$$ as the probabilities of the student passing in A,B,C respectively.\n\nLet $$p+q+r=\\frac xy$$ for coprime positive integers $$x,y$$, find the value of $$x-y$$.\n\n\u00d7","date":"2017-10-19 07:27:54","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\": 0.8944158554077148, \"perplexity\": 496.1061286645804}, \"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-2017-43\/segments\/1508187823255.12\/warc\/CC-MAIN-20171019065335-20171019085335-00116.warc.gz\"}"} | null | null |
Jean-Jacques Grasset (c.1769 – 25 August 1839) was a French classical violinist.
He was born in Paris about 1769, and was a pupil of Isidore Bertheaume. After several years' obligatory service in the army, he soon became well-known on his return. On the death of Pierre Gaviniès in 1800 he was appointed professor of the violin at the Conservatoire de Paris. Soon afterwards he succeeded Antonio Bartolomeo Bruni as chef d'orchestre at the Italian Opera, holding the post until 1829, when he retired from public life. He died in Paris in 1839.
He published three Concertos for the Violin, five books of Violin-Duos, and a Sonata for Piano and Violin.
References
Attribution
External links
1769 births
1839 deaths
Musicians from Paris
18th-century French male classical violinists
19th-century French male classical violinists
Academic staff of the Conservatoire de Paris | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 9,701 |
Q: How to escape "this" problem in a javascript class? get_O3(e)
{
e.preventDefault();
let station = document.getElementById(e.target.id);
let lon = station.getAttribute('lon');
let lat = station.getAttribute('lat');
let code_station = station.getAttribute('code');
this.get_previsions(lon, lat, "O3").bind(this).then((data) =>
{
console.log(data);
});
}
I have a "this" problem, when i call the function get_previsions i get the error :
Uncaught TypeError: this.get_previsions is not a function.
It might be because of the (e) parameter because when i do a console.log(this) it returns the target. I would like that this == my class.
Thanks for the help
A: At any given point you can check what the current this reference is pointing to by doing the 4 following rules:
*
*New: Was the function called using new then the this points to the new instance.
*Explicit Binding: Was the function called using Function#call, Function#apply or Function#bind
*Implicit Binding: Was the function called by its owner? (i.e. owner.foo() or owner["bar"]())
*Default Rule: If none of the other rules happen then this is set to the window object if the script is running in "use strict" mode otherwise undefined.
Event-listeners call a function using Explicit binding (callBack.call(target, ...)) so the this reference gets set to the target. To change the this reference you either need to wrap it and call it implicitly or use Function#bind.
Implicit call Example (+ closure):
var something = {
foo: function() {
var self = this;
addEventListener("click", function(e) {
self.bar(e);
});
},
bar: function() {
}
};
Explicit call Example (Function#bind):
var something = {
foo: function() {
addEventListener("click", this.bar.bind(this));
},
bar: function() {
}
};
A: I'm assuming you have a class defined similar to
class thing {
get_O3(e) { ... },
get_previsions() { ... }
}
There are a few options for you. First option, you can bind all functions to this in the constructor:
class thing {
constructor () {
this.get_03 = this.get03.bind(this);
this.get_previsions = this.get_previsions.bind(this);
}
get_O3(e) { ... },
get_previsions() { ... }
}
This can get awkward, especially if you have many functions. You can write a helper bindAll function, but a less awkward/verbose solution is to use a factory method instead, bypassing this altogether:
function makeThing {
const thing = {
get_O3(e) {
...
thing.get_previsions();
},
get_previsions() { ... }
};
return thing;
}
Eric Elliot on Medium has some good reading on the topic if you want to get more in depth.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,516 |
Both Jim Lamastra and Rob Reddy qualified for the Ironman 70.3 World Championships in Clearwater tomorrow.
Whoa!! This race is REALLY cool. Incredible pro field. I'm standing in transition with Jim now. Great vibe! Very professionally done, big international field.
Track Jim and Rob here. Rob is bib number 1051, and Jim is 513.
Hammer down, guys! Good luck!
11:44: Jim has finished! His final time was 4:07, which looks like it's good for 3rd in his age group. Rob is still on course.
12:50: Rob has finished! He crossed the line in a smokin' 4:53. Congrats to both of you on finishing the World Championships!!
Love it!!! Nice work, guys.
Where was the sub 4 Jim? Slacker. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,084 |
Q: Transfer a form name containing an id to jquery I want js to get the fields of a form, located in a modal. The id of my modal is set this way: th:id="myModal+${employee.id}".
And the id of my form, this way: <form th:id="update+${employee.id}" class="form-horizontal" role="form">. I want this form id to be recuperated in my js and what I have done will not work.
I have posted both my modal and JS code below:
<button type="button" class="btn btn-default" data-toggle="modal" th:attr="data-target='#myModal'+${employee.id}">
Update Modal
</button>
<div class="modal fade" th:id="myModal+${employee.id}" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close">
<span aria-hidden="true">×</span>
</button>
<h4 class="modal-title" id="myModalLabel">
Modal title
</h4>
</div>
<form th:id="update+${employee.id}" class="form-horizontal" role="form">
<div class="modal-body">
<input type="text" name="fname" id="fname" />
<input type="text" name="id" th:value="${employee.id}" id="id" />
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">
Close
</button>
<input type="button" class="btn btn-default" value="Update" id="update" th:onclick="'javascript:updateEmp();'" />
</div>
</form>
</div>
</div>
</div>
function updateEmp() {
var location = "http://localhost:8080/update";
console.log($("#update+id"));
var employeeForm = $("#update+id").serializeJSON();
console.log(employeeForm);
var employeeData = JSON.stringify(employeeForm);
console.log(employeeData);
$.ajax({
url: location,
type: 'POST',
data: employeeData, //not included for GET
//dataType: 'json',
contentType: "application/json; charset=utf-8",
success: function (results) {}
And I don't really know how to do this.
A: Since you are using jQuery, bind event using it. You can add a CSS class and the bind event using Class Selector $('.class')
HTML
<input type="button"
class="btn btn-default updateButton"
value="Update" />
Script
$(function(){
$(document).on('click', ".updateButton", function(){
//Get form ans serialize it
var employeeForm = $(this).closest('form').serializeJSON();
//Rest of the code
});
});
A: HTML
<input type="button" class="btn btn-default" value="Update" id="update" th:onclick="'javascript:updateEmp(this);'" />
JS
function updateEmp(button) {
$.ajax({
url: location,
type: 'POST',
data: $(button).parent('form').serialize(), //not included for GET
//dataType: 'json',
contentType: "application/json; charset=utf-8",
success: function (results) {}
})
}
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,029 |
"The nature of human beings and the question of their ultimate origin"
A discussion between Richard Dawkins and the Archbishop of Canterbury The Sheldonian Theatre, Oxford, Thursday, 23 Feburary 2012 4.00-5.30pmhttp://fsmevents.com/sophiaeuropa/
Breaking News: General Aladeen has been banned from attending the Zionists Oscars
Stung by bad publicity, the Academy Of Motion Picture Arts & Zionists is trying to find a way out of its Oscars standoff with General Aladeen.http://www.deadline.com/2012/02/sacha-baron-cohen-oscars-2012-banned-dic..."Admiral General...
30000 vuotta ikiroudassa ollut kasvi kukkii jälleen
A plant that last flowered when woolly mammoths roamed the plains is back in bloom. Biologists have resurrected a 30,000-year-old plant, cultivating it...
Something From Nothing? Richard Dawkins & Lawrence Krauss [HD] 02-04-12
Join critically-acclaimed author and evolutionary biologist Richard Dawkins and world-renowned theoretical physicist and author Lawrence Krauss as they discuss biology, cosmology, religion, and a host of other topics.
Fuck Bay City Rollers!
Siinä missä Hurriganes oli Suomen ensimmäinen suuri rockbändi, oli Hurriganesin Cisse Häkkinen myös maamme ensimmäinen todellinen rocktähti. Eli unohtakaa Eric fucking Faulkner!Ja sitten jööttiä...
Big Beautiful Woman - AC/DC - Whole Lotta Rosie
The song is about an obese Tasmanian woman, Rosie, with whom the singer (Bon Scott) had a one night stand at the Freeway Gardens Motel in North Melbourne. In addition to pointing out the woman's size, the singer finds her to be one of the most...
Fly Me To Ramallah On Delta Airlines
It appears that someone gave Delta a ring and well, the rest is history...http://www.youtube.com/watch?feature=player_embedded&v=D_tp3Lp4z_I
Koiran Vastaisku - Das Auto mainos episode 2
Koira: I can't do it, R2. I can't go on alone.Obi-Wan: Yoda and I will always be with you.Koira: Ben! Why didn't you tell me? You told me that Darth Vader betrayed and murdered my father.Obi-Wan: Your father... was seduced by the Dark Side of the... | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,213 |
Meet duendita, a Spiritual Soul Singer Who Worships at the Altar of Self-Love
The New York City artist talks about how she uses community (and weed) to combat oppression and sorrow in this Rising interview.
Photos by James Emmerman
by Vrinda Jagota
Pop/R&B
Candace Camacho and I meet on one of the coldest days of the year—the kind of day where it takes minutes to un-hunch your shoulders and regain feeling in your fingers. We burrow into our seats at Bahia, an El Salvadorian restaurant in Williamsburg, Brooklyn, where surreal images of nature, animals, and praying people line the walls. The casual spot is bubbling with chatter and pop music. When a Spanish cover of Donna Summer's "Heaven Knows" plays, Camacho lights up and starts humming along. She comes here often to celebrate and plan projects with her friends, and her order—three bean pupusas and peppermint tea—is a familiar one.
Eating well, and eating vegan food especially, is important to her. It's one of the ways she expresses her spirituality, a throughline in the mesmerizing soul music she makes as duendita. Camacho grew up going to church, and still does occasionally, but she now thinks of her spirituality as a set of personalized practices that includes drawing with colored pencils, smoking weed, and paying attention to what she puts in her body. She details some of her favorite dishes: sweet plantain salad with blueberries, vegan grilled cheese with soup. She didn't always have access to good food. "I used to eat boxed mashed potatoes and things that had artificial shit," she says, "not because I wanted to, but usually that's all my parents could afford."
duendita: "Magdalena" (via SoundCloud)
After growing up in Queens, Camacho attended New York University's Clive Davis Institute of Recorded Music. The 23-year-old recently released her debut record, direct line to My Creator, which also served as her college senior thesis project. The 10-track collection is a celebration of the divine in herself and the people she loves, expertly sequenced and intentionally self-contained. It begins with a dial tone and ends with the sound of a phone being hung up. What plays in between is a conversation with God and with those closest to her; on the bouncy highlight "pray," she sings, "I pray to my God and my mom because they're both the same."
Camacho's stage name was inspired by the Spanish poet Federico García Lorca's concept of duende—that inexplicable, mysterious feeling when a piece of art makes the hair on your arms stand up. This sense of wonder radiates throughout direct line, her voice pirouetting from clear, open tones to deep, rounded wails. Camacho sings about heartache, loneliness, and death while situating love, tenderness, and a relationship with something bigger than herself as antidotes. Camacho, who is Puerto Rican-American and Afro-Latinx, also addresses the pain of racial and gendered oppression while uplifting Black women and celebrating her ancestry through prayer and love. Despite her hardships, she finds salvation in God, in the feeling of sunlight on her face, in a future full of daughters.
"I'm very loving and full of joy, but I'm also depressed all the time," she explains. "Maybe everything that has happened in the world has just created this divided person. I often think, Colonization has divided me this way. I cannot recognize myself. I do not fit in America."
During dinner, Camacho gets emotional telling me about how she was a sad child who often felt she couldn't express herself. Her mother encouraged her to be creative, she says, "not so that I could be something, but so I could be OK."
Camacho produced direct line alongside her best friend, Ezra O.S.T., and voices from those in her circle show up multiple times—even the sound of a pal's dog flapping his ears can be heard in the mix. It's all a reflection of the close-knit sense of camaraderie that drives her work and life. "I'm just the person who brings cake to your house and smokes you out forever," she says.
The music video for her song "blue hands," a condemnation of racialized police brutality, speaks to the way she fosters community. The single-take clip, which was filmed in Camacho's bedroom, begins with a close up of her face but slowly zooms out to reveal that she is surrounded by other Black women, including her younger sister. A sign behind her reads, "Say Her Name." The scene is strikingly intimate, a nod to the spaces and people that offer solace from the world's injustices. "Collaboration is, in a way, my reparations," she says.
After finishing our meal, Camacho and I sit at our table for hours, our conversation unspooling slowly. At one point, she pulls out her colored pencils, and we start doodling. I draw random lines, and she begins shading in an amorphous brown swath that eventually becomes a portrait of me. Speaking in shapes and squiggles, it feels comfortable, like being around an old friend.
Pitchfork: On your song "pray," you sing, "I pray for the girls who feel lonely." What do you do when you feel lonely?
duendita: When I'm lonely, I don't really try to cure it. I spend a lot of time alone, and I don't see it as an ailment anymore. The line is for my younger self, who felt ugly and lonely. I grew up feeling rejected and different from everybody else. Me and my grandma look exactly the same, but she's mean. I'm like, "Listen, we look the same, so stop calling me ugly. We actually should be besties, and you should be giving me all the cheat codes. Stop putting me down." But that's colonization. She has been taught to hate herself and to hate her Blackness.
When I was young I couldn't speak up and say, "I don't like this." It was very hard for me to say what was wrong for me and my body. But gaining communication skills changed my life. I feel like I was born last year. I'm more confident, in a sense where I'm relying on my intuition. So that line is the center of a larger feeling that that song is radiating, and that I'm still discovering today.
When I first made "pray," the Ableton session was called "Bipolar Prayer," and, with every lyric, the first line and the second line contradict each other. It talks about how I'm feeling sad and then how I'm feeling happy. And then it lists things you can do to prevent those feelings.
direct line is full of candid sounds that feel intimate and bespoke. Why is it important to you to capture those unique noises?
It all starts in those moments where I'm like, "OK, what does that sound like?" [lifts and drops silverware onto the table] That's what interests me. I went to L.A. last week and was staying in this bougie-ass hotel, but they had a clay pot in the corner. I was looking for something to stretch across the pot to make a drum so I could make some samples. I couldn't find anything, but I had a condom, so I stretched it with my fingers and played it over the pot. It was giving me all kinds of different tones and pitches. I had a microphone just hanging on my neck. That's the kind of stuff I like to do.
We recorded "thunder" in a room all together live. It wasn't orchestrated in a big studio. I got my friends together. It doesn't always have to be super pristine, although I do sometimes wish it were. When I listen to direct line, sometimes it's hard, because I'm like, "Ugh, this doesn't stand next to SZA." You know? Even though I know I shouldn't compare. I don't mean to degrade my work. It's just hard when you're making artifacts and you're becoming at once.
duendita: "thunder" (via SoundCloud)
Aside from music, what else connects you to your creator?
Nature more than anything. I birdwatch a lot. In a lot of my demos, you can hear birds in the background. Being present with nature really makes me feel like, "Wow, I am part of this and am equally as beautiful." When it's hard for me to tell myself that, I just go outside and I see it. It's everywhere and it's undeniable. I once went two hours outside of Berlin, to the border of Germany and Poland, and hung out with birds. I rolled two joints but didn't smoke any of it because I felt the birds would have been mad at me somehow.
Do you think of yourself as a creator?
Not as a god, but I am a creator of something. But I am also a creation, and that's my favorite part; it's not all on me. I also think about how being alive is not a consensual experience. Late at night, when things get really weird, I ask my friends, like, "You would tell me if you asked to be born, right? And if you find out that you did ask, let me know as soon as possible, because I want to know if I agreed to this shit."
Looking forward, what are you hoping to do?
I want to see all the birds and record new music. I have a lot of songs. Go on tour. I want to put out a mixtape. And I've been thinking a lot about how having a career in music gives me an opportunity to have the same economic power and brand equity that others have had for years—it doesn't give it to me now, because I'm broke y'all, but it could. And I'm really interested in having a global conversation and building that slowly, because it's a really long life. I'm not interested in going viral or letting this last for just a moment. I'm not doing this shit like anyone else is doing it. I'm the one making the art. I have the control and the power and the friends and the data. I want to start a company and just publish books by Black women. We're going to get it poppin'. This should have been done a long time ago.
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Charles de Moreau (8. prosince 1758 Rimaucourt, département Haute-Marne , 3. listopadu 1840 Leopoldstadt) byl francouzský architekt a malíř, více než 10 let žijící ve Vídni a známý především svými projekty v Rakousku, Uhrách i v českých zemích.
Život
Opakovaně se v literatuře uvádí jako Moreauovo rodiště Paříž, protože je zapsáno v úmrtní matrice. Roku 2014 však bylo prameny doloženo, že se narodil ve vesnici Rimaucourt, v départementu Haute-Marne Nejdříve vystudoval v Paříži architekturu u Jeana Truffauta. Malířem se vyškolil rovněž v Paříži, v ateliéru slavného portrétisty Jacquesa-Louise Davida. Vyučoval pak několik let malbu v Louvru. Kariéru architekta zahájil ve Francii.
Již roku 1803 však s manželkou Adelaide a dvěma dětmi přesídlil do Vídně, kde se usadil na předměstí Alsergrund, zakoupil rybí tržiště a podle vlastního projektu si tam dal postavit budovu Dianiných lázní (Dianabad). Vstoupil do služeb knížete Mikuláše II. Esterházyho jako jeho dvorní architekt. Proslavil se monumentální přestavbou jeho rodového sídla v Eisenstadtu. Na návrh knížete Metternicha byl roku 1812 jmenován členem Akademie výtvarných umění ve Vídni a od Vídeňského kongresu v roce 1815 se věnoval především projektům pro objednavatele z Rakouské monarchie. V jeho ateliéru se vyučil například moravský architekt Josef Esch. K stáru opět projevil zájem o obchodní podnikání, v Leopoldstadtu zakoupil trh s rybami a přestěhoval se tam. Za své zásluhy byl povýšen na rytíře čestné legie.
Zemřel v 82 letech na mozkovou mrtvici a byl pohřben na vídeňském předměstském hřbitově ve Währingenu.
Dílo (výběr)
přestavba zámku Esterházy v Eisenstadtu (1805–1815), s projektem parku se sloupovým chrámem kněžny Leopoldiny (Leopoldinentempel, se sochou kněžny od Antonia Canovy ), terénními úpravami, vodotrysky, grottou a glorietem
rozšíření knížecího sídla Esterházyů ve Vídni, Naglergasse 9 (1805–1820)
Palác knížat Lubomirskich ve Vídni, Mölker Bastei (1806–1809); zbořen
Apollonův sál ve Vídni, Zieglergasse 17, (1808)
Dianina lázeň ve Vídni (1808–1810)
Pohřební kostel rodiny Esterházyů, Nagyganna, okres Veszprém (1808–1818)
Palác Pálffy ve Vídni, Wallnerstraße 6, (1809–1813)
Přestavba zámku Esterházy, Csákvár (1810–1814)
Palác rodiny Erdödy ve Vídni, (1812); 1956 zbořen
Hala Metternichovy zahrady ve Vídni, Rennweg (1814)
Hlavní budova Technické university ve Vídni, Karlsplatz, (1816–1818)
Nový zámek Werianda Windischgrätze ve Vintířově (1817-1823)
Prodejní prostory porcelánky ve Vídni, Porzellangasse 52, (1818); 1867 přestavěny
Palác Gentz ve Vídni, Währinger Straße 169–171, (1819); po roce 1918 přestavěn
Lázně Frauenbad a Karolinenbad, Baden (1820–1821); 1876 silně přestavěny
Palác knížete Šternberka ve Vídni, Ungargasse 43, (1820–1821), na půdorysu písmene U, zahrada zanikla pozdější uliční zástavbou
Rakouská národní banka (Oesterreichische Nationalbank) ve Vídni, Herrengasse 17, (1821–1823)
Odkazy
Reference
Literatura
Richard H. Kastner: Der Architekt Karl (Charles) Moreau; in: Wiener Geschichtsblätter, Heft 4/2014, s. 277–304.
Externí odkazy
Österreichisches Biographisches Lexikon
Rakouští architekti
Francouzští architekti
Francouzští malíři
Narození v Haute-Marne
Narození v roce 1758
Úmrtí v Rakousku
Úmrtí v roce 1840
Muži
Narození 8. prosince
Úmrtí 3. listopadu
Rytíři Řádu čestné legie | {
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Hyundai Cars & Prices in Ghana (2022)
PGadmin January 4, 2022 Automobile No Comments
The world is revolving and it is safe to say that the automobile world is also on the move and in this part of the world, the Hyundai brand is giving that a go. There is a high possibility you didn't know about the Hyundai brand some decades ago. In fact, one would only rate Hyundai cars among the "upcoming" automobile brands some ten years ago. The development and evolution of this brand have been remarkable. While the likes of Toyota and Honda can share the bragging right when it comes to customer base and quantity of cars produced each year, it is Hyundai that seems to be taking the market by storm with absolutely remarkable cars and sedans. Do you know what the coolest part is? These vehicles come at some amazingly cheap prices compared to their qualities.
Since making their way into the West African market over 15 years ago as a cheap and simple brand, Hyundai has developed into one of the most sought-after automobile brands in Ghana today.
With consistent unique and top-quality products pumped into the market every year, it is a matter of time before the brand starts getting the ovations they deserve. The latest models of Elantra, Accent, Azira, Grandi10, and Sonata are just simple definitions of perfection and simplicity. If you are a lover of comfortable, classy, and stylish cars, then be sure to fall in love with Hyundai products after your first ride. From the wonderfully crafted layout to the well-designed interior, the brand produces a touch of excellence all through. Although they don't have the most commanding of presence, even their SUVs, the products are one of those automobiles that will make you look twice whenever they pass by you.
From their small, mid-sized, and medium vehicles and hatches to their SUVs and sporty sedans, Ghanaians have embraced this brand with love and great joy. The fact that spare parts of models of this brand can be quite expensive and hard to come by in Ghana has not reduced its rate of demand thanks to its amazing ad user-friendly features and designs.
Here, we will take a look at the prices of the most popular models of the brand in Ghana in 2018. Please note that the variations in the prices are due to certain calculated factors including custom and clearance tariff for new and foreign used vehicles, condition of vehicle, and fuel mileage for locally used versions.
Prices of Hyundai Vehicles in Ghana
Hyundai Elantra Prices in Ghana
Hyundai Elantra is one of the most popular Hyundai models in Ghana. It is rated among the company's best products in recent years. Formerly called Lantra, the name was changed in 2001 also adding to its style, classiness, and elegance. Elantra is one of those vehicles you see and fall in love with. The latest model is fantastic and a joy to behold. It boasts amazing features including adaptive cruise control, lane departure intervention, forward collision mitigation, a new base 2.0-liter engine, a turbocharged 1.4-liter engine tuned for fuel economy, and a sporty 1.6-liter turbocharged four-cylinder designed to raise pulses. Let us take a look at how much they go for in Ghana in the current market.
Brand new Hyundai Elantra (2016 -2017 models) can be purchased for between ¢110, 000 – ¢160, 000 in Ghana.
Fairly used Hyundai Elantra
2006 – 2009: ¢55, 000 – ¢90, 000
2014 – 2017: ¢65, 000 – ¢120, 000
Hyundai Accent Prices in Ghana
If you desire a small classy car that is compact, sleek, and efficient, then Hyundai Accent is one of the best cars in that category. Apart from the fact that the car boast a unique layout, the interior is something to savor. Although this remains the brand's lightest and most fragile model, Accent still boasts of high-class features. This affordable Hyundai brand, available as either a hatch or sedan, is powered by a 1.4-liter four-cylinder engine. The sedan is available in base SE and Value Edition trims, while the hatchback comes in SE and Sport variations. Accent is among the fastest cars in its range and is among the Sportier Hyundai models available in Ghana today. It is silent and soft on the road. Handling is OK and you will not stress on maintenance. Accent is well worth the price it goes for in Ghana today, or even more.
Brand new Hyundai Accent (2016 -2017 models) can be purchased for between ¢90, 000 – ¢120, 000 in Ghana.
Fairly used Hyundai Accent
Hyundai Sonata Prices in Ghana
Apart from Elantra and maybe Accent, Hyundai Sonata is arguably the most popular Hyundai vehicle in Ghana. You will be right to categorize this road magic as the Camry of Hyundai Motor Company. To date, Sonata is regarded as the most successful Hyundai model, sitting just a little below Genesis as the most high-spec and technical vehicle of the brand. Sonata is a mid-sized family car that is decent enough to fit every occasion and slides into various environments. They are great for official purposes and pose well for personal use. The latest version features a big new grille, angrier headlights, and boot-lid with taillights to make the model look sportier, saucier, and noticeably shorter. Except for the difficulty in getting its spare parts like any other Hyundai vehicle, Sonata is overall decent and gives you almost everything you want in a medium car.
Brand new Hyundai Sonata (2017 – 2018 models) can be purchased for between ¢150, 000 – ¢190, 000 in Ghana.
Fairly used Hyundai Sonata
Other Hyundai Models include
Brand New: between ¢200, 000 – ¢250, 000
Fairly used
2013 – 2016: ¢140, 000 – ¢180, 000
Brand New: between ¢65, 000 – ¢75, 000
See Prices of Other Cars in Ghana
Suzuki Alto Prices in Ghana (2022)
Hyundai i10 Prices in Ghana (2022)
Toyota Cars & Prices in Ghana (2022)
Nissan Car Prices in Ghana (2022)
Bicycle Prices in Ghana (2022)
Kia Picanto Prices in Ghana (2022)
Toyota Corolla Prices in Ghana (2022)
Prices of Range Rover Sport in Ghana (2022)
KIA Car Prices in Ghana (2022)
Honda Car Prices in Ghana (2022)
Aboboyaa Prices in Ghana (2022) | {
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